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Using The Practice of Statistics, Fifth Edition, for Advanced Placement (AP®) Statistics
(The percents in parentheses reflect coverage on the AP® exam.)
Topic Outline for AP® Statistics
from the College Board’s AP ® Statistics Course Description
The Practice of Statistics, 5th ed.
Chapter and Section references
I. Exploring data: describing patterns and departures from patterns (20%–30%)
A. Constructing and interpreting graphical displays of distributions of univariate data (dotplot,
stemplot, histogram, cumulative frequency plot)
1. Center and spread
2. Clusters and gaps
3. Outliers and unusual features
4. Shape
B. Summarizing distributions of univariate data
1. Measuring center: median, mean
2. Measuring spread: range, interquartile range, standard deviation
3. Measuring position: quartiles, percentiles, standardized scores (z-scores)
4. Using boxplots
5. The effect of changing units on summary measures
C. Comparing distributions of univariate data (dotplots, back-to-back stemplots, parallel boxplots)
1. Comparing center and spread
2. Comparing clusters and gaps
3. Comparing outliers and unusual features
4. Comparing shape
D. Exploring bivariate data
1. Analyzing patterns in scatterplots
2. Correlation and linearity
3. Least-squares regression line
4. Residual plots, outliers, and influential points
5. Transformations to achieve linearity: logarithmic and power transformations
E. Exploring categorical data
1. Frequency tables and bar charts
2. Marginal and joint frequencies for two-way tables
3. Conditional relative frequencies and association
4. Comparing distributions using bar charts
Dotplot, stemplot, histogram 1.2;
Cumulative frequency plot 2.1
1.2
1.2
1.2
1.2
1.3 and 2.1
1.3
1.3
Quartiles 1.3; percentiles and z-scores 2.1
1.3
2.1
Dotplots and stemplots 1.2;
boxplots 1.3
1.2 and 1.3
1.2 and 1.3
1.2 and 1.3
1.2 and 1.3
Chapter 3 and Section 12.2
3.1
3.1
3.2
3.2
12.2
Sections 1.1, 5.2, 5.3
1.1 (we call them bar graphs)
Marginal 1.1; joint 5.2
1.1 and 5.3
1.1
II. Sampling and experimentation: planning and conducting a study (10%–15%)
A. Overview of methods of data collection
1. Census
2. Sample survey
3. Experiment
4. Observational study
B. Planning and conducting surveys
1. Characteristics of a well-designed and well-conducted survey
2. Populations, samples, and random selection
3. Sources of bias in sampling and surveys
4. Sampling methods, including simple random sampling, stratified random sampling,
and cluster sampling
C. Planning and conducting experiments
1. Characteristics of a well-designed and well-conducted experiment
2. Treatments, control groups, experimental units, random assignments, and replication
3. Sources of bias and confounding, including placebo effect and blinding
4. Completely randomized design
5. Randomized block design, including matched pairs design
D. Generalizability of results and types of conclusions that can be drawn from ­
observational studies, experiments, and surveys
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Sections 4.1 and 4.2
4.1
4.1
4.2
4.2
Section 4.1
4.1
4.1
4.1
4.1
Section 4.2
4.2
4.2
4.2
4.2
4.2
Section 4.3
12/9/13 5:51 PM
Using The Practice of Statistics, Fifth Edition, for Advanced Placement (AP®) Statistics
(The percents in parentheses reflect coverage on the AP® exam.)
Topic Outline for AP® Statistics
from the College Board’s AP ® Statistics Course Description
The Practice of Statistics, 5th ed.
Chapter and Section references
III. Anticipating patterns: exploring random phenomena using probability and simulation (20%–30%)
A. Probability
1. Interpreting probability, including long-run relative frequency interpretation
2. “Law of large numbers” concept
3. Addition rule, multiplication rule, conditional probability, and independence
4. Discrete random variables and their probability distributions, including binomial and geometric
5. Simulation of random behavior and probability distributions
6. Mean (expected value) and standard deviation of a random variable, and linear
transformation of a random variable
B. Combining independent random variables
1. Notion of independence versus dependence
2. Mean and standard deviation for sums and differences of independent random variables
C. The Normal distribution
1. Properties of the Normal distribution
2. Using tables of the Normal distribution
3. The Normal distribution as a model for measurements
D. Sampling distributions
1. Sampling distribution of a sample proportion
2. Sampling distribution of a sample mean
3. Central limit theorem
4. Sampling distribution of a difference between two independent sample proportions
5. Sampling distribution of a difference between two independent sample means
6. Simulation of sampling distributions
7. t distribution
8. Chi-square distribution
Chapters 5 and 6
5.1
5.1
Addition rule 5.2; other three topics 5.3
Discrete 6.1; Binomial
and geometric 6.3
5.1
Mean and standard deviation 6.1;
Linear transformation 6.2
Section 6.2
6.2
6.2
Section 2.2
2.2
2.2
2.2
Chapter 7; Sections 8.3,
10.1, 10.2, 11.1
7.2
7.3
7.3
10.1
10.2
7.1
8.3
11.1
IV. Statistical inference: estimating population parameters and testing hypotheses (30%–40%)
A. Estimation (point estimators and confidence intervals)
1. Estimating population parameters and margins of error
2. Properties of point estimators, including unbiasedness and variability
3. Logic of confidence intervals, meaning of confidence level and confidence intervals,
and properties of confidence intervals
4. Large-sample confidence interval for a proportion
5. Large-sample confidence interval for a difference between two proportions
6. Confidence interval for a mean
7. Confidence interval for a difference between two means (unpaired and paired)
8. Confidence interval for the slope of a least-squares regression line
B. Tests of significance
1. Logic of significance testing, null and alternative hypotheses; P-values; one-and two-sided tests;
concepts of Type I and Type II errors; concept of power
2. Large-sample test for a proportion
3. Large-sample test for a difference between two proportions
4. Test for a mean
5. Test for a difference between two means (unpaired and paired)
6. Chi-square test for goodness of fit, homogeneity of proportions, and independence
(one- and two-way tables)
7. Test for the slope of a least-squares regression line
Starnes-Yates5e_FM_endpp.hr.indd 2
Chapter 8 plus parts of Sections 9.3, 10.1,
10.2, 12.1
8.1
8.1
8.1
8.2
10.1
8.3
Paired 9.3; unpaired 10.2
12.1
Chapters 9 and 11 plus parts of Sections
10.1, 10.2, 12.1
9.1; power in 9.2
9.2
10.1
9.3
Paired 9.3; unpaired 10.2
Chapter 11
12.1
12/9/13 5:51 PM
For the AP® Exam
The Practice
of Statistics
f i f t h E D ITI O N
AP® is a trademark registered by the College Board, which was not involved in the production of, and does not endorse, this product.
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Publisher: Ann Heath
Assistant Editor: Enrico Bruno
Editorial Assistant: Matt Belford
Development Editor: Donald Gecewicz
Executive Marketing Manager: Cindi Weiss
Photo Editor: Cecilia Varas
Photo Researcher: Julie Tesser
Art Director: Diana Blume
Cover Designers: Diana Blume, Rae Grant
Text Designer: Patrice Sheridan
Cover Image: Joseph Devenney/Getty Images
Senior Project Editor: Vivien Weiss
Illustrations: Network Graphics
Production Manager: Susan Wein
Composition: Preparé
Printing and Binding: RR Donnelley
TI-83™, TI-84™, TI-89™, and TI-Nspire screen shots are used with permission of the publisher:
© 1996, Texas Instruments Incorporated.
TI-83™, TI-84™, TI-89™, and TI-Nspire Graphic Calculators are registered trademarks of Texas Instruments
Incorporated.
Minitab is a registered trademark of Minitab, Inc.
Microsoft© and Windows© are registered trademarks of the Microsoft Corporation in the United States
and other countries.
Fathom Dynamic Statistics is a trademark of Key Curriculum, a McGraw-Hill Education Company.
M&M’S is a registered trademark of Mars, Incorporated and its affiliates. This trademark is used with
permission. Mars, Incorporated is not associated with Macmillan Higher Education. Images printed with
permission of Mars, Incorporated.
Library of Congress Control Number: 2013949503
ISBN-13: 978-1-4641-0873-0
ISBN-10: 1-4641-0873-0
© 2015, 2012, 2008, 2003, 1999 by W. H. Freeman and Company
First printing 2014
All rights reserved
Printed in the United States of America
W. H. Freeman and Company
41 Madison Avenue
New York, NY 10010
Houndmills, Basingstoke RG21 6XS, England
www.whfreeman.com
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For the AP® Exam
The Practice
of Statistics
f i f t h E D ITI O N
Daren S. Starnes
The Lawrenceville School
Josh Tabor
Canyon del Oro High School
Daniel S. Yates
Statistics Consultant
David S. Moore
Purdue University
W. H. Freeman and Company/BFW
New York
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Contents
About the Authors vi
To the Student xii
Overview: What Is Statistics? xxi
1 Exploring Data
5 Probability: What Are the Chances?
xxxii
Introduction: Data Analysis: Making Sense of Data 2
1.1 Analyzing Categorical Data 7
1.2 Displaying Quantitative Data with Graphs 25
1.3 Describing Quantitative Data with Numbers 48
Free Response AP® Problem, Yay! 74
Chapter 1 Review 74
Chapter 1 Review Exercises 76
Chapter 1 AP® Statistics Practice Test 78
2 Modeling Distributions of Data
Introduction 84
2.1 Describing Location in a Distribution 85
2.2 Density Curves and Normal Distributions 103
Free Response AP® Problem, Yay! 134
Chapter 2 Review 134
Chapter 2 Review Exercises 136
Chapter 2 AP® Statistics Practice Test 137
3 Describing Relationships
140
344
Introduction 346
6.1 Discrete and Continuous Random Variables 347
6.2 Transforming and Combining Random Variables 363
6.3 Binomial and Geometric Random Variables 386
Free Response AP® Problem, Yay! 414
Chapter 6 Review 415
Chapter 6 Review Exercises 416
Chapter 6 AP® Statistics Practice Test 418
7 Sampling Distributions
Introduction 142
3.1 Scatterplots and Correlation 143
3.2 Least-Squares Regression 164
Free Response AP® Problem, Yay! 199
Chapter 3 Review 200
Chapter 3 Review Exercises 202
Chapter 3 AP® Statistics Practice Test 203
4 Designing Studies
Introduction 288
5.1 Randomness, Probability, and Simulation 289
5.2 Probability Rules 305
5.3 Conditional Probability and Independence 318
Free Response AP® Problem, Yay! 338
Chapter 5 Review 338
Chapter 5 Review Exercises 340
Chapter 5 AP® Statistics Practice Test 342
6 Random Variables
82
286
420
Introduction 422
7.1 What Is a Sampling Distribution? 424
7.2 Sample Proportions 440
7.3 Sample Means 450
Free Response AP® Problem, Yay! 464
Chapter 7 Review 465
Chapter 7 Review Exercises 466
Chapter 7 AP® Statistics Practice Test 468
Cumulative AP® Practice Test 2
470
206
Introduction 208
4.1 Sampling and Surveys 209
4.2 Experiments 234
4.3 Using Studies Wisely 266
Free Response AP® Problem, Yay! 275
Chapter 4 Review 276
Chapter 4 Review Exercises 278
Chapter 4 AP® Statistics Practice Test 279
Cumulative AP® Practice Test 1
282
8 Estimating with Confidence
474
Introduction 476
8.1 Confidence Intervals: The Basics 477
8.2 Estimating a Population Proportion 492
8.3 Estimating a Population Mean 507
Free Response AP® Problem, Yay! 530
Chapter 8 Review 531
Chapter 8 Review Exercises 532
Chapter 8 AP® Statistics Practice Test 534
iv
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Section 2.1 Scatterplots and Correlations
9 Testing a Claim
536
Introduction 538
9.1 Significance Tests: The Basics 539
9.2 Tests about a Population Proportion 554
9.3 Tests about a Population Mean 574
Free Response AP® Problem, Yay! 601
Chapter 9 Review 602
Chapter 9 Review Exercises 604
Chapter 9 AP® Statistics Practice Test 605
10 Comparing Two Populations or Groups
13 Analysis of Variance
14 Multiple Linear Regression
15 Logistic Regression
Photo Credits
608
11 Inference for Distributions
676
Introduction 678
11.1 Chi-Square Tests for Goodness of Fit 680
11.2 Inference for Two-Way Tables 697
Free Response AP® Problem, Yay! 730
Chapter 11 Review 731
Chapter 11 Review Exercises 732
Chapter 11 AP® Statistics Practice Test 734
12 More about Regression
PC-1
Notes and Data Sources
Introduction 610
10.1 Comparing Two Proportions 612
10.2 Comparing Two Means 634
Free Response AP® Problem, Yay! 662
Chapter 10 Review 662
Chapter 10 Review Exercises 664
Chapter 10 AP® Statistics Practice Test 666
Cumulative AP® Practice Test 3
669
of Categorical Data
Additional topics on CD-ROM and at
http://www.whfreeman.com/tps5e
N/DS-1
Solutions
S-1
Appendices
A-1
Appendix A: About the AP® Exam A-1
Appendix B: TI-Nspire Technology Corners
A-3
Formulas for AP® Statistics Exam
F-1
Tables
T-1
Table A: Standard Normal Probabilities T-1
Table B: t Distribution Critical Values T-3
Table C: Chi-Square Distribution
Critical Values T-4
Table D: Random Digits T-5
Glossary/Glosario G-1
Index I-1
Technology Corners Reference Back Endpaper
Inference Summary Back Endpaper
736
Introduction 738
12.1 Inference for Linear Regression 739
12.2 Transforming to Achieve Linearity 765
Free Response AP® Problem, Yay! 793
Chapter 12 Review 794
Chapter 12 Review Exercises 795
Chapter 12 AP® Statistics Practice Test 797
Cumulative AP® Practice Test 4
800
v
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About the Authors
Daren S. Starnes is Mathematics Department Chair and
holds the Robert S. and Christina Seix Dow Distinguished Master
Teacher Chair in Mathematics at The Lawrenceville School near
Princeton, New Jersey. He earned his MA in Mathematics from
the University of Michigan and his BS in Mathematics from the
University of North Carolina at Charlotte. Daren is also an alumnus of the North Carolina School of Science and Mathematics.
Daren has led numerous one-day and weeklong AP® Statistics institutes for new and experienced AP® teachers, and he has been a
Reader, Table Leader, and Question Leader for the AP® Statistics
exam since 1998. Daren is a frequent speaker at local, state, regional, national, and international conferences. For two years, he
served as coeditor of the Technology Tips column in the NCTM
journal The Mathematics Teacher. From 2004 to 2009, Daren
served on the ASA/NCTM Joint Committee on the Curriculum
in Statistics and Probability (which he chaired in 2009). While
on the committee, he edited the Guidelines for Assessment and
Instruction in Statistics Education (GAISE) pre-K–12 report and
coauthored (with Roxy Peck) Making Sense of Statistical Studies,
a capstone module in statistical thinking for high school students.
Daren is also coauthor of the popular text Statistics Through
Applications, First and Second Editions.
Josh Tabor has enjoyed teaching general and AP® Statistics
to high school students for more than 18 years, most recently at
his alma mater, Canyon del Oro High School in Oro Valley, Arizona. He received a BS in Mathematics from Biola University, in
La Mirada, California. In recognition of his outstanding work as
an educator, Josh was named one of the five finalists for Arizona
Teacher of the Year in 2011. He is a past member of the AP®
Statistics Development Committee (2005–2009), as well as an experienced Table Leader and Question Leader at the AP® Statistics
Reading. Each year, Josh leads one-week AP® Summer Institutes
and one-day College Board workshops around the country and
frequently speaks at local, national, and international conferences.
In addition to teaching and speaking, Josh has authored articles in
The Mathematics Teacher, STATS Magazine, and The Journal of
Statistics Education. He is the author of the Annotated Teacher’s
Edition and Teacher’s Resource Materials for The Practice of Statistics 4e and 5e, along with the Solutions Manual for The Practice
of Statistics 5e. Combining his love of statistics and love of sports,
Josh teamed with Christine Franklin to write Statistical Reasoning
in Sports, an innovative textbook for on-level statistics courses.
Daniel S. Yates taught AP® Statistics in the Electronic Classroom (a distance-learning facility) affiliated with Henrico County
Public Schools in Richmond, Virginia. Prior to high school teaching, he was on the mathematics faculty at Virginia Tech and
Randolph-Macon College. He has a PhD in Mathematics Education from Florida State University. Dan received a College Board/
Siemens Foundation Advanced Placement Teaching Award in
2000.
David S. Moore is Shanti S. Gupta Distinguished Professor of
Statistics (Emeritus) at Purdue University and was 1998 President of
the American Statistical Association. David is an elected fellow of the
American Statistical Association and of the Institute of Mathematical
Statistics and an elected member of the International Statistical Institute. He has served as program director for statistics and probability
at the National Science Foundation. He is the author of influential
articles on statistics education and of several leading textbooks.
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Section 2.1 Scatterplots and Correlations
Content Advisory Board and Supplements Team
Jason Molesky, Lakeville Area Public School, Lakeville, MN
Media Coordinator—Worked Examples, PPT lectures, Strive for a
5 Guide
Jason has served as an AP® Statistics Reader and Table Leader
since 2006. After teaching AP® Statistics for eight years and developing the FRAPPY system for AP® exam preparation, Jason moved
into administration. He now serves as the Director of Program
Evaluation and Accountability, overseeing the district’s research
and evaluation, continuous improvement efforts, and assessment
programs. Jason also provides professional development to statistics teachers across the United States and maintains the “Stats
Monkey” Web site, a clearinghouse for AP® Statistics resources.
Tim Brown, The Lawrenceville School, Lawrenceville, NJ
Content Advisor, Test Bank, TRM Tests and Quizzes
Tim first piloted an AP® Statistics course the year before the
first exam was administered. He has been an AP® Reader since
1997 and a Table Leader since 2004. He has taught math and statistics at The Lawrenceville School since 1982 and currently holds
the Bruce McClellan Distinguished Teaching Chair.
Doug Tyson, Central York High School, York, PA
Exercise Videos, Learning Curve
Doug has taught mathematics and statistics to high school
and undergraduate students for 22 years. He has taught AP® Statistics for 7 years and served as an AP® Reader for 4 years. Doug
is the co-author of a curriculum module for the College Board,
conducts student review sessions around the country, and gives
workshops on teaching statistics.
Paul Buckley, Gonzaga College High School, Washington, DC
Exercise Videos
Paul has taught high school math for 20 years and AP® Statistics for 12 years. He has been an AP® Statistics Reader for six years
and helps to coordinate the integration of new Readers (Acorns)
into the Reading process. Paul has presented at Conferences for
AP®, NCTM, NCEA (National Catholic Education Association)
and JSEA (Jesuit Secondary Education Association).
Leigh Nataro, Moravian Academy, Bethlehem, PA
Technology Corner Videos
Leigh has taught AP® Statistics for nine years and has served
as an AP® Statistics Reader for the past four years. She enjoys
the challenge of writing multiple-choice questions for the College Board for use on the AP® Statistics exam. Leigh is a National
Board Certified Teacher in Adolescence and Young Adulthood
Mathematics and was previously named a finalist for the Presidential Award for Excellence in Mathematics and Science Teaching
in New Jersey.
Ann Cannon, Cornell College, Mount Vernon, IA
Content Advisor, Accuracy Checker
Ann has served as Reader, Table Leader, and Question Leader for the AP® Statistics exam for the past 13 years. She has taught
introductory statistics at the college level for 20 years and is very
active in the Statistics Education Section of the American Statistical Association, serving on the Executive Committee for two
3-year terms. She is co-author of STAT2: Building Models for a
World of Data (W. H. Freeman and Company).
Michel Legacy, Greenhill School, Dallas, TX
Content Advisor, Strive for a 5 Guide
Michael is a past member of the AP® Statistics Development
Committee (2001–2005) and a former Table Leader at the Reading. He currently reads the Alternate Exam and is a lead teacher at
many AP® Summer Institutes. Michael is the author of the 2007
College Board AP® Statistics Teacher’s Guide and was named the
Texas 2009–2010 AP® Math/Science Teacher of the Year by the
Siemens Corporation.
James Bush, Waynesburg University, Waynesburg, PA
Learning Curve, Media Reviewer
James has taught introductory and advanced courses in Statistics for over 25 years. He is currently a Professor of Mathematics at
Waynesburg University and is the recipient of the Lucas Hathaway
Teaching Excellence Award. James has served as an AP® Statistics
Reader for the past seven years and conducts many AP® Statistics
preparation workshops.
Beth Benzing, Strath Haven High School, Wallingford/
Swarthmore School District, Wallingford, PA
Activities Videos
Beth has taught AP® Statistics for 14 years and has served as
a Reader for the AP® Statistics exam for the past four years. She
serves as Vice President on the board for the regional affiliate for
NCTM in the Philadelphia area and is a moderator for an on-line
course, Teaching Statistics with Fathom.
Heather Overstreet, Franklin County High School,
Rocky Mount, VA
TI-Nspire Technology Corners
Heather has taught AP® Statistics for nine years and has
served as an AP® Statistics Reader for the past six years. While
working with Virginia Advanced Study Strategies, a program for
promoting AP® math, science, and English courses in Virginia
High Schools, she led many AP® Statistics Review Sessions and
served as a Laying the Foundation trainer of teachers of pre-AP®
math classes.
vii
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Acknowledgments
Teamwork: It has been the secret to the successful evolution of The Practice of Statistics (TPS) through
its fourth and fifth editions. We are indebted to each and every member of the team for the time,
energy, and passion that they have invested in making our collective vision for TPS a reality.
To our team captain, Ann Heath, we offer our utmost gratitude. Managing a revision project of this scope is a Herculean task! Ann has a knack for troubleshooting thorny issues with an
uncanny blend of forthrightness and finesse. She assembled an all-star cast to collaborate on the
fifth edition of TPS and trusted each of us to deliver an excellent finished product. We hope
you’ll agree that the results speak for themselves. Thank you, Ann, for your unwavering support,
patience, and friendship throughout the production of these past two editions.
Truth be told, we had some initial reservations about adding a Development Editor to our
team starting with the fourth edition of TPS. Don Gecewicz quickly erased our doubts. His keen
mind and sharp eye were evident in the many probing questions and insightful suggestions he
offered at all stages of the project. Thanks, Don, for your willingness to push the boundaries of
our thinking.
Working behind the scenes, Enrico Bruno busily prepared the manuscript chapters for production. He did yeoman’s work in ensuring that the changes we intended were made as planned.
For the countless hours that Enrico invested sweating the small stuff, we offer him our sincere
appreciation.
We are deeply grateful to Patrice Sheridan and to Diana Blume for their aesthetic contributions to the eye-catching design of TPS 5e. Our heartfelt thanks also go to Vivien Weiss and
Susan Wein at W. H. Freeman for their skillful oversight of the production process. Patti Brecht
did a superb job copyediting a very complex manuscript.
A special thank you goes to our good friends on the high school sales and marketing staff
at Bedford, Freeman, and Worth (BFW) Publishers. We feel blessed to have such enthusiastic
professionals on our extended team. In particular, we want to thank our chief cheerleader, Cindi
Weiss, for her willingness to promote The Practice of Statistics at every opportunity.
We cannot say enough about the members of our Content Advisory Board and Supplements
Team. This remarkable group is a veritable who’s who of the AP® Statistics community. We’ll
start with Ann Cannon, who once again reviewed the statistical content of every chapter. Ann
also checked the solutions to every exercise in the book. What a task! More than that, Ann offered us sage advice about virtually everything between the covers of TPS 5e. We are so grateful
to Ann for all that she has done to enhance the quality of The Practice of Statistics over these
past two editions.
Jason Molesky, aka “Stats Monkey,” greatly expanded his involvement in the fifth edition by
becoming our Media Coordinator in addition to his ongoing roles as author of the Strive for a 5
Guide and creator of the PowerPoint presentations for the book. Jason also graciously loaned us
his Free Response AP® Problem, Yay! (FRAPPY!) concept for use in TPS 5e. We feel incredibly
fortunate to have such a creative, energetic, and deeply thoughtful person at the helm as the
media side of our project explodes in many new directions at once.
Jason is surrounded by a talented media team. Doug Tyson and Paul Buckley have expertly
recorded the worked exercise videos. Leigh Nataro has produced screencasts for the Technology
Corners on the TI-83/84, TI-89, and TI-Nspire. Beth Benzing followed through on her creative
suggestion to produce “how to” screencasts for teachers to accompany the Activities in the book.
James Bush capably served as our expert reviewer for all of the media elements and is now partnering with Doug Tyson on a new Learning Curve media component. Heather Overstreet once
viii
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Section 2.1 Scatterplots and Correlations
again compiled the TI-Nspire Technology Corners in Appendix B. We wish to thank the entire
media team for their many contributions.
Tim Brown, my Lawrenceville colleague, has been busy creating many new, high-quality
assessment items for the fifth edition. The fruits of his labors are contained in the revised Test
Bank and in the quizzes and tests that are part of the Teacher’s Resource Materials. We are especially thankful that Tim was willing to compose an additional cumulative test for Chapters 2
through 12.
We offer our continued thanks to Michael Legacy for composing the superb questions in
the Cumulative AP® Practice Tests and the Strive for a 5 Guide. Michael’s expertise as a twotime former member of the AP® Statistics Test Development Committee is invaluable to us.
Although Dan Yates and David Moore both retired several years ago, their influence lives
on in TPS 5e. They both had a dramatic impact on our thinking about how statistics is best
taught and learned through their pioneering work as textbook authors. Without their early efforts, The Practice of Statistics would not exist!
Thanks to all of you who reviewed chapters of the fourth and fifth editions of TPS and offered your constructive suggestions. The book is better as a result of your input.
—Daren Starnes and Josh Tabor
A final note from Daren: It has been a privilege for me to work so closely with Josh Tabor on
TPS 5e. He is a gifted statistics educator; a successful author in his own right; and a caring parent, colleague, and friend. Josh’s influence is present on virtually every page of the book. He
also took on the thankless task of revising all of the solutions for the fifth edition in addition to
updating his exceptional Annotated Teacher’s Edition. I don’t know how he finds the time and
energy to do all that he does!
The most vital member of the TPS 5e team for me is my wonderful wife, Judy. She has read
page proofs, typed in data sets, and endured countless statistical conversations in the car, in airports, on planes, in restaurants, and on our frequent strolls. If writing is a labor of love, then I am
truly blessed to share my labor with the person I love more than anyone else in the world. Judy,
thank you so much for making the seemingly impossible become reality time and time again.
And to our three sons, Simon, Nick, and Ben—thanks for the inspiration, love, and support that
you provide even in the toughest of times.
A final note from Josh: When Daren asked me to join the TPS team for the fourth edition,
I didn’t know what I was getting into. Having now completed another full cycle with TPS 5e, I
couldn’t have imagined the challenge of producing a textbook—from the initial brainstorming
sessions to the final edits on a wide array of supplementary materials. In all honesty, I still don’t
know the full story. For taking on all sorts of additional tasks—managing the word-by-word revisions to the text, reviewing copyedits and page proofs, encouraging his co-author, and overseeing just about everything—I owe my sincere thanks to Daren. He has been a great colleague and
friend throughout this process.
I especially want to thank the two most important people in my life. To my wife Anne, your
patience while I spent countless hours working on this project is greatly appreciated. I couldn’t
have survived without your consistent support and encouragement. To my daughter Jordan, I
look forward to being home more often and spending less time on my computer when I am
there. I also look forward to when you get to use TPS 7e in about 10 years. For now, we have a
lot of fun and games to catch up on. I love you both very much.
ix
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Acknowledgments
Fifth Edition Survey
Participants and Reviewers
Blake Abbott, Bishop Kelley High School, Tulsa, OK
Maureen Bailey, Millcreek Township School District, Erie, PA
Kevin Bandura, Lincoln County High School, Stanford, KY
Elissa Belli, Highland High School, Highland, IN
Jeffrey Betlan, Yough School District, Herminie, PA
Nancy Cantrell, Macon County Schools, Franklin, NC
Julie Coyne, Center Grove HS, Greenwood, IN
Mary Cuba, Linden Hall, Lititz, PA
Tina Fox, Porter-Gaud School, Charleston, SC
Ann Hankinson, Pine View, Osprey, FL
Bill Harrington, State College Area School District,
State College, PA
Ronald Hinton, Pendleton Heights High School, Pendleton, IN
Kara Immonen, Norton High School, Norton, MA
Linda Jayne, Kent Island High School, Stevensville, MD
Earl Johnson, Chicago Public Schools, Chicago, IL
Christine Kashiwabara, Mid-Pacific Institute, Honolulu, HI
Melissa Kennedy, Holy Names Academy, Seattle, WA
Casey Koopmans, Bridgman Public Schools, Bridgman, MI
David Lee, SPHS, Sun Prairie, WI
Carolyn Leggert, Hanford High School, Richland, WA
Jeri Madrigal, Ontario High School, Ontario, CA
Tom Marshall, Kents Hill School, Kents Hill, ME
Allen Martin, Loyola High School, Los Angeles, CA
Andre Mathurin, Bellarmine College Preparatory, San Jose, CA
Brett Mertens, Crean Lutheran High School, Irvine, CA
Sara Moneypenny, East High School, Denver, CO
Mary Mortlock, The Harker School, San Jose, CA
Mary Ann Moyer, Hollidaysburg Area School District,
Hollidaysburg, PA
Howie Nelson, Vista Murrieta HS, Murrieta, CA
Shawnee Patry, Goddard High School, Wichita, KS
Sue Pedrick, University High School, Hartford, CT
Shannon Pridgeon, The Overlake School, Redmond, WA
Sean Rivera, Folsom High, Folsom, CA
Alyssa Rodriguez, Munster High School, Munster, IN
Sheryl Rodwin, West Broward High School, Pembroke Pines, FL
Sandra Rojas, Americas HS, El Paso, TX
Christine Schneider, Columbia Independent School,
Boonville, MO
Amanda Schneider, Battle Creek Public Schools, Charlotte, MI
Steve Schramm, West Linn High School, West Linn, OR
Katie Sinnott, Revere High School, Revere, MA
Amanda Spina, Valor Christian High School, Highlands Ranch,
CO
Julie Venne, Pine Crest School, Fort Lauderdale, FL
Dana Wells, Sarasota High School, Sarasota, FL
Luke Wilcox, East Kentwood High School, Grand Rapids, MI
Thomas Young, Woodstock Academy, Putnam, CT
Fourth Edition Focus Group
Participants and Reviewers
Gloria Barrett, Virginia Advanced Study Strategies,
Richmond, VA
David Bernklau, Long Island University, Brookville, NY
Patricia Busso, Shrewsbury High School, Shrewsbury, MA
Lynn Church, Caldwell Academy, Greensboro, NC
Steven Dafilou, Springside High School, Philadelphia, PA
Sandra Daire, Felix Varela High School, Miami, FL
Roger Day, Pontiac High School, Pontiac, IL
Jared Derksen, Rancho Cucamonga High School,
Rancho Cucamonga, CA
Michael Drozin, Munroe Falls High School, Stow, OH
Therese Ferrell, I. H. Kempner High School, Sugar Land, TX
Sharon Friedman, Newport High School, Bellevue, WA
Jennifer Gregor, Central High School, Omaha, NE
Julia Guggenheimer, Greenwich Academy, Greenwich, CT
Dorinda Hewitt, Diamond Bar High School,
Diamond Bar, CA
Dorothy Klausner, Bronx High School of Science, Bronx, NY
Robert Lochel, Hatboro-Horsham High School, Horsham, PA
Lynn Luton, Duchesne Academy of the Sacred Heart,
Houston, TX
Jim Mariani, Woodland Hills High School, Greensburgh, PA
Stephen Miller, Winchester Thurston High School,
Pittsburgh, PA
Jason Molesky, Lakeville Area Public Schools, Lakeville, MN
Mary Mortlock, Harker School, San Jose, CA
Heather Nichols, Oak Creek High School, Oak Creek, WI
Jamis Perrett, Texas A&M University, College Station, TX
Heather Pessy, Mount Lebanon High School, Pittsburgh, PA
Kathleen Petko, Palatine High School, Palatine, IL
Todd Phillips, Mills Godwin High School, Richmond, VA
Paula Schute, Mount Notre Dame High School, Cincinnati, OH
Susan Stauffer, Boise High School, Boise, ID
Doug Tyson, Central York High School, York, PA
Bill Van Leer, Flint High School, Oakton, VA
Julie Verne, Pine Crest High School, Fort Lauderdale, FL
x
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Section 2.1 Scatterplots and Correlations
Steve Willot, Francis Howell North High School, St. Charles, MO
Jay C. Windley, A. B. Miller High School, Fontana, CA
Reviewers of previous editions:
Christopher E. Barat, Villa Julie College, Stevenson, MD
Jason Bell, Canal Winchester High School, Canal Winchester, OH
Zack Bigner, Elkins High School, Missouri City, TX
Naomi Bjork, University High School, Irvine, CA
Robert Blaschke, Lynbrook High School, San Jose, CA
Alla Bogomolnaya, Orange High School, Pepper Pike, OH
Andrew Bowen, Grand Island Central School District,
Grand Island, NY
Jacqueline Briant, Bishop Feehan High School, Attleboro, MA
Marlys Jean Brimmer, Ridgeview High School, Bakersfield, CA
Floyd E. Brown, Science Hill High School, Johnson City, TN
James Cannestra, Germantown High School, Germantown, WI
Joseph T. Champine, King High School, Corpus Christi, TX
Jared Derksen, Rancho Cucamonga High School, Rancho
Cucamonga, CA
George J. DiMundo, Coast Union High School, Cambria, CA
Jeffrey S. Dinkelmann, Novi High School, Novi, MI
Ronald S. Dirkse, American School in Japan, Tokyo, Japan
Cynthia L. Dishburger, Whitewater High School, Fayetteville, GA
Michael Drake, Springfield High School, Erdenheim, PA
Mark A. Fahling, Gaffney High School, Gaffney, SC
David Ferris, Noblesville High School, Noblesville, IN
David Fong, University High School, Irvine, CA
Terry C. French, Lake Braddock Secondary School, Burke, VA
Glenn Gabanski, Oak Park and River Forest High School,
Oak Park, IL
Jason Gould, Eaglecrest High School, Centennial, CO
Dr. Gene Vernon Hair, West Orange High School,
Winter Garden, FL
Stephen Hansen, Napa High School, Napa, CA
Katherine Hawks, Meadowcreek High School, Norcross, GA
Gregory D. Henry, Hanford West High School, Hanford, CA
Duane C. Hinders, Foothill College, Los Altos Hills, CA
Beth Howard, Saint Edwards, Sebastian, FL
Michael Irvin, Legacy High School, Broomfield, CO
Beverly A. Johnson, Fort Worth Country Day School,
Fort Worth, TX
Matthew L. Knupp, Danville High School, Danville, KY
Kenneth Kravetz, Westford Academy, Westford, MA
Lee E. Kucera, Capistrano Valley High School,
Mission Viejo, CA
Christina Lepi, Farmington High School, Farmington, CT
Jean E. Lorenson, Stone Ridge School of the Sacred Heart,
Bethesda, MD
Thedora R. Lund, Millard North High School, Omaha, NE
Philip Mallinson, Phillips Exeter Academy, Exeter, NH
Dru Martin, Ramstein American High School,
Ramstein, Germany
Richard L. McClintock, Ticonderoga High School,
Ticonderoga, NY
Louise McGuire, Pattonville High School,
Maryland Heights, MO
Jennifer Michaelis, Collins Hill High School, Suwanee, GA
Dr. Jackie Miller, Ohio State University
Jason M. Molesky, Lakeville South High School, Lakeville, MN
Wayne Nirode, Troy High School, Troy, OH
Heather Pessy, Mount Lebanon High School, Pittsburgh, PA
Sarah Peterson, University Preparatory Academy, Seattle, WA
Kathleen Petko, Palatine High School, Palatine, IL
German J. Pliego, University of St. Thomas
Stoney Pryor, A&M Consolidated High School,
College Station, TX
Judy Quan, Alameda High School, Alameda, CA
Stephanie Ragucci, Andover High School, Andover, MA
James M. Reeder, University School, Hunting Valley, OH
Joseph Reiken, Bishop Garcia Diego High School,
Santa Barbara, CA
Roger V. Rioux, Cape Elizabeth High School, Cape Elizabeth,
ME
Tom Robinson, Kentridge Senior High School, Kent, WA
Albert Roos, Lexington High School, Lexington, MA
Linda C. Schrader, Cuyahoga Heights High School, Cuyahoga
Heights, OH
Daniel R. Shuster, Royal High School, Simi Valley, CA
David Stein, Paint Branch High School, Burtonsville, MD
Vivian Annette Stephens, Dalton High School, Dalton, GA
Charles E. Taylor, Flowing Wells High School, Tucson, AZ
Reba Taylor, Blacksburg High School, Blacksburg, VA
Shelli Temple, Jenks High School, Jenks, OK
David Thiel, Math/Science Institute, Las Vegas, NV
William Thill, Harvard-Westlake School, North Hollywood, CA
Richard Van Gilst, Westminster Christian Academy,
St. Louis, MO
Joseph Robert Vignolini, Glen Cove High School, Glen Cove, NY
Ira Wallin, Elmwood Park Memorial High School,
Elmwood Park, NJ
Linda L. Wohlever, Hathaway Brown School, Shaker Heights, OH
xi
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To the Student
Statistical Thinking and You
The purpose of this book is to give you a working knowledge of the big ideas of statistics and
of the methods used in solving statistical problems. Because data always come from a realworld context, doing statistics means more than just manipulating data. The Practice of Statistics
(TPS), Fifth Edition, is full of data. Each set of data has some brief background to help you
understand what the data say. We deliberately chose contexts and data sets in the examples and
exercises to pique your interest.
TPS 5e is designed to be easy to read and easy to use. This book is written by current high
school AP® Statistics teachers, for high school students. We aimed for clear, concise explanations and a conversational approach that would encourage you to read the book. We also tried
to enhance both the visual appeal and the book’s clear organization in the layout of the pages.
Be sure to take advantage of all that TPS 5e has to offer. You can learn a lot by reading the
text, but you will develop deeper understanding by doing Activities and Data Explorations and
answering the Check Your Understanding questions along the way. The walkthrough guide on
pages xiv–xx gives you an inside look at the important features of the text.
You learn statistics best by doing statistical problems. This book offers many different types
of problems for you to tackle.
•
•
•
•
Section Exercises include paired odd- and even-numbered problems that test the
same skill or concept from that section. There are also some multiple-choice questions to help prepare you for the AP® exam. Recycle and Review exercises at the
end of each exercise set involve material you studied in previous sections.
Chapter Review Exercises consist of free-response questions aligned to specific
learning objectives from the chapter. Go through the list of learning objectives
summarized in the Chapter Review and be sure you can say “I can do that” to each
item. Then prove it by solving some problems.
The AP® Statistics Practice Test at the end of each chapter will help you prepare
for in-class exams. Each test has 10 to 12 multiple-choice questions and three freeresponse problems, very much in the style of the AP® exam.
Finally, the Cumulative AP® Practice Tests after Chapters 4, 7, 10, and 12 provide
challenging, cumulative multiple-choice and free-response questions like ones you
might find on a midterm, final, or the AP® Statistics exam.
The main ideas of statistics, like the main ideas of any important subject, took a long time to
discover and take some time to master. The basic principle of learning them is to be persistent.
Once you put it all together, statistics will help you make informed decisions based on data in
your daily life.
xii
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Section 2.1 Scatterplots and Correlations
TPS and AP® Statistics
The Practice of Statistics (TPS) was the first book written specifically for the Advanced Placement (AP®) Statistics course. Like the previous four editions, TPS 5e is organized to closely follow the AP® Statistics Course Description. Every item on the College Board’s “Topic Outline”
is covered thoroughly in the text. Look inside the front cover for a detailed alignment guide. The
few topics in the book that go beyond the AP® syllabus are marked with an asterisk (*).
Most importantly, TPS 5e is designed to prepare you for the AP® Statistics exam. The entire
author team has been involved in the AP® Statistics program since its early days. We have more
than 80 years’ combined experience teaching introductory statistics and more than 30 years’
combined experience grading the AP® exam! Two of us (Starnes and Tabor) have served as
Question Leaders for several years, helping to write scoring rubrics for free-response questions.
Including our Content Advisory Board and Supplements Team (page vii), we have two former
Test Development Committee members and 11 AP® exam Readers.
TPS 5e will help you get ready for the AP® Statistics exam throughout the course by:
•
•
•
•
Using terms, notation, formulas, and tables consistent with those found on the
AP® exam. Key terms are shown in bold in the text, and they are defined in the
Glossary. Key terms also are cross-referenced in the Index. See page F-1 to find
“Formulas for the AP® Statistics Exam” as well as Tables A, B, and C in the back of
the book for reference.
Following accepted conventions from AP® exam rubrics when presenting model
solutions. Over the years, the scoring guidelines for free-response questions have
become fairly consistent. We kept these guidelines in mind when writing the
solutions that appear throughout TPS 5e. For example, the four-step State-PlanDo-Conclude process that we use to complete inference problems in Chapters 8
through 12 closely matches the four-point AP® scoring rubrics.
Including AP® Exam Tips in the margin where appropriate. We place exam
tips in the margins and in some Technology Corners as “on-the-spot” reminders
of common mistakes and how to avoid them. These tips are collected and summarized in Appendix A.
Providing hundreds of AP® -style exercises throughout the book. We even added
a new kind of problem just prior to each Chapter Review, called a FRAPPY (Free
Response AP® Problem, Yay!). Each FRAPPY gives you the chance to solve an
AP®-style free-response problem based on the material in the chapter. After you
finish, you can view and critique two example solutions from the book’s Web site
(www.whfreeman.com/tps5e). Then you can score your own response using a rubric provided by your teacher.
Turn the page for a tour of the text. See how to use the book to realize success in the course and
on the AP® exam.
xiii
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143
section 3.1 scatterplots and correlation
increments starting with 135 cm. Refer to the sketch in the margin
for comparison.
160
5. Plot each point from your class data table as accurately as you can
on the graph. Compare your graph with those of your group members.
Height (cm)
READ THE TEXT and use the book’s features
to help you grasp the big ideas.
155
150
6. As a group, discuss what the graph tells you about the relationship
between hand span and height. Summarize your observations in a
sentence or two.
145
140
7. Ask your teacher for a copy of the handprint found at the scene
and the math department roster. Which math teacher does your group
believe is the “prime suspect”? Justify your answer with appropriate
statistical evidence.
135
15
15.5
16
16.5
17
Hand span (cm)
3.1
Read the LEARNING
OBJECTIVES at the
beginning of each section.
Focus on mastering these
skills and concepts as you
work through the chapter.
17.5
Scatterplots and Correlation
145
section 3.1 scatterplots and correlation
WHAT You WILL LEARn
By the end of the section, you should be able to:
• Identify explanatory and response variables in situations
Displaying
Relationships: Scatterplots
where one variable helps to explain or influences the other.
Mean Math score
• Interpret the correlation.
• Understand the basic properties of correlation,
625
The
most useful
relationship between
• Make
a scatterplot
to graph
displayfor
thedisplaying
relationshipthe
between
including how the correlation is influenced by outliers.
quantitative
variables is a scatterplot. Figure 3.2 shows a
600
twotwo
quantitative
variables.
• Use technology to calculate correlation.
ofethe
graduates in each state
168
C H• A PDescribe
T scatterplot
E R 3the direction,
D
s cpercent
r
i b iand
nofghigh
r eschool
l at
575
form,
strength
of ai o n s h i p s • Explain why association does not imply causation.
who took the SAT and the state’s mean SAT Math score
in a
relationship displayed in a scatterplot and identify outli550
recent year. We think that “percent taking” will help explain
ers in a scatterplot.
“mean score.” SoA“percent
is the explanatory
negativetaking”
price doesn’t
make muchvariable
sense in this context. Look again at Figure
525
and “mean score”
the response
variable.
want
to see
3.8. Aistruck
with 300,000
milesWe
driven
is far
outside the set of x values for our data.
500
how mean score
when percent
taking changes,
somiles driven and price remains linWe changes
can’t say whether
the relationship
between
475
Most
statistical
studies
examine
data
on more
onemiles
variable.
we put percentear
taking
(theextreme
explanatory
variable)
on the
horiat such
values.
Predicting
price
for
a truck
with than
300,000
drivenFortunately,
analysis
of
several-variable
data
builds
on the
the data
toolsshow.
we used to examine individual
450
zontal axis. Each
point
represents
a
single
state.
In
Colorado,
is an extrapolation of the relationship beyond what
0
90
10
20
30
40
50
60
70
80
variables.
Thetheir
principles
that guide
for example, 21% took the
SAT, and
mean SAT
Math our work also remain the same:
Percent taking SAT
• the
Plot
the data, then
score was 570. Find 21 on
x (horizontal)
axisadd
andnumerical
570 on summaries.
Often, using the regression
DEFInItIon:
Extrapolation
theliney (vertical) axis.
Colorado
appears
the point
(21, and
570).departures from those patterns.
FIGURE 3.2 Scatterplot of the
• Look
for as
overall
patterns
to make a prediction for x = 0 is
mean SAT Math score in each state
Extrapolation
is the use
of a aregression
line for pattern,
predictionuse
faraoutside
the interval
• When
there’s
regular overall
simplified
model to describe it.
an extrapolation. That’s why the y
against the percent of that state’s
of values of the explanatory variable x used to obtain the line. Such predictions are
intercept isn’t always statistically
DEFInItIon:
Scatterplot
high school graduates who took the
meaningful.
often
not
accurate.
SAT. The dotted lines intersect at
A scatterplot shows the relationship between two quantitative variables measured
the point (21, 570), the values for
on the same individuals. The values of one variable
horizontal
Weappear
think on
thatthecar
weight axis,
helpsand
explain accident deaths and that smoking influColorado.
Few relationships are linear for all values of the explanatory variable. autio
the values of the other variable appear on
the vertical
Each individual
the data
encesaxis.
life expectancy.
Ininthese
relationships, the two variables play different roles.
Don’t make predictions using values of x that are much larger or much
appears as a point in the graph.
Accident death rate and life expectancy are the response variables of interest. Car
smaller than those that actually appear in your data.
weight and number of cigarettes smoked are the explanatory variables.
Scan the margins for
the purple notes, which
represent the “voice of
the teacher” giving helpful
hints for being successful
in the course.
Here’s a helpful way to remember: the
eXplanatory variable goes on the x
axis.
Look for the boxes
with the blue bands.
Some explain how to
make graphs or set
up calculations while
others recap important
concepts.
EXAMPLE
!
n
c
Explanatory and Response Variables
Take note of the green
DEFINITION boxes
that explain important
vocabulary. Flip back to
them to review key terms
and their definitions.
Always plot the explanatory variable, if there is one, on the horizontal axis (the
x axis) of a scatterplot. As a reminder, we usually
call the explanatory
DEFInItIon:
Response variable
variable,x explanatory variable
Your understanding
and the response variable y. If thereCheCk
is no explanatory-response
distinction, either
Some data were
collected on
the weight
of a male
white laboratory
rat for
first 25 weeks
A response
variable
measures
an outcome
of a study.
An the
explanatory
variable
variable can go on the horizontal axis.
155
section
3.1shows
scatterplots and correlation
after its birth.may
A scatterplot
of the
weight
grams)inand
time since
birth
(in weeks)
helpFor
explain
orproblems,
predict(in
changes
a response
variable.
We used computer software to produce
Figure
3.2.
some
you’ll
a fairly strong, positive linear relationship. The linear regression equation weight = 100 +
be expected to make scatterplots by40(time)
hand. Here’s
how
to
do
it.
models the data fairly well.
1. What is the slope of the regression line?Standardized
Explain what itvalues have no units—in this example, they are no longer measured in points.
How to MakE a ScattERplotmeans in context.
–
Somethe
people
like to writeExplain
the
To standardize
2. What’s
y intercept?
what it means
in context. the number of wins, we use y = 8.08 and sy = 3.34. For
correlation formula as
12 − 8.08
1. Decide which variable should3.goPredict
on each
theaxis.
rat’s weight after 16 weeks. Show
your
work.
= 1.17. Alabama’s number of wins (12) is 1.17 stanAlabama, zy =
1
zx zy
r = you ∙
3.34
2.
Label and scale your axes. 4. Should
Starnes-Yates5e_c03_140-205hr2.indd 143
9/30/13 4:44 PM
n − 1 use this line to predict the rat’s weight at
dard deviations
above the mean number of wins for SEC teams. When we mulage
2
years?
Use
the
equation
to
make
the
prediction
and
3. Plot individual data values.
to emphasize the product of
thisare
team’s
think about
the reasonableness
of the result.tiply
(There
454 two z-scores, we get a product of 1.2636. The correlation r is an
standardized
scores in the calculation.
“average” of the products of the standardized scores for all the teams. Just
grams in a pound.)
The following example illustrates the process of constructing a scatterplot.
as in the case of the standard deviation sx, the average here divides by one fewer
than the number of individuals. Finishing the calculation reveals that r = 0.936
for the SEC teams.
Watch for CAUTION
ICONS. They alert you
to common mistakes
that students make.
Residuals and the Least-Squares
SEC
Football
What does correlation measure? The Fathom screen shots below proTHINK
Make connections and Regression Line
vide more detail. At the left is a scatterplot of the SEC football data with two lines
Making a scatterplot
ABOUT
ITcases, added—a
vertical
line at the
group’s
In most
no line will
pass exactly
through
allmean
the points per game and a horizontal line
deepen Atyour
understanding
at the mean
number
of wins
of the
group. Most of the points fall in the upper-right
the end of the 2011 college football season, thepoints
University
Alabama
in a of
scatterplot.
Because
we use
the line
to predict
or lower-left
the graph.
That is, teams with above-average points
defeatedon
Louisiana
State deviation
University
for the national
championship.
In- errors“quadrants”
y from
x, the prediction
we make areoferrors
in y, the
by reflecting
the Vertical
from the line
peringame
tend to have
above-average
numbers of wins, and teams with belowterestingly, both of these teams were from the Southeastern
Conference
vertical direction
the scatterplot.
A good
regression line
questions
asked
in the
THINK
average
pointsofper
numbers of wins that are below average.
(SEC).
Here are
averageRegression
number line
of points scored
per the
game
and
nummakes
vertical
deviations
the game
points tend
from to
thehave
line as
1
This confirms the positive association between the variables.
ber of wins for each of the twelve
teams
in the SECsmall
that season.
as possible.
ŷ = 38,257
2 0.1629x
ABOUT IT passages.
Below
on the right
is aFord
scatterplot
the standardized scores. To get this graph,
Figure 3.9 shows
a scatterplot
of the
F-150 of
data
45,000
Price (in dollars)
40,000
35,000
30,000
25,000
20,000
Data point
15,000
10,000
5000
Starnes-Yates5e_c03_140-205hr2.indd 145
0
20,000 40,000
183
section
3.2 we
least-squares
regression
transformed
bothprediction
the x- and errors
the y-values
with
a regression
line
added. The
are by subtracting their mean and dividing by their standard deviation. As we saw in Chapter 2, standardizing a data set
converts the mean to 0 and the standard deviation to 1. That’s why the vertical and
variables and their correlation. Exploring this method will highlight an important
FIGURE 3.9 Scatterplot
of the
FordinF-150
data with a regression
horizontal
lines
the right-hand
graph are both at 0.
relationship between the correlation
and
the
slope oflinea should
least-squares
line added. A good
regression
make the regression
prediction
60,000 80,000 100,000 120,000 140,000 160,000
line—and reveal why we include
the word
“regression”
in theasexpression
“leasterrors (shown
as bold
vertical4:44
segments)
small as possible.
9/30/13
PM
Miles driven
squares regression line.”
How to calcUlatE tHE lEaSt-SqUaRES REGRESSIon lInE
Read the AP® EXAM
TIPS. They give advice on
how to be successful on
the AP® exam.
ap® ExaM tIp The formula
sheet for the AP® exam uses
Starnes-Yates5e_c03_140-205hr2.indd 168
different notation for these
sy
equations: b1 = r and
sx
b0 = y– − b1 x– . That’s because
the least-squares line is written
as y^ = b0 + b1x . We prefer our
simpler versions without the
subscripts!
and y intercept
9/30/13 4:44 PM
sy
b = r Notice that all the products of the standardized values will be positive—not
sx
surprising, considering the strong positive association between the variables. What
if
there
was a negative association between two variables? Most of the points would
a = y–be
− in
bx–the upper-left and lower-right “quadrants” and their z-score products would
be negative, resulting in a negative correlation.
The formula for the y intercept comes from the fact that the least-squares regression line always passes through the point (x– , y– ). You discovered this in Step 4 of the
Activity on page 170. Substituting (x– , y– ) into the equation y^ = a + bx produces
the equation y– = a + bx– . Solving this equation for a gives the equation shown in
How correlation behaves is more important than the details of the formula. Here’s
the definition box, a = y– − bx– .
what you
in order to interpret correlation correctly.
To see how these formulas work in practice,
let’sneed
looktoatknow
an example.
Facts about Correlation
xiv
Starnes-Yates5e_fm_i-xxiii_hr.indd 14
We have data on an explanatory variable x and a response variable y for n
individuals. From the data, calculate the means x– and y– and the standard deviations sx and sy of the two variables and their correlation r. The least-squares
regression line is the line y^ = a + bx with slope
EXAMPLE
Using Feet to Predict Height
Calculating the least-squares regression line
11/20/13 7:43 PM
538
426
CHAPTER 7
CHAPTER 9
TesTing a Claim
Sampling DiStributionS
Introduction
To make sense of sampling variability, we ask, “What would happen if we took
many samples?” Here’s how to answer that question:
• Take a large number of samples from the same population.
• Calculate the statistic (like the sample mean x– or sample proportion p^ ) for
each sample.
• Make a graph of the values of the statistic.
• Examine the distribution displayed in the graph for shape, center, and spread,
as well as outliers or other unusual features.
The following Activity gives you a chance to see sampling variability in action.
LEARN STATISTICS BY DOING STATISTICS
Confidence intervals are one of the two most common types of statistical inference.
Use a confidence interval when your goal is to estimate a population parameter.
The second common type of inference, called significance tests, has a different
goal: to assess the evidence provided by data about some claim concerning a
parameter. Here is an Activity that illustrates the reasoning of statistical tests.
Activity
Activity
MATERIAlS:
200 colored chips, including
100 of the same color; large
bag or other container
I’m a Great Free-Throw Shooter!
MATERIAlS:
Reaching for Chips
A basketball player claims to make 80% of the free throws that he attempts. We
think he might be exaggerating. To test this claim, we’ll ask him to shoot some free
throws—virtually—using The Reasoning of a Statistical Test applet at the book’s
Web site.
1. Go to www.whfreeman.com/tps5e and launch the applet.
Computer with Internet
access and projection
capability
Before class, your teacher will prepare a population of 200 colored chips, with 100
PLET
AP
having the same color (say, red). The parameter is the actual proportion p of red
chips in the population: p = 0.50. In this Activity, you will investigate sampling
variability by taking repeated random samples of size 20 from the population.
1. After your teacher has mixed the chips thoroughly, each student in the class
should take a sample of 20 chips and note the sample proportion p^ of red chips.
When finished, the student should return all the chips to the bag, stir them up,
and pass the bag to the next student.
Note: If your class has fewer than 25 students, have some students take two
157
section 3.1 scatterplots and correlation
samples.
^
andcourse,
plot this
2. Each student should record the p-value in a chart on the board Of
even giving means, standard deviations, and the correlation for “state
value on a class dotplot. Label the graph scale from 0.10 to 0.90SAT
withMath
tick marks
scores” and “percent taking” will not point out the clusters in Figure 3.2.
spaced 0.05 units apart.
Numerical summaries complement plots of data, but they do not replace them.
3. Describe what you see: shape, center, spread, and any outliers or other un144
CHAPTER 3
D e s c r i b i n g r e l at i o n s h i p s
usual features.
EXAMPLE
You will often see explanatory variables
Scoring
Skaters
It is easiest toFigure
identify explanatory
and response variables when we actually
When Mr. Caldwell’s called
class independent
did the “Reaching
for Chips” Activity, his 35 stuvariables and
specify
values
of one variable
see the
howwhole
it affectsstory
another variable. For instance,
Why
correlation
doesn’ttotell
dents produced the graphresponse
shown in
Figurecalled
7.1. dependent
Here’s what the class
said
about its
variables
2. Set the
to take 25 shots.researchers
Click “Shoot.”
How
many of
the 25 shots did
to study the effect of alcohol
on applet
body temperature,
gave
several
difdistribution of p^ -values. variables. Because the words
Until a scandal at the 2002
Olympics
brought
change,
figure
skating
was
scored
the player
make?Then
Do you
have
enough
data
tochange
decide whether
ferent
amounts
of from
alcohol
to
mice.
they
measured
theWe
in eachthe player’s claim
“independent”
andis“dependent”
have
by judges
on apeak
scale
0.0
to
6.0.
The
scores
were
often
controversial.
have
Shape:
The graph
roughly symmetric
with
a single
is valid? minutes later. In this
mouse’s
temperature
the scoresbody
awarded
by two judges,15
Pierre and Elena, for many skaters. How well do
at 0.5.other meanings in statistics, we won’t
3.
“Shoot”
again
for
25
more
shots.
Keep
they amount
agree? Weof
calculate
thatClick
correlation
between
their
scores
is r =
0.9. doing
But this until you are
use them here.
case,
alcohol
isthethe
explanatory
variable,
player
makes less than 80% of his shots or that the
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Center: The mean of our sample proportions
is
0.499.
This
the change
mean
of Pierre’s
scoresconvinced
is 0.8 pointeither
lower
than
Elena’s
mean.
and
in
body
temperature
isthat
thethe
response
^
player’s
claim
is
true.
How
large
a
sample
of
shots
did you need to make your
p
is the balance point of the distribution. variable. When we don’t specify the values of either
decision?
These facts don’t contradict
each other. They simply give different kinds of inforfiGURe 7.1 Dotplot of sample proportions obtained by the
Spread: The standard deviation of our sample
proportions
is
variable
both
variables,
there
may
mation. but
The just
meanobserve
scores4.show
that“Show
Pierretrue
awards
lower
scores
than Elena.
But Was your conclusion
Click
probability”
to reveal
the truth.
35 students in Mr. Caldwell’s class.
0.112. The values of p^ are typically about
0.112
from
orbecause
may away
not
be
explanatory
response
Pierre
gives
everycorrect?
skaterand
a score
about 0.8varipoint lower than Elena does,
the mean.
the correlation
the same
number to all values of either x
ables.
Whetherremains
therehigh.
are Adding
depends
on how
If time permits,
a newthe
shooter
repeat
or plan
y doesto
notuse
change
the5.correlation.
If both choose
judges score
same and
skaters,
the Steps 2 through 4. Is it
you
the data.
Outliers: There are no obvious outliers or other unusual features.
easier to because
tell thatPierre
the player
is exaggerating
whenperhis actual proportion of free
competition is scored consistently
and Elena
agree on which
throws
made
is
closer
to
0.8
or
farther
from
0.8?
Of course, the class only took 35 different simple random samples
of 20are
chips.
formances
better than others. The high r shows their agreement. But if Pierre
There are many, many possible SRSs of size 20 from a population scores
of sizesome
200 (about
skaters and Elena others, we should add 0.8 point to Pierre’s scores to
1.6 · 1027, actually). If we took every one of those possible samples, calculated
arrive at athe
fairvalue
comparison.
of p^ for each, and graphed all those p^ -values, then we’d have a sampling distribution.
Every chapter begins with a hands-on ACTIVITY
that introduces the content of the chapter. Many
of these activities involve collecting data and
drawing conclusions from the data. In other
activities, you’ll use dynamic applets to explore
statistical concepts.
EXAMPLE
Linking SAT Math and Critical
Reading Scores
DATA EXPLORATION The SAT essay: Is longer better?
DATA EXPLORATIONS
ask you to play the role of
data detective. Your goal
is to answer a puzzling,
real-world question by
examining data graphically
and numerically.
Explanatory
response?
Following the debutor
of the
new SAT Writing test in March 2005, Dr. Les Perelman
Starnes-Yates5e_c09_536-607hr.indd 538
10/15/13 10:43 AM
from the Massachusetts Institute of Technology stirred controversy by reporting,
Julie
asks, “Can I predict
a state’s mean SAT Math score if I know its mean SAT
10/10/13
4:59 PM of what a student wrote, the longer the essay, the
“It appeared to me that
regardless
Critical
Reading
to know
the
mean SATpredictor
Math and Critical
higher the
score.” score?”
He went Jim
on towants
say, “I have
neverhow
found
a quantifiable
Reading
scores
this year
50 states
are related
to each
in 25 years
of grading
that in
wasthe
anywhere
as strong
as this one.
If youother.
just graded
Starnes-Yates5e_c07_420-473hr.indd 426
them based on length without ever reading them, you’d be right over 90 percent
Problem:
each
student,
variable
and the
response
The
table
belowidentify
shows the explanatory
data that Dr.
Perelman
used
to drawvariable
his if possible.
of the time.”3For
conclusions.4
solution:
Julie is treating the mean SAT Critical Reading score as the explanatory variable and
the mean SAT Math score
asofthe
response
variable.
Jim is
simply
interested in exploring the relationLength
essay
and score
for a sample
of SAT
essays
shipWords:
between 460
the two
there is
variable.
422variables.
402 For
365him,357
278no clear
236 explanatory
201
168or response
156
133
Score:
6
6
5
5
6
5
4
4
Words:
114
108
100
403
401
388
320
258
Score:
2
1
1
5
6
6
5
Words:
67
697
387
355
337
325
272
4
3
2
4
4
3
2
150
135
For
236 Practice
189
128Try Exercise
1
Score:
1
6 the6 goal 5is to 5show4 that 4changes
2
In
many studies,
in3 one or more explanatory
variables
actually
cause
changeswrite
in aaresponse
variable.
However,
other explanatoryDoes this
mean that
if students
lot, they are
guaranteed
high scores?
response
relationships
don’t involve
direct
causation.
the alcohol
and
Carry out
your own analysis
of the data.
How
would youInrespond
to each
of mice study,
Dr. Perelman’s
claims?a change in body temperature. But there is no cause-and-effect
alcohol
actually causes
relationship between SAT Math and Critical Reading scores. Because the scores are
closely related, we can still use a state’s mean SAT Critical Reading score to predict its
mean Math score. We will learn how to make such predictions in Section 3.2.
Starnes-Yates5e_c03_140-205hr2.indd 157
CHECK YOUR UNDERSTANDING
questions appear throughout the
section. They help you to clarify definitions, concepts, and procedures.
Be sure to check your answers in the
back of the book.
CheCk Your understanding
9/30/13 4:44 PM
Identify the explanatory and response variables in each setting.
1. How does drinking beer affect the level of alcohol in people’s blood? The legal limit
for driving in all states is 0.08%. In a study, adult volunteers drank different numbers of
cans of beer. Thirty minutes later, a police officer measured their blood alcohol levels.
2. The National Student Loan Survey provides data on the amount of debt for recent
college graduates, their current income, and how stressed they feel about college debt. A
sociologist looks at the data with the goal of using amount of debt and income to explain
the stress caused by college debt.
xv
Starnes-Yates5e_c03_140-205hr2.indd 144
Starnes-Yates5e_fm_i-xxiii_hr.indd 15
9/30/13 4:44 PM
11/20/13 7:43 PM
EXAMPLES: Model statistical problems
and how to solve them
CHAPTER 3
D e s c r i b i n g r e l at i o n s h i p s
You will often see explanatory variables
called independent variables and
response variables called dependent
variables. Because the words
“independent” and “dependent” have
other meanings in statistics, we won’t
use them here.
EXAMPLE
It is easiest to identify explanatory and response variables when we actually
specify values of one variable to see how it affects another variable. For instance,
to study the effect of alcohol on body temperature, researchers gave several different amounts of alcohol to mice. Then they measured the change in each
mouse’s body temperature 15 minutes later. In this
case, amount of alcohol is the explanatory variable,
and change in body temperature is the response
variable. When we don’t specify the values of either
variable but just observe both variables, there may
or may not be explanatory and response variables. Whether there are depends on how
you plan to use the data.
Exercises
section 3.1
The red number box next
to the exercise directs you
back to the page in the
section where the model
example appears.
Read through each example,
and then try out the concept
yourself by working the FOR
PRACTICE exercise in the
Section Exercises.
Need extra help? Examples
and exercises marked with
Linking SAT Math and Critical
the PLAY ICON
are
Reading Scores
supported by short video clips
Explanatory or response?
prepared by experienced AP®
Julie asks, “Can I predict a state’s mean SAT Math score if I know its mean SAT
teachers. The video guides
Critical Reading score?” Jim wants to know how the mean SAT Math and Critical
Reading scores this year in the 50 states are related to each other.
185
section 3.2 least-squares regression
you through each
step in the
Problem: For each student, identify the explanatory variable and the response variable if possible.
example
and
solution
and
solution: Julie is treating the mean SAT Critical Reading
score
as r.
theIn
explanatory
variablethe
andslope is equal to the correlation. The Fathom
means
b=
other words,
2
the mean SAT Math score as the response variable. Jim isscreen
simply interested
in
exploring
the
relationshot confirms these results. It shows that r = 0.49,
so r =you
!0.49
= 0.7,
gives
extra
help when you
ship between the two variables. For him, there is no clear explanatory
or response
variable.
159
section
approximately
the same
value3.1
as thescatterplots
slope of 0.697.and correlation
need
it.
For Practice Try Exercise 1
In many studies, the goal is to show that changes in one or more explanatory
variables actually cause changes in a response variable. However, other explanatoryresponse relationships don’t involve direct causation. In the alcohol and mice study,
alcohol actually causes a change in body temperature. But there is no cause-and-effect
1. coralbetween
reefs How
changes
in water
how
relationship
SATsensitive
Math andtoCritical
Reading
scores. Because the4.scores
aremuch gas? Joan is concerned about the
arestill
coral
findSAT
out,Critical
measure
amount
related, we can
use reefs?
a state’sTo
mean
Reading score to predict
its of energy she uses to heat her home. The
pg closely
144 temperature
mean the
Math
score. of
Wecorals
will learn
how to make
suchthe
predictions
growth
in aquariums
where
water in Section 3.2.graph below plots the mean number of cubic feet of
temperature is controlled at different levels. Growth is
gas per day that Joan used each month against the
measured by weighing the coral before and after the
average temperature that month (in degrees FahrenCheCk
Your
understanding
experiment.
What
are the explanatory and response
heit) for one heating season.
Identifyvariables?
the explanatory
and
response
variables
in
each
setting.
Are they categorical or quantitative?
1000
Putting It All Together: Correlation
and Regression
It is now usual to remove only the tumor and nearby
1. How does drinking beer affect the level of alcohol in people’s blood? The legal limit
2.driving
treating
Early
on, adult
the most
common
for
in all breast
states iscancer
0.08%. In
a study,
volunteers
drank different numbers of
800
forminutes
breast cancer
was removal
of the breast.
cans oftreatment
beer. Thirty
later, a police
officer measured
their blood alcohol levels.
Gas consumed (cubic feet)
144
2. The National Student Loan Survey provides data on the amount of debt for recent
600
Inchange
Chapter
introduced
a four-step
process for organizing a statistics problem.
nodes,
followed
by radiation.
Thestressed
in 1,
collegelymph
graduates,
their
current income,
and how
they
feelwe
about
college debt.
A
Herethat
is of
another
example
four-step process in action.
sociologist
looks
the to
data
with the
goal ofexperiment
using amount
debt and
income of
to the
explain
policy
wasatdue
a large
medical
com400
the stress
caused
debt. Some breast cancer patients,
pared
the by
twocollege
treatments.
chosen at random, were given one or the other treatment.
The patients were closely followed to see how long they
lived following surgery. What are the explanatory and
response variables? Are they categorical or quantitative?
EXAMPLE
3.
4-STEP EXAMPLES: By reading the 4-Step Examples and
mastering the special “StatePlan-Do-Conclude” framework,
you can develop good problemsolving skills and your ability to
tackle more complex problems
like those on the AP® exam.
Gesell Scores
25
30
35
40
45
50
55
60
11
10
cHIlD
9
aGE
8
1
15
7
2
26
3
10
6
5
4
9
5
15
2
6
20
1
7
18
4
3
60
70
80
90
100
110
120
130
140
IQ score
(a) Does the plot show a positive or negative association
between the variables? Why does this make sense?
(b) What is the form of the relationship? Is it very strong?
Starnes-Yates5e_c03_140-205hr2.indd 185
Explain your answers.
Starnes-Yates5e_fm_i-xxiii_hr1.indd 16
20
4
0
xvi
0
TEP
iQ and grades Do studentsSwith
higher
IQ test scores
Temperature (degrees Fahrenheit)
Putting
it all together
tend to do better in school? The figure below shows
Does the age at which a child begins to
talk4:44
predict
a later score on a test of mental
9/30/13
PM
(a) Does the plot show
a positive
or negative association
a scatterplot of IQ and school grade point
average
ability?
A study of the development of young children recorded the age in months
between the variables? Why does this make sense?
(GPA) for all 78 seventh-grade studentsatin
a
rural
which each of 21 children spoke their first word and their Gesell Adaptive Score,
midwestern school. (GPA was recordedthe
onresult
a 12-point
data appear inIsthe
tablestrong?
beof an aptitude(b)
test What
taken is
much
later.16
the form
ofThe
the relationship?
it very
scale with A+ = 12, A = 11, A− = 10,low,
B+along
= 9,with
. . . , a scatterplot,Explain
residual your
plot, answers.
and computer output. Should we use a
5
D− = 1, and F = 0.)
linear model to predict a child’s Gesell score from his or her age at first word? If so,
(c) Explain
how accurate will our predictions
be?what the point at the bottom right of the plot
12
represents.
Grade point average
Starnes-Yates5e_c03_140-205hr2.indd 144
200
(c) At the bottom of the plot are several points that we
might call outliers. One student in particular has a
very low GPA despite an average IQ score. What are
Age5.(months)
first word and
Gesell score
heavyatbackpacks
Ninth-grade
students at the Webb
Schools aGE
go on a backpacking
trip each
cHIlD
ScoRE
cHIlD
aGEfall. Students
ScoRE are
divided
into
groups of15
size 8 by
95
8
11 hiking100
11selecting
102names
from
leaving, students
and
71
9 a hat.8 Before 104
16
10 their backpacks
100
are
one hiking
83
10 weighed.
20 The data
94 here are
17 from 12
105group
in a recent year. Make a scatterplot by hand that shows
91
11
7
113
18
42
57
how backpack weight relates to body weight.
ScoRE
pg 145
102
12
9
87
13
10
93
14
11
Body weight (lb):
Backpack weight (lb):
6.
96
83
19
120
187
109
26
30
26
84
20
17
11
121
103
131
165
24
29
35
21
10
86
158
116
31
28
100
bird colonies One of nature’s patterns connects the
percent of adult birds in a colony that return from
the previous year and the number of new adults that
join the colony. Here are data for 13 colonies of
sparrowhawks:6
9/30/13 4:45 PM
Percent return: 74 66 81 52 73 62 52 45 62 46 60 46 38
new adults:
5
6
8 11 12 15 16 17 18 18 19 20 20
Make a scatterplot by hand that shows how the number
12/2/13 11:15 AM
EXERCISES: Practice makes perfect!
192
CHAPTER 3
section 3.2
Start by reading the
SECTION SUMMARY to be
sure that you understand
the key concepts.
Most of the exercises
are paired, meaning that
odd- and even-numbered
problems test the same skill
or concept. If you answer an
assigned problem incorrectly,
try to figure out your mistake.
Then see if you can solve the
paired exercise.
Summary
A regression line is a straight line that describes how a response variable y changes
as an explanatory variable x changes. You can use a regression line to predict the
value of y for any value of x by substituting this x into the equation of the line.
• The slope b of a regression line y^ = a + bx is the rate at which the predicted
response y^ changes along the line as the explanatory variable x changes. Specifically, b is the predicted change in y when x increases by 1 unit.
• The y intercept a of a regression line y^ = a + bx is the predicted response y^
when the explanatory variable x equals 0. This prediction is of no statistical
use unless x can actually take values near 0.
• Avoid extrapolation, the use of a regression line for prediction using values
193
section 3.2 least-squares regression
of the explanatory variable outside the range of the data from which the line
was calculated.
• The most common method of fitting a line to a scatterplot is least squares. The
least-squares regression line is the straight line y^ = a + bx that minimizes the
sum of the squares of the vertical distances of the observed points from the line.
• You can examine the fit of a regression line by studying the residuals, which
arethe
thesame
differences
between
the observed
predicted
values
of y.
Befigure
on the
35. What’s my line? You use
bar of soap
to
in and
Joan’s
midwestern
home.
The
below shows
lookout
for patterns
in when
the residual plot,
thatwith
a linear
model line
shower each morning. The
bar weighs
80 grams
thewhich
originalindicate
scatterplot
the least-squares
it is new. Its weight goesmay
down
bybe
6 grams
per day on
added. The equation of the least-squares line is
not
appropriate.
average. What is the
of the regression
line of the residuals
y^ =s1425
− 19.87x.
• equation
The standard
deviation
measures
the typical size of the
197
section 3.2 least-squares regression
for predicting weight from days of use?
900
prediction errors (residuals) when using the regression
line.
36. What’s my line? An eccentric professor believes that
800
of the variation in the
• The coefficient of determination r2 is the fraction
a child with IQ 100 should have a reading test score
700
husbands and wives Refer toresponse
Exercisevariable
61.
If so,
stumps
shouldregression
produce more
beetle
is accounted
formore
by least-squares
on the
ex-larvae.
of 50 and predicts that reading score
shouldthat
increase
24
600
Here are the data:
2
planatory
variable.
Find rby
and
interpret
this value
in context.
1 point
for every
additional
point
of IQ. What
500
is thedata,
equation
of Interpret
the
regression line
for
• professor’s
Thethis
least-squares
regression
line of y on x is the line with slope b = r(sy/sx)
For these
s = 1.2.
value.
2 4002 1 3 3 4 3 1 – 2– 5 1 3
predicting reading scoreand
from
IQ?
a = y– − bx–. ThisStumps:
line always passes
through the point (x , y ).
intercept
the stock market Refer to Exercise 62.
Beetle larvae: 10 300
30 12 24 36 40 43 11 27 56 18 40
37. gas mileage We expect
a car’s highway
gas
mileagemust be interpreted with caution. Plot the data to be
•
Correlation
and
regression
200
2
Stumps:
2
1
2 2 35 1 401 45
4 150 2 55 1 604
Find rtoand
interpret
value
in
context.
be related
to this
its city
gas
mileage.
for all 1198
sure
that theData
relationship
is roughly linear and to detect30 outliers.
Also look
for inlarvae: 25 8 21 14Temperature
16 6(degrees
54 Fahrenheit)
9 13 14 50
vehicles
the8.3.
government’s
recent
Fuel Economy
fluential
observations,
individualBeetle
points
that substantially change
the
correlation
For these
data,ins =
Interpret
this
value.
Guide give the regression
line:
predicted
highway
(a)
Identify
the
slope
of
the
line
and
explain
what
Can
we use
a linearfor
model
to predict the
of it
or the regression
line. Outliers in x are
often
influential
the regression
line.number
Will impg
bomb
the final?
We(city
expect
that students
= 4.62
+ 1.109
mpg).
thisthe
setting.
beetlemeans
larvae in
from
number of stumps? If so, how
who do well on the midterm
exam of
in all,
a course
will not to conclude that there is a cause-and-effect rela• Most
be careful
accurate
will our
be?the
Follow
the four-step
(a) What’s
the well
slopeon
of this
Interpret
this value
in context.
Identify
the predictions
y intercept of
line. Explain
why it’s
usually
also do
theline?
final
exam.between
Gary
Smith
tionship
two variables just(b)
because
they
are
strongly associated.
•
section 3.2
63.
(a)
(b)
64.
(a)
(b)
65.
Exercises
Gas consumed (cubic feet)
Practice! Work the
EXERCISES assigned by
your teacher. Compare
your answers to those in
the Solutions Appendix at
the back of the book. Short
solutions to the exercises
numbered in red are found in
the appendix.
D e s c r i b i n g r e l at i o n s h i p s
process.
risky to use this value as a prediction.
What’sCollege
the y intercept?
why
the value
of(b)
Pomona
looked atExplain
the exam
scores
of of the
intercept
not statistically
meaningful.
(c)and
Usecalories
the regression
line to of
predict
theinamount
all 346
studentsiswho
took his statistics
class over a
68. Fat
The number
calories
a food of
STEP
22
natural gas
use in
a monththe
withamount
an average
10-year
period.
Assumehighway
that both
the midterm
item depends
on Joan
manywill
factors,
including
(c) Find
the predicted
mileage
for a carand
that gets
temperature
of 30°F.
final exam
were
of fat in
the item. The
data below show the amount
16 miles
perscored
gallonout
in of
the100
city.points.
of
fat
(in
grams)
and
the
number
of
calories
in
7
41. acid rain Refer to Exercise
39. Would it bebeef
appropri(a) State
equation
of the
least-squares
38. the
iQ and
reading
scores
Data onregression
the IQ testline
scores
25
sandwiches
at McDonalds.
ate to use
the regression line to predict pH after 1000
if eachand
student
scored
same
the midterm
and
reading
test the
scores
foron
a group
of fifth-grade
3.2 TECHNOLOGY
months?
Justify
your
answer.
the
final.
children give the following regression line: predicted Sandwich
Fat
Calories
CORNERs
reading
score = −33.4
+ predicting
0.882(IQ score).
® how much gas? Refer to Exercise 40. Would it be
42.
(b)TI-nspire
The actual
least-squares
line for
final29
550
Big Mac
Instructions
in Appendix
B; HP Prime instructions
on the book’s
Web site
appropriate
to
use
the
regression
line
to
predict Joan’s
®
exam
score
y
from
midterm-exam
score
x
was
(a) What’s the slope of this line? Interpret this value in
26
520
Quarter Pounder with Cheese
natural-gas® consumption in a future month with an
y^ = 46.6
+ 0.41x. Predict the score of a student who
context.
with Cheeseof 65°F?page
42 171
Double Quarter
Pounder
8.scored
least-squares
linesa student
on the calculator
average
temperature
Justify
your 750
answer.
50 on theregression
midterm and
who scored
(b) What’s the y intercept? Explain why the value of the
Hamburger
9 175 250
9.100
residual
on the calculator
on the plots
midterm.
43. least-squares idea The tablepage
below
gives a small
intercept is not statistically meaningful.
Cheeseburger
12 two lines
300 fits the
set of data. Which of the following
199
section
3.2 least-squares
regression
(c) Explain
how
answers
to partscore
(b) illustrate
regres(c) Find
theyour
predicted
reading
for a child
with an
Double Cheeseburger
440the leastdata better: y^ = 1 − x or y^ = 23
3 − 2x? Use
sion toIQthe
mean.
score
of 90.
answer. (Note:
McDouble squares criterion to justify your19
390 Neither
it’s
early
expect
that
a baseball
player
39.still
acid
rain
Researchers
studying
acid
rain who
measured
of these two lines is the least-squares regression line
0.93.
(b)66.r2 will
increase,
s We
will
stay the
same.
battingofaverage
in the in
first
month of wilderness
the
Can we
use
a linear
model to predict the number of
pg
166 a high
2 has
the
acidity
precipitation
a
Colorado
for
these
data.)
6.4.
(c) r will
increase,
s will
decrease.
season
willfor
also
have
a high batting
average
rest of
calories from the amount of fat? If so, how accurate will
area
150
consecutive
weeks.
Aciditythe
is measured
Can’t tell without seeing the data.
(d) r2 will
stay theUsing
same,66
s will
stayLeague
the same.
−1 Follow1 the four-step
x:
1
3
5
the season.
Major
playersThe
our predictions
be?
process.
by pH. Lower
pH
values
show Baseball
higher acidity.
23
Starnes-Yates5e_c03_140-205hr2.indd
192
9/30/13 4:45 PM
2 from the
2010
season,
least-squares
regression
In addition to the regression line, the report on the
(e) r will stay
the
same,
s will adecrease.
researchers
observed
a linear pattern
over time.
−1 measure
−5
y: diabetes
2 People
0 with1 diabetes
69. Managing
line was
calculated
to
predict
rest-of-season
batting
Mumbai measurements says that r 2 = 0.95. This
They reported that the regression line pH = 5.43 −
their
fasting
plasma
glucose
(FPG;
measured
in
units
Exercises
79 0.0053(weeks)
and
80 refer
to the
following
setting.
19
average
y from
first-month
batting
average
x. Note: A
44. least-squares idea In Exercise 40, the line drawn
suggests that
fit the
data well.
of milligrams per milliliter) after fasting for at least
player’s
batting
average
is thethe
proportion
of times at
on the scatterplot is the least-squares regression line.
In its recent
Fuel
Economy
Guide,
Environmental
although arm span and height are correlated, arm
8 hours. Another measurement, made at regular
(a)that
Identify
the
slope
of the line
and explain
what
it
bat
he
gets
a
hit.
A
batting
average
over
0.300
is
Explain the meaning of the phrase “least-squares” to
Protection Agency gives data on 1152 vehicles. There are
span does not predict height very accurately.
medical checkups, is called HbA. This is roughly
meansvery
in this
setting.
considered
good
in
Major with
League
Joan, who knows very little about statistics.
a number
of outliers,
mainly
vehicles
veryBaseball.
poor gas
the
percent
of red blood cells that have a glucose
height increases by !0.95 = 0.97 cm for each ad(b)
Identify the
intercepthowever,
of the line
and
explain what it
mileage.
If we
theyoutliers,
the
combined
45. acidattached.
rain In the
acid rainaverage
study ofexposure
Exercise to
39, the
(a) State
theignore
equation
of the least-squares
regression
line
molecule
It measures
ditional centimeter of arm span.
means
in
this
setting.
city andif highway
gas had
mileage
of thebatting
other 1120
or so
pgglucose
169 actual
pH
measurement
Week 50
was
5.08. Find
each player
the same
average
theverest of
over
a period
of severalfor
months.
The
table
95% of the relationship between height and arm span
hicles isthe
approximately
Normal
with
mean
18.7
miles
per
(c)
According
to
the
regression
line,
what
was
the
pH
at
interpret
residual
for FPG
this week.
season as he did in the first month of the season.
belowand
gives
data onthe
both
HbA and
for 18 diabetis accounted for by the regression line.
gallon (mpg)the
andend
standard
of this deviation
study? 4.3 mpg.
ics
months
had
diabetes
46.fivehow
muchafter
gas?they
Refer
tocompleted
Exercise 40.aDuring
March,
(b) The actual equation of the least-squares regression line
95% of the variation in height is accounted for by the
27
79. inismy
chevrolet
(2.2)
The
Chevrolet
Malibu
with
education
class.
40.
how
much
gas?
In
Exercise
4
(page
159),
we
examthe average temperature was 46.4°F and Joan used 490
y^ = 0.245 + 0.109x. Predict the rest-of-season batting
regression line.
a four-cylinder
has ahad
combined
ofmonthly
ined
relationship
between
themileage
average
cubic feet of gas per day. Find and interpret the residual
average
for the
a engine
player
who
a 0.200 gas
batting
average
95% of the height measurements are accounted for by
25the
mpg.
What
percent
of
all vehicles
worse
gas
temperature
and
the
amount
natural
gas
consumed
forHbA
this month.
FPG
HbA
FPG
first
month
of the
season
and for have
aofplayer
who
had
a
the regression line.
mileage
than the
Malibu?
Subject
(%)
(mg/mL)
Subject
(%)
(mg/mL)
0.400 batting
average
the first month of the season.
One child in the Mumbai study had height 59 cm
80. the top 10% (2.2) How high must a vehicle’s gas
1
6.1
141
10
8.7
172
(c) Explain how your answers to part (b) illustrate regresand arm span 60 cm. This child’s residual is
mileage be in order to fall in the top 10% of all
2
6.3
158
11
9.4
200
sion to the mean.
vehicles? (The distribution omits a few high outliers,
−3.2 cm.
(c) −1.3 cm.
(e) 62.2 cm.
3
6.4
112
12
10.4
271
67.mainly
beavers
andgas-electric
beetles Dovehicles.)
beavers benefit beetles?
hybrid
−2.2 cm.
(d) 3.2 cm.
STEP
4
6.8
153
13
10.6
103
9/30/13 4:45 PM
Starnes-Yates5e_c03_140-205hr2.indd
193
Researchers
out 23
circular(1.1)
plots, Researchers
each 4 meters
81. Marijuana
andlaid
traffic
accidents
Suppose that a tall child with arm span 120 cm and
5
7.0
134
14
10.7
172
diameter,
ininterviewed
an area where
cutting
in in
New
Zealand
907beavers
driverswere
at age
21.
height 118 cm was added to the sample used in this
down
cottonwood
trees.
In
each
plot,
they
counted
the
6
7.1
95
15
10.7
359
They had data on traffic accidents and they asked the
study. What effect will adding this child have on the
number
of stumps
fromuse.
trees
cut are
by beavers
drivers
about
marijuana
Here
data onand
the the
7
7.5
96
16
11.2
145
correlation and the slope of the least-squares regresnumberofofaccidents
clusters ofcaused
beetleby
larvae.
Ecologists
think
numbers
these
drivers
at
age
8
7.7
78
17
13.7
147
sion line?
pg 185 that the new sprouts from stumps are more tender 29
19, broken down by marijuana use at the same age: than
9
7.9
148
18
19.3
255
Correlation will increase, slope will increase.
other cottonwood growth, so that beetles prefer them.
Marijuana use per year
Correlation will increase, slope will stay the same.
never 1–10 times 11–50 times 51 ∙ times
Correlation will increase, slope will decrease.
Drivers
452
229
70
156
Correlation will stay the same, slope will stay the same.
Accidents caused
59
36
15
50
Correlation will stay the same, slope will increase.
9/30/13 4:45 PM
Starnes-Yates5e_c03_140-205hr2.indd 197
(a) Make a graph that displays the accident rate for each
Suppose that the measurements of arm span and
class. Is there evidence of an association between
height were converted from centimeters to meters
marijuana use and traffic accidents?
by dividing each measurement by 100. How will this
conversion affect the values of r2 and s?
(b) Explain why we can’t conclude that marijuana use
2
causes accidents.
r will increase, s will increase.
4
Look for icons that appear
next(c)
to selected problems.
(d)
(e) will guide you to
They
75.
• an Example that models
(a)the problem.
•
•
•
videos that provide stepby-step instructions for
(c)
solving the problem.
(b)
(d)
earlier sections on which
the problem draws (here,
76.
Section 2.2).
(e)
(a)
examples with the
4-Step State-Plan-DoConclude way of solving
problems.
(b)
77.
(a)
(b)
(c)
(d)
(e)
78.
(a)
Starnes-Yates5e_fm_i-xxiii_hr1.indd 17
4
xvii
12/2/13 11:15 AM
200
CHAPTER 3
D e s c r i b i n g r e l at i o n s h i p s
and observed how many hours each flower continued to
look fresh. A scatterplot of the data is shown below.
240
230
Freshness (h)
220
210
200
190
180
170
160
(c) Calculate and interpret the residual for the flower
that had 2 tablespoons of sugar and looked fresh for
204 hours.
(d) Suppose that another group of students conducted
a similar experiment using 12 flowers, but included different varieties in addition to carnations.
Would you expect the value of r2 for the second
group’s data to be greater than, less than, or about
the same as the value of r2 for the first group’s data?
Explain.
After you finish, you can view two example solutions on the book’s
Web site (www.whfreeman.com/tps5e). Determine whether you
think each solution is “complete,” “substantial,” “developing,” or
“minimal.” If the solution is not complete, what improvements would
you suggest to the student who wrote it? Finally, your teacher will
provide you with a scoring rubric. Score your response and note
what, if anything, you would do differently to improve your own
score.
REVIEW and PRACTICE for quizzes and tests
0
1
2
3
Sugar (tbsp)
(a) Briefly describe the association shown in the
scatterplot.
(b) The equation of the least-squares regression line
for these data is y^ = 180.8 + 15.8x. Interpret the
slope of the line in the context of the study.
Chapter Review
Section 3.1: Scatterplots and Correlation
In this section, you learned how to explore the relationship
between two quantitative variables. As with distributions of
a single variable, the first step is always to make a graph. A
scatterplot is the appropriate type of graph to investigate associations between two quantitative variables. To describe a
scatterplot, be sure to discuss four characteristics: direction,
form, strength, and outliers. The direction of an association might be positive, negative, or neither. The form of
an association can be linear or nonlinear. An association is
strong if it closely follows a specific form. Finally, outliers
are any points that clearly fall outside the pattern of the rest
of the data.
The correlation r is a numerical summary that describes
the direction and strength of a linear association. When
r > 0, the association is positive, and when r < 0, the
association is negative. The correlation will always take
values between −1 and 1, with r = −1 and r = 1 indicating a perfectly linear relationship. Strong linear associations have correlations near 1 or −1, while weak linear
relationships have correlations near 0. However, it isn’t
Starnes-Yates5e_c03_140-205hr2.indd 200
Use the WHAT DID YOU
LEARN? table to guide you
to model examples and exercises to verify your mastery of
each Learning Objective.
202
possible to determine the form of an association from
only the correlation. Strong nonlinear relationships can
have a correlation close to 1 or a correlation close to 0,
depending on the association. You also learned that outliers
can greatly
affect
value of
theofcorrelation
The
difference
between
thethe
observed
value
y and the and
that value
correlation
does anot
implyResiduals
causation.
predicted
of y is called
residual.
are That
the keyis, we
can’t assumealmost
that changes
in one
variable
cause
changes
to understanding
everything
in this
section.
To find
in the other
justregression
because line,
theyfind
have
correlathe equation
of the variable,
least-squares
thealine
tion closethe
to 1sum
or –1.
that minimizes
of the squared residuals. To see if a
linear model is appropriate, make a residual plot. If there is
Sectionpattern
3.2: Least-Squares
Regression
no leftover
in the residual
plot, you know the model
In this section,
howfitstothe
use
least-squares
is appropriate.
To assessyou
howlearned
well a line
data,
calculate regressiondeviation
lines as ofmodels
for relationships
the standard
the residuals
s to estimatebetween
the size vari2
ables
that
have
a
linear
association.
It
is
of a typical prediction error. You can also calculate rimportant
, which to
understand the difference between the actual data and
the model used to describe the data. For example, when
you are interpreting the slope of a least-squares regression
line, describe the predicted change in the y variable. To
emphasize that the model only provides predicted values, least-squares regression lines are always expressed in
terms of y^ instead of y.
Section
3.1
144
R3.4
3.1
145, 148
R3.4
Describe the direction, form, and strength of a relationship
displayed in a scatterplot and recognize outliers in a scatterplot.
3.1
147, 148
R3.1
Interpret the correlation.
3.1
152
R3.3, R3.4
Understand the basic properties of correlation, including how the
correlation is influenced by outliers.
3.1
152, 156, 157
R3.1, R3.2
Use technology to calculate correlation.
3.1
Activity on 152, 171
R3.4
Explain why association does not imply causation.
3.1
Discussion on 156, 190
R3.6
Interpret the slope and y intercept of a least-squares regression line.
3.2
166
R3.2, R3.4
Use the least-squares regression line to predict y for a given x.
Explain the dangers of extrapolation.
3.2
167, Discussion on 168
(for extrapolation)
R3.2, R3.4, R3.5
Calculate and interpret residuals.
3.2
169
R3.3, R3.4
Explain the concept of least squares.
3.2
Discussion on 169
R3.5
Determine the equation of a least-squares regression line using
technology or computer output.
3.2
Technology Corner on
171, 181
R3.3, R3.4
Construct and interpret residual plots to assess whether a linear
model is appropriate.
3.2
Discussion on 175, 180
R3.3, R3.4
9/30/13 4:45 PM
3.2
180
R3.3, R3.5
Describe how the slope, y intercept, standard deviation of the
residuals, and r 2 are influenced by outliers.
3.2
Discussion on 188
R3.1
Find the slope and y intercept of the least-squares regression
line from the means and standard deviations of x and y and their
correlation.
3.2
183
R3.5
r3.3 stats teachers’ cars A random sample of AP® Statistics teachers was asked to report the age (in years)
and mileage of their primary vehicles. A scatterplot
of the data, a least-squares regression printout, and
a residual plot are provided below.
Coef
3704
12188
S = 20870.5
A
SE Coef
8268
1492
R-Sq = 83.7%
T
0.45
8.17
P
0.662
0.000
140,000
120,000
30
100,000
80,000
60,000
40,000
20,000
10
0
0
0
100
200
300
400
500
600
(a) Describe the association shown in the scatterplot.
xviii
(b) Point A is the hippopotamus. What effect does this
point have on the correlation, the equation of the
least-squares regression line, and the standard deviation of the residuals?
(c) Point B is the Asian elephant. What effect does this
point have on the correlation, the equation of the
least-squares regression line, and the standard deviation of the residuals?
r3.2 penguins diving A study of king penguins looked
for a relationship between how deep the penguins
Starnes-Yates5e_fm_i-xxiii_hr.indd 18
dive to seek food and how long they stay under
31
2
4
6
8
10
12
8
10
12
Tackle the CHAPTER REVIEW
EXERCISES for practice in
solving problems that test
concepts from throughout the
chapter.
Age
700
Gestation (days)
9/30/13 4:45 PM
60,000
50,000
40,000
30,000
20,000
10,000
0
10,000
20,000
30,000
Residual
0
201
R-Sq(adj) = 82.4%
160,000
Mileage
Life span (years)
B
20
Relevant Chapter
Review Exercise(s)
Make a scatterplot to display the relationship between two
quantitative variables.
r3.1 born to be old? Is there a relationship between the
gestational period (time from conception to birth)
of an animal and its average life span? The figure
Predictor
shows a scatterplot of the gestational period and
avConstant201
Starnes-Yates5e_c03_140-205hr2.indd
Age
erage life span for 43 species of animals.30
40
Related Example
on Page(s)
Identify explanatory and response variables in situations where one
variable helps to explain or influences the other.
Chapter 3 Chapter Review Exercises
These exercises are designed to help you review the important
ideas and methods of the chapter.
summary statistics (the means and standard deviations of
two variables and their correlation). As with the correlation,
the equation of the least-squares regression line and the values of s and r2 can be greatly influenced by outliers, so be
sure to plot the data and note any unusual values before
making any calculations.
what Did You learn?
Learning Objective
Interpret the standard deviation of the residuals and r 2 and use
these values to assess how well the least-squares regression line
D e s c r i b i n g r e l at i o n s h i p s
models the relationship between two variables.
CHAPTER 3
Review the CHAPTER
SUMMARY to be sure that
measures the fraction of the variation in the y variable that is
you
understand
the key
accounted for by its linear
relationship
with the x variable.
You also learned how to obtain the equation of a leastinoutput
eachandsection.
squares regression line concepts
from computer
from
0
2
4
6
Age
(a) Give the equation of the least-squares regression
line for these data. Identify any variables you use.
(b) One teacher reported that her 6-year-old car had
65,000 miles on it. Find and interpret its residual.
11/20/13 7:44 PM
dicting when the cherry trees will bloom from the
temperature. Which variable did you choose as the
explanatory variable? Explain.
(b) Use technology to calculate the correlation and the
equation of the least-squares regression line. Interpret the correlation, slope, and y intercept of the line
in this setting.
(c) Suppose that the average March temperature this year
was 8.2°C. Would you be willing to use the equation
in part (b) to predict the date of first bloom? Explain.
(a)
(b)
(c)
(d)
(d) Calculate and interpret the residual for the year when the
average March temperature was 4.5°C. Show your work.
(e) Use technology to help construct a residual plot. Describe what you see.
r3.5 What’s my grade? In Professor Friedman’s economics course, the correlation between the students’ total
scores prior to the final examination and their finalexamination scores is r = 0.6. The pre-exam totals for
all students in the course have mean 280 and stanDesignin
g deviation
s t u D i e30.
s The final-exam scores have mean
dard
75 and standard deviation 8. Professor Friedman has
r3.6
®
the exam was 300. He decides to predict her finalexam score from her pre-exam total.
Find the equation for the appropriate least-squares
regression line for Professor Friedman’s prediction.
Use the least-squares regression line to predict Julie’s
final-exam score.
Explain the meaning of the phrase “least squares” in
the context of this question.
Julie doesn’t think this method accurately predicts
how well she did on the final exam. Determine r2.
Use this result to argue that her actual score could
have been much higher (or much lower) than the
predicted value.
calculating achievement The principal of a high
school read a study that reported a high correlation
between the number of calculators owned by high
school students and their math achievement. Based
on this study, he decides to buy each student at his
school two calculators, hoping to improve their
math achievement. Explain the flaw in the principal’s reasoning.
and the AP Exam
284
CHAPTER 4
AP1.11 You are planning an experiment to determine
AP1.13 The frequency table below summarizes the times
the effect of the brand of gasoline and the weight
in the last month that patients at the emergency
of a car on gas mileage measured in miles per
room of a small-city hospital waited to receive
gallon. You will use a single test car, adding
medical attention.
weights so that its total weight is 3000, 3500, or
®
waiting time
frequency
4000 pounds. The car will drive on a test track
Less than 10 minutes
5
at each weight using each of Amoco, Marathon,
At least 10 but less than 20 minutes
24
gasoline.
is ithe
way to
C H A P T E Rand
4 Speedway
Desig
n i n gWhich
s tSection
uD
e sI:best
282
Multiple choice Select the best answer for each question.
At least 20 but less than 30 minutes
45
® organize the study?
199
section 3.2 least-squares regression
t3.1 Aand
school
guidance counselor examines
theless
number
alcoholic
(a) Start with 3000 pounds and Amoco
run the
At least 30 but
than 40 minutes
38 beverages for each of 11 regions in Great
of extracurricular
activities
that students
do
and
their
Britain was recorded. A scatterplot of spending on
car on
the test
track. Then
do 3500 and
4000
(b) Why was one adult
chosen
at random
in each
of caffeine
normally
ingested
by
that
subject
in
one
At least 40 but less than 50 minutes
19
grade
point
average.
The
guidance
counselor
says,
alcohol
versus
spending
on
tobacco
is
shown
below.
pounds.
to Marathon and go through theday. During the other study period, the subjects were
household to respond
to Change
the survey?
least 50 but less
than 60 minutes
(c) 0.93.that theAt correlation
(b) r2 will
increase,
will stay the same.
“The
evidence
indicates
between
Which 7of the following
statements
is strue?
three weightscould
in order.
to Speedway
given
placebos. The order in which
each subject
(c) Explain how undercoverage
leadThen
to biaschange
in
At least 60a but
less than
70 minutes
2
the more.
number ofreceived
extracurricular
activities
student
par(d)
6.4.
(c) r2 will increase, s will decrease.
and
do
the
three
weights
in
order
once
the
two
types
of
capsules
was
randomized.
6.5
this sample survey.
2
ticipates
in
and
his
or
her
grade
point
average
is
close
the following
represents
were of
restricted
during
each of thepossible values
Can’t diets
tellWhich
without
seeing
the
data.
(d) r will stay the same, s will stay the same.
(b)start
Start
with
and Amoco and run the The(e)subjects’
t4.14 Many people
their
day3000
with apounds
jolt of caffeine
6.0
toMarathon
zero.” A correct
interpretation
ofofthis
the
and
mean
times
study
Atfor
theto
end
eachstatement
2-day
study
period,on
car on
theMost
test track.
Then
75.periods.
In addition
themedian
regression
line,
thewaiting
report
the for the
(e) r2 will stay the same, s will decrease.
from coffee or a soft
drink.
experts
agreechange
that towould
be thatsubjects were evaluated
2 in which
emergency
room
last
month?
using
a
tapping
task
and
then
to
Speedway
without
changing
the
Mumbai measurements says that r = 0.95. This 5.5
people who take in large amounts of caffeine each
(a)
active
students
tend
to instructed
be(a)students
with
poor
grades,
they
were
to
press
a
button
200
times
as
fast
weight.
Then
add
weights
to
get
3500
pounds
and
Exercises 79 and 80 refer to the following setting.
median
=
27
minutes
and
mean
=
24
minutes
suggests
that
day may suffer from physical withdrawal symptoms
66
viceorder.
versa.
5.0
as they could.
go through
the amounts
three gasolines
in theand
same
(b) median
= 28
andcorrelated,
mean = 30arm
minutes
if they stop ingesting
their usual
of caffeine.
In its recent Fuel Economy Guide, the Environmental
(a) although
arm span
andminutes
height are
(b)
students
with
good
grades
tend
to
be
students
who
Then
change
to
4000
pounds
and
do
the
three
(a) How and
why
was
blocking
used
in the
design
of = 35 minutes
Researchers recruited 11 volunteers who were cafProtection Agency gives data on 1152 vehicles. There are
(c)
median
= 31
minutes
and
mean
span
does
not
predict
height
very
accurately.
4.5
are
not
involved
in
many
extracurricular
activities,
gasolines
in
order
again.
this
experiment?
feine dependent and who were willing to take part in
a number of outliers, mainly vehicles with very poor gas
(d) increases
median =
minutes
andcm
mean
39 adminutes
(b) height
by 35
!0.95
= 0.97
for =
each
andthe
vice
versa.
(c) Choose
a gasolineThe
at random,
and run
car(b)
with
a caffeine withdrawal
experiment.
experiment
Why did researchers randomize the order in which
mileage. If we ignore the outliers, however, the combined
ditional
centimeter
ofactivities
arm
span.
(e)extracurricular
median = 45
minutes
and mean = 46 minutes
students
in many
are
this
gasoline
at 3000,
3500,
and (c)
4000
poundsinvolved
in subjects
was conducted on
two
2-day periods
that
occurred
received
the two treatments?
3.0 city and
3.5 highway
4.0gas mileage
4.5 of the other 1120 or so ve(c)
95%
of
the
relationship
between
height
and
arm
span
just
as
likely
to
get
good
grades
as
bad
grades;
the
same
is
AP1.14
Boxplots
of
two
data
sets
are
shown.
order.
Choose
one
of
the
two
remaining
gasolines
one week apart. During one of the 2-day periods, each
hiclesTobacco
is approximately Normal with mean 18.7 miles per
(c) Could this experiment have been carried out in a
is accounted
for by the regression
line.
true forthen
students
involved
in
few
extracurricular
activities.
and again run
car at 3000,
subject was givenatarandom
capsule containing
the the
amount
gallon
(mpg)
and
standard
deviation 4.3 mpg.
double-blind manner? Explain.
(a) The observation (4.5, 6.0) is an outlier.
3500, then 4000 pounds. Do the(d)
same
with
thelinear(d)
95% of the
variation
in height
is accounted
there
is no
relationship
between
number
of activPlotfor
1 by the
79. in of
myachevrolet
(2.2) The beChevrolet Malibu with
(b)
There
is clear evidence
negative association
last gasoline.
regression
ities and grade point average
for line.
students at this school.
a four-cylinder
engine has a combined gas mileage of
tween spending on alcohol
and tobacco.
(d) There are nine combinations of weight
and
gasoline.
Plot 2 for by
(e) involvement in (e)
many
extracurricular
activities and are accounted
95%
of the height measurements
mpg. What line
percent
allplot
vehicles have worse gas
(c)
The equation of the 25
least-squares
for of
this
Run the car several times using each of
these
combigood
grades
go hand the
in hand.
regression line.
mileage
than
the Malibu?
would be approximately
y^ = 10
− 2x.
nations. Make all these runs in random order.
76.
One
child
in
the
Mumbai
study
had
height
59
cm
(d) The correlation for
r =(2.2)
0.99. How high must a vehicle’s gas
80.these
thedata
top is10%
t3.2 The British government conducts regular surveys
Randomly
select
amount
of for
weight
and a brand
and arm
span
60
This
child’s
residual
Section i: Multiple(e)
Choice
Choose
thean
best
answer
Questions
AP1.1
to AP1.14.
be incorner
orderof
tothe
fallplot
in the
of household
spending.
TheBased
average
weekly
house(e)is The below
observation
in themileage
lower-right
is top 10% of all
oncm.
the
boxplots,
which
statement
is
of gasoline, and run the car on the test
track.
vehicles? line.
(The distribution omits a few high outliers,
hold
spending
(in
pounds)
on
tobacco
products
and
influential
for
the
least-squares
(a)
−3.2
cm.
(c)
−1.3
cm.
(e)
62.2
cm.
true?
AP1.1 You look at real Repeat
estate ads
housesa in
Sarasota,
thisfor
process
total
of 30 times. AP1.4 For a certain experiment, the available experimenmainly hybrid gas-electric vehicles.)
Therats,
range
of both
plotsare
is female
about the same.
cm.
3.2 four
cm.
Florida. Many houses range from $200,000 to
tal (b)
units−2.2
are(a)
eight
of(d)
which
AP1.12
A linear
regression
performed
$400,000
in price.
The few
houses was
on the
water, using the five (F1,
F3,
andmeans
are
male
(M1,
M2,
M3,
(b)F4)
The
of both
plots
approximately
77.F2,Suppose
that
afour
tall child
with
armare
span
120 cm andequal. 81. Marijuana and traffic accidents (1.1) Researchers
following
data
points:
A(2,
22),
B(10,
4),
C(6,
14),
in New Zealand interviewed 907 drivers at age 21.
however, have prices up to $15 million. Which of
M4). There
are
to cm
be
four
treatment
groups,
A,used
B, inPlot
height
118
was added
to the
sample
this1.
(c)
Plot
2 contains
more
data
points
than
D(14, 2), best
E(18,
−4). The
They had data on traffic accidents and they asked the
the following statements
describes
theresidual
distribu-for which of the
C, and study.
D. If aWhat
randomized
block
design
used,have on the
effect
will
adding
thisischild
(d)
The
medians
are
approximately
equal.
Starnes-Yates5e_c03_140-205hr2.indd
203
9/30/13
PM on the
five points
has the largest absolute value?
drivers about marijuana use. Here
are4:45
data
tion of home prices
in Sarasota?
with thecorrelation
experimental
unitsslope
blocked
gender,
and
the
of thebyleast-squares
regres(e)
Plot
1
is
more
symmetric
than
Plot
2.
numbers of accidents caused by these drivers at age
(a) Ais most
(b) likely
B
(c)
C to(d)
which of
theline?
following assignments of treatments
sion
(a) The distribution
skewed
the D
left, (e) E
® age:29
19, broken down by marijuana use at the same
is impossible?
and the mean is greater than the median.
(a) Correlation will increase, slope will increase.
(a) A S (F1, M1), B S (F2, M2),
(b) The distribution is most likely skewed to the left,
Marijuana use per year
(b)
Correlation
will
increase,
slope
will
stay
the
same.
free
Response
Show all your work. Indicate
clearly
the methods
CS
(F3, M3),
D S (F4,you
M4)use, because you will be graded
and theSection
mean isii:less
than
the median.
never 1–10 times 11–50 times 51 ∙ times
(c) Correlation
will increase, slope will decrease.
on the correctness of your methods as well as on (b)
the accuracy
and completeness
A S (F1, M2),
B S (F2, M3), of your results and explanations.
(c) The distribution is roughly symmetric with a few
Drivers
452
229
70
156
(d)(F3,
Correlation
will
stayM1)
the same, slope will stay the same.
CS
M4), D S
(F4,
high outliers, and the mean is approximately equal
59
36
15
50
AP1.15 The manufacturer of exercise machines for fitness
theStwo
Noteslope
that will
higher
scores indicate Accidents caused
(e)(F1,
Correlation
will
stay
the same,
increase.
to the median.
(c) A S
M2), B
(F3,machines.
F2),
designed
newright,
elliptical machinesC S
gains
in fitness. of arm span and
(a) Make a graph that displays the accident rate for each
78.(F4,
Suppose
that
measurements
(d) The distributioncenters
is mosthas
likely
skewedtwo
to the
M1), larger
D
Sthe
(M3,
M4)
are meant
to median.
increase cardiovascular fitclass. Is there evidence of an association between
height
to meters
and the mean isthat
greater
than the
Machine
A centimetersMachine
B
(d) A S (F4,
M1),were
B Sconverted
(F2,
M3),from
ness. The two machines are being tested on 30
marijuana use and traffic accidents?
by dividing
each
by
(e) The distribution is most likely skewed to the right,
M2), D S
(F1,measurement
M4)
0 100. 2How will this
volunteers at a fitness center near the company’s C S (F3,
2
conversion affect the values of r and s?
and the mean is less than the median.
(b) Explain why we can’t conclude that marijuana use
(e) A S (F4,2 M1), B S (F1, M4),5 4
1
0
headquarters. The volunteers are randomly ascauses accidents.
(a)(F3,
r will
increase,
s will
increase.
M2),
D S (F2,
M3)
AP1.2 A child is 40 inches
tall,towhich
her at theand
90thuse it daily for C S
signed
one ofplaces
the machines
Chapter 3 AP® Statistics Practice Test
Alcohol
Each chapter concludes with
an AP STATISTICS
PRACTICE TEST. This test
includes about 10 AP -style
multiple-choice questions and
3 free-response questions.
Cumulative AP® Practice Test 1
876320
2
Four CUMULATIVE AP TESTS
simulate the real exam. They
are placed after Chapters 4, 7,
10, and 12. The tests expand
in length and content coverage
from the first through the
fourth.
159
percentile of all two
children
of similar
age. The
heights
months.
A measure
of cardiovascular
fitness For a biology project, you9measure
7 4 1 1 the weight
3
2489
AP1.5
in
for children of this
age form an approximately
is administered
at the start of the experiment and grams (g) and the tail length in millimeters (mm) of
61
4
257
Normal distribution
with
a
mean
of
38
inches.
Based
again at the end. The following table contains thea group of mice. The equation of the least-squares
5
359
on this information,
what is in
thethe
standard
deviation
differences
two scores
(Afterof– Before) for line for predicting tail length from weight
is
the heights of all children of this age?
predicted tail length = 20 + 3 × weight
(a) 0.20 inches (c) 0.65 inches (e) 1.56 inches
Which of the following is not correct?
®
(b) 0.31 inches (d) 1.21 inches
(a) The slope is 3, which indicates that a mouse’s
AP1.3 A large set of test scores has mean 60 and standard
weight should increase by about 3 grams for each
®
The following problem is modeled after actual AP Statistics exam
Two statistics students went to a flower shop and randeviation 18. If each score is doubled, and then 5 is
additional millimeter of tail length.
Starnes-Yates5e_c04_206-285hr.indd 284
9/17/13 4:44 PM
free response questions. Your task is to generate a complete, condomly selected 12 carnations. When they got home, the
subtracted from the result, the mean and standard
(b) The
predicted
of a mouse that weighs
cise
response tail
in 15length
minutes.
students prepared 12 identical vases with exactly the same
deviation of the new scores are
38 grams is 134 millimeters.
amount of water in each vase. They put one tablespoon of
(a) mean 115; std. dev. 31. (d) mean 120; std. dev. 31.
(c) By Directions:
looking at the
equation
the least-squares
line,
Show
all yourofwork.
Indicate clearly
the methods
sugar in 3 vases, two tablespoons of sugar in 3 vases, and
(b) mean 115; std. dev. 36. (e) mean 120; std. dev. 36.
youyou
canuse,
seebecause
that theyou
correlation
between
weight
will be scored
on the
correctness of your
three tablespoons of sugar in 3 vases. In the remaining
200
C isH positive.
AasP on
T Ethe
R accuracy
3
D e s c r i b i n g r e l at i o n s h i p s
(c) mean 120; std. dev. 6.
andmethods
tail length
as well
and completeness of your
3 vases, they put no sugar. After the vases were prepared,
results and explanations.
the students randomly assigned 1 carnation to each vase
FRAPPY! Free Response AP Problem, Yay!
Learn how to answer free-response
questions successfully by working
the FRAPPY! The Free Response
AP® Problem, Yay! that comes
just before the Chapter Review in
every chapter.
and observed how many hours each flower continued to
look fresh. A scatterplot of the data is shown below.
Starnes-Yates5e_c04_206-285hr.indd 282
9/17/13 4:44 PM
240
230
Starnes-Yates5e_c03_140-205hr2.indd
199
Freshness (h)
220
210
200
190
180
170
160
0
1
2
3
Sugar (tbsp)
(a) Briefly describe the association shown in the
scatterplot.
(b) The equation of the least-squares regression line
for these data is y^ = 180.8 + 15.8x. Interpret the
slope of the line in the context of the study.
(c) Calculate and interpret the residual for the flower
that had 2 tablespoons of sugar and looked fresh for
204 hours.
(d) Suppose that another group of students conducted
a similar experiment using 12 flowers, but in-9/30/13
cluded different varieties in addition to carnations.
Would you expect the value of r2 for the second
group’s data to be greater than, less than, or about
the same as the value of r2 for the first group’s data?
Explain.
4:45 PM
After you finish, you can view two example solutions on the book’s
Web site (www.whfreeman.com/tps5e). Determine whether you
think each solution is “complete,” “substantial,” “developing,” or
“minimal.” If the solution is not complete, what improvements would
you suggest to the student who wrote it? Finally, your teacher will
provide you with a scoring rubric. Score your response and note
what, if anything, you would do differently to improve your own
score.
xix
Chapter Review
Starnes-Yates5e_fm_i-xxiii_hr.indd 19
11/20/13 7:44 PM
Use Technology to discover and analyze
Use technology as a tool for discovery and analysis. TECHNOLOGY CORNERS
give step-by-step instructions for using the TI-83/84 and TI-89 calculator.
192
CHAPTER 3
D e s c r i b i n g r e l at i o n s h i p s
Instructions for the TI-Nspire are in an end-of-book appendix. HP Prime
instructions are on the book’s Web site and in the e-Book.
150
CHAPTER 3
D e s c r i b i n g r e l at i o n s h i p s
Summary
section 3.2
You can access video instructions for
the Technology Corners through the
e-Book or on the book’s Web site.
A regression line is a straight line that describes how a response variable y changes
as an explanatory variable x changes. You can use a regression line to predict the
value of y for any value of x by substituting this x into the equation of the line.
TI-nspire instructions in Appendix B; HP Prime instructions on the book’s Web•site The slope b of a regression line y^ = a + bx is the rate at which the predicted
response y^ changes along the line as the explanatory variable x changes. Specifically,
Making scatterplots with technology is much easier than constructing them by hand. We’ll use the SEC football
data b is the predicted change in y when x increases by 1 unit.
from page 146 to show how to construct a scatterplot on a TI-83/84 or TI-89.
• The y intercept a of a regression line y^ = a + bx is the predicted response y^
• Enter the data values into your lists. Put the points per game in L1/list1 and the number of wins in L2/list2. when the explanatory variable x equals 0. This prediction is of no statistical
• Define a scatterplot in the statistics plot menu (press F2 on the TI-89). Specify the settings shown below. use unless x can actually take values near 0.
• Avoid extrapolation, the use of a regression line for prediction using values
of the explanatory variable outside the range of the data from which the line
was calculated.
• The most common method of fitting a line to a scatterplot is least squares. The
least-squares regression line is the straight line y^ = a + bx that minimizes the
sum of the squares of the vertical distances of the observed points from the line.
• You can examine the fit of a regression line by studying the residuals, which
are the differences between the observed and predicted values of y. Be on the
• Use ZoomStat (ZoomData on the TI-89) to obtain a graph. The calculator will set the window dimensions automatically
lookout for patterns in the residual plot, which indicate that a linear model
by looking at the values in L1/list1 and L2/list2.
may not be appropriate.
• The standard deviation of the residuals s measures the typical size of the
prediction errors (residuals) when using the regression line.
• The coefficient of determination r2 is the fraction of the variation in the
response variable that is accounted for by least-squares regression on the explanatory variable.
The least-squares regression line of y on x is the line with slope b = r(sy/sx)
196
CHAPTER 3
D e s c r i b i n g r e l at i o n s h i p s•
andonto
intercept a = y– − bx–. This line always passes through the point (x–, y–).
Notice that there are no scales on the axes and that the axes are not labeled. If you copy a scatterplot from your calculator
your paper, make sure that you scale and label the axes.
• Correlation and regression must be interpreted with caution. Plot the data to be
20
brain thatsure
responds
to physical
pain goes
up as distress
that the
relationship
is roughly
linear and to detect outliers. Also look for infrom social
exclusion
goes up. A scatterplot shows
ap® ExaM10tIp If you are asked to make a scatterplot on a free-response question,
be sure
to
fluential
observations, individual points that substantially change the correlation
a moderately
strong, linear relationship. The figure
label and scale both axes. Don’t just copy an unlabeled calculator graph directly
onto yourorpaper.
the regression line. Outliers in x are often influential for the regression line.
below shows Minitab regression output for these data.
0
• Most of all, be careful not to conclude that there is a cause-and-effect rela10
tionship between two variables just because they are strongly associated.
•
ScattERplotS on tHE calcUlatoR
Residual
7. TECHNOLOGY
CORNER
Measuring Linear Association: Correlation
20
Some people refer to r as the
“correlation coefficient.”
c
Find the Technology Corners easily by
consulting the summary table at the
end of each section or the complete
table inside the back cover of the book.
59. Merlins breeding Exercise
13 (page 160) gives data on192
Starnes-Yates5e_c03_140-205hr2.indd
the number of breeding pairs of merlins in an isolated
area in each of seven years and the percent of males who
returned the next year. The data show that the percent
returning is lower after successful breeding seasons and
that the relationship is roughly linear. The figure below
shows Minitab regression output for these data.
Regression Analysis: Percent return versus Breeding pairs
Predictor
Constant
Breeding pairs
S = 7.76227
xx
Starnes-Yates5e_fm_i-xxiii_hr1.indd 20
page 171
page 175
relationship with social distress score?
Explain what each of these values means in this setting.
(c) Use the information in the figure to find the correla58.DEFInItIon:
Do heavier people
burn
more
energy?
Refer
to
Exercorrelation r
tion r between social distress score and brain activity.
cises 48 and 50. For the regression you performed earlier,
do you know whether the sign of r is + or −?
The
correlation
the direction
and
strength of the linear How
relationship
r2 =
0.768 and sr=measures
95.08. Explain
what each
of these
between
two quantitative
variables.
(d) Interpret the value of s in this setting.
values means
in this setting.
pg 181
Starnes-Yates5e_c03_140-205hr2.indd 150
!
n
A scatterplot displays the direction, form, and strength of the relationship between
two quantitative10variables.
11 12 Linear
13 14 relationships
15 16 17 are particularly important because a
straight line is a simple pattern
Agethat is quite common. A linear relationship is strong
if the points lie close to a straight line and weak if they are widely scattered about a
(a) Calculate and interpret the residual for the student
3.2
TECHNOLOGY
line.
Unfortunately, our eyes are not good
judges
of how strong a linear relationship is.
who was 141 cm tall at age 10.
(a) same
Whatdata,
is the equation of the least-squares regression
CORNERs
The two scatterplots in Figure 3.5 (on the facing
page) show the
autio
(b)
Is
a
linear
model
appropriate
for
these
data?
Explain.
line for predicting
brain activity from social distress
but the graph on the right is drawn smaller inTI-nspire
a large field.
The right-hand
Instructions
in Appendix B; HP Prime instructions on the book’s Web site
score? Use the equation to predict brain activity for
(c)
Interpret
s.
graph
seemsthe
to value
show of
a stronger
linear relationship.
distress score 2.0.
2
it’s value
easy of
to rbe
scales or by the social
amount
of space
(d) Because
Interpret the
. fooled by different
8. least-squares regression lines on the calculator
What percent
of the variation in brain activity among
around
the cloud
of points
in a 47
scatterplot,
we
need to use (b)
a numerical
measure
57.
bird colonies
Refer
to Exercises
and 49. For9.
the
residual plots on these
the calculator
subjects
is accounted for by the straight-line
to supplement
graph. earlier,
correlation
is and
the smeasure
regression youthe
performed
r2 = 0.56
= 3.67. we use.
Coef
266.07
-6.650
SE Coef
52.15
1.736
R-Sq = 74.6%
T
5.10
-3.83
P
0.004
0.012
R-Sq(adj) = 69.5%
(a) What is the equation of the least-squares regression
line for predicting the percent of males that return
from the number of breeding pairs? Use the equation to predict the percent of returning males after a
season with 30 breeding pairs.
(b) What percent of the year-to-year variation in percent
of returning males is accounted for by the straightline relationship with number of breeding pairs the
previous year?
(c) Use the information in the figure to find the correlation r between percent of males that return and
number of breeding pairs. How do you know whether
the sign of r is + or −?
(d) Interpret the value of s in this setting.
60. Does social rejection hurt? Exercise 14 (page 161)
gives data from a study that shows that social exclusion
causes “real pain.” That is, activity in an area of the
61. husbands and wives The mean height of married
American women in their early twenties is 64.5 inches
and the standard deviation is 2.5 inches. The mean
height of married men the same age is 68.5 inches, with
standard deviation 2.7 inches. The correlation between
4:44 PMis about r = 0.5.
the heights of husbands9/30/13
and wives
pg 183
9/30/13 4:45 PM
(a) Find the equation of the least-squares regression line
for predicting a husband’s height from his wife’s height
for married couples in their early 20s. Show your work.
Other types of software displays,
including Minitab, Fathom, and applet
screen captures, appear throughout
help
youthink
learn
tobehavread
62. thethe
stockbook
marketto
Some
people
that the
ior of the stock market in January predicts its behavior
many
differentvariable
kinds of
for and
the restinterpret
of the year. Take
the explanatory
x to be the percent change in a stock market index in
output.
January
and the response variable y to be the change
(b) Suppose that the height of a randomly selected wife
was 1 standard deviation below average. Predict the
height of her husband without using the least-squares
line. Show your work.
in the index for the entire year. We expect a positive
correlation between x and y because the change
during January contributes to the full year’s change.
Calculation from data for an 18-year period gives
x– = 1.75% sx = 5.36% y– = 9.07%
sy = 15.35% r = 0.596
(a) Find the equation of the least-squares line for predicting
full-year change from January change. Show your work.
(b) Suppose that the percent change in a particular January was 2 standard deviations above average. Predict
the percent change for the entire year, without using
the least-squares line. Show your work.
12/2/13 11:15 AM
Overview
What Is Statistics?
Does listening to music while studying help or hinder learning? If an athlete fails a drug test, how sure
can we be that she took a banned substance? Does having a pet help people live longer? How well do SAT
scores predict college success? Do most people recycle? Which of two diets will help obese children lose
more weight and keep it off? Should a poker player go “all in” with pocket aces? Can a new drug help
people quit smoking? How strong is the evidence for global warming?
These are just a few of the questions that statistics can help answer. But what is statistics? And why
should you study it?
Statistics Is the Science of Learning from Data
Data are usually numbers, but they are not “just numbers.” Data are numbers with a context. The number
10.5, for example, carries no information by itself. But if we hear that a family friend’s new baby weighed 10.5
pounds at birth, we congratulate her on the healthy size of the child. The context engages our knowledge
about the world and allows us to make judgments. We know that a
baby weighing 10.5 pounds is quite large, and that a human baby is
unlikely to weigh 10.5 ounces or 10.5 kilograms. The context makes
the number meaningful.
In your lifetime, you will be bombarded with data and statistical information. Poll results, television ratings, music sales,
gas prices, unemployment rates, medical study outcomes, and
standardized test scores are discussed daily in the media. Using
data effectively is a large and growing part of most professions. A
solid understanding of statistics will enable you to make sound,
data-based decisions in your career and everyday life.
Data Beat Personal Experiences
It is tempting to base conclusions on your own experiences or the experiences of those you know. But our
experiences may not be typical. In fact, the incidents that stick in our memory are often the unusual ones.
Do cell phones cause brain cancer?
Italian businessman Innocente Marcolini developed a brain tumor at age 60. He also talked on a cellular
phone up to 6 hours per day for 12 years as part of his job. Mr. Marcolini’s physician suggested that the
brain tumor may have been caused by cell-phone use. So Mr. Marcolini decided to file suit in
the Italian court system. A court ruled in his favor in October 2012.
Several statistical studies have investigated the link between cell-phone use and brain
cancer. One of the largest was conducted by the Danish Cancer Society. Over 350,000 residents of Denmark were included in the study. Researchers compared the brain-cancer rate for
the cell-phone users with the rate in the general population. The result: no statistical difference in brain-cancer rates.1 In fact, most studies have produced similar conclusions. In spite
of the evidence, many people (like Mr. Marcolini) are still convinced that cell phones can
cause brain cancer.
In the public’s mind, the compelling story wins every time. A statistically literate person
knows better. Data are more reliable than personal experiences because they systematically
describe an overall picture rather than focus on a few incidents.
xxi
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Where the Data Come from Matters
Are you kidding me?
The famous advice columnist Ann Landers once asked her readers, “If you had it to do over
again, would you have children?” A few weeks later, her column was headlined “70% OF
PARENTS SAY KIDS NOT WORTH IT.” Indeed, 70% of the nearly 10,000 parents who
wrote in said they would not have children if they could make the choice again. Do you
believe that 70% of all parents regret having children?
You shouldn’t. The people who took the trouble to write Ann Landers are not representative of all parents. Their letters showed that many of them were angry with their children.
All we know from these data is that there are some unhappy parents out there. A statistically
designed poll, unlike Ann Landers’s appeal, targets specific people chosen in a way that gives
all parents the same chance to be asked. Such a poll showed that 91% of parents would have
children again.
Where data come from matters a lot. If you are careless about how you get your data, you
may ­announce 70% “No” when the truth is close to 90% “Yes.”
Who talks more—women or men?
According to Louann Brizendine, author of The Female Brain, women say nearly three times as many
words per day as men. Skeptical researchers devised a study to test this claim. They used electronic devices
to record the talking patterns of 396 university students from Texas, Arizona, and Mexico. The device was
programmed to record 30 seconds of sound every 12.5 minutes without the carrier’s knowledge. What
were the results?
According to a published report of the study in Scientific American, “Men showed a slightly wider
variability in words uttered. . . . But in the end, the sexes came out just about even in the daily averages:
women at 16,215 words and men at 15,669.”2 When asked where she got her figures, Brizendine admitted
that she used unreliable sources.3
The most important information about any statistical study is how the data were produced. Only
carefully designed studies produce results that can be trusted.
Always Plot Your Data
Yogi Berra, a famous New York Yankees baseball player known for his unusual quotes, had this to say:
“You can observe a lot just by watching.” That’s a motto for learning from data. A carefully chosen graph
is often more instructive than a bunch of numbers.
Do people live longer in wealthier countries?
The Gapminder Web site, www.gapminder.org, provides loads of data on the health and well-being of the
world’s inhabitants. The graph on the next pages displays some data from Gapminder.4 The individual
points represent all the world’s nations for which data are available. Each point shows the income per
person and life expectancy in years for one country.
We expect people in richer countries to live longer. The overall pattern of the graph does show
this, but the relationship has an interesting shape. Life expectancy rises very quickly as personal income
increases and then levels off. People in very rich countries like the United States live no longer than
people in poorer but not extremely poor nations. In some less wealthy countries, people live longer
than in the United States. Several other nations stand out in the graph. What’s special about each of
these countries?
xxii
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11/20/13 7:44 PM
80
Life expectancy
United States
Qatar
America
70
East Asia & Pacific
Europe & Central Asia
Gabon
Middle East & North Africa
60
South Asia
South Africa
Sub-Saharan Africa
Equatorial Guinea
50
Graph of the life
expectancy of people
in many nations against
each nation’s income
per person in 2012.
00
,0
90
00
,0
80
00
,0
70
00
,0
60
00
,0
50
00
,0
40
00
,0
30
00
,0
20
10
,0
00
Botswana
Income per person in 2012
Variation Is Everywhere
Individuals vary. Repeated measurements on the same individual vary. Chance outcomes—like spins of
a roulette wheel or tosses of a coin—vary. Almost everything varies over time. Statistics provides tools for
understanding variation.
Have most students cheated on a test?
Researchers from the Josephson Institute were determined to find out. So they surveyed about 23,000
students from 100 randomly selected schools (both public and private) nationwide. The question they
asked was “How many times have you cheated during a test at school in the past year?” Fifty-one percent
said they had cheated at least once.5
If the researchers had asked the same question of all high school students, would exactly 51% have
answered “Yes”? Probably not. If the Josephson Institute had selected a different sample of about 23,000
students to respond to the survey, they would probably have gotten a different estimate. Variation is everywhere!
Fortunately, statistics provides a description of how the sample results will vary in relation to the actual population percent. Based on the sampling method that this study used, we can say that the estimate
of 51% is very likely to be within 1% of the true population value. That is, we can be quite confident that
between 50% and 52% of all high school students would say that they have cheated on a test.
Because variation is everywhere, conclusions are uncertain. Statistics gives us a language for talking about uncertainty that is understood by statistically literate people everywhere.
xxiii
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Chapter
1
Introduction
Data Analysis:
Making Sense of Data
2
Section 1.1
Analyzing Categorical Data
7
Section 1.2
Displaying Quantitative
Data with Graphs
25
Section 1.3
Describing Quantitative
Data with Numbers
48
Free Response
AP® Problem, YAY!
74
Chapter 1 Review
74
Chapter 1 Review Exercises
76
Chapter 1 AP® Statistics
Practice Test
78
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11/13/13 1:06 PM
Exploring Data
case study
Do Pets or Friends Help Reduce Stress?
If you are a dog lover, having your dog with you may reduce your stress level. Does having a friend with
you reduce stress? To examine the effect of pets and friends in stressful situations, researchers recruited 45
women who said they were dog lovers. Fifteen women were assigned at random to each of three groups:
to do a stressful task alone, with a good friend present, or with their dogs present. The stressful task was
to count backward by 13s or 17s. The woman’s average heart rate during the task was one measure of the
effect of stress. The table below shows the data.1
Average heart rates during stress with a pet (P), with a friend (F), and for the control group (C)
Group
Rate
Group
Rate
Group
Rate
Group
Rate
P
69.169
P
68.862
C
84.738
C
75.477
F
99.692
C
87.231
C
84.877
C
62.646
P
70.169
P
64.169
P
58.692
P
70.077
C
80.369
C
91.754
P
79.662
F
88.015
C
87.446
C
87.785
P
69.231
F
81.600
P
75.985
F
91.354
C
73.277
F
86.985
F
83.400
F
100.877
C
84.523
F
92.492
F
102.154
C
77.800
C
70.877
P
72.262
P
86.446
P
97.538
F
89.815
P
65.446
F
80.277
P
85.000
F
98.200
C
90.015
F
101.062
F
76.908
C
99.046
F
97.046
P
69.538
Based on the data, does it appear that the presence of a pet
or friend reduces heart rate during a stressful task? In this
chapter, you’ll develop the tools to help answer this question.
1
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2
CHAPTER 1
E x p l o r i n g Data
Introduction
What You Will Learn
•
Data Analysis: Making Sense
of Data
By the end of the section, you should be able to:
Identify the individuals and variables in a set of data.
•
Classify variables as categorical or quantitative.
Statistics is the science of data. The volume of data available to us is overwhelming. For example, the Census Bureau’s American Community Survey collects
data from 3,000,000 housing units each year. Astronomers work with data on tens
of millions of galaxies. The checkout scanners at Walmart’s 10,000 stores in 27
countries record hundreds of millions of transactions every week.
In all these cases, the data are trying to tell us a story—about U.S. households,
objects in space, or Walmart shoppers. To hear what the data are saying, we need
to help them speak by organizing, displaying, summarizing, and asking questions.
That’s data analysis.
Individuals and Variables
Any set of data contains information about some group of individuals. The characteristics we measure on each individual are called variables.
Definition: Individuals and variables
Individuals are the objects described by a set of data. Individuals may be people,
animals, or things.
A variable is any characteristic of an individual. A variable can take different values
for different individuals.
A high school’s student data base, for example, includes data about every currently enrolled student. The students are the individuals described by the data
set. For each individual, the data contain the values of variables such as age,
gender, grade point average, homeroom, and grade level. In practice, any set of
data is accompanied by background information that helps us understand the
data. When you first meet a new data set, ask yourself the following questions:
1. Who are the individuals described by the data? How many individuals are there?
2. What are the variables? In what units are the variables recorded? Weights, for example, might be recorded in grams, pounds, thousands of pounds, or kilograms.
We could follow a newspaper reporter’s lead and extend our list of questions
to include Why, When, Where, and How were the data produced? For now, we’ll
focus on the first two questions.
Some variables, like gender and grade level, assign labels to individuals that place
them into categories. Others, like age and grade point average (GPA), take numerical values for which we can do arithmetic. It makes sense to give an average GPA for
a group of students, but it doesn’t make sense to give an “average” gender.
Starnes-Yates5e_c01_xxiv-081hr3.indd 2
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3
Introduction Data Analysis: Making Sense of Data
Definition: Categorical variable and quantitative variable
A categorical variable places an individual into one of several groups or categories.
A quantitative variable takes numerical values for which it makes sense to find an
average.
EXAMPLE
c
Not every variable that takes number values is quantitative. Zip code is autio
one ­example. Although zip codes are numbers, it doesn’t make sense to
talk about the average zip code. In fact, zip codes place individuals (people or dwellings) into categories based on location. Some variables—such as gender, race, and o­ ccupation—are categorical by nature. Other categorical variables
are created by grouping values of a quantitative variable into classes. For instance,
we could classify people in a data set by age: 0–9, 10–19, 20–29, and so on.
The proper method of analysis for a variable depends on whether it is categorical or quantitative. As a result, it is important to be able to distinguish these two
types of variables. The type of data determines what kinds of graphs and which
numerical summaries are appropriate.
!
n
AP® EXAM TIP ​If you learn
to distinguish categorical from
quantitative variables now, it
will pay big rewards later. You
will be expected to analyze
categorical and quantitative
variables correctly on the AP®
exam.
Census at School
Data, individuals, and variables
CensusAtSchool is an international project that collects data about primary and
secondary school students using surveys. Hundreds of thousands of students from
Australia, Canada, New Zealand, South Africa, and the United Kingdom have
taken part in the project since 2000. Data from the surveys are available at the
project’s Web site (www.censusatschool.com). We used the site’s “Random Data
Selector” to choose 10 Canadian students who completed the survey in a recent
year. The table below displays the data.
There is at least one suspicious
value in the data table. We doubt
that the girl who is 166 cm tall really
has a wrist circumference of 65 mm
(about 2.6 inches). Always look to be
sure the values make sense!
Starnes-Yates5e_c01_xxiv-081hr3.indd 3
Language
spoken
Handed
Height
(cm)
Male
1
Right
175
180
In person
Ontario
Female
1
Right
162.5
160
In person
Alberta
Male
1
Right
178
174
Facebook
Ontario
Male
2
Right
169
160
Cell phone
Ontario
Female
2
Right
166
65
In person
Nunavut
Male
1
Right
168.5
160
Text messaging
Ontario
Female
1
Right
166
165
Cell phone
Ontario
Male
4
Left
157.5
147
Text Messaging
Ontario
Female
2
Right
150.5
187
Text Messaging
Ontario
Female
1
Right
171
180
Text Messaging
Province
Gender
Saskatchewan
Wrist
Preferred
circum. (mm) communication
11/13/13 1:06 PM
4
CHAPTER 1
E x p l o r i n g Data
Problem:
(a) Who are the individuals in this data set?
(b) What variables were measured? Identify each as categorical or quantitative.
(c) Describe the individual in the highlighted row.
We’ll see in Chapter 4 why choosing
at random, as we did in this
example, is a good idea.
To make life simpler, we sometimes
refer to “categorical data” or
“quantitative data” instead of
identifying the variable as categorical
or quantitative.
Solution:
(a) The individuals are the 10 randomly selected Canadian students who participated in the
CensusAtSchool survey.
(b) The seven variables measured are the province where the student lives (categorical), gender
(categorical), number of languages spoken (quantitative), dominant hand (categorical), height (quantitative), wrist circumference (quantitative), and preferred communication method (categorical).
(c) ​This student lives in Ontario, is male, speaks four languages, is left-handed, is 157.5 cm tall
(about 62 inches), has a wrist circumference of 147 mm (about 5.8 inches), and prefers to communicate via text messaging.
For Practice Try Exercise 3
Most data tables follow the format shown in the example—each row is an individual, and each column is a variable. Sometimes the individuals are called cases.
A variable generally takes values that vary (hence the name “variable”!).
Categorical variables sometimes have similar counts in each category and sometimes don’t. For instance, we might have expected similar numbers of males
and ­females in the CensusAtSchool data set. But we aren’t surprised to see that
most students are right-handed. Quantitative variables may take values that are
very close ­together or values that are quite spread out. We call the pattern of variation of a variable its distribution.
Definition: Distribution
The distribution of a variable tells us what values the variable takes and how often
it takes these values.
Section 1.1 begins by looking at how to describe the distribution of a single categorical variable and then examines relationships between categorical variables.
Sections 1.2 and 1.3 and all of Chapter 2 focus on describing the distribution of
a quantitative variable. Chapter 3 investigates relationships between two quantitative variables. In each case, we begin with graphical displays, then add numerical
summaries for a more complete description.
How to Explore Data
•
•
Begin by examining each variable by itself. Then move on to study relationships among the variables.
Start with a graph or graphs. Then add numerical summaries.
Check Your Understanding
Jake is a car buff who wants to find out more about the vehicles that students at his
school drive. He gets permission to go to the student parking lot and record some data.
Later, he does some research about each model of car on the Internet. Finally, Jake
Starnes-Yates5e_c01_xxiv-081hr3.indd 4
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Introduction Data Analysis: Making Sense of Data
5
makes a spreadsheet that includes each car’s model, year, color, number of cylinders,
gas mileage, weight, and whether it has a navigation system.
1. Who are the individuals in Jake’s study?
2.​What variables did Jake measure? Identify each as categorical or quantitative.
From Data Analysis to Inference
Sometimes, we’re interested in drawing conclusions that go beyond the data at
hand. That’s the idea of inference. In the CensusAtSchool example, 9 of the 10
randomly selected Canadian students are right-handed. That’s 90% of the sample.
Can we conclude that 90% of the population of Canadian students who participated in CensusAtSchool are right-handed? No.
If another random sample of 10 students was selected, the percent who are
right-handed might not be exactly 90%. Can we at least say that the actual population value is “close” to 90%? That depends on what we mean by “close.”
The following Activity gives you an idea of how statistical inference works.
ACTIVITY
MATERIALS:
Bag with 25 beads (15 of
one color and 10 of another)
or 25 identical slips of paper
(15 labeled “M” and 10
labeled “F”) for each student
or pair of students
Hiring discrimination—it just won’t fly!
An airline has just finished training 25 pilots—15 male and 10 female—to ­become
captains. Unfortunately, only eight captain positions are available right now. Airline managers announce that they will use a lottery to determine which pilots will
fill the available positions. The names of all 25 pilots will be written on identical
slips of paper. The slips will be placed in a hat, mixed thoroughly, and drawn out
one at a time until all eight captains have been identified.
A day later, managers announce the results of the lottery. Of the 8 captains
chosen, 5 are female and 3 are male. Some of the male pilots who weren’t selected
suspect that the lottery was not carried out fairly. One of these pilots asks your
statistics class for advice about whether to file a grievance with the pilots’ union.
The key question in this possible discrimination case seems to
be: Is it plausible (believable) that these results happened just by
chance? To find out, you and your classmates will simulate the
lottery process that airline managers said they used.
1. Mix the beads/slips thoroughly. Without looking, remove
8 beads/slips from the bag. Count the number of female pilots
selected. Then return the beads/slips to the bag.
2. Your teacher will draw and label a number line for a class dotplot. On the graph, plot the number of females you got in Step 1.
3. Repeat Steps 1 and 2 if needed to get a total of at least 40
simulated lottery results for your class.
4. Discuss the results with your classmates. Does it seem believable that airline
managers carried out a fair lottery? What advice would you give the male pilot
who contacted you?
5. Would your advice change if the lottery had chosen 6 female (and 2 male)
pilots? What about 7 female pilots? Explain.
Starnes-Yates5e_c01_xxiv-081hr3.indd 5
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6
CHAPTER 1
E x p l o r i n g Data
Our ability to do inference is determined by how the data are produced.
Chapter 4 discusses the two main methods of data production—sampling and
experiments—and the types of conclusions that can be drawn from each. As the
Activity illustrates, the logic of inference rests on asking, “What are the chances?”
Probability, the study of chance behavior, is the topic of Chapters 5 through 7.
We’ll introduce the most common inference techniques in Chapters 8 through 12.
Introduction
Summary
•
•
•
Introduction
1.
2.
3.
pg 3
A data set contains information about a number of individuals. Individuals
may be people, animals, or things. For each individual, the data give values
for one or more variables. A variable describes some characteristic of an individual, such as a person’s height, gender, or salary.
Some variables are categorical and others are quantitative. A categorical variable assigns a label that places each individual into one of several groups, such as
male or female. A quantitative variable has numerical values that measure some
characteristic of each individual, such as height in centimeters or salary in dollars.
The distribution of a variable describes what values the variable takes and
how often it takes them.
Exercises
Protecting wood How can we help wood surfaces
resist weathering, especially when restoring historic
wooden buildings? In a study of this question, researchers prepared wooden panels and then exposed
them to the weather. Here are some of the variables
recorded: type of wood (yellow poplar, pine, cedar);
type of water repellent (solvent-based, water-based);
paint thickness (millimeters); paint color (white, gray,
light blue); weathering time (months). Identify each
variable as categorical or quantitative.
Medical study variables Data from a medical study
contain values of many variables for each of the people
who were the subjects of the study. Here are some of
the variables recorded: gender (female or male); age
(years); race (Asian, black, white, or other); smoker
(yes or no); systolic blood pressure (millimeters of mercury); level of calcium in the blood (micrograms per
milliliter). Identify each as categorical or quantitative.
A class survey Here is a small part of the data set that
describes the students in an AP® Statistics class. The
data come from anonymous responses to a questionnaire filled out on the first day of class.
Starnes-Yates5e_c01_xxiv-081hr3.indd 6
The solutions to all exercises numbered in red are
found in the Solutions Appendix, starting on page S-1.
Homework
time (min)
Favorite
music
Pocket
change
(cents)
Gender
Hand
Height
(in.)
F
L
65
200
Hip-hop
50
M
L
72
30
Country
35
M
R
62
95
Rock
35
F
L
64
120
Alternative
0
M
R
63
220
Hip-hop
0
F
R
58
60
F
R
67
150
Alternative
Rock
76
215
(a) What individuals does this data set describe?
(b) What variables were measured? Identify each as categorical or quantitative.
(c) Describe the individual in the highlighted row.
4.
Coaster craze Many people like to ride roller coasters. Amusement parks try to increase attendance by
building exciting new coasters. The following table
displays data on several roller coasters that were
opened in a recent year.2
11/13/13 1:07 PM
7
Section 1.1 Analyzing Categorical Data
Roller coaster
Type
Height
(ft)
Design
Speed
(mph)
Duration
(s)
Weight
(lb)
Age
(yr)
Travel
to work
(min)
Wild Mouse
Steel
49.3
Sit down
28
70
Terminator
Wood
95
Sit down
50.1
Manta
Steel
140
Flying
56
Gender
Income
last
year ($)
180
187
66
0
Ninth grade
1
24,000
155
158
66
54
n/a
High school grad
2
0
10
Assoc. degree
2
11,900
School
Prowler
Wood
102.3
Sit down
51.2
150
176
Diamondback
Steel
230
Sit down
80
180
339
37
10
Assoc. degree
1
6000
91
27
10
Some college
2
30,000
(a)​What individuals does this data set describe?
155
18
n/a
High school grad
2
0
(b) What variables were measured? Identify each as categorical or quantitative.
213
38
15
Master’s degree
2
125,000
(c)​Describe the individual in the highlighted row.
194
40
0
High school grad
1
800
221
18
20
High school grad
1
2500
193
11
n/a
Fifth grade
1
0
5.
Ranking colleges Popular magazines rank colleges
and universities on their “academic quality” in serving undergraduate students. Describe two categorical
variables and two quantitative variables that you
might record for each institution.
6. Students and TV You are preparing to study the
television-viewing habits of high school students. Describe two categorical variables and two quantitative
variables that you might record for each student.
Multiple choice: Select the best answer.
Exercises 7 and 8 refer to the following setting. At the
Census Bureau Web site www.census.gov, you can view
detailed data collected by the American Community
Survey. The following table includes data for 10 people
chosen at random from the more than 1 million people
in households contacted by the survey. “School” gives the
highest level of education completed.
1.1
What You Will Learn
•
•
•
7.
The individuals in this data set are
(a)​households.
(b)​people.
(c)​adults.
(d) ​120 variables.
(e) ​columns.
8.
This data set contains
(a)​7 variables, 2 of which are categorical.
(b)​7 variables, 1 of which is categorical.
(c) 6 variables, 2 of which are categorical.
(d) 6 variables, 1 of which is categorical.
(e) None of these.
Analyzing Categorical Data
By the end of the section, you should be able to:
Display categorical data with a bar graph. Decide if it
would be appropriate to make a pie chart.
Identify what makes some graphs of categorical data
deceptive.
Calculate and display the marginal distribution of a
categorical variable from a two-way table.
•
•
Calculate and display the conditional distribution of a
categorical variable for a particular value of the other
categorical variable in a two-way table.
Describe the association between two categorical variables by comparing appropriate conditional
distributions.
The values of a categorical variable are labels for the categories, such as “male”
and “female.” The distribution of a categorical variable lists the categories and
gives either the count or the percent of individuals who fall within each category.
Here’s an example.
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8
CHAPTER 1
EXAMPLE
E x p l o r i n g Data
Radio Station Formats
Distribution of a categorical variable
The radio audience rating service Arbitron places U.S radio stations into categories
that describe the kinds of programs they broadcast. Here are two different tables
showing the distribution of station formats in a recent year:3
Frequency table
Format
Relative frequency table
Count of stations
Format
Percent of stations
Adult contemporary
1556
Adult contemporary
Adult standards
1196
Adult standards
8.6
Contemporary hit
4.1
Contemporary hit
569
11.2
Country
2066
Country
14.9
News/Talk/Information
2179
News/Talk/Information
15.7
Oldies
1060
Oldies
Religious
2014
Religious
7.7
14.6
Rock
869
Rock
6.3
Spanish language
750
Spanish language
5.4
Other formats
Total
1579
13,838
Other formats
11.4
Total
99.9
In this case, the individuals are the radio stations and the variable being measured
is the kind of programming that each station broadcasts. The table on the left,
which we call a frequency table, displays the counts (frequencies) of stations in
each format category. On the right, we see a relative frequency table of the data
that shows the percents (relative frequencies) of stations in each format category.
It’s a good idea to check data for consistency. The counts should add to 13,838,
the total number of stations. They do. The percents should add to 100%. In
fact, they add to 99.9%. What happened? Each percent is rounded to the nearest tenth. The exact percents would add to 100, but the rounded percents only
come close. This is roundoff error. Roundoff errors don’t point to mistakes in
our work, just to the effect of rounding off results.
Bar Graphs and Pie Charts
Columns of numbers take time to read. You can use a pie chart or a bar graph to
display the distribution of a categorical variable more vividly. Figure 1.1 illustrates
both displays for the distribution of radio stations by format.
Pie charts show the distribution of a categorical variable as a “pie” whose slices
are sized by the counts or percents for the categories. A pie chart must include
all the categories that make up a whole. In the radio station example, we needed
the “Other formats” category to complete the whole (all radio stations) and allow
us to make a pie chart. Use a pie chart only when you want to emphasize each
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9
Section 1.1 Analyzing Categorical Data
This bar has height 14.9%
because 14.9% of the
radio stations have a
“Country” format.
This slice occupies 14.9% of the pie
because 14.9% of the radio stations
have a “Country” format.
Contemporary hit
14
Percent of stations
Country
16
Adult
standards
Adult
contemporary
News/Talk
Other
Oldies
12
10
8
6
4
2
(a)
Spanish
(b)
r
th
e
ish
O
an
oc
k
R
Rock
Sp
on
t
A
dS
ta
n
C
on
H
it
C
ou
nt
N
r
ew y
s/T
al
k
O
ld
ie
R
s
el
ig
io
us
0
A
dC
Religious
Radio station format
FIGURE 1.1 ​(a) Pie chart and (b) bar graph of U.S. radio stations by format.
category’s relation to the whole. Pie charts are awkward to make by hand, but
technology will do the job for you.
Bar graphs are also called bar charts.
EXAMPLE
Bar graphs represent each category as a bar. The bar heights show the category
counts or percents. Bar graphs are easier to make than pie charts and are also
easier to read. To convince yourself, try to use the pie chart in Figure 1.1 to estimate the percent of radio stations that have an “Oldies” format. Now look at the
bar graph—it’s easy to see that the answer is about 8%.
Bar graphs are also more flexible than pie charts. Both graphs can display the
distribution of a categorical variable, but a bar graph can also compare any set of
quantities that are measured in the same units.
Who Owns an MP3 Player?
Choosing the best graph to display the data
Portable MP3 music players, such as the Apple iPod, are popular—but not equally
popular with people of all ages. Here are the percents of people in various age
groups who own a portable MP3 player, according to an Arbitron survey of 1112
randomly selected people.4
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10
CHAPTER 1
E x p l o r i n g Data
Age group (years)
Percent owning an MP3 player
12 to 17
54
18 to 24
30
25 to 34
30
35 to 54
13
55 and older
5
Problem:
50
(a) Make a well-labeled bar graph to display the data. Describe what you see.
(b) ​Would it be appropriate to make a pie chart for these data? Explain.
40
Solution:
30
(a) We start by labeling the axes: age group goes on the horizontal axis, and percent
who own an MP3 player goes on the vertical axis. For the vertical scale, which is
measured in percents, we’ll start at 0 and go up to 60, with tick marks for every 10.
Then for each age category, we draw a bar with height corresponding to the percent of
survey respondents who said they have an MP3 player. Figure 1.2 shows the completed bar graph. It appears that MP3 players are more popular among young people
and that their popularity generally decreases as the age category increases.
(b) ​Making a pie chart to display these data is not appropriate because each percent
in the table refers to a different age group, not to parts of a single whole.
% who own mp3 player
60
20
10
0
12–17 18–24 25–34 35–54
55+
Age group (years)
FIGURE 1.2 ​Bar graph comparing the percents of several age groups who own portable
MP3 players.
For Practice Try Exercise
15
Graphs: Good and Bad
Bar graphs compare several quantities by comparing the heights of bars that represent the quantities. Our eyes, however, react to the area of the bars as well as to
their height. When all bars have the same width, the area (width × height) varies
in proportion to the height, and our eyes receive the right impression. When you
draw a bar graph, make the bars equally wide.
Artistically speaking, bar graphs are a bit dull. It is tempting to replace the bars
with pictures for greater eye appeal. Don’t do it! The following example shows why.
EXAMPLE
Who Buys iMacs?
Beware the pictograph!
When Apple, Inc., introduced the iMac, the company wanted to know whether
this new computer was expanding Apple’s market share. Was the iMac mainly being bought by previous Macintosh owners, or was it being purchased by first-time
computer buyers and by previous PC users who were switching over? To find out,
Apple hired a firm to conduct a survey of 500 iMac customers. Each customer was
categorized as a new computer purchaser, a previous PC owner, or a previous Macintosh owner. The table summarizes the survey results.5
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11
Section 1.1 Analyzing Categorical Data
Previous ownership
300
200
Count
Percent (%)
None
85
17.0
PC
60
12.0
Macintosh
355
71.0
Total
500
100.0
Problem:
100
None
Windows
(a) Here’s a clever graph of the data that uses pictures instead of the more
traditional bars. How is this graph misleading?
(b) Two possible bar graphs of the data are shown below. Which one could be
considered deceptive? Why?
Macintosh
Percent
Previous computer
80
80
70
70
60
60
50
Percent
Number of buyers
400
40
30
40
30
20
20
10
0
50
None
PC
Macintosh
Previous computer
10
None
PC
Macintosh
Previous computer
Solution:
(a) Although the heights of the pictures are accurate, our eyes respond to the area of the pictures. The
pictograph makes it seem like the percent of iMac buyers who are former Mac owners is at least ten times
higher than either of the other two categories, which isn’t the case.
(b) The bar graph on the right is misleading. By starting the vertical scale at 10 instead of 0, it looks
like the percent of iMac buyers who previously owned a PC is less than half the percent who are first-time
computer buyers. We get a distorted impression of the relative percents in the three categories.
17
autio
!
n
There are two important lessons to be learned from this example:
(1) beware the pictograph, and (2) watch those scales.
c
For Practice Try Exercise
Two-Way Tables and Marginal Distributions
We have learned some techniques for analyzing the distribution of a single categorical variable. What do we do when a data set involves two categorical variables?
We begin by examining the counts or percents in various categories for one of the
variables. Here’s an example to show what we mean.
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12
CHAPTER 1
EXAMPLE
E x p l o r i n g Data
I’m Gonna Be Rich!
Two-way tables
A survey of 4826 randomly selected young adults (aged 19 to 25) asked, “What do
you think the chances are you will have much more than a middle-class income
at age 30?” The table below shows the responses.6
Young adults by gender and chance of getting rich
Gender
Male
Total
96
98
194
Some chance but probably not
426
286
712
A 50-50 chance
696
720
1416
A good chance
663
758
1421
Almost certain
486
597
1083
2367
2459
4826
Opinion
Almost no chance
Total
Female
This is a two-way table because it describes two categorical variables, gender and
opinion about becoming rich. Opinion is the row variable because each row in the
table describes young adults who held one of the five opinions about their chances.
Because the opinions have a natural order from “Almost no chance” to “Almost
certain,” the rows are also in this order. Gender is the column variable. The entries
in the table are the counts of individuals in each opinion-by-gender class.
How can we best grasp the information contained in the two-way table above?
First, look at the distribution of each variable separately. The distribution of a categorical variable says how often each outcome occurred. The “Total” column
at the right of the table contains the totals for each of the rows. These row totals
give the distribution of opinions about becoming rich in the entire group of 4826
young adults: 194 thought that they had almost no chance, 712 thought they had
just some chance, and so on. (If the row and column totals are missing, the first
thing to do in studying a two-way table is to calculate them.) The distributions of
opinion alone and gender alone are called marginal distributions because they
appear at the right and bottom margins of the two-way table.
Definition: Marginal distribution
The marginal distribution of one of the categorical variables in a two-way table of
counts is the distribution of values of that variable among all individuals described by
the table.
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Section 1.1 Analyzing Categorical Data
13
Percents are often more informative than counts, especially when we are comparing groups of different sizes. We can display the marginal distribution of opinions in percents by dividing each row total by the table total and converting to a
percent. For instance, the percent of these young adults who think they are almost
certain to be rich by age 30 is
almost certain total 1083
=
= 0.224 = 22.4%
table total
4826
EXAMPLE
I’m Gonna Be Rich!
Examining a marginal distribution
Problem:
(a) ​Use the data in the two-way table to calculate the marginal distribution (in percents) of
opinions.
(b) ​Make a graph to display the marginal distribution. Describe what you see.
Solution:
(a) ​We can do four more calculations like the one shown above to obtain the marginal distribution of
opinions in percents. Here is the complete distribution.
Response
Percent
Almost no chance
194
= 4.0%
4826
30
Some chance
712
= 14.8%
4826
20
A 50–50 chance
1416
= 29.3%
4826
10
A good chance
1421
= 29.4%
4826
Almost certain
1083
= 22.4%
4826
Percent
Chance of being wealthy by age 30
0
Almost Some 50–50 Good Almost
none chance chance chance certain
Survey response
FIGURE 1.3 ​Bar graph showing the marginal
distribution of opinion about chance of being
rich by age 30.
(b) ​Figure 1.3 is a bar graph of the distribution of opinion among these young adults.
It seems that many young adults are optimistic about their future income. Over 50%
of those who responded to the survey felt that they had “a good chance” or were
“almost certain” to be rich by age 30.
For Practice Try Exercise
19
Each marginal distribution from a two-way table is a distribution for a single
categorical variable. As we saw earlier, we can use a bar graph or a pie chart to
display such a distribution.
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14
CHAPTER 1
E x p l o r i n g Data
Check Your Understanding
Country
Superpower
U.K.
U.S.
Fly
54
45
Freeze time
52
44
Invisibility
30
37
Superstrength
20
23
Telepathy
44
66
A random sample of 415 children aged 9 to 17 from the United Kingdom and the United
States who completed a CensusAtSchool survey in a recent year was selected. Each student’s
country of origin was recorded along with which superpower they would most like to have:
the ability to fly, ability to freeze time, invisibility, superstrength, or telepathy (ability to read
minds). The data are summarized in the table.7
1. Use the two-way table to calculate the marginal distribution (in percents) of
superpower preferences.
2. Make a graph to display the marginal distribution. Describe what you see.
Relationships between Categorical
Variables: Conditional Distributions
The two-way table contains much more information than the two marginal distributions of opinion alone and gender alone. Marginal distributions tell us nothing
about the relationship between two variables. To describe a relationship between
two categorical variables, we must calculate some well-chosen percents from the
counts given in the body of the table.
Young adults by gender and chance of getting rich
Gender
Female
Male
Total
96
98
194
Some chance but probably not
426
286
712
A 50-50 chance
696
720
1416
A good chance
663
758
1421
Almost certain
486
597
1083
2367
2459
4826
Opinion
Almost no chance
Total
Conditional distribution of opinion among women
Response
Percent
Almost no chance
96
= 4.1%
2367
Some chance
426
= 18.0%
2367
A 50-50 chance
696
= 29.4%
2367
A good chance
663
= 28.0%
2367
Almost certain
486
= 20.5%
2367
Starnes-Yates5e_c01_xxiv-081hr3.indd 14
We can study the opinions of women alone by looking only at the
“Female” column in the two-way table. To find the percent of young
women who think they are almost certain to be rich by age 30, divide
the count of such women by the total number of women, the column
total:
women who are almost certain
486
=
= 0.205 = 20.5%
column total
2367
Doing this for all five entries in the “Female” column gives the conditional distribution of opinion among women. See the table in
the margin. We use the term “conditional” because this distribution
describes only young adults who satisfy the condition that they are
female.
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Section 1.1 Analyzing Categorical Data
15
Definition: Conditional distribution
A conditional distribution of a variable describes the values of that variable among
individuals who have a specific value of another variable. There is a separate conditional distribution for each value of the other variable.
Now let’s examine the men’s opinions.
EXAMPLE
I’m Gonna Be Rich!
Calculating a conditional distribution
Problem: Calculate the conditional distribution of opinion among the young men.
Solution: To find the percent of young men who think they are almost certain to be rich by age
30, divide the count of such men by the total number of men, the column total:
men who are almost certain
597
=
= 24.3%
column total
2459
If we do this for all five entries in the “Male” column, we get the conditional distribution shown in the
table.
Conditional distribution of opinion among men
Response
Percent
Almost no chance
98
= 4.0%
2459
Some chance
286
= 11.6%
2459
A 50-50 chance
720
= 29.3%
2459
A good chance
758
= 30.8%
2459
Almost certain
597
= 24.3%
2459
For Practice Try Exercise
21
There are two sets of conditional distributions for any two-way table: one for the
column variable and one for the row variable. So far, we have looked at the conditional distributions of opinion for the two genders. We could also examine the five
conditional distributions of gender, one for each of the five opinions, by looking
separately at the rows in the original two-way table. For instance, the conditional
distribution of gender among those who responded “Almost certain” is
Female
486
= 44.9%
1083
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Male
597
= 55.1%
1083
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16
CHAPTER 1
E x p l o r i n g Data
That is, of the young adults who said they were almost certain to be rich by age 30, 44.9% were female and 55.1% were
male.
Because the variable “gender” has only two categories, comparing the five conditional distributions amounts to comparing
the percents of women among young adults who hold each
opinion. Figure 1.4 makes this comparison in a bar graph. The
bar heights do not add to 100%, because each bar represents a
different group of people.
70
Percent of women
in the opinion group
60
50
40
30
20
10
0
Almost Some A 50–50 A good Almost
none
chance chance chance certain
Opinion
FIGURE 1.4 ​Bar graph comparing the percents of
females among those who hold each opinion about
their chance of being rich by age 30.
THINK
ABOUT IT
Which conditional distributions should we compare? Our goal
all along has been to analyze the relationship between gender and opinion about
chances of becoming rich for these young adults. We started by examining the
conditional distributions of opinion for males and females. Then we looked at the
conditional distributions of gender for each of the five opinion categories. Which
of these two gives us the information we want? Here’s a hint: think about whether
changes in one variable might help explain changes in the other. In this case, it
seems reasonable to think that gender might influence young adults’ opinions
about their chances of getting rich. To see whether the data support this idea, we
should compare the conditional distributions of opinion for women and men.
Software will calculate conditional distributions for you. Most programs allow
you to choose which conditional distributions you want to compute.
1.Technology
Corner
Analyzing two-way tables
Figure 1.5 presents the two conditional distributions of
opinion, for women and for men, and also the marginal
distribution of opinion for all of the young adults. The
distributions agree (up to rounding) with the results in the
last two examples.
FIGURE 1.5 ​Minitab output for the two-way table of young adults
by gender and chance of being rich, along with each entry as a
percent of its column total. The “Female” and “Male” columns give
the conditional distributions of opinion for women and men, and the
“All” column shows the marginal distribution of opinion for all these
young adults.
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Section 1.1 Analyzing Categorical Data
17
Putting It All Together: Relationships
Between Categorical Variables
Now it’s time to complete our analysis of the relationship between gender and
opinion about chances of becoming rich later in life.
Example
Women’s and Men’s Opinions
Conditional distributions and relationships
Problem: Based on the survey data, can we conclude that young men and women differ in their
opinions about the likelihood of future wealth? Give appropriate evidence to support your answer.
solution: We suspect that gender might influence a young adult’s opinion about the chance of getting rich. So we’ll compare the conditional distributions of response for men alone and for women alone.
Response
Chance of being wealthy by age 30
40
Female
Male
Percent
30
20
10
0
Almost Some 50–50 Good Almost
none chance chance chance certain
Opinion
FIGURE 1.6 ​Side-by-side bar graph comparing the
opinions of males and females.
100
Almost certain
Percent
90
80
A good chance
70
A 50–50 chance
60
Some chance
50
Almost none
40
30
20
Percent of Females
Percent of Males
Almost no chance
96
= 4.1%
2367
98
= 4.0%
2459
Some chance
426
= 18.0%
2367
286
= 11.6%
2459
A 50-50 chance
696
= 29.4%
2367
720
= 29.3%
2459
A good chance
663
= 28.0%
2367
758
= 30.8%
2459
Almost certain
486
= 20.5%
2367
597
= 24.3%
2459
We’ll make a side-by-side bar graph to compare the opinions of males and
females. Figure 1.6 displays the completed graph.
Based on the sample data, men seem somewhat more optimistic about
their future income than women. Men were less likely to say that they have
“some chance but probably not” than women (11.6% vs. 18.0%). Men were
more likely to say that they have “a good chance” (30.8% vs. 28.0%) or are
“almost certain” (24.3% vs. 20.5%) to have much more than a middle-class
income by age 30 than women were.
For Practice Try Exercise
25
We could have used a segmented bar graph to compare
the distributions of male and female responses in the previous
example. Figure 1.7 shows the completed graph. Each bar has
five segments—one for each of the opinion categories. It’s fairly
difficult to compare the percents of males and females in each
category because the “middle” segments in the two bars start
at different locations on the vertical axis. The side-by-side bar
graph in Figure 1.6 makes comparison easier.
10
0
Female
Opinion
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Male
FIGURE 1.7 ​Segmented bar graph comparing the opinions of males and females.
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18
CHAPTER 1
E x p l o r i n g Data
Both graphs provide evidence of an association between gender and opinion
about future wealth in this sample of young adults. Men more often rated their
chances of becoming rich in the two highest categories; women said “some chance
but probably not” much more frequently.
Definition: Association
We say that there is an association between two variables if knowing the value of
one variable helps predict the value of the other. If knowing the value of one variable
does not help you predict the value of the other, then there is no association between
the variables.
Can we say that there is an association between gender and opinion in the population of young adults? Making this determination requires formal inference, which
will have to wait a few chapters.
THINK
ABOUT IT
What does “no association” mean? Figure 1.6 (page 17) suggests
an association between gender and opinion about future wealth for young
adults. Knowing that a young adult is male helps us predict his opinion: he is
more likely than a female to say “a good chance” or “almost certain.” What
would the graph look like if there was no association between the two variables? In that case, knowing that a young adult is male would not help us
predict his opinion. He would be no more or less likely than a female to say “a
good chance” or “almost certain” or any of the other possible responses. That
is, the conditional distributions of opinion about becoming rich would be the
same for males and females. The segmented bar graphs for the two genders
would look the same, too.
Check Your Understanding
U.K.
U.S.
Fly
54
45
Freeze time
52
44
Invisibility
30
37
Superstrength
20
23
Telepathy
44
66
1. Find the conditional distributions of superpower preference among students from
the United Kingdom and the United States.
2. Make an appropriate graph to compare the conditional distributions.
3. Is there an association between country of origin and superpower preference? Give
appropriate evidence to support your answer.
Starnes-Yates5e_c01_xxiv-081hr3.indd 18
autio
!
n
There’s one caution that we need to offer: even a strong association between two categorical variables can be influenced by other variables lurking
in the background. The Data Exploration that follows gives you a chance to
explore this idea using a famous (or infamous) data set.
c
Country
Superpower
Let’s complete our analysis of the data on superpower preferences from the previous Check Your Understanding (page 14). Here is the two-way table of counts once
again.
11/13/13 1:07 PM
19
Section 1.1 Analyzing Categorical Data
DATA EXPLORATION A Titanic disaster
In 1912 the luxury liner Titanic, on its first voyage across the Atlantic,
struck an iceberg and sank. Some passengers got off the ship in lifeboats,
but many died. The two-way table below gives information about adult
passengers who lived and who died, by class of travel.
Class of Travel
Survival status
First class
Second class
Third class
197
122
94
167
151
476
Lived
Died
Here’s another table that displays data on survival status by gender and
class of travel.
Class of Travel
First class
Second class
Third class
Female
Male
Female
Male
Female
Male
Lived
140
57
80
14
76
75
Died
4
118
13
154
89
387
Survival status
The movie Titanic, starring Leonardo DiCaprio and Kate Winslet, suggested
the following:
• First-class passengers received special treatment in boarding the lifeboats,
while some other passengers were prevented from doing so (especially thirdclass passengers).
• Women and children boarded the lifeboats first, followed by the men.
1. What do the data tell us about these two suggestions? Give appropriate
graphical and numerical evidence to support your answer.
2. How does gender affect the relationship between class of travel and survival
status? Explain.
Section 1.1
Summary
•
•
•
Starnes-Yates5e_c01_xxiv-081hr3.indd 19
The distribution of a categorical variable lists the categories and gives the
count (frequency) or percent (relative frequency) of individuals that fall
within each category.
Pie charts and bar graphs display the distribution of a categorical variable.
Bar graphs can also compare any set of quantities measured in the same
units. When examining any graph, ask yourself, “What do I see?”
A two-way table of counts organizes data about two categorical variables measured for the same set of individuals. Two-way tables are often used to summarize large amounts of information by grouping outcomes into categories.
11/13/13 1:07 PM
20
CHAPTER 1
E x p l o r i n g Data
•
•
•
The row totals and column totals in a two-way table give the marginal distributions of the two individual variables. It is clearer to present these distributions as percents of the table total. Marginal distributions tell us nothing
about the relationship between the variables.
There are two sets of conditional distributions for a two-way table: the distributions of the row variable for each value of the column variable, and the
distributions of the column variable for each value of the row variable. You
may want to use a side-by-side bar graph (or possibly a segmented bar graph)
to display conditional distributions.
There is an association between two variables if knowing the value of one variable helps predict the value of the other. To see whether there is an association
between two categorical variables, compare an appropriate set of conditional
distributions. Remember that even a strong association between two categorical variables can be influenced by other variables.
1.1 T echnology
Corner
TI-Nspire instructions in ApTI-
1. Analyzing two-way tables
Section 1.1
9.
Exercises
Cool car colors The most popular colors for cars
and light trucks change over time. Silver passed
green in 2000 to become the most popular color
worldwide, then gave way to shades of white in 2007.
Here is the distribution of colors for vehicles sold in
North America in 2011.8
Color
Percent of vehicles
White
23
Black
18
Silver
16
Gray
13
Red
10
Blue
9
Brown/beige
5
Yellow/gold
3
Green
2
(a) What percent of vehicles had colors other than those
listed?
(b) Display these data in a bar graph. Be sure to label
your axes.
Starnes-Yates5e_c01_xxiv-081hr3.indd 20
page 16
(c) Would it be appropriate to make a pie chart of these
data? Explain.
10. Spam Email spam is the curse of the Internet. Here
is a compilation of the most common types of spam:9
Type of spam
Percent
Adult
19
Financial
20
Health
7
Internet
7
Leisure
6
Products
25
Scams
9
Other
??
(a) ​What percent of spam would fall in the “Other”
category?
(b) Display these data in a bar graph. Be sure to label
your axes.
(c) Would it be appropriate to make a pie chart of these
data? Explain.
11/13/13 1:07 PM
21
Section 1.1 Analyzing Categorical Data
11. Birth days Births are not evenly distributed across the
days of the week. Here are the average numbers of babies born on each day of the week in the United States
in a recent year:10
Day
Births
Sunday
7374
Monday
11,704
Tuesday
13,169
Wednesday
13,038
Thursday
13,013
Friday
12,664
Saturday
14. Which major? About 1.6 million first-year students
enroll in colleges and universities each year. What do
they plan to study? The pie chart displays data on the
percents of first-year students who plan to major in several discipline areas.13 About what percent of first-year
students plan to major in business? In social science?
Technical
Other
Arts/
humanities
Biological
sciences
Professional
8459
Business
(a) Present these data in a well-labeled bar graph. Would
it also be correct to make a pie chart?
(b) Suggest some possible reasons why there are fewer
births on weekends.
12. Deaths among young people Among persons aged
15 to 24 years in the United States, the leading causes
of death and number of deaths in a recent year were
as follows: accidents, 12,015; homicide, 4651; suicide,
4559; cancer, 1594; heart disease, 984; congenital
defects, 401.11
Social science
Education
Physical sciences
Engineering
15. Buying music online Young people are more likely
than older folk to buy music online. Here are the
percents of people in several age groups who bought
music online in a recent year:14
pg 9
(a) Make a bar graph to display these data.
Age group
(b) To make a pie chart, you need one additional piece of
information. What is it?
12 to 17 years
24%
18 to 24 years
21%
25 to 34 years
20%
35 to 44 years
16%
45 to 54 years
10%
55 to 64 years
3%
65 years and over
1%
13. Hispanic origins Below is a pie chart prepared by the
Census Bureau to show the origin of the more than
50 million Hispanics in the United States in 2010.12
About what percent of Hispanics are Mexican? Puerto
Rican?
Percent Distribution of Hispanics by Type: 2010
Bought music online
(a) Explain why it is not correct to use a pie chart to
display these data.
(b) Make a bar graph of the data. Be sure to label your axes.
Puerto Rican
Mexican
Cuban
Central
American
South
American
Other
Hispanic
Comment: You see that it is hard to determine numbers from a pie chart. Bar graphs are much easier to
use. (The Census Bureau did include the percents in
its pie chart.)
Starnes-Yates5e_c01_xxiv-081hr3.indd 21
16. The audience for movies Here are data on the
percent of people in several age groups who attended
a movie in the past 12 months:15
Age group
Movie attendance
18 to 24 years
83%
25 to 34 years
73%
35 to 44 years
68%
45 to 54 years
60%
55 to 64 years
47%
65 to 74 years
32%
75 years and over
20%
(a)​Display these data in a bar graph. Describe what
you see.
11/13/13 1:07 PM
E x p l o r i n g Data
Car
Key:
pg 13
Lower
9
43
Starnes-Yates5e_c01_xxiv-081hr3.indd 22
ol
d
25
/g
7
0
w
The same
5
lo
29
10
n
No
20
Percent
15
en
Yes
Higher
20
w
Think quality is
U.S.
Europe
25
re
Buy recycled filters?
30
Ye
l
19. Attitudes toward recycled products Recycling is supposed to save resources. Some people think recycled
products are lower in quality than other products, a
fact that makes recycling less practical. People who
use a recycled product may have different opinions
from those who don’t use it. Here are data on attitudes
toward coffee filters made of recycled paper from a
sample of people who do and don’t buy these filters:16
23. Popular colors—here and there Favorite vehicle
colors may differ among countries. The side-by-side bar
graph shows data on the most popular colors of cars in
a recent year for the United States and Europe. Write a
few sentences comparing the two distributions.
ro
(b) M
​ ake a new graph that isn’t misleading. What do
you conclude about the relationship between eating
oatmeal and cholesterol reduction?
G
(a) ​How is this graph misleading?
/b
Week 1 Week 2 Week 3 Week 4
ge
196
y
198
ed
200
22. Smoking by students and parents Refer to Exercise
20. Calculate three conditional distributions of
students’ smoking behavior: one for each of the three
parental smoking categories. Describe the relationship between the smoking behaviors of students and
their parents in a few sentences.
R
202
ei
204
B
206
400
e
Cholesterol
208
1380
416
ra
210
1823
188
21. Attitudes toward recycled products Exercise 19
gives data on the opinions of people who have and
have not bought coffee filters made from recycled
paper. To see the relationship between opinion and
experience with the product, find the conditional
distributions of opinion (the response variable) for
buyers and nonbuyers. What do you conclude?
er
Representative cholesterol point drop
1168
pg 15
lv
18. Oatmeal and cholesterol Does eating oatmeal
reduce cholesterol? An advertisement included the
following graph as evidence that the answer is “Yes.”
Both
parents
smoke
(a) H
​ ow many students are described in the two-way
table? What percent of these students smoke?
(b) ​Give the marginal distribution (in percents) of parents’
smoking behavior, both in counts and in percents.
Si
(b) ​Make a new graph that isn’t misleading.
One
parent
smokes
Student smokes
rl
(a) ​How is this graph misleading?
Neither
parent
smokes
Student does not smoke
ck
STATE COLLEGE BUS SERVICE
Walk
la
= 1 Cycler
= 2 Cars
= 7 Bus takers
= 2 Walkers
STATE COLLEGE BUS SERVICE
B
STATE COLLEGE BUS SERVICE
pe
a
Bus
te
/
Mode of transport
Cycle
20. Smoking by students and parents Here are data
from a survey conducted at eight high schools on
smoking among students and their parents:17
lu
17. Going to school Students in a high school statistics
class were given data about the main method of transportation to school for a group of 30 students. They
produced the pictograph shown.
pg 10
(a)​How many people does this table describe? How many
of these were buyers of coffee filters made of recycled
paper?
(b) Give the marginal distribution (in percents) of opinion
about the quality of recycled filters. What percent
of the people in the sample think the quality of the
recycled product is the same or higher than the quality
of other filters?
G
(b) W
​ ould it be correct to make a pie chart of these data?
Why or why not?
(c) ​A movie studio wants to know what percent of the total audience for movies is 18 to 24 years old. Explain
why these data do not answer this question.
B
CHAPTER 1
W
hi
22
Color
11/13/13 1:07 PM
23
Section 1.1 Analyzing Categorical Data
24. Comparing car colors Favorite vehicle colors may
differ among types of vehicle. Here are data on the
most popular colors in a recent year for luxury cars
and for SUVs, trucks, and vans.
Color
Luxury cars (%)
Black
22
13
Silver
16
16
White pearl
14
1
Gray
12
13
White
11
25
Blue
7
10
Red
7
11
Yellow/gold
6
1
Green
3
4
Beige/brown
2
6
Low anger
53
110
27
190
No CHD
3057
4621
606
8284
Total
3110
4731
633
8474
SUVs, trucks, vans (%)
Do these data support the study’s conclusion about
the relationship between anger and heart disease?
Give appropriate evidence to support your answer.
Multiple choice: Select the best answer for Exercises 27
to 34.
Exercises 27 to 30 refer to the following setting. The
National Survey of Adolescent Health interviewed several
thousand teens (grades 7 to 12). One question asked was
“What do you think are the chances you will be married
in the next ten years?” Here is a two-way table of the
responses by gender:18
25. Snowmobiles in the park Yellowstone National Park
surveyed a random sample of 1526 winter visitors
to the park. They asked each person whether they
owned, rented, or had never used a snowmobile. Respondents were also asked whether they belonged to
an environmental organization (like the Sierra Club).
The two-way table summarizes the survey responses.
pg 17
Environmental Club
Total
No
Yes
Never used
445
212
657
Snowmobile renter
497
77
574
Snowmobile owner
279
16
295
1221
305
1526
Do these data suggest that there is an association
between environmental club membership and
snowmobile use among visitors to Yellowstone National Park? Give appropriate evidence to support
your answer.
26. Angry people and heart disease People who get angry easily tend to have more heart disease. That’s the
conclusion of a study that followed a random sample
of 12,986 people from three locations for about four
years. All subjects were free of heart disease at the
beginning of the study. The subjects took the Spielberger Trait Anger Scale test, which measures how
prone a person is to sudden anger. Here are data for
the 8474 people in the sample who had normal blood
pressure. CHD stands for “coronary heart disease.”
This includes people who had heart attacks and those
who needed medical treatment for heart disease.
Starnes-Yates5e_c01_xxiv-081hr3.indd 23
Total
High anger
CHD
(a) Make a graph to compare colors by vehicle type.
(b) Write a few sentences describing what you see.
Total
Moderate anger
Female
Male
Almost no chance
119
103
Some chance, but probably not
150
171
A 50-50 chance
447
512
A good chance
735
710
Almost certain
1174
756
27. The percent of females among the respondents was
(a) 2625.
(c) ​about 46%. (e) ​None of these.
(b)​4877.
(d) ​about 54%.
28.
(a)
(b)
(c)
(d)
Your percent from the previous exercise is part of
​the marginal distribution of females.
the marginal distribution of gender.
the marginal distribution of opinion about marriage.
the conditional distribution of gender among adolescents with a given opinion.
(e) ​the conditional distribution of opinion among adolescents of a given gender.
29. What percent of females thought that they were
almost certain to be married in the next ten years?
(a) About 16%
(c) About 40%
(e) About 61%
(b) About 24%
(d) About 45%
30.
(a)
(b)
(c)
Your percent from the previous exercise is part of
the marginal distribution of gender.
the marginal distribution of opinion about marriage.
the conditional distribution of gender among adolescents with a given opinion.
(d) the conditional distribution of opinion among adolescents of a given gender.
(e) the conditional distribution of “Almost certain”
among females.
11/13/13 1:07 PM
24
CHAPTER 1
E x p l o r i n g Data
31. For which of the following would it be inappropriate
to display the data with a single pie chart?
(a) The distribution of car colors for vehicles purchased
in the last month.
(b) The distribution of unemployment percentages for
each of the 50 states.
(c) The distribution of favorite sport for a sample of 30
middle school students.
(d) The distribution of shoe type worn by shoppers at a
local mall.
(e) The distribution of presidential candidate preference
for voters in a state.
(d) The Mavericks won a higher proportion of games
when scoring at least 100 points (43/47) than when
they scored fewer than 100 points (14/35).
(e) The combination of scoring 100 or more points and
winning the game occurred more often (43 times)
than any other combination of outcomes.
34. The following partially complete two-way table shows
the marginal distributions of gender and handedness
for a sample of 100 high school students.
Male
Right
The subjects are not listed in the correct order.
This distribution should be displayed with a pie chart.
The vertical axis should show the percent of students.
The vertical axis should start at 0 rather than 100.
The foreign language bar should be broken up by
language.
33. In the 2010–2011 season, the Dallas Mavericks won
the NBA championship. The two-way table below
displays the relationship between the outcome of
each game in the regular season and whether the
Mavericks scored at least 100 points.
100 or more points
Fewer than 100 points
Total
Win
43
14
57
Loss
4
21
25
Total
47
35
82
Which of the following is the best evidence that there
is an association between the outcome of a game and
whether or not the Mavericks scored at least 100 points?
(a) The Mavericks won 57 games and lost only 25 games.
(b) The Mavericks scored at least 100 points in 47 games
and fewer than 100 points in only 35 games.
(c) The Mavericks won 43 games when scoring at least
100 points and only 14 games when scoring fewer
than 100 points.
Starnes-Yates5e_c01_xxiv-081hr3.indd 24
40
60
100
If there is no association between gender and handedness for the members of the sample, which of the
following is the correct value of x?
20.
30.
36.
45.
Impossible to determine without more information.
ts
ar
e
Fi
n
F
la ore
ng ig
ua n
ge
ng
E
S
st ocia
ud l
ie
s
lis
h
e
nc
ie
Sc
Number of students
Favorite subject
(a)
(b)
(c)
(d)
(e)
10
Total
(a)
(b)
(c)
(d)
(e)
h
at
M
90
x
Left
32. The following bar graph shows the distribution of
favorite subject for a sample of 1000 students. What is
the most serious problem with the graph?
280
260
240
220
200
180
160
140
120
100
Total
Female
35. Marginal distributions aren’t the whole story Here
are the row and column totals for a two-way table with
two rows and two columns:
a
b
50
c
d
50
60
40
100
Find two different sets of counts a, b, c, and d for the
body of the table that give these same totals. This
shows that the relationship between two variables cannot be obtained from the two individual distributions
of the variables.
36. Fuel economy (Introduction) Here is a small part
of a data set that describes the fuel economy (in miles
per gallon) of model year 2012 motor vehicles:
Make and
model
Vehicle type
Transmission Number of City Highway
type
cylinders mpg
mpg
Aston Martin Two-seater
Vantage
Manual
8
14
20
Honda Civic
Hybrid
Subcompact
Automatic
4
44
44
Toyota Prius
Midsize
Automatic
4
51
48
Chevrolet
Impala
Large
Automatic
6
18
30
(a) ​What are the individuals in this data set?
(b) ​What variables were measured? Identify each as categorical or quantitative.
11/13/13 1:07 PM
25
Section 1.2 Displaying Quantitative Data with Graphs
Displaying Quantitative Data
with Graphs
1.2
What You Will Learn
By the end of the section, you should be able to:
•
Make and interpret dotplots and stemplots of quantitative data.
•
•
Describe the overall pattern (shape, center, and spread) of
a distribution and identify any major departures from the
pattern (outliers).
•
•
Identify the shape of a distribution from a graph
as roughly symmetric or skewed.
Make and interpret histograms of quantitative data.
Compare distributions of quantitative data using dotplots, stemplots, or histograms.
To display the distribution of a categorical variable, use a bar graph or a pie chart.
How can we picture the distribution of a quantitative variable? In this section, we
present several types of graphs that can be used to display quantitative data.
Dotplots
One of the simplest graphs to construct and interpret is a dotplot. Each data value
is shown as a dot above its location on a number line. We’ll show how to make a
dotplot using some sports data.
EXAMPLE
Gooooaaaaallllll!
How to make a dotplot
How good was the 2012 U.S. women’s soccer team? With players like Abby Wambach,
Megan Rapinoe, and Hope Solo, the team put on an impressive showing en route to
winning the gold medal at the 2012 Olympics in London. Here are data on the number of goals scored by the team in the 12 months prior to the 2012 Olympics.19
1
3
1
14
13
4
3
4
2
5
2
0
1
3
4
3
4
2
4
3
1
2
4
2
4
Here are the steps in making a dotplot:
• Draw a horizontal axis (a number line) and label it with the variable name. In
this case, the variable is number of goals scored.
•
0
2
4
6
Scale the axis. Start by looking at the minimum and maximum values of the
variable. For these data, the minimum number
of goals scored was 0, and the maximum was
14. So we mark our scale from 0 to 14, with tick
marks at every whole-number value.
8
10
12
14
Number of goals scored
FIGURE 1.8 ​A dotplot of goals scored by the U.S. women’s soccer
team in 2012.
Starnes-Yates5e_c01_xxiv-081hr3.indd 25
• Mark a dot above the location on the horizontal axis corresponding to each data value.
Figure 1.8 displays a completed dotplot for the
soccer data.
11/13/13 1:07 PM
26
CHAPTER 1
E x p l o r i n g Data
Making a graph is not an end in itself. The purpose of a graph is to help us
­understand the data. After you make a graph, always ask, “What do I see?” Here is
a general strategy for interpreting graphs of quantitative data.
How to Examine the Distribution of a Quantitative Variable
In any graph, look for the overall pattern and for striking departures from
that pattern.
•
•
You can describe the overall pattern of a distribution by its shape,
center, and spread.
An important kind of departure is an outlier, an individual value that
falls outside the overall pattern.
You’ll learn more formal ways of describing shape, center, and spread and identifying outliers soon. For now, let’s use our informal understanding of these ideas
to examine the graph of the U.S. women’s soccer team data.
Shape: The dotplot has a peak at 4, a single main cluster of dots between 0 and 5,
and a large gap between 5 and 13. The main cluster has a longer tail to the left of
the peak than to the right. What does the shape tell us? The U.S. women’s soccer
team scored between 0 and 5 goals in most of its games, with 4 being the most
common value (known as the mode).
Center: The “midpoint” of the 25 values shown in the graph is the 13th value if
we count in from either end. You can confirm that the midpoint is at 3. What does
this number tell us? In a typical game during the 2012 season, the U.S. women’s
soccer team scored about 3 goals.
When describing a distribution of
quantitative data, don’t forget your
SOCS (shape, outliers, center, spread)!
Spread: The data vary from 0 goals scored to 14 goals scored.
EXAMPLE
Are You Driving a Gas Guzzler?
Outliers: The games in which the women’s team scored 13 goals and 14 goals
clearly stand out from the overall pattern of the distribution. So we label them
as possible outliers. (In Section 1.3, we’ll establish a procedure for determining
whether a particular value is an outlier.)
Interpreting a dotplot
The Environmental Protection Agency (EPA) is in charge of determining
and reporting fuel economy ratings for cars (think of those large window
stickers on a new car). For years, consumers complained that their actual
gas mileages were noticeably lower than the values reported by the EPA. It
seems that the EPA’s tests—all of which are done on computerized devices
to ensure consistency—did not consider things like outdoor temperature,
use of the air conditioner, or realistic acceleration and braking by drivers. In 2008 the EPA changed the method for measuring a vehicle’s fuel
economy to try to give more accurate e­stimates.
The following table displays the EPA estimates of highway gas mileage in
miles per gallon (mpg) for a sample of 24 model year 2012 midsize cars.20
Starnes-Yates5e_c01_xxiv-081hr3.indd 26
11/13/13 1:07 PM
27
Section 1.2 Displaying Quantitative Data with Graphs
Model
mpg
Model
mpg
Model
mpg
Acura RL
24
Dodge Avenger
30
Mercedes-Benz E350
30
Audi A8
28
Ford Fusion
25
Mitsubishi Galant
30
Bentley Mulsanne
18
Hyundai Elantra
40
Nissan Maxima
26
BMW 550I
23
Jaguar XF
23
Saab 9-5 Sedan
28
Buick Lacrosse
27
Kia Optima
34
Subaru Legacy
31
Cadillac CTS
27
Lexus ES 350
28
Toyota Prius
48
Chevrolet Malibu
33
Lincoln MKZ
27
Volkswagen Passat
31
Chrysler 200
30
Mazda 6
31
Volvo S80
26
Figure 1.9 shows a dotplot of the data:
FIGURE 1.9 Dotplot displaying
EPA estimates of highway gas
mileage for model year 2012
midsize cars.
20
25
30
35
HwyMPG
40
45
Problem: Describe the shape, center, and spread of the distribution. Are there any outliers?
Solution:
The 2012 Nissan Leaf, an electric
car, got an EPA estimated 92 miles
per gallon on the highway. With
the U.S. government’s plan to raise
the fuel economy standard to an
average of 54.5 mpg by 2025, even
more alternative-fuel vehicles like
the Leaf will have to be developed.
Shape: The dotplot has a peak at 30 mpg and a main cluster of values from 23 to 34 mpg. There are
large gaps between 18 and 23, 34 and 40, 40 and 48 mpg.
Center: The midpoint of the 24 values shown in the graph is 28. So a typical model year 2012
midsize car in the sample got about 28 miles per gallon on the highway.
Spread: The data vary from 18 mpg to 48 mpg.
Outliers: We see two midsize cars with unusually high gas mileage ratings: the Hyundai Elantra
(40 mpg) and the Toyota Prius (48 mpg). The Bentley Mulsanne stands out for its low gas mileage
rating (18 mpg). All three of these values seem like clear outliers.
For Practice Try Exercise
39
Describing Shape
When you describe a distribution’s shape, concentrate on the main features. Look
for major peaks, not for minor ups and downs in the graph. Look for clusters of
values and obvious gaps. Look for potential outliers, not just for the smallest and
largest observations. Look for rough symmetry or clear skewness.
Skewed
to the
left!
For his own safety, which way
should Mr. Starnes go “skewing”?
Starnes-Yates5e_c01_xxiv-081hr3.indd 27
Definition: Symmetric and skewed distributions
A distribution is roughly symmetric if the right and left sides of the graph are approximately mirror images of each other.
A distribution is skewed to the right if the right side of the graph (containing the half
of the observations with larger values) is much longer than the left side. It is skewed
to the left if the left side of the graph is much longer than the right side.
11/13/13 1:07 PM
28
CHAPTER 1
E x p l o r i n g Data
c
EXAMPLE
!
n
For brevity, we sometimes say “left-skewed” instead of “skewed to the left” and
“right-skewed” instead of “skewed to the right.” We could also describe a distribution with a long tail to the left as “skewed toward negative values” or “negatively
skewed” and a distribution with a long right tail as “positively skewed.”
The direction of skewness is the direction of the long tail, not the direc- autio
tion where most observations are clustered. See the drawing in the margin
on page 27 for a cute but corny way to help you keep this straight.
Die Rolls and Quiz Scores
Describing shape
Figure 1.10 displays dotplots for two different sets of quantitative data. Let’s practice
describing the shapes of these distributions. Figure 1.10(a) shows the results of rolling a pair of fair, six-sided dice and finding the sum of the up-faces 100 times. This
distribution is roughly symmetric. The dotplot in Figure 1.10(b) shows the scores
on an AP® Statistics class’s first quiz. This distribution is skewed to the left.
(a)
(b)
2
4
6
8
10
12
Dice rolls
70
75
80
85
90
95
100
Score
FIGURE 1.10 ​Dotplots displaying different shapes: (a) roughly symmetric; (b) skewed to the left.
Although the dotplots in the previous example have different shapes, they do
have something in common. Both are unimodal, that is, they have a single peak:
the graph of dice rolls at 7 and the graph of quiz scores at 90. (We don’t count
minor ups and downs in a graph, like the “bumps” at 9 and 11 in the dice rolls
dotplot, as “peaks.”) Figure 1.11 is a dotplot of the duration (in minutes) of 220
FIGURE 1.11 ​Dotplot displaying
duration (in minutes) of Old Faithful
eruptions. This graph has a bimodal
shape.
Starnes-Yates5e_c01_xxiv-081hr3.indd 28
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Section 1.2 Displaying Quantitative Data with Graphs
29
eruptions of the Old Faithful geyser. We would describe this distribution’s shape
as roughly symmetric and bimodal because it has two clear peaks: one near 2
minutes and the other near 4.5 minutes. (Although we could continue the pattern
with “trimodal” for three peaks and so on, it’s more common to refer to distributions with more than two clear peaks as multimodal.)
THINK
ABOUT IT
What shape will the graph have? Some variables have distributions
with predictable shapes. Many biological measurements on individuals from the
same species and gender—lengths of bird bills, heights of young women—have
roughly symmetric distributions. Salaries and home prices, on the other hand,
usually have right-skewed distributions. There are many moderately priced houses, for example, but the few very expensive mansions give the distribution of house
prices a strong right skew. Many distributions have irregular shapes that are neither symmetric nor skewed. Some data show other patterns, such as the two peaks
in Figure 1.11. Use your eyes, describe the pattern you see, and then try to explain
the pattern.
Check Your Understanding
The Fathom dotplot displays data on the number of siblings reported by each student in
a statistics class.
1. Describe the shape of the distribution.
2. Describe the center of the distribution.
3. Describe the spread of the distribution.
4. Identify any potential outliers.
Comparing Distributions
Some of the most interesting statistics questions involve comparing two or more
groups. Which of two popular diets leads to greater long-term weight loss? Who
texts more—males or females? Does the number of people living in a household
differ among countries? As the following example suggests, you should always
discuss shape, center, spread, and possible outliers whenever you compare distributions of a quantitative variable.
Starnes-Yates5e_c01_xxiv-081hr3.indd 29
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30
CHAPTER 1
EXAMPLE
E x p l o r i n g Data
Household Size: U.K. versus
South Africa
Comparing distributions
How do the numbers of people living in households in the United Kingdom (U.K.)
and South Africa compare? To help answer this question, we used CensusAtSchool’s
“Random Data Selector” to choose 50 students from each country. ­Figure 1.12 is a
dotplot of the household sizes reported by the survey respondents.
Problem: Compare the distributions of household size for these two countries.
Solution:
South Africa
Shape: The distribution of household size for the U.K. sample is roughly symmetric and unimodal,
while the distribution for the South Africa sample is skewed to the right and unimodal.
Center: Household sizes for the South African students tended to be larger than for the U.K.
students. The midpoints of the household sizes for the two groups are 6 people and 4 people,
respectively.
Spread: The household sizes for the South African students vary more (from 3 to 26 people) than
for the U.K. students (from 2 to 6 people).
Outliers: There don’t appear
to be any outliers in the U.K.
distribution. The South African
distribution seems to have two
outliers in the right tail of the
distribution—students who
reported living in households
with 15 and 26 people.
Place
0
5
10
15
20
25
30
25
30
Household size
U.K.
AP® EXAM TIP ​When
comparing distributions of
quantitative data, it’s not
enough just to list values
for the center and spread of
each distribution. You have
to explicitly compare these
values, using words like
“greater than,” “less than,”
or “about the same as.”
FIGURE 1.12 ​Dotplots of
household size for random
samples of 50 students from
the United Kingdom and
South Africa.
0
5
10
15
20
Household size
For Practice Try Exercise
43
Notice that we discussed the distributions of household size only for the two
samples of 50 students in the previous example. We might be interested in w
­ hether
the sample data give us convincing evidence of a difference in the population distributions of household size for South Africa and the United Kingdom. We’ll have
to wait a few chapters to decide whether we can reach such a conclusion, but our
Starnes-Yates5e_c01_xxiv-081hr3.indd 30
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31
Section 1.2 Displaying Quantitative Data with Graphs
ability to make such an inference later will be helped by the fact that the students
in our samples were chosen at random.
Stemplots
Another simple graphical display for fairly small data sets is a stemplot (also called
a stem-and-leaf plot). Stemplots give us a quick picture of the shape of a distribution while including the actual numerical values in the graph. Here’s an example
that shows how to make a stemplot.
EXAMPLE
How Many Shoes?
Making a stemplot
How many pairs of shoes does a typical teenager have? To find out, a group of
AP® Statistics students conducted a survey. They selected a random sample of 20
female students from their school. Then they recorded the number of pairs of shoes that each respondent reported having. Here
are the data:
50
13
26
50
26
13
31
34
57
23
19
30
24
49
22
13
23
15
38
51
Here are the steps in making a stemplot. Figure 1.13 displays the
process.
• Separate each observation into a stem, consisting of all but
the final digit, and a leaf, the final digit. Write the stems in a vertical column with the smallest at the top, and draw a vertical line
at the right of this column. Do not skip any stems, even if there is
no data value for a particular stem. For these data, the
tens digits are the stems, and the ones digits are the
1
1 93335
1 33359
leaves. The stems run from 1 to 5.
Key: 4|9 represents
2
2 664233 2 233466 a female student
• Write each leaf in the row to the right of its stem.
3
3 1840
3 0148
who reported having
4
4 9
4 9
For example, the female student with 50 pairs of shoes
49 pairs of shoes.
5
5 0701
5 0017
would have stem 5 and leaf 0, while the student with
Stems
Add leaves
Order leaves
Add a key
31 pairs of shoes would have stem 3 and leaf 1.
• Arrange the leaves in increasing order out from the
FIGURE 1.13 ​Making a stemplot of the shoe data. (1) Write
stem.
the stems. (2) Go through the data and write each leaf on the
• Provide a key that explains in context what the stems
proper stem. (3) Arrange the leaves on each stem in order out
and leaves represent.
from the stem. (4) Add a key.
The AP® Statistics students in the previous example also collected data from a
random sample of 20 male students at their school. Here are the numbers of pairs
of shoes reported by each male in the sample:
14 7 6 5 12 38 8 7
10 11 4 5 22 7 5 10
Starnes-Yates5e_c01_xxiv-081hr3.indd 31
10 10
35 7
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32
CHAPTER 1
E x p l o r i n g Data
What would happen if we tried the same approach as before: using the first digits
as stems and the last digits as leaves? The completed stemplot is shown in Figure
1.14(a). What shape does this distribution have? It is difficult to tell with so few
stems. We can get a better picture of male shoe ownership by splitting stems.
In Figure 1.14(a), the values from 0 to 9 are placed on the “0” stem. Figure 1.14(b)
shows another stemplot of the same data. This time, values having leaves 0 through
4 are placed on one stem, while values ending in 5 through 9 are placed on another
stem. Now we can see the single peak, the cluster of values between 4 and 14, and
the large gap between 22 and 35 more clearly.
What if we want to compare the number of pairs of shoes that males and females
have? That calls for a back-to-back stemplot with common stems. The leaves on
each side are ordered out from the common stem. Figure 1.15 is a back-to-back
stemplot for the male and female shoe data. Note that we have used the split stems
from Figure 1.14(b) as the common stems. The values on the right are the male data
from Figure 1.14(b). The values on the left are the female data, ordered out from
the stem from right to left. We’ll ask you to compare these two distributions shortly.
0
1
2
3
4555677778
0000124
2
58
(a)
0
0
1
1
2
2
3
3
4
555677778
0000124
2
58
(b)
Key: 2|2 represents
a male student with
22 pairs of shoes.
Notice that we include
this stem even though
it contains no data.
FIGURE 1.14 ​Two stemplots showing the male shoe
data. Figure 1.14(b) improves on the stemplot of
Figure 1.14(a) by splitting stems.
•
•
•
Instead of rounding, you can also
truncate (remove one or more digits)
when data have too many digits. The
teacher’s salary of $42,549 would
truncate to $42,000.
•
Females
333
95
4332
66
410
8
9
100
7
0
0
1
1
2
2
3
3
4
4
5
5
Males
4
555677778
0000124
2
Key: 2|2 represents
a male student with
22 pairs of shoes.
58
FIGURE 1.15 ​Back-to-back stemplot
comparing numbers of pairs of shoes for
male and female students at a school.
Here are a few tips to consider before making a stemplot:
Stemplots do not work well for large data sets, where each stem must hold a
large number of leaves.
There is no magic number of stems to use, but five is a good minimum. Too
few or too many stems will make it difficult to see the distribution’s shape.
If you split stems, be sure that each stem is assigned an equal number of possible leaf digits (two stems, each with five possible leaves; or five stems, each
with two possible leaves).
You can get more flexibility by rounding the data so that the final digit after
rounding is suitable as a leaf. Do this when the data have too many digits.
For example, in reporting teachers’ salaries, using all five digits (for example,
$42,549) would be unreasonable. It would be better to round to the nearest
thousand and use 4 as a stem and 3 as a leaf.
Check Your Understanding
1. Use the back-to-back stemplot in Figure 1.15 to write a few sentences comparing the
number of pairs of shoes owned by males and females. Be sure to address shape, center,
spread, and outliers.
Starnes-Yates5e_c01_xxiv-081hr3.indd 32
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33
Section 1.2 Displaying Quantitative Data with Graphs
6
7
8
9
10
11
12
13
14
15
16
8
8
79
08
15566
012223444457888999
01233333444899
02666
23
8
Key: 8|8 represents a state
in which 8.8% of residents
are 65 and older.
Multiple choice: Select the best answer for Questions 2 through 4.
Here is a stemplot of the percents of residents aged 65 and older in the 50 states and the
District of Columbia. The stems are whole percents and the leaves are tenths of a percent.
2. The low outlier is Alaska. What percent of Alaska residents are 65 or older?
(a) 0.68 (b) ​6.8 (c) ​8.8 (d) ​16.8 (e) ​68
3.​Ignoring the outlier, the shape of the distribution is
(a) skewed to the right.
(b)​​skewed to the left.
(c) skewed to the middle.
(d) ​​bimodal.
(e) ​roughly symmetric.
4. The center of the distribution is close to
(a) 11.6%. (b) 12.0%. (c) ​12.8%. (d) ​13.3%. (e) ​6.8% to 16.8%.
Histograms
Quantitative variables often take many values. A graph of the distribution is clearer
if nearby values are grouped together. One very common graph of the distribution
of a quantitative variable is a histogram. Let’s look at how to make a histogram
using data on foreign-born residents in the United States.
Example
Foreign-Born Residents
Making a histogram
What percent of your home state’s residents were born outside the U
­ nited States?
A few years ago, the country as a whole had 12.5% foreign-born residents, but the
states varied from 1.2% in West Virginia to 27.2% in California. The following table
presents the data for all 50 states.21 The individuals in this data set are the states. The
variable is the percent of a state’s residents who are foreign-born. It’s much easier
to see from a graph than from the table how your state compared with other states.
Starnes-Yates5e_c01_xxiv-081hr3.indd 33
StatePercent
StatePercent
StatePercent
Alabama
Alaska
Arizona
Arkansas
California
Colorado
Connecticut
Delaware
Florida
Georgia
Hawaii
Idaho
Illinois
Indiana
Iowa
Kansas
Kentucky
Louisiana
Maine
Maryland
Massachusetts
Michigan
Minnesota
Mississippi
Missouri
Montana
Nebraska
Nevada
New Hampshire
New Jersey
New Mexico
New York
North Carolina
North Dakota
Ohio
Oklahoma
Oregon
Pennsylvania
Rhode Island
South Carolina
South Dakota
Tennessee
Texas
Utah
Vermont
Virginia
Washington
West Virginia
Wisconsin
Wyoming
2.8
7.0
15.1
3.8
27.2
10.3
12.9
8.1
18.9
9.2
16.3
5.6
13.8
4.2
3.8
6.3
2.7
2.9
3.2
12.2
14.1
5.9
6.6
1.8
3.3
1.9
5.6
19.1
5.4
20.1
10.1
21.6
6.9
2.1
3.6
4.9
9.7
5.1
12.6
4.1
2.2
3.9
15.9
8.3
3.9
10.1
12.4
1.2
4.4
2.7
11/13/13 1:07 PM
34
CHAPTER 1
E x p l o r i n g Data
Here are the steps in making a histogram:
• Divide the data into classes of equal width. The data in the table vary from 1.2
to 27.2, so we might choose to use classes of width 5, beginning at 0:
0–5
5–10
10–15
15–20
20–25
25–30
But we need to specify the classes so that each individual falls into exactly one class.
For instance, what if exactly 5.0% of the residents in a state were born outside the
United States? Because a value of 0.0% would go in the 0–5 class, we’ll agree to place
a value of 5.0% in the 5–10 class, a value of 10.0% in the 10–15 class, and so on. In
reality, then, our classes for the percent of foreign-born residents in the states are
0 to <5 5 to <10 10 to <15 15 to <20 20 to <25 25 to <30
• Find the count (frequency) or percent (relative frequency) of individuals in each
class. Here are a frequency table and a relative frequency table for these data:
Frequency table
Class
Notice that the frequencies add
to 50, the number of individuals
(states) in the data, and that the
relative frequencies add to 100%.
Count
Percent
20
0 to <5
40
5 to <10
13
5 to <10
26
10 to <15
9
10 to <15
18
15 to <20
5
15 to <20
10
20 to <25
2
20 to <25
4
25 to <30
1
25 to <30
2
Total
100
50
• Label and scale your axes and draw the histogram. Label the horizontal axis
with the variable whose distribution you are displaying. That’s the percent of
a state’s residents who are foreign-born. The scale on the horizontal axis runs
from 0 to 30 because that is the span of the classes we chose. The vertical axis
contains the scale of counts or percents. Each bar represents a class. The base
of the bar covers the class, and the bar height is the class frequency or relative
frequency. Draw the bars with no horizontal space between them unless a class
is empty, so that its bar has height zero.
FIGURE 1.16 ​(a) Frequency histogram and (b) relative frequency
histogram of the distribution
of the percent of foreign-born
residents in the 50 states.
Figure 1.16(a) shows a completed frequency histogram; Figure 1.16(b) shows a
completed relative frequency histogram. The two graphs look identical except for
the vertical scales.
This bar has height 13
because 13 states have
between 5.0% and 9.9%
foreign-born residents.
20
15
10
5
0
(b)
This bar has height 26
because 26% of states
have between 5.0% and
9.9% foreign-born
residents.
40
Percent of states
(a)
Number of states
Class
0 to <5
Total
30
20
10
0
0
5
10
15
20
25
Percent of foreign-born residents
Starnes-Yates5e_c01_xxiv-081hr3.indd 34
Relative frequency table
30
0
5
10
15
20
25
30
Percent of foreign-born residents
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35
Section 1.2 Displaying Quantitative Data with Graphs
What do the histograms in Figure 1.16 tell us about the percent of foreign-born
residents in the states? To find out, we follow our familiar routine: describe the
pattern and look for any departures from the pattern.
Shape: The distribution is skewed to the right and unimodal. Most states have fewer than 10% foreign-born residents, but several states have much higher percents.
To find the center, remember that we’re
looking for the value having 25 states
with smaller percents foreign-born and
25 with larger.
Center: From the graph, we see that the midpoint would fall somewhere in the
5.0% to 9.9% class. (Arranging the values in the table in order of size shows that
the midpoint is 6.1%.)
Spread: The percent of foreign-born residents in the states varies from less than
5% to over 25%.
Outliers: We don’t see any observations outside the overall pattern of the distribution.
Figure 1.17 shows (a) a frequency histogram and (b) a relative frequency histogram
of the same distribution, with classes half as wide. The new classes are 0–2.4, 2.5–4.9,
and so on. Now California, at 27.2%, stands out as a potential outlier in the right tail.
The choice of classes in a histogram can influence the appearance of a distribution.
Histograms with more classes show more detail but may have a less clear pattern.
(a)
(b)
16
30
25
12
Percent of states
Number of states
14
10
8
6
4
20
15
10
5
2
0
0
0
5
10
15
20
25
Percent of foreign-born residents
30
0
5
10
15
20
25
30
Percent of foreign-born residents
FIGURE 1.17 ​(a) Frequency histogram and (b) relative frequency histogram of the distribution of
the percent of foreign-born residents in the 50 states, with classes half as wide as in Figure 1.16.
Here are some important things to consider when you are constructing a
­histogram:
• Our eyes respond to the area of the bars in a histogram, so be sure to choose
classes that are all the same width. Then area is determined by height, and all
classes are fairly represented.
• There is no one right choice of the classes in a histogram. Too few classes
will give a “skyscraper” graph, with all values in a few classes with tall bars.
Too many will produce a “pancake” graph, with most classes having one or
no observations. Neither choice will give a good picture of the shape of the
distribution. Five classes is a good minimum.
THINK
ABOUT IT
Starnes-Yates5e_c01_xxiv-081hr3.indd 35
What are we actually doing when we make a histogram?
The dotplot on the left below shows the foreign-born resident data. We grouped
the data values into classes of width 5, beginning with 0 to <5, as indicated by the
dashed lines. Then we tallied the number of values in each class. The dotplot on
11/13/13 1:07 PM
36
CHAPTER 1
E x p l o r i n g Data
the right shows the results of that process. Finally, we drew bars of the appropriate
height for each class to get the completed histogram shown.
i
5
10
15
20
25
Percent of foreign-born residents
0
30
5
10
15
20
Percent of foreign-born residents
20
25
30
i
Number of states
0
15
10
5
0
5
10
15
20
25
30
Percent of foreign-born residents
PLET
AP
2.
T echnology
Corner
Statistical software and graphing calculators will choose the classes for you. The
default choice is a good starting point, but you should adjust the classes to suit
your needs. To see what we’re talking about, launch the One-Variable Statistical
Calculator applet at the book’s Web site, www.whfreeman.com/tps5e. Select the
“Percent of foreign-born residents” data set, and then click on the “Histogram”
tab. You can change the number of classes by dragging the horizontal axis with
your mouse or by entering different values in the boxes above the graph. By doing
so, it’s easy to see how the choice of classes affects the histogram. Bottom line: Use
your judgment in choosing classes to display the shape.
Histograms on the calculator
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
TI-83/84TI-89
1. ​Enter the data for the percent of state residents born outside the United States in your Statistics/List Editor.
•
Press STAT and choose Edit...
•
Press APPS and select Stats/List Editor.
•
Type the values into list L1.
•
Type the values into list1.
Starnes-Yates5e_c01_xxiv-081hr3.indd 36
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Section 1.2 Displaying Quantitative Data with Graphs
37
2. ​Set up a histogram in the Statistics Plots menu.
•
Press 2nd Y= (STAT PLOT).
•
Press F2 and choose Plot Setup...
•
Press ENTER or 1 to go into Plot1.
•
With Plot1 highlighted, press F1 to define.
Set Hist.
Bucket
Width to 5.
•
Adjust the settings as shown.
•
Adjust the settings as shown.
3. ​Use ZoomStat (ZoomData on the TI-89) to let the calculator choose ­classes and make a histogram.
•
Press ZOOM and choose ZoomStat.
•
Press TRACE and
◀
▶
•
to examine the classes. •
Press F5 (ZoomData).
Press F3 (Trace) and
◀
▶
to examine the classes.
Note the calculator’s
unusual choice of
classes.
4. ​Adjust the classes to match those in Figure 1.16, and then graph the histogram.
•
Press WINDOW and enter the values
shown below.
•
Press ◆ F2 (WINDOW) and enter the values shown
below.
•
Press GRAPH .
•
Press
•
Press TRACE and
◀
▶
to examine the classes. •
◆ F3
(GRAPH).
Press F3 (Trace) and
◀
▶
to examine the classes.
5. ​See if you can match the histogram in Figure 1.17.
AP® EXAM TIP ​If you’re asked to make a graph on a free-response question, be sure
to label and scale your axes. Unless your calculator shows labels and scaling, don’t just
transfer a calculator screen shot to your paper.
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38
CHAPTER 1
E x p l o r i n g Data
Check Your Understanding
Many people believe that the distribution of IQ scores follows a “bell curve,” like the one
shown in the margin. But is this really how such scores are distributed? The IQ scores of
60 fifth-grade students chosen at random from one school are shown below.22
145 139
101 142
123 94
106 124
117 90
102 108
126
134
100
115
103
110
122
124
136
133
114
128
125
112
109
116
139
114
130 96
109 134
131 117
102 127
101 122
112 114
110 118 118
113 81 113
110 127 124
117 109 137
105 97 89
102 82 101
1.​Construct a histogram that displays the distribution of IQ scores effectively.
2. Describe what you see. Is the distribution bell-shaped?
Using Histograms Wisely
c
Histogram !
n
We offer several cautions based on common mistakes students make when using
histograms.
1.​
Don’t confuse histograms and bar graphs. Although histograms re- autio
semble bar graphs, their details and uses are different. A histogram
displays the distribution of a quantitative variable. The horizontal
axis of a histogram is marked in the units of measurement for the variable.
A bar graph is used to display the distribution of a categorical variable or to
compare the sizes of different quantities. The horizontal axis of a bar graph
identifies the categories or quantities being compared. Draw bar graphs with
blank space between the bars to separate the items being compared. Draw
histograms with no space, to show the equal-width classes. For comparison,
here is one of each type of graph from previous examples.
Bar graph
16
14
Percent of stations
30
20
15
10
5
12
10
8
6
4
2
th
er
Radio station format
autio
!
n
2. U
​ se percents instead of counts on the vertical axis when comparing
distributions with different numbers of observations. Mary was interested in comparing the reading levels of a medical journal and an
c
Starnes-Yates5e_c01_xxiv-081hr3.indd 38
O
N
ew
R
oc
k
Sp
an
ish
ry
s/T
al
k
O
ld
ie
R
s
el
ig
io
us
it
ou
nt
A
Percent of foreign-born residents
C
30
ta
25
on
H
20
C
15
on
t
10
dS
5
dC
0
n
0
0
A
Percent of states
25
11/13/13 1:07 PM
39
Section 1.2 Displaying Quantitative Data with Graphs
airline magazine. She counted the number of letters in the first 400 words
of an article in the medical journal and of the first 100 words of an article in
the airline magazine. Mary then used Minitab statistical software to produce
the histograms shown in Figure 1.18(a). This figure is misleading—it compares frequencies, but the two samples were of very different sizes (100 and
400). Using the same data, Mary’s teacher produced the histograms in Figure
1.18(b). By using relative frequencies, this figure provides an accurate comparison of word lengths in the two samples.
(a)
(b)
JLength
MLength
90
70
20
60
Percent
Frequency
MLength
25
80
50
40
15
10
30
20
5
10
0
JLength
30
0
2
4
6
8 10 12 14 0
2
4
6
0
8 10 12 14
0
2
4
6
8 10 12 14 0
2
4
6
8 10 12 14
FIGURE 1.18 ​Two sets of histograms comparing word lengths in articles from a journal and from an airline magazine. In (a), the
vertical scale uses frequencies. The graph in (b) fixes this problem by using percents on the vertical scale.
c
4
5
6
First name length
!
n
3. Just because a graph looks nice doesn’t make it a meaningful display of autio
data. The students in a small statistics class recorded the number of
letters in their first names. One student entered the data into an Excel
spreadsheet and then used Excel’s “chart maker” to produce the graph shown
below left. What kind of graph is this? It’s a bar graph that compares the raw data
values. But first-name length is a quantitative variable, so a bar graph is not an appropriate way to display its distribution. The dotplot on the right is a much better
choice.
7
Check Your Understanding
About 1.6 million first-year students enroll in colleges and universities each year. What
do they plan to study? The graph on the next page displays data on the percents of firstyear students who plan to major in several discipline areas.23
Starnes-Yates5e_c01_xxiv-081hr3.indd 39
11/13/13 1:07 PM
CHAPTER 1
E x p l o r i n g Data
20
15
10
5
0
B
us
in
Pr
es
of
s
es
sio
na
l
A
O
rt
s/h
th
e
um
r
an
So
iti
ci
al
es
sc
ie
nc
es
E
du
ca
tio
E
ng
B
n
io
in
lo
e
er
gi
ca
in
g
Ph l sc
ie
ys
nc
ic
al
es
sc
ie
nc
es
Te
ch
ni
ca
l
Percent of students who plan to major
40
Field of study
1.​Is this a bar graph or a histogram? Explain.
2.​Would it be correct to describe this distribution as right-skewed? Why or why not?
Section 1.2
Summary
•
•
•
•
•
Starnes-Yates5e_c01_xxiv-081hr3.indd 40
You can use a dotplot, stemplot, or histogram to show the distribution of a
quantitative variable. A dotplot displays individual values on a number line.
Stemplots separate each observation into a stem and a one-digit leaf. Histograms plot the counts (frequencies) or percents (relative frequencies) of
values in equal-width classes.
When examining any graph, look for an overall pattern and for notable
departures from that pattern. Shape, center, and spread describe the overall
pattern of the distribution of a quantitative variable. Outliers are observations
that lie outside the overall pattern of a distribution. Always look for outliers
and try to explain them. Don’t forget your SOCS!
Some distributions have simple shapes, such as symmetric, skewed to the
left, or skewed to the right. The number of modes (major peaks) is another
aspect of overall shape. So are distinct clusters and gaps. Not all distributions
have a simple overall shape, especially when there are few observations.
When comparing distributions of quantitative data, be sure to compare
shape, center, spread, and possible outliers.
Remember: histograms are for quantitative data; bar graphs are for categorical data. Also, be sure to use relative frequency histograms when comparing
data sets of different sizes.
11/13/13 1:07 PM
41
Section 1.2 Displaying Quantitative Data with Graphs
1.2 T ECHNOLOGY
CORNER
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
2. Histograms on the calculator
page 36
Exercises
Section 1.2
37. Feeling sleepy? Students in a college statistics class
responded to a survey designed by their teacher. One
of the survey questions was “How much sleep did you
get last night?” Here are the data (in hours):
9
5
6
6
8
11
6
6
8
3
8
6
6
6
6.5
10
6
7
7
8
9
4.5
4
9
3
7
4
7
(b) O
​ verall, 205 countries participated in the 2012 Summer Olympics, of which 54 won at least one gold
medal. Do you believe that the sample of countries
listed in the table is representative of this larger population? Why or why not?
39. U.S. women’s soccer—2012 Earlier, we examined
data on the number of goals scored by the U.S. women’s soccer team in games played in the 12 months
prior to the 2012 Olympics. The dotplot below
displays the goal differential for those same games,
computed as U.S. score minus opponent’s score.
pg 26
(a) M
​ ake a dotplot to display the data.
(b) ​Describe the overall pattern of the distribution and
any departures from that pattern.
38. Olympic gold! The following table displays the total
number of gold medals won by a sample of countries
in the 2012 Summer Olympic Games in London.
Country
Sri Lanka
China
Vietnam
Great Britain
Gold
medals
0
38
0
29
Country
Gold
medals
Thailand
0
Kuwait
0
Bahamas
1
Kenya
2
Norway
2
Trinidad and Tobago
1
Romania
2
Greece
0
Switzerland
2
Mozambique
0
Armenia
0
Kazakhstan
7
Netherlands
6
Denmark
2
India
0
Latvia
1
Georgia
1
Czech Republic
4
Kyrgyzstan
0
Hungary
8
Costa Rica
0
Sweden
1
Brazil
3
Uruguay
0
Uzbekistan
1
United States
46
(a) M
​ ake a dotplot to display these data. Describe the
overall pattern of the distribution and any departures
from that pattern.
Starnes-Yates5e_c01_xxiv-081hr4.indd 41
(a) ​Explain what the dot above −1 represents.
(b) W
​ hat does the graph tell us about how well the team
did in 2012? Be specific.
40. Fuel efficiency In an earlier example, we examined
data on highway gas mileages of model year 2012
midsize cars. The following dotplot shows the difference (highway – city) in EPA mileage ratings for each
of the 24 car models from the earlier example.
-2
0
6
2
4
Difference (highway – city)
8
12
10
(a) E
​ xplain what the dot above 12 represents.
(b) ​What does the graph tell us about fuel economy in the
city versus on the highway for these car models? Be
specific.
11/20/13 6:23 PM
42
CHAPTER 1
E x p l o r i n g Data
41. Dates on coins ​Suppose that you and your friends
emptied your pockets of coins and recorded the year
marked on each coin. The distribution of dates would
be skewed to the left. Explain why.
of the data is shown below. Compare the three distributions. Critics claim that supermarkets tend to put sugary
kids’ cereals on lower shelves, where the kids can see
them. Do the data from this study support this claim?
42. Phone numbers The dotplot below displays the last
digit of 100 phone numbers chosen at random from
a phone book. Describe the shape of the distribution.
Does this shape make sense to you? Explain.
0
2
4
6
8
Last digit
43. Creative writing Do external rewards—things like
money, praise, fame, and grades—promote creativity?
Researcher Teresa Amabile designed an experiment
to find out. She recruited 47 experienced creative
writers who were college students and divided them
into two groups using a chance process (like drawing names from a hat). The students in one group
were given a list of statements about external reasons
(E) for writing, such as public recognition, making
­money, or pleasing their parents. Students in the other group were given a list of statements about internal
reasons (I) for writing, such as expressing yourself and
enjoying playing with words. Both groups were then
instructed to write a poem about laughter. Each student’s poem was rated separately by 12 different poets
using a creativity scale.24 These ratings were averaged
to obtain an overall creativity score for each poem.
Dotplots of the two groups’ creativity scores are
shown below. Compare the two distributions. What
do you conclude about whether external rewards
promote creativity?
pg 30
Reward
E
0
5
10
15
20
25
30
Average rating
I
0
5
10
15
20
25
30
Average rating
44. Healthy cereal? Researchers collected data on 77
brands of cereal at a local supermarket.25 For each
brand, the sugar content (grams per serving) and the
shelf in the store on which the cereal was located (1 =
bottom, 2 = middle, 3 = top) were recorded. A dotplot
Starnes-Yates5e_c01_xxiv-081hr3.indd 42
45. Where do the young live? Below is a stemplot of
the percent of residents aged 25 to 34 in each of the
50 states. As in the stemplot for older residents (page
33), the stems are whole percents, and the leaves are
tenths of a percent. This time, each stem has been
split in two, with values having leaves 0 through 4
placed on one stem, and values ending in 5 through 9
placed on another stem.
11
11
12
12
13
13
14
14
15
15
16
44
66778
0134
666778888
0000001111444
7788999
0044
567
11
0
(a) ​Why did we split stems?
(b) Give an appropriate key for this stemplot.​
(c) D
​ escribe the shape, center, and spread of the distribution. Are there any outliers?
46. Watch that caffeine! The U.S. Food and Drug
Administration (USFDA) limits the amount of caffeine
in a 12-ounce can of carbonated beverage to 72 milligrams. That translates to a maximum of 48 milligrams
of caffeine per 8-ounce serving. Data on the caffeine
content of popular soft drinks (in milligrams per
8-ounce serving) are displayed in the stemplot below.
1
2
2
3
3
4
4
556
033344
55667778888899
113
55567778
33
77
(a) ​Why did we split stems?
(b) ​Give an appropriate key for this graph.
(c) D
​ escribe the shape, center, and spread of the distribution. Are there any outliers?
11/13/13 1:07 PM
43
Section 1.2 Displaying Quantitative Data with Graphs
47. El Niño and the monsoon It appears that El Niño,
the periodic warming of the Pacific Ocean west of
South America, affects the monsoon rains that are essential for agriculture in India. Here are the monsoon
rains (in millimeters) for the 23 strong El Niño years
between 1871 and 2004:26
(a) E
​ xamine the data. Why are you not surprised that
most responses are multiples of 10 minutes? Are there
any responses you consider suspicious?
(b) Make a back-to-back stemplot to compare the two samples. Does it appear that women study more than men
(or at least claim that they do)? Justify your answer.
628 669 740 651 710 736 717 698 653 604 781 784
50. Basketball playoffs Here are the numbers of points
scored by teams in the California Division I-AAA high
school basketball playoffs in a single day’s games:27
(a) T
​ o make a stemplot of these rainfall amounts, round
the data to the nearest 10, so that stems are hundreds
of millimeters and leaves are tens of millimeters.
Make two stemplots, with and without splitting the
stems. Which plot do you prefer? Why?
(b) D
​ escribe the shape, center, and spread of the
distribution.
(c) T
​ he average monsoon rainfall for all years from 1871
to 2004 is about 850 millimeters. What effect does El
Niño appear to have on monsoon rains?
48. Shopping spree A marketing consultant observed
50 consecutive shoppers at a supermarket. One variable of interest was how much each shopper spent in
the store. Here are the data (in dollars), arranged in
increasing order:
3.11 8.88 9.26
18.36 18.43 19.27
24.58 25.13 26.24
36.37 38.64 39.16
50.39 52.75 54.80
10.81
19.50
26.26
41.02
59.07
12.69
19.54
27.65
42.97
61.22
13.78
20.16
28.06
44.08
70.32
15.23
20.59
28.08
44.67
82.70
15.62
22.22
28.38
45.40
85.76
17.00
23.04
32.03
46.69
86.37
17.39
24.47
34.98
48.65
93.34
(a) R
​ ound each amount to the nearest dollar. Then make
a stemplot using tens of dollars as the stems and dollars as the leaves.
(b) M
​ ake another stemplot of the data by splitting stems.
Which of the plots shows the shape of the distribution
better?
(c) W
​ rite a few sentences describing the amount of
money spent by shoppers at this supermarket.
49. Do women study more than men? We asked the
students in a large first-year college class how many
minutes they studied on a typical weeknight. Here are
the responses of random samples of 30 women and 30
men from the class:
Women
180 120
120 180
150 120
200 150
120 
60
90 240
180
120
180
180
120
180
Starnes-Yates5e_c01_xxiv-081hr3.indd 43
Men
360
240
180
150
180
115
240 90 120 
30 
90 200
170 90 45 30 120 75
150
150 120 
60 240 300
180
240 
60 120 
60 
30
180 30 230 120 
95 150
120 0 200 120 120 180
71 38 52 47
55 53 76 65 77 63 65 63 68
54 64 62 87
47 64 56 78 64 58 51 91 74
71 41 67 62 106 46
On the same day, the final scores of games in Division V-AA were
98
45
67
44
74
60
96
54
92
72
93
46
98
67
62
37
37
36
69
44
86
66
66
58
(a) C
​ onstruct a back-to-back stemplot to compare the
points scored by the 32 teams in the Division I-AAA
playoffs and the 24 teams in the Division V-AA playoffs.
(b) ​Write a few sentences comparing the two distributions.
51. Returns on common stocks The return on a stock
is the change in its market price plus any dividend
payments made. Total return is usually expressed as a
percent of the beginning price. The figure below shows
a histogram of the distribution of the monthly returns
for all common stocks listed on U.S. markets over a
273-month period.28 The extreme low outlier represents
the market crash of October 1987, when stocks lost 23%
of their value in one month.
80
Number of months
790 811 830 858 858 896 806 790 792 957 872
60
40
20
0
–25
–20
–15
–10
–5
0
5
10
15
Monthly percent return on common stocks
(a) Describe the overall shape of the distribution of
monthly returns.
(b) ​What is the approximate center of this distribution?
(c) ​Approximately what were the smallest and largest
monthly returns, leaving out the outliers?
(d) ​A return less than zero means that stocks lost ­value
in that month. About what percent of all months had
returns less than zero?
11/13/13 1:07 PM
44
CHAPTER 1
E x p l o r i n g Data
Percent of Shakespeare’s words
52. Shakespeare The histogram below shows the distribution of lengths of words used in Shakespeare’s
plays.29 Describe the shape, center, and spread of this
distribution.
25
(a) M
​ ake a histogram of the data using classes of width 2,
starting at 0.
20
(b) D
​ escribe the shape, center, and spread of the distribution. Which countries are outliers?
15
Carbon dioxide emissions
(metric tons per person)
10
5
0
1
2
3
4
5
6
7
8
9
10
11
12
Number of letters in word
23.6
LA
25.1
OH
22.1
AK
17.7
ME
22.3
OK
20.0
AZ
25.0
MD
30.6
OR
21.8
AR
20.7
MA
26.6
PA
25.0
CA
26.8
MI
23.4
RI
22.3
CO
23.9
MN
22.0
SC
22.9
CT
24.1
MS
24.0
SD
15.9
DE
23.6
MO
22.9
TN
23.5
FL
25.9
MT
17.6
TX
24.6
GA
27.3
NE
17.7
UT
20.8
HI
25.5
NV
24.2
VT
21.2
ID
20.1
NH
24.6
VA
26.9
IL
27.9
NJ
29.1
WA
25.2
IN
22.3
NM
20.9
WV
25.6
IA
18.2
NY
30.9
WI
20.8
KS
18.5
NC
23.4
WY
17.9
KY
22.4
ND
15.5
DC
29.2
(a) M
​ ake a histogram of the travel times using classes of
width 2 minutes, starting at 14 minutes. That is, the
first class is 14 to 16 minutes, the second is 16 to 18
minutes, and so on.
(b) T
​ he shape of the distribution is a bit irregular. Is it
closer to symmetric or skewed? Describe the center
and spread of the distribution. Are there any outliers?
54. Carbon dioxide emissions Burning fuels in power
plants and motor vehicles emits carbon dioxide
Starnes-Yates5e_c01_xxiv-081hr3.indd 44
Country
CO2
Country
CO2
Algeria
2.6
Mexico
3.7
Argentina
3.6
Morocco
1.4
Australia
18.4
Myanmar
0.2
0.3
Nepal
0.1
1.8
Nigeria
0.4
17.0
Pakistan
0.8
Bangladesh
53. Traveling to work How long do people travel each
day to get to work? The following table gives the
average travel times to work (in minutes) for workers
in each state and the District of Columbia who are at
least 16 years old and don’t work at home.30
AL
(CO2), which contributes to global warming. The
table below displays CO2 emissions per person from
countries with populations of at least 20 million.31
Brazil
Canada
China
3.9
Peru
1.0
Colombia
1.3
Philippines
0.9
Congo
0.2
Poland
7.8
Egypt
2.0
Romania
Ethiopia
0.1
Russia
10.8
France
6.2
Saudi Arabia
13.8
Germany
9.9
South Africa
7.0
Ghana
0.3
Spain
7.9
India
1.1
Sudan
0.3
Indonesia
1.6
Tanzania
0.1
Iran
6.0
Thailand
3.3
Iraq
2.9
Turkey
3.0
Italy
7.8
Ukraine
6.3
Japan
9.5
United Kingdom
Kenya
0.3
United States
Korea, North
3.3
Uzbekistan
4.2
Korea, South
9.3
Venezuela
5.4
Malaysia
5.5
Vietnam
1.0
4.2
8.8
19.6
55. DRP test scores There are many ways to measure
the reading ability of children. One frequently used
test is the Degree of Reading Power (DRP). In a
research study on third-grade students, the DRP was
administered to 44 students.32 Their scores were:
40
47
52
47
26
19
25
35
39
26
35
48
14
35
35
22
42
34
33
33
18
15
29
41
25
44
34
51
43
40
41
27
46
38
49
14
27
31
28
54
19
46
52
45
Make a histogram to display the data. Write a paragraph describing the distribution of DRP scores.
11/13/13 1:07 PM
45
Section 1.2 Displaying Quantitative Data with Graphs
56. Drive time Professor Moore, who lives a few miles
outside a college town, records the time he takes to
drive to the college each morning. Here are the times
(in minutes) for 42 consecutive weekdays:
Philadelphia Phillies won the World Series. Maybe
the Yankees didn’t spend enough money that year.
The graph below shows histograms of the salary distributions for the two teams during the 2008 season.
Why can’t you use this graph to effectively compare
the team payrolls?
8.25 7.83 8.30 8.42 8.50 8.67 8.17 9.00 9.00 8.17 7.92
9.00 8.50 9.00 7.75 7.92 8.00 8.08 8.42 8.75 8.08 9.75
8.33 7.83 7.92 8.58 7.83 8.42 7.75 7.42 6.75 7.42 8.50
8.67 10.17 8.75 8.58 8.67 9.17 9.08 8.83 8.67 6
4
34
18
42
658
35
81
43
370
36
185
44
92
37
420
45
50
38
749
46
21
39
1073
47
4
40
1079
48
1
(a) ​Make a histogram of this distribution.
(b) D
​ escribe the shape, center, and spread of the chest
measurements distribution. Why might this information
be useful?
59. Paying for championships Does paying high salaries
lead to more victories in professional sports? The
New York Yankees have long been known for having
Major League Baseball’s highest team payroll. And
over the years, the team has won many championships. This strategy didn’t pay off in 2008, when the
Starnes-Yates5e_c01_xxiv-081hr3.indd 45
20 0
00
00
40 0
00
00
60 0
00
00
80 0
00
0
10 00
00
00
12 00
00
00
14 00
00
00
00
50 0
00
0
10 00
00
00
15 00
00
00
20 00
00
00
25 00
00
00
30 00
00
00
35 00
00
00
00
14
8
00
32
00
00
00
00
00
24
00
00
00
32
00
00
Salary
00
0
00
2
0
00
4
2
00
6
4
16
6
10
00
0
8
12
00
10
0
12
80
Frequency
16
14
00
Frequency
18
16
24
934
Phillies 2008
18
00
41
Salary
Yankees 2008
00
3
3
60. Paying for championships Refer to Exercise 59.
Here is another graph of the 2008 salary distributions
for the Yankees and the Phillies. Write a few sentences comparing these two distributions.
80
33
4
Salary
00
Count
5
0
0
Chest size
6
1
0
16
Count
7
2
2
0
Chest size
Frequency
8
(a) M
​ ake a histogram of this distribution. Describe its
shape, center, and spread.
58. Chest out, Soldier! In 1846, a published paper provided chest measurements (in inches) of 5738 Scottish
militiamen. The table below summarizes the data.34
8
10
Length:
1 2 3
4
5 6 7 8 9 10 11 12 13 14 15
Percent: 3.6 14.8 18.7 16.0 12.5 8.2 8.1 5.9 4.4 3.6 2.1 0.9 0.6 0.4 0.2
(b) H
​ ow does the distribution of lengths of words used in
Popular Science compare with the similar distribution for
Shakespeare’s plays in Exercise 52? Look in particular at
short words (2, 3, and 4 letters) and very long words (more
than 10 letters).
9
00
57. The statistics of writing style Numerical data can
distinguish different types of writing and, sometimes,
even individual authors. Here are data on the percent
of words of 1 to 15 letters used in articles in Popular
Science magazine:33
Phillies 2008
12
Frequency
Make a histogram to display the data. Write a paragraph describing the distribution of Professor Moore’s
drive times.
Yankees 2008
Salary
61. Birth months Imagine asking a random sample of 60
students from your school about their birth months.
Draw a plausible graph of the distribution of birth
months. Should you use a bar graph or a histogram to
display the data?
62. Die rolls Imagine rolling a fair, six-sided die 60
times. Draw a plausible graph of the distribution of
die rolls. Should you use a bar graph or a histogram
to display the data?
63. Who makes more? A manufacturing company is
reviewing the salaries of its full-time employees below
the executive level at a large plant. The clerical staff
is almost entirely female, while a majority of the production workers and technical staff is male. As a result, the distributions of salaries for male and female
employees may be quite different. The following
table gives the frequencies and relative frequencies
for women and men.
11/13/13 1:07 PM
46
CHAPTER 1
E x p l o r i n g Data
Women
Salary ($1000)
Number
%
10–15
89
11.8
26
15–20
192
25.4
221
9.0
20–25
236
31.2
677
27.6
Number
%
1.1
25–30
111
14.7
823
33.6
30–35
86
11.4
365
14.9
35–40
25
3.3
182
7.4
40–45
11
1.5
91
3.7
45–50
3
0.4
33
1.3
50–55
2
0.3
19
0.8
55–60
0
0.0
11
0.4
60–65
0
0.0
0
0.0
65–70
1
0.1
3
0.1
756
100.1
2451
99.9
Total
66. Population pyramids Refer to Exercise 65. Here
is a graph of the projected population distribution
for China in the year 2050. Describe what the
graph suggests about China’s future population. Be
specific.
Men
Male
60
(a) Explain why the total for women is greater than 100%.
(b) Make histograms for these data, choosing the vertical
scale that is most appropriate for comparing the two
distributions.
(c) Write a few sentences comparing the salary distributions for men and women.
64. Comparing AP® scores The table below gives the
distribution of grades earned by students taking the
AP® Calculus AB and AP® Statistics exams in 2012.35
Grade
No. of
exams
5
4
3
2
1
Calculus AB 266,994
67,394 45,523 46,526 27,216 80,335
Statistics
19,267 32,521 39,355 27,684 35,032
153,859
48
China 2050
36
24
12
100
95
90
85
80
75
70
65
60
55
50
45
40
35
30
25
20
15
10
5
0
0 0
Female
12
24
Population (in millions)
36
48
60
67. Student survey A survey of a large high school class
asked the following questions:
(i) Are you female or male? (In the data, male = 0,
female = 1.)
(ii) Are you right-handed or left-handed? (In the data,
right = 0, left = 1.)
(iii) What is your height in inches?
(iv) How many minutes do you study on a typical
weeknight?
The figure below shows graphs of the student responses, in scrambled order and without scale markings.
Which graph goes with each variable? Explain your
­reasoning.
(a) Make an appropriate graphical display to compare the grade
distributions for AP® Calculus AB and AP® Statistics.
(b) Write a few sentences comparing the two distributions
of exam grades.
65. Population pyramids A population pyramid is a
­helpful graph for examining the distribution of a
country’s population. Here is a population pyramid for
Vietnam in the year 2010. Describe what the graph tells
you about Vietnam’s population that year. Be specific.
Male
5
Vietnam 2010
4
3
2
1
100
95
90
85
80
75
70
65
60
55
50
45
40
35
30
25
20
15
10
5
0
0 0
(b)
(c)
(d)
Female
68. Choose a graph What type of graph or graphs would
you make in a study of each of the following issues at
your school? Explain your choices.
(a) Which radio stations are most popular with students?
1
Population (in millions)
Starnes-Yates5e_c01_xxiv-081hr3.indd 46
(a)
2
3
4
5
(b) How many hours per week do students study?
(c) How many calories do students consume per day?
11/13/13 1:07 PM
47
Section 1.2 Displaying Quantitative Data with Graphs
69. Here are the amounts of money (cents) in coins carried by 10 students in a statistics class: 50, 35, 0, 97,
76, 0, 0, 87, 23, 65. To make a stemplot of these data,
you would use stems
(a)
(b)
(c)
(d)
(e)
70.
0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
0, 2, 3, 5, 6, 7, 8, 9.
0, 3, 5, 6, 7.
00, 10, 20, 30, 40, 50, 60, 70, 80, 90.
None of these.
The histogram below shows the heights of 300 randomly selected high school students. Which of the
following is the best description of the shape of the
distribution of heights?
(c) It would be better if the values from 34 to 50 were
deleted on the horizontal axis so there wouldn’t be a
large gap.
(d) There was one state with a value of exactly 33%.
(e) About half of the states had percents between 24%
and 28%.
14
12
Number of states
Multiple choice: Select the best answer for Exercises 69 to 74.
10
8
6
4
2
0
20
24
28
32
36
40
44
48
52
Percent of women over
age 15 who never married
(a)
(b)
(c)
(d)
(e)
71.
(a)
(b)
(c)
(d)
(e)
Roughly symmetric and unimodal
Roughly symmetric and bimodal
Roughly symmetric and multimodal
Skewed to the left
Skewed to the right
You look at real estate ads for houses in Naples, Florida. There are many houses ranging from $200,000
to $500,000 in price. The few houses on the water,
however, have prices up to $15 million. The distribution of house prices will be
skewed to the left.
roughly symmetric.
skewed to the right.
unimodal.
too high.
72. The following histogram shows the distribution of the
percents of women aged 15 and over who have never
married in each of the 50 states and the District of
Columbia. Which of the following statements about
the histogram is correct?
(a) The center of the distribution is about 36%.
(b) There are more states with percents above 32 than
there are states with percents less than 24.
Starnes-Yates5e_c01_xxiv-081hr3.indd 47
73. When comparing two distributions, it would be best
to use relative frequency histograms rather than frequency histograms when
(a) the distributions have different shapes.
(b) the distributions have different spreads.
(c) the distributions have different centers.
(d) the distributions have different numbers of observations.
(e) at least one of the distributions has outliers.
74. Which of the following is the best reason for choosing a stemplot rather than a histogram to display the
distribution of a quantitative variable?
(a) Stemplots allow you to split stems; histograms don’t.
(b) Stemplots allow you to see the values of individual
­observations.
(c) Stemplots are better for displaying very large sets of data.
(d) Stemplots never require rounding of values.
(e) Stemplots make it easier to determine the shape of a
distribution.
75. Baseball players (Introduction) Here is a small part
of a data set that describes Major League Baseball
players as of opening day of the 2012 season:
Player
Team
Position
Age Height Weight
Salary
Rodriguez,
Alex
Yankees
Infielder
37
6-3
225
29,000,000
Gonzalez,
Adrian
Dodgers
Infielder
30
6-2
225
21,000,000
Cruz,
Nelson
Rangers
Outfielder
32
6-2
240
5,000,000
Lester,
Jon
Red Sox
Pitcher
28
6-4
240
7,625,000
Strasburg,
Stephen Nationals Pitcher
24
6-4
220
3,000,000
11/13/13 1:07 PM
48
CHAPTER 1
E x p l o r i n g Data
(a) What individuals does this data set describe?
(b) In addition to the player’s name, how many variables
does the data set contain? Which of these variables
are categorical and which are quantitative?
77. Risks of playing soccer (1.1) A study in Sweden
looked at former elite soccer players, people who had
played soccer but not at the elite level, and people of
the same age who did not play soccer. Here is a twoway table that classifies these individuals by whether
or not they had arthritis of the hip or knee by their
mid-fifties:37
76. I love my iPod! (1.1) The rating service Arbitron
asked adults who used several high-tech devices and
services whether they “loved” using them. Below is a
graph of the percents who said they did.36
(a) Summarize what this graph tells you in a sentence or
two.
(b) Would it be appropriate to make a pie chart of these
data? Why or why not?
Non-Elite
Did not play
Arthritis
10
9
24
No arthritis
61
206
548
(a) What percent of the people in this study were elite
soccer players? What percent had arthritis?
(b) What percent of the elite soccer players had arthritis?
What percent of those who had arthritis were elite soccer players?
50
Percent of users who love it
Elite
40
78. Risks of playing soccer (1.1) Refer to Exercise 77.
We suspect that the more serious soccer players have
more arthritis later in life. Do the data confirm this
suspicion? Give graphical and numerical evidence to
support your answer.
30
20
10
V
TV
yT
le
ab
C
Pa
y
rr
P3
R
B
la
ck
be
M
io
D
V
ad
Sa
tr
TV
nd
D
ba
H
B
ro
ad
iP
od
0
High-tech device or service
1.3
What You Will Learn
•
•
•
Describing Quantitative Data
with Numbers
By the end of the section, you should be able to:
Calculate measures of center (mean, median).
Calculate and interpret measures of spread (range, IQR,
standard deviation).
Choose the most appropriate measure of center and
spread in a given setting.
•
•
•
Identify outliers using the 1.5 × IQR rule.
Make and interpret boxplots of quantitative data.
Use appropriate graphs and numerical summaries to
compare distributions of quantitative variables.
How long do people spend traveling to work? The answer may depend on where
they live. Here are the travel times in minutes for 15 workers in North Carolina,
chosen at random by the Census Bureau:38
30
Starnes-Yates5e_c01_xxiv-081hr3.indd 48
20
10
40
25
20
10
60
15
40
5
30
12
10
10
11/13/13 1:07 PM
Section 1.3 Describing Quantitative Data with Numbers
49
We aren’t surprised that most people estimate their travel time in multiples of
5 minutes. Here is a stemplot of these data:
0
1
2
3
4
5
6
5
000025
005
Key: 2|5 is a NC
00
worker who travels 25
00
minutes to work.
0
The distribution is single-peaked and right-skewed. The longest travel time
(60 minutes) may be an outlier. Our main goal in this section is to describe the
center and spread of this and other distributions of quantitative data with numbers.
Measuring Center: The Mean
The most common measure of center is the ordinary arithmetic average, or mean.
Definition: The mean x
To find the mean x– (pronounced “x-bar”) of a set of observations, add their values
and divide by the number of observations. If the n observations are x1, x2, . . . , xn,
their mean is
sum of observations x1 + x2 + c+ xn
=
x– =
n
n
or, in more compact notation,
x– =
∙x
i
n
∙
The (capital Greek letter sigma) in the formula for the mean is short for “add
them all up.” The subscripts on the observations xi are just a way of keeping the n
observations distinct. They do not necessarily indicate order or any other special
facts about the data.
Actually, the notation x– refers to the mean of a sample. Most of the time, the data
we’ll encounter can be thought of as a sample from some larger population. When
we need to refer to a population mean, we’ll use the symbol m (Greek letter mu,
pronounced “mew”). If you have the entire population of data available, then you
calculate m in just the way you’d expect: add the values of all the observations, and
divide by the number of observations.
EXAMPLE
Travel Times to Work in North Carolina
Calculating the mean
Here is a stemplot of the travel times to work for the sample of 15 North ­Carolinians.
0
1
2
3
4
5
6
Starnes-Yates5e_c01_xxiv-081hr3.indd 49
5
000025
005
Key: 2|5 is a NC
00
worker who travels 25
00
minutes to work.
0
11/13/13 1:07 PM
50
CHAPTER 1
E x p l o r i n g Data
Problem:
(a) ​Find the mean travel time for all 15 workers.
(b) Calculate the mean again, this time excluding the person who reported a 60-minute travel time
to work. What do you notice?
Solution:
(a) The mean travel time for the sample of 15 North Carolina workers is
x– =
∙x = x
i
n
1
+ x2 + c+ xn 30 + 20 + c+ 10 337
=
=
= 22.5 minutes
n
15
15
(b) If we leave out the longest travel time, 60 minutes, the mean for the remaining 14 people is
∙x = x
+ x2 + c + xn 277
=
= 19.8 minutes
n
n
14
This one observation raises the mean by 2.7 minutes.
x– =
i
1
For Practice Try Exercise
79
c
THINK
ABOUT IT
!
n
The previous example illustrates an important weakness of the mean autio
as a ­measure of center: the mean is sensitive to the influence of extreme
observations. These may be outliers, but a skewed distribution that has
no outliers will also pull the mean toward its long tail. Because the mean cannot
resist the influence of extreme observations, we say that it is not a resistant measure of center.
What does the mean mean? ​A group of elementary schoolchildren was
asked how many pets they have. Here are their responses, arranged from lowest
to ­highest:39
1
3
4
4
4
5
7
8
9
What’s the mean number of pets for this group of children? It’s
x– =
sum of observations 1 + 3 + 4 + 4 + 4 + 5 + 7 + 8 + 9
=
= 5 pets
n
9
But what does that number tell us? Here’s one way to look at it: if every child in
the group had the same number of pets, each would have 5 pets. In other words,
the mean is the “fair share” value.
The mean tells us how large each data value would be if the total were split
equally among all the observations. The mean of a distribution also has a physical
interpretation, as the following Activity shows.
Starnes-Yates5e_c01_xxiv-081hr3.indd 50
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Section 1.3 Describing Quantitative Data with Numbers
Activity
51
Mean as a “balance point”
MATERIALS:
In this Activity, you’ll investigate an interesting property of the mean.
Foot-long ruler, pencil, and
5 pennies per group of 3 to
4 students
1. Stack all 5 pennies above the 6-inch mark on your ruler. Place your pencil
under the ruler to make a “seesaw” on a desk or table. Move the pencil until the
ruler balances. What is the relationship between the location of the pencil and
the mean of the five data values: 6, 6, 6, 6, 6?
2. Move one penny off the stack to the 8-inch mark on your
ruler. Now move one other penny so that the ruler balances
again without moving the pencil. Where did you put the
other penny? What is the mean of the five data values represented by the pennies now?
3. Move one more penny off the stack to the 2-inch mark
on your ruler. Now move both remaining pennies from the
6-inch mark so that the ruler still balances with the pencil in
the same location. Is the mean of the data values still 6?
4. Do you see why the mean is sometimes called the “balance point” of a ­distribution?
Measuring Center: The Median
In Section 1.2, we introduced the median as an informal measure of center that
describes the “midpoint” of a distribution. Now it’s time to offer an official “rule”
for calculating the median.
Definition: The median
The median is the midpoint of a distribution, the number such that about half the observations are smaller and about half are larger. To find the median of a distribution:
1. Arrange all observations in order of size, from smallest to largest.
2.If the number of observations n is odd, the median is the center observation in
the ordered list.
3.If the number of observations n is even, the median is the average of the two
center observations in the ordered list.
Medians require little arithmetic, so they are easy to find by hand for small sets
of data. Arranging even a moderate number of values in order is tedious, however,
so finding the median by hand for larger sets of data is unpleasant.
Starnes-Yates5e_c01_xxiv-081hr3.indd 51
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52
CHAPTER 1
EXAMPLE
E x p l o r i n g Data
Travel Times to Work in North Carolina
Finding the median when n is odd
What is the median travel time for our 15 North Carolina workers? Here are the
data arranged in order:
5 10 10 10 10 12 15 20 20 25 30 30 40 40 60
The count of observations n = 15 is odd. The bold 20 is the center observation in
the ordered list, with 7 observations to its left and 7 to its right. This is the median,
20 minutes.
The next example shows you how to find the median when there is an even
number of data values.
EXAMPLE
Stuck in Traffic
Finding the median when n is even
People say that it takes a long time to get to work in New York State due to the
heavy traffic near big cities. What do the data say? Here are the travel times in
minutes of 20 randomly chosen New York workers:
10
15
30 5
20 85
25
15
40
65
20
15
10
60
15
60
30
40
20
45
Problem:
(a) Make a stemplot of the data. Be sure to include a key.
(b) Find the median by hand. Show your work.
Solution:
0
1
2
3
4
5
6
7
8
0
1
2
3
4
5
6
5
005 5 5 5
0005
00
005
005
Key: 4|5 is a
New York worker
who reported a
45-minute travel
time to work.
5
5
000025
005
Key: 2|5 is a NC
00
worker who travels 25
00
minutes to work.
0
Starnes-Yates5e_c01_xxiv-081hr3.indd 52
(a) Here is a stemplot of the data. The stems indicate10 minutes and the leaves indicate
minutes.
(b) Because there is an even number of data values, there is no center observation. There is a center
pair—the bold 20 and 25 in the stemplot—which have 9 observations before them and 9 after
them in the ordered list. The median is the average of these two observations:
20 + 25
= 22.5 minutes
2
For Practice Try Exercise 81
Comparing the Mean and the Median
Our discussion of travel times to work in North Carolina illustrates an important difference between the mean and the median. The median travel time (the midpoint
of the distribution) is 20 minutes. The mean travel time is higher, 22.5 minutes. The
mean is pulled toward the right tail of this right-skewed distribution. The median,
11/13/13 1:07 PM
Section 1.3 Describing Quantitative Data with Numbers
PLET
AP
53
unlike the mean, is resistant. If the longest travel time were 600 minutes rather than
60 minutes, the mean would increase to more than 58 minutes but the median
would not change at all. The outlier just counts as one observation above the center,
no matter how far above the center it lies. The mean uses the actual value of each
observation and so will chase a single large observation upward.
You can compare the behavior of the mean and median by using the Mean and
Median applet at the book’s Web site, www.whfreeman.com/tps5e.
Comparing the Mean and Median
The mean and median of a roughly symmetric distribution are close t­ ogether.
If the distribution is exactly symmetric, the mean and median are exactly the
same. In a skewed distribution, the mean is usually farther out in the long tail
than is the median.40
The mean and median measure center in different ways, and both are useful.
THINK
ABOUT IT
Should we choose the mean or the median? ​Many economic vari-
ables have distributions that are skewed to the right. College tuitions, home prices, and personal incomes are all right-skewed. In Major League Baseball (MLB),
for instance, most players earn close to the minimum salary (which was $480,000
in 2012), while a few earn more than $10 million. The median salary for MLB
players in 2012 was about $1.08 million—but the mean salary was about $3.44
million. Alex Rodriguez, Prince Fielder, Joe Mauer, and several other highly paid
superstars pull the mean up but do not affect the median.
Reports about incomes and other strongly skewed distributions usually give
the median (“midpoint”) rather than the mean (“arithmetic average”). However,
a county that is about to impose a tax of 1% on the incomes of its residents cares
about the mean income, not the median. The tax revenue will be 1% of total
­income, and the total is the mean times the number of residents.
Check Your Understanding
0
1
2
3
4
5
6
7
8
5
005555
0005
Key: 4|5 is a
00
New York worker
005
who reported a
005
45-minute travel
time to work.
5
Here, once again, is the stemplot of travel times to work for 20 randomly selected New
Yorkers. Earlier, we found that the median was 22.5 minutes.
1. Based only on the stemplot, would you expect the mean travel time to be less than,
about the same as, or larger than the median? Why?
2. Use your calculator to find the mean travel time. Was your answer to Question 1
correct?
3. Would the mean or the median be a more appropriate summary of the center of this
distribution of drive times? Justify your answer.
Measuring Spread:
Range and Interquartile Range (IQR )
A measure of center alone can be misleading. The mean annual temperature in
San Francisco, California, is 57°F—the same as in Springfield, Missouri. But the
wardrobe needed to live in these two cities is very different! That’s because daily
Starnes-Yates5e_c01_xxiv-081hr3.indd 53
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54
CHAPTER 1
Note that the range of a data set
is a single number that represents
the distance between the maximum
and the minimum value. In everyday
language, people sometimes say
things like, “The data values range
from 5 to 85.” Be sure to use the term
range correctly, now that you know its
statistical definition.
E x p l o r i n g Data
temperatures vary a lot more in Springfield than in San Francisco. A useful numerical
description of a distribution requires both a measure of center and a measure of spread.
The simplest measure of variability is the range. To compute the range of a
quantitative data set, subtract the smallest value from the largest value. For the
New York travel time data, the range is 85 − 5 = 80 minutes. The range shows
the full spread of the data. But it depends on only the maximum and minimum
values, which may be outliers.
We can improve our description of spread by also looking at the spread of the
middle half of the data. Here’s the idea. Count up the ordered list of observations,
starting from the minimum. The first quartile Q1 lies one-quarter of the way up
the list. The second quartile is the median, which is halfway up the list. The third
quartile Q3 lies three-quarters of the way up the list. These quartiles mark out the
middle half of the distribution. The interquartile range (IQR) measures the range
of the middle 50% of the data. We need a rule to make this idea exact. The process
for calculating the quartiles and the IQR uses the rule for finding the median.
How to Calculate the Quartiles Q1 and Q3 and the Interquartile
Range (IQR )
To calculate the quartiles:
1. Arrange the observations in increasing order and locate the median in
the ordered list of observations.
2. The first quartile Q1 is the median of the observations that are to the
left of the median in the ordered list.
3. The third quartile Q3 is the median of the observations that are to
the right of the median in the ordered list.
The interquartile range (IQR) is defined as
IQR = Q3 − Q1
Be careful in locating the quartiles when several observations take the same
numerical value. Write down all the observations, arrange them in order,
and apply the rules just as if they all had distinct values.
Let’s look at how this process works using a familiar set of data.
EXAMPLE
Travel Times to Work in North Carolina
Calculating quartiles
Our North Carolina sample of 15 workers’ travel times, arranged in increasing
order, is
5
10 10 10 10 12 15 20 20 25 30 30 40 40 60
Median
Q 1 is the median
of the values to the
left of the median.
Starnes-Yates5e_c01_xxiv-081hr3.indd 54
Q 3 is the median
of the values to the
right of the median.
11/13/13 1:07 PM
55
Section 1.3 Describing Quantitative Data with Numbers
c
!
n
There is an odd number of observations, so the median is the middle one, the
bold 20 in the list. The first quartile is the median of the 7 observations to the left
of the median. This is the 4th of these 7 observations, so Q1 = 10 minutes (shown
in blue). The third quartile is the median of the 7 observations to the right of the
median, Q3 = 30 minutes (shown in green). So the spread of the middle
autio
50% of the travel times is IQR = Q3 − Q1 = 30 − 10 = 20 minutes.
Be sure to leave out the overall median when you locate the quartiles.
The quartiles and the interquartile range are resistant because they are not
a­ ffected by a few extreme observations. For example, Q3 would still be 30 and the
IQR would still be 20 if the maximum were 600 rather than 60.
EXAMPLE
Stuck in Traffic Again
Finding and interpreting the IQR
In an earlier example, we looked at data on travel times to work for 20 randomly
selected New Yorkers. Here is the stemplot once again:
0
1
2
3
4
5
6
7
8
5
005555
0005
Key: 4|5 is a
00
New York worker
005
who reported a
005
45-minute travel
time to work.
5
Problem: Find and interpret the interquartile range (IQR).
Solution: We begin by writing the travel times arranged in increasing order:
5 10 10 15 15 15 15 20 20 20 25 30 30 40 40 45 60 60 65 85
Q 1 = 15
Median = 22.5
Q 3 = 42.5
There is an even number of observations, so the median lies halfway between the middle pair. Its value
is 22.5 minutes. (We marked the location of the median by |.) The first quartile is the median of the
10 observations to the left of 22.5. So it’s the average of the two bold 15s: Q1 = 15 minutes. The
third quartile is the median of the 10 observations to the right of 22.5. It’s the average of the bold
numbers 40 and 45: Q3 = 42.5 minutes. The interquartile range is
IQR = Q3 − Q1 = 42.5 − 15 = 27.5 minutes
Interpretation: The range of the middle half of travel times for the New Yorkers in the sample is
27.5 minutes.
For Practice Try Exercise 89a
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56
CHAPTER 1
E x p l o r i n g Data
Identifying Outliers
In addition to serving as a measure of spread, the interquartile range (IQR) is used
as part of a rule of thumb for identifying outliers.
Definition: The 1.5 × IQR rule for outliers
Call an observation an outlier if it falls more than 1.5 × IQR above the third quartile
or below the first quartile.
0
1
2
3
4
5
6
7
8
5
005555
0005
Key: 4|5 is a
00
New York worker
005
who reported a
005
45-minute travel
time to work.
Does the 1.5 × IQR rule identify any outliers for the New York travel time
data? In the previous example, we found that Q1 = 15 minutes, Q3 = 42.5 minutes, and IQR = 27.5 minutes. For these data,
1.5 × IQR = 1.5(27.5) = 41.25
Any values not falling between
5
Q1 − 1.5 × IQR = 15 − 41.25 = −26.25
Q3 + 1.5 × IQR = 42.5 + 41.25 = 83.75
and
are flagged as outliers. Look again at the stemplot: the only outlier is the longest travel
time, 85 minutes. The 1.5 × IQR rule suggests that the three next-longest travel times
(60 and 65 minutes) are just part of the long right tail of this skewed distribution.
Example
Travel Times to Work in North Carolina
Identifying outliers
Earlier, we noted the influence of one long travel time of 60 minutes in our sample of 15 North Carolina workers.
0
1
2
3
4
5
6
5
000025
005
Key: 2|5 is a NC
00
worker who travels 25
00
minutes to work.
0
Problem: Determine whether this value is an outlier.
Solution: Earlier, we found that Q1 = 10 minutes, Q3 = 30 minutes, and IQR = 20
minutes. To check for outliers, we first calculate
1.5 × IQR = 1.5(20) = 30
By the 1.5 × IQR rule, any value greater than
Q3 + 1.5 × IQR = 30 + 30 = 60
or less than
Q1 − 1.5 × IQR = 10 − 30 = −20
would be classified as an outlier. The maximum value of 60 minutes is not quite large enough to be an
outlier because it falls right on the upper cutoff value.
For Practice Try Exercise 89b
Whenever you find outliers in your data, try to find an explanation for them.
Sometimes the explanation is as simple as a typing error, like typing 10.1 as 101.
Sometimes a measuring device broke down or someone gave a silly response, like
the student in a class survey who claimed to study 30,000 minutes per night. (Yes,
Starnes-Yates5e_c01_xxiv-081hr3.indd 56
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Section 1.3 Describing Quantitative Data with Numbers
AP® EXAM TIP ​You may be
asked to determine whether a
quantitative data set has any
outliers. Be prepared to state
and use the rule for identifying
outliers.
57
that really happened.) In all these cases, you can simply remove the outlier from your
data. When outliers are “real data,” like the long travel times of some New York workers, you should choose measures of center and spread that are not greatly a­ ffected by
the outliers.
The Five-Number Summary and Boxplots
The smallest and largest observations tell us little about the distribution as a whole,
but they give information about the tails of the distribution that is missing if we
know only the median and the quartiles. To get a quick summary of both center
and spread, use all five numbers.
Definition: The five-number summary
The five-number summary of a distribution consists of the smallest observation,
the first quartile, the median, the third quartile, and the largest observation, written in
order from smallest to largest. That is, the five-number summary is
Minimum Q1 Median Q3 Maximum
These five numbers divide each distribution roughly into quarters. About 25%
of the data values fall between the minimum and Q1, about 25% are between Q1
and the median, about 25% are between the median and Q3, and about 25% are
between Q3 and the maximum.
The five-number summary of a distribution leads to a new graph, the boxplot
(sometimes called a box-and-whisker plot).
How to Make a Boxplot
•
•
•
•
A central box is drawn from the first quartile (Q1) to the third quartile (Q3).
A line in the box marks the median.
Lines (called whiskers) extend from the box out to the smallest and largest observations that are not outliers.
Outliers are marked with a special symbol such as an asterisk (*).
Here’s an example that shows how to make a boxplot.
Example
Home Run King
Making a boxplot
Barry Bonds set the major league record by hitting 73 home runs in a single
season in 2001. On ­August 7, 2007, Bonds hit his 756th career home run, which
broke Hank Aaron’s longstanding record of 755. By the end of the 2007 season
when Bonds retired, he had increased the total to 762. Here are data on the number of home runs that Bonds hit in each of his 21 complete seasons:
16
40
Starnes-Yates5e_c01_xxiv-081hr3.indd 57
25
37
24
34
19
49
33
73
25
46
34
45
46
45
37
26
33 42
28
11/13/13 1:07 PM
58
CHAPTER 1
E x p l o r i n g Data
Problem: Make a boxplot for these data.
Solution: Let’s start by ordering the data values so that we can find the five-number summary.
16 19 24 25 25 26 28 33 33 34 34 37 37 40 42 45 45 46 46 49 73
Min Q1 = 25.5 min
16
15
Q1
25.5
20
25
Med
34
30
35 40
45
50
55
60 65
Max
N ow we check for outliers. Because IQR = 45 − 25.5 = 19.5, by
the 1.5 × IQR rule, any value greater than Q3 + 1.5 × IQR = 45 +
1.5 × 19.5 = 74.25 or less than Q1 − 1.5 × IQR = 25.5 − 1.5
× 19.5 = −3.75 would be classified as an outlier. So there are no
outliers in this data set. Now we are ready to draw the boxplot. See
the finished graph at left.
max
73
Q3
45
Q3 = 45 Median 70 75
Number of home runs hit in a season by Barry Bonds
For Practice Try Exercise 91
THINK
ABOUT IT
What are we actually doing when we make a boxplot? The top
dotplot shows Barry Bonds’s home run data. We have marked the first quartile, the
median, and the third quartile with blue lines. The process of testing for outliers
with the 1.5 × IQR rule is shown in visual form. Because there are no outliers,
we draw the whiskers to the maximum and minimum data values, as shown in the
finished boxplot at right.
1.5 IQR 29.25
Lower cutoff for outliers
Upper cutoff for outliers
IQR 19.5
Q1 25.5 Med 34 Q3 45
0
10
20
Q1 25.5
30
40
Home runs
Med 34
50
60
Q3 45
Min 16
0
Travel time to work (minutes)
90
*
80
An outlier is any point
more than 1.5 box lengths
below Q 1 or above Q 3 .
70
Whisker ends at last
data value that is not
an outlier, 65.
60
50
40
30
The interquartile range
is the length of the box.
20
10
0
North Carolina
New York
FIGURE 1.19 ​Boxplots comparing the travel times to work
of samples of workers in North Carolina and New York.
Starnes-Yates5e_c01_xxiv-081hr3.indd 58
10
70
Max 73
20
30
40
Home runs
50
60
70
Figure 1.19 shows boxplots (this time, they are oriented
vertically) comparing travel times to work for the samples of
workers from North Carolina and New York. We will identify
outliers as isolated points in the graph (like the * for the maximum value in the New York data set).
Boxplots show less detail than histograms or stemplots, so
they are best used for side-by-side comparison of more than
one distribution, as in Figure 1.19. As always, be sure to discuss shape, center, spread, and outliers as part of your comparison. For the travel time to work data:
Shape: We see from the graph that both distributions are
right-skewed. For both states, the distance from the minimum
to the median is much smaller than the distance from the
median to the maximum.
Center: It appears that travel times to work are generally a
bit longer in New York than in North Carolina. The median,
both quartiles, and the maximum are all larger in New York.
11/13/13 1:07 PM
Section 1.3 Describing Quantitative Data with Numbers
59
Spread: Travel times are also more variable in New York, as shown by the lengths
of the boxes (the IQR) and the range.
Outliers: Earlier, we showed that the maximum travel time of 85 minutes is an
outlier for the New York data. There are no outliers in the North Carolina sample.
Check Your Understanding
The 2011 roster of the Dallas Cowboys professional football team included 8 offensive
linemen. Their weights (in pounds) were
310
1.
2.
3.
4.
3. T ECHNOLOGY
CORNER
307
345
324
305
301
290
307
Find the five-number summary for these data by hand. Show your work.
Calculate the IQR. Interpret this value in context.
Determine whether there are any outliers using the 1.5 × IQR rule.
Draw a boxplot of the data.
​Making calculator boxplots
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
The TI-83/84 and TI-89 can plot up to three boxplots in the same viewing window. Let’s use the calculator to make parallel
boxplots of the travel time to work data for the samples from North Carolina and New York.
1.​Enter the travel time data for North Carolina in L1/list1 and for New York in L2/list2.
2.​Set up two statistics plots: Plot1 to show a boxplot of the North Carolina data and Plot2 to show a boxplot of the New
York data. The setup for Plot1 is shown below. When you define Plot2, be sure to change L1/list1 to L2/list2.
TI-83/84TI-89
Note: The calculator offers two types of boxplots: one that shows outliers and one that doesn’t. We’ll always use the
type that identifies outliers.
3. Use the calculator’s Zoom feature to display the parallel boxplots. Then Trace to view the five-number summary.
TI-83/84TI-89
•
Press ZOOM and select ZoomStat.
•
Press F5 (ZoomData).
•
Press TRACE .
•
Press F3 (Trace).
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60
CHAPTER 1
E x p l o r i n g Data
Measuring Spread: The Standard Deviation
The five-number summary is not the most common numerical description of a
distribution. That distinction belongs to the combination of the mean to measure
center and the standard deviation to measure spread. The standard deviation and
its close relative, the variance, measure spread by looking at how far the observations are from their mean. Let’s explore this idea using a simple set of data.
EXAMPLE
How Many Pets?
Investigating spread around the mean
In the Think About It on page 50, we examined data on the number of pets owned
by a group of 9 children. Here are the data again, arranged from lowest to highest:
1
3
4
4
4
5
7
8
9
Earlier, we found the mean number of pets to be x = 5. Let’s look at where the
observations in the data set are relative to the mean.
deviation = –4
x=1
0
1
deviation = 2
x=5
2
3
4
5
x=7
6
7
8
9
Mean = balance point
Number of pets
FIGURE 1.20 Dotplot of the
pet data with the mean and
two of the deviations marked.
Figure 1.20 displays the data in a dotplot, with the
mean clearly marked. The data value 1 is 4 units below the mean. We say that its deviation from the mean
is −4. What about the data value 7? Its deviation is
7 − 5 = 2 (it is 2 units above the mean). The arrows
in the figure mark these two deviations from the mean.
The deviations show how much the data vary about
their mean. They are the starting point for calculating
the variance and standard deviation.
The table below shows the deviation from the mean (xi − x) for each value in the
data set. Sum the deviations from the mean. You should get 0, because the mean
is the balance point of the distribution. Because the sum of the deviations from
the mean will be 0 for any set of data, we need another way to calculate spread
around the mean.
Observations
xi
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Deviations
xi − x–
Squared deviations
(xi − x– )2
1
1 − 5 = −4
(−4)2 = 16
3
3 − 5 = −2
(−2)2 = 4
4
4 − 5 = −1
(−1)2 = 1
4
4 − 5 = −1
(−1)2 = 1
4
4 − 5 = −1
(−1)2 = 1
5
5−5=0
02 = 0
7
7−5=2
22 = 4
8
8−5=3
32 = 9
9
9−5=4
42 = 16
sum = 0
sum = 52
11/13/13 1:07 PM
61
Section 1.3 Describing Quantitative Data with Numbers
How can we fix the problem of the positive and negative deviations canceling
out? We could take the absolute value of each deviation. Or we could square the
deviations. For mathematical reasons beyond the scope of this book, statisticians
choose to square rather than to use absolute values.
We have added a column to the table that shows the square of each deviation
(xi − x– )2. Add up the squared deviations. Did you get 52? Now we compute the
average squared deviation—sort of. Instead of dividing by the number of observations n, we divide by n − 1:
“average” squared deviation =
16 + 4 + 1 + 1 + 1 + 0 + 4 + 9 + 16 52
=
= 6.5
9−1
8
This value, 6.5, is called the variance.
Because we squared all the deviations, our units are in “squared pets.” That’s no
good. We’ll take the square root to get back to the correct units—pets. The resulting value is the standard deviation:
standard deviation = "variance = "6.5 = 2.55 pets
This 2.55 is the “typical” distance of the values in the data set from the mean. In this
case, the number of pets typically varies from the mean by about 2.55 pets.
As you can see, the “average” in the standard deviation calculation is found in
a rather unexpected way. Why do we divide by n − 1 instead of n when calculating the variance and standard deviation? The answer is complicated but will be
revealed in Chapter 7.
Definition: The standard deviation sx and variance s x2
The standard deviation sx measures the typical distance of the values in a distribution from the mean. It is calculated by finding an average of the squared deviations and then taking the square root. This average squared deviation is called the
variance. In symbols, the variance sx2 is given by
s 2x
(x1 − x–)2 + (x2 − x–)2 + c + (xn − x–)2
1
=
=
n−1
n−1
∙(x − x–)
i
2
and the standard deviation is given by
sx =
1
Ån − 1
∙(x − x–)
i
2
Here’s a brief summary of the process for calculating the standard deviation.
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62
CHAPTER 1
E x p l o r i n g Data
How to Find the Standard Deviation
To find the standard deviation of n observations:
1. ​Find the distance of each observation from the mean and square each of
these distances.
2. ​Average the distances by dividing their sum by n − 1.
3. ​The standard deviation sx is the square root of this average squared distance:
sx =
1
Ån − 1
∙(x − x)
i
2
Many calculators report two standard deviations. One is usually labeled sx,
the symbol for the standard d
­ eviation of a population. This standard deviation is
calculated by dividing the sum of squared deviations by n instead of n − 1 before
taking the square root. If your data set consists of the entire population, then it’s
appropriate to use sx. Most often, the data we’re examining come from a sample.
In that case, we should use sx.
c
Starnes-Yates5e_c01_xxiv-081hr3.indd 62
!
n
More important than the details of calculating sx are the properties that describe the usefulness of the standard deviation:
• sx measures spread about the mean and should be used only when the mean is
chosen as the measure of center.
• sx is always greater than or equal to 0. sx = 0 only when there is no variability. This happens only when all observations have the same value. Otherwise,
sx > 0. As the observations become more spread out about their mean, sx gets
larger.
• sx has the same units of measurement as the original observations. For example,
if you measure metabolic rates in calories, both the mean x and the standard
deviation sx are also in calories. This is one reason to prefer sx to the variance
s2x, which is in squared calories.
• Like the mean x, sx is not resistant. A few outliers can make sx very large. The
use of squared deviations makes sx even more sensitive than x to a few ­extreme
observations. For example, the standard deviation of the travel times
for the 15 North Carolina workers is 15.23 minutes. If we omit the autio
maximum value of 60 minutes, the standard deviation drops to 11.56
minutes.
11/13/13 1:07 PM
Section 1.3 Describing Quantitative Data with Numbers
63
Check Your Understanding
The heights (in inches) of the five starters on a basketball team are 67, 72, 76, 76, and 84.
1.​Find the mean. Show your work.
2.​Make a table that shows, for each value, its deviation from the mean and its squared
deviation from the mean.
3.​Show how to calculate the variance and standard deviation from the values in your
table.
4.​Interpret the standard deviation in this setting.
Numerical Summaries with Technology
Graphing calculators and computer software will calculate numerical summaries
for you. That will free you up to concentrate on choosing the right methods and
interpreting your results.
4. T echnology
Corner
Computing numerical summaries
with technology
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
Let’s find numerical summaries for the travel times of North Carolina and New York workers from the previous Technology Corner (page 59). We’ll start by showing you the necessary calculator techniques and then look at output from
computer software.
I. One-variable statistics on the calculator If you haven’t done so already, enter the North Carolina data in L1/list1
and the New York data in L2/list2.
1. Find the summary statistics for the North Carolina travel times.
TI-83/84TI-89
(CALC); choose 1-VarStats.
•
Press F4 (Calc); choose 1-Var Stats.
OS 2.55 or later: In the dialog box, press 2nd
1 (L1) and ENTER to specify L1 as the List.
Leave FreqList blank. Arrow down to Calculate
and press ENTER . Older OS: Press 2nd 1
(L1) and ENTER .
•
Type list1 in the list box. Press ENTER .
•
Press
Press STAT
▼
▶
to see the rest of the one-variable statistics for North Carolina.
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64
CHAPTER 1
E x p l o r i n g Data
2. ​Repeat Step 1 using L2/list2 to find the summary statistics for the New York travel times.
II. Output from statistical software We used Minitab statistical software to produce descriptive statistics for the
New York and North Carolina travel time data. Minitab allows you to choose which numerical summaries are
included in the output.
Descriptive Statistics: Travel time to work
Variable
N
Mean
StDev
Minimum
Q1
Median
Q3
Maximum
NY Time
20
31.25
21.88
5.00
15.00
22.50
43.75
85.00
NC Time
15
22.47
15.23
5.00
10.00
20.00
30.00
60.00
THINK
ABOUT IT
What’s with that third quartile? ​Earlier, we saw that the quartiles of the
New York travel times are Q1 = 15 and Q3 = 42.5. Look at the Minitab output in
the Technology Corner. Minitab says that Q3 = 43.75. What happened? Minitab
and some other software use different rules for locating quartiles. Results from the
various rules are always close to each other, so the differences are rarely important
in practice. But because of the slight difference, Minitab wouldn’t identify the
maximum value of 85 as an outlier by the 1.5 × IQR rule.
Choosing Measures of Center and Spread
We now have a choice between two descriptions of the center and spread of a
distribution: the median and IQR, or x and sx. Because x and sx are sensitive to
extreme observations, they can be misleading when a distribution is strongly
skewed or has outliers. In these cases, the median and IQR, which are both
resistant to extreme values, provide a better summary. We’ll see in the next
chapter that the mean and standard deviation are the natural measures of center
and spread for a very important class of symmetric distributions, the Normal
distributions.
Choosing Measures of Center and Spread
The median and IQR are usually better than the mean and standard deviation for describing a skewed distribution or a distribution with strong outliers. Use x and sx only for reasonably symmetric distributions that don’t have
outliers.
Starnes-Yates5e_c01_xxiv-081hr3.indd 64
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65
Section 1.3 Describing Quantitative Data with Numbers
c
!
n
Remember that a graph gives the best overall picture of a distribution. autio
Numerical measures of center and spread report specific facts about a distribution, but they do not describe its entire shape. Numerical summaries
do not highlight the presence of multiple peaks or clusters, for example. Always
plot your data.
Organizing a Statistics Problem
As you learn more about statistics, you will be asked to solve more complex problems. Although no single strategy will work on every problem, it can be helpful to
have a general framework for organizing your thinking. Here is a four-step process
you can follow.
STEP
4
How to Organize a StatisticS Problem: A Four-Step Process
State: What’s the question that you’re trying to answer?
Plan: How will you go about answering the question? What statistical techniques does this problem call for?
To keep the four steps straight, just
remember: Statistics Problems Demand
Consistency!
Do: Make graphs and carry out needed calculations.
Conclude: Give your conclusion in the setting of the real-world problem.
Many examples and exercises in this book will tell you what to do—construct a
graph, perform a calculation, interpret a result, and so on. Real statistics problems
don’t come with such detailed instructions. From now on, you will encounter
some examples and exercises that are more realistic. They are marked with the
four-step icon. Use the four-step process as a guide to solving these problems, as
the following example illustrates.
EXAMPLE
Who Texts More—Males or Females?
Putting it all together
STEP
4
For their final project, a group of AP® Statistics students wanted to
compare the texting habits of males and females. They asked a random
sample of students from their school to record the number of text messages sent
and received over a two-day period. Here are their data:
Males:127 44 28
Females: 112 203 102
83
0
54 379
6 78 6
5 213 73
305 179 24 127 65 41
20 214 28 11
27 298 6 130
0
What conclusion should the students draw? Give appropriate evidence to support
your answer.
STATE: Do males and females at the school differ in their texting habits?
PLAN: We’ll begin by making parallel boxplots of the data about males and females. Then we’ll
calculate one-variable statistics. Finally, we’ll compare shape, center, spread, and outliers for the
two distributions.
Starnes-Yates5e_c01_xxiv-081hr3.indd 65
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66
CHAPTER 1
E x p l o r i n g Data
DO: Figure 1.21 is a sketch of the boxplots we got from our calculator. The table below shows
numerical summaries for males and females.
Male
Female
x
sx
Min
Q1
Med
Q3
Max
IQR
62.4
128.3
71.4
116.0
0
0
6
34
28
107
83
191
214
379
77
157
Due to the strong skewness and outliers, we’ll use the median and IQR instead
of the mean and standard deviation when comparing center and spread.
Shape: Both distributions are strongly right-skewed.
Center: Females typically text more than males. The median number of
texts for females (107) is about four times as high as for males (28). In
Males
fact, the median for the females is above the third quartile for the males.
0 6 28 83 127
213 214
This indicates that over 75% of the males texted less than the “typical”
Females
­(median) female.
0 34
107
191
379
Spread: There is much more variation in texting among the females than
the males. The IQR for females (157) is about twice the IQR for males (77).
0
100
200
300
400
Outliers: There are two outliers in the male distribution: students who
Number of text messages in 2-day period
reported 213 and 214 texts in two days. The female distribution has no
FIGURE 1.21 Parallel boxplots of the texting data.
outliers.
Conclude: The data from this survey project give very strong evidence that male and female texting habits differ considerably at the school. A typical female sends and receives about
79 more text messages in a two-day period than a typical male. The males as a group are also
much more consistent in their texting frequency than the females.
These two values appeared as
one dot in the calculator graph.
We found them both by tracing.
For Practice Try Exercise
105
Now it’s time for you to put what you have learned into practice in the following Data Exploration.
DATA EXPLORATION Did Mr. Starnes stack his class?
Mr. Starnes teaches AP® Statistics, but he also does the class scheduling for the
high school. There are two AP® Statistics classes—one taught by Mr. Starnes and
one taught by Ms. McGrail. The two teachers give the same first test to their classes
and grade the test together. Mr. Starnes’s students earned an average score that
was 8 points higher than the average for Ms. McGrail’s class. Ms. McGrail wonders whether Mr. Starnes might have “adjusted” the class rosters from the computer scheduling program. In other words, she thinks he might have “stacked” his
class. He denies this, of course.
To help resolve the dispute, the teachers collect data on the cumulative grade
point averages and SAT Math scores of their students. Mr. Starnes provides the
GPA data from his computer. The students report their SAT Math scores. The following table shows the data for each student in the two classes. Note that the two
data values in each row come from a single student.
Starnes-Yates5e_c01_xxiv-081hr3.indd 66
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Section 1.3 Describing Quantitative Data with Numbers
Starnes GPA
Starnes SAT-M
McGrail GPA
McGrail SAT-M
2.9
670
2.9
620
2.86
520
3.3
590
2.6
570
3.98
650
3.6
710
2.9
600
3.2
600
3.2
620
2.7
590
3.5
680
3.1
640
2.8
500
3.085
570
2.9
502.5
3.75
710
3.95
640
3.4
630
3.1
630
3.338
630
2.85
580
3.56
670
2.9
590
3.8
650
3.245
600
3.2
660
3.0
600
3.1
510
3.0
620
2.8
580
2.9
600
3.2
600
67
Did Mr. Starnes stack his class? Give appropriate graphical and numerical
­evidence to support your conclusion.
AP® EXAM TIP ​Use statistical terms carefully and correctly on the AP® exam. Don’t say
“mean” if you really mean “median.” Range is a single number; so are Q1, Q3, and IQR. Avoid
colloquial use of language, like “the outlier skews the mean.” Skewed is a shape. If you misuse
a term, expect to lose some credit.
case closed
Do pets or friends help reduce stress?
Refer to the chapter-opening Case Study (page 1). You will now use
what you have learned in this chapter to analyze the data.
1.
2.
3.
4.
5.
Starnes-Yates5e_c01_xxiv-081hr3.indd 67
Construct an appropriate graph for comparing the heart rates of
the women in the three groups.
Calculate numerical summaries for each group’s data. Which measures of center and spread would you choose to compare? Why?
Determine if there are any outliers in each of the three groups.
Show your work.
Write a few sentences comparing the distributions of heart rates
for the women in the three groups.
Based on the data, does it appear that the presence of a pet or friend
reduces heart rate during a stressful task? Justify your answer.
11/13/13 2:09 PM
68
CHAPTER 1
Section 1.3
E x p l o r i n g Data
Summary
•
•
•
•
•
•
•
•
•
•
A numerical summary of a distribution should report at least its center and
its spread, or variability.
The mean –x and the median describe the center of a distribution in different ways. The mean is the average of the observations, and the median is the
midpoint of the values.
When you use the median to indicate the center of a distribution, describe its
spread using the quartiles. The first quartile Q1 has about one-fourth of the
observations below it, and the third quartile Q3 has about three-fourths of
the observations below it. The interquartile range (IQR) is the range of the
middle 50% of the observations and is found by IQR = Q3 − Q1.
An extreme observation is an outlier if it is smaller than Q1 − (1.5 × IQR) or
larger than Q3 + (1.5 × IQR).
The five-number summary consisting of the median, the quartiles, and the
maximum and minimum values provides a quick overall description of a distribution. The median describes the center, and the IQR and range describe
the spread.
Boxplots based on the five-number summary are useful for comparing distributions. The box spans the quartiles and shows the spread of the middle
half of the distribution. The median is marked within the box. Lines extend
from the box to the smallest and the largest observations that are not outliers.
Outliers are plotted as isolated points.
The variance s2x and especially its square root, the standard deviation sx, are common measures of spread about the mean. The standard
deviation sx is zero when there is no variability and gets larger as the spread
increases.
The median is a resistant measure of center because it is relatively unaffected by extreme observations. The mean is nonresistant. Among measures
of spread, the IQR is resistant, but the standard deviation and range are not.
The mean and standard deviation are good descriptions for roughly symmetric distributions without outliers. They are most useful for the Normal
distributions introduced in the next chapter. The median and IQR are a better description for skewed distributions.
Numerical summaries do not fully describe the shape of a distribution.
Always plot your data.
1.3 T echnology
Corners
TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site.
3. Making calculator boxplots
4. Computing numerical summaries with technology
Starnes-Yates5e_c01_xxiv-081hr3.indd 68
page 59
page 63
11/13/13 1:07 PM
69
Section 1.3 Describing Quantitative Data with Numbers
Section 1.3
Exercises
79. Quiz grades Joey’s first 14 quiz grades in a marking
period were
pg 49
86
87
84
76
91
96
75
82
78
90
80
98
74
93
Calculate the mean. Show your work.
80. Cowboys The 2011 roster of the Dallas Cowboys
professional football team included 7 defensive linemen. Their weights (in pounds) were 321, 285, 300,
285, 286, 293, and 298. Calculate the mean. Show
your work.
81. Quiz grades Refer to Exercise 79.
(a) Find the median by hand. Show your work.
(b) Suppose Joey has an unexcused absence for the 15th
52 quiz, and he receives a score of zero. Recalculate the
mean and the median. What property of measures of
center does this illustrate?
82. Cowboys Refer to Exercise 80.
(a) Find the median by hand. Show your work.
(b) Suppose the heaviest lineman had weighed 341
pounds instead of 321 pounds. How would this
change affect the mean and the median? What property of measures of center does this illustrate?
83. Incomes of college grads According to the Census
Bureau, the mean and median income in a recent
year of people at least 25 years old who had a bachelor’s degree but no higher degree were $48,097 and
$60,954. Which of these numbers is the mean and
which is the median? Explain your reasoning.
84. House prices The mean and median selling prices
of existing single-family homes sold in July 2012 were
$263,200 and $224,200.41 Which of these numbers is
the mean and which is the median? Explain how you
know.
85. Baseball salaries Suppose that a Major League Baseball team’s mean yearly salary for its players is $1.2
million and that the team has 25 players on its active
roster. What is the team’s total annual payroll? If you
knew only the median salary, would you be able to
answer this question? Why or why not?
86. Mean salary? Last year a small accounting firm paid
each of its five clerks $22,000, two junior accountants
$50,000 each, and the firm’s owner $270,000. What is
the mean salary paid at this firm? How many of the employees earn less than the mean? What is the median
salary? Write a sentence to describe how an unethical
recruiter could use statistics to mislead prospective
employees.
Starnes-Yates5e_c01_xxiv-081hr3.indd 69
(a) Estimate the mean and median of the distribution.
Explain your method clearly.
(b) If you wanted to argue that shorter domain names were
more popular, which measure of center would you
choose—the mean or the median? Justify your answer.
88. Do adolescent girls eat fruit? We all know that fruit
is good for us. Below is a histogram of the number of
servings of fruit per day claimed by 74 seventeen-yearold girls in a study in Pennsylvania.42
15
Number of subjects
pg
87. Domain names When it comes to Internet domain
names, is shorter better? According to one ranking
of Web sites in 2012, the top 8 sites (by number of
“hits”) were google.com, youtube.com, wikipedia.org,
yahoo.com, amazon.com, ebay.com, craigslist.org, and
facebook.com. These familiar sites certainly have short
domain names. The histogram below shows the domain name lengths (in number of letters in the name,
not including the extensions .com and .org) for the 500
most popular Web sites.
10
5
0
0
1
2
3
4
5
6
7
8
Servings of fruit per day
(a) With a little care, you can find the median and the
quartiles from the histogram. What are these numbers? How did you find them?
(b) Estimate the mean of the distribution. Explain your
method clearly.
11/13/13 1:07 PM
70
pg 55
pg 56
pg
CHAPTER 1
E x p l o r i n g Data
89. Quiz grades Refer to Exercise 79.
(a) Find and interpret the interquartile range (IQR).
(b) Determine whether there are any outliers. Show your
work.
90. Cowboys Refer to Exercise 80.
(a) Find and interpret the interquartile range (IQR).
(b) Determine whether there are any outliers. Show your
work.
91. Don’t call me In a September 28, 2008, article titled
57 “Letting Our Fingers Do the Talking,” the New York
Times reported that Americans now send more text
messages than they make phone calls. According to a
study by Nielsen Mobile, “Teenagers ages 13 to 17 are
by far the most prolific texters, sending or receiving
1742 messages a month.” Mr. Williams, a high school
statistics teacher, was skeptical about the claims in
the article. So he collected data from his first-period
statistics class on the number of text messages and
calls they had sent or received in the past 24 hours.
Here are the texting data:
0
7
8 118
1
72
29
0
25
92
8
52
5
14
1
3
25
3
98
44
9 0 26
5 42
(a) Make a boxplot of these data by hand. Be sure to
check for outliers.
(b) Explain how these data seem to contradict the claim
in the article.
92. Acing the first test Here are the scores of Mrs. Liao’s
students on their first statistics test:
(b) Can we draw any conclusion about the preferences
of all students in the school based on the data from
Mr. Williams’s statistics class? Why or why not?
94. Electoral votes To become president of the United
States, a candidate does not have to receive a majority
of the popular vote. The candidate does have to win a
majority of the 538 electoral votes that are cast in the
Electoral College. Here is a stemplot of the number of
electoral votes for each of the 50 states and the District
of Columbia.
0
0
1
1
2
2
3
3
4
4
5
5
Key: 1|5 is a
state with 15
electoral votes.
5
(a) Make a boxplot of these data by hand. Be sure to
check for outliers.
(b) Which measure of center and spread would you use
to summarize the distribution—the mean and standard deviation or the median and IQR? Justify your
answer.
95. Comparing investments Should you put your money into a fund that buys stocks or a fund that invests in
real estate? The boxplots compare the daily returns
(in percent) on a “total stock market” fund and a
real estate fund over a one-year period.43
Daily percent return
93 93 87.5 91 94.5 72 96 95 93.5 93.5 73
82 45 88 80 86 85.5 87.5 81 78 86 89
92 91 98 85 82.5 88 94.5 43
(a) Make a boxplot of the test score data by hand. Be sure
to check for outliers.
(b) How did the students do on Mrs. Liao’s first test?
Justify your answer.
93. Texts or calls? Refer to Exercise 91. A boxplot of the
difference (texts – calls) in the number of texts and
calls for each student is shown below.
3333333344444
55555666777788999
0000111123
5557
011
7
14
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
–0.5
–1.0
–1.5
–2.0
–2.5
–3.0
–3.5
–4.0
Stocks
Real estate
Type of investment
–20
0
20
40
60
80
100
120
Difference (texts – calls)
(a) Do these data support the claim in the article about
texting versus calling? Justify your answer with appropriate evidence.
Starnes-Yates5e_c01_xxiv-081hr3.indd 70
(a) Read the graph: about what were the highest and lowest daily returns on the stock fund?
(b) Read the graph: the median return was about the
same on both investments. About what was the median return?
(c) What is the most important difference between the
two distributions?
11/13/13 1:08 PM
71
Section 1.3 Describing Quantitative Data with Numbers
Total income (thousands of dollars)
96. Income and education level Each March, the
Bureau of Labor Statistics compiles an Annual
Demographic Supplement to its monthly Current Population Survey.44 Data on about 71,067
individuals between the ages of 25 and 64 who
were employed full-time were collected in one of
these surveys. The boxplots below compare the
distributions of income for people with five levels of
education. This figure is a variation of the boxplot
idea: because large data sets often contain very
extreme observations, we omitted the individuals
in each category with the top 5% and bottom 5% of
incomes. Write a brief description of how the distribution of income changes with the highest level of
education reached. Give specifics from the graphs
to support your statements.
300
250
(c)
Do you think it’s safe to conclude that the mean
amount of sleep for all 30 students in this class is
close to 8 hours? Why or why not?
99.
Shopping spree The figure displays computer output
for data on the amount spent by 50 grocery shoppers.
(a)
What would you guess is the shape of the distribution based only on the computer output? Explain.
Interpret the value of the standard deviation.
Are there any outliers? Justify your answer.
(b)
(c)
100. C-sections Do male doctors perform more cesarean
sections (C-sections) than female doctors? A study
in Switzerland examined the number of cesarean
sections (surgical deliveries of babies) performed in
a year by samples of male and female doctors. Here
are summary statistics for the two distributions:
x
200
150
100
50
Male
doctors
41.333 20.607
Female
doctors
19.1
(a)
0
Not
HS grad
HS
grad
Some
college
Bachelor’s Advanced
97. Phosphate levels The level of various substances in
the blood influences our health. Here are measurements of the level of phosphate in the blood of a patient, in milligrams of phosphate per deciliter of blood,
made on 6 consecutive visits to a clinic: 5.6, 5.2, 4.6,
4.9, 5.7, 6.4. A graph of only 6 observations gives little
information, so we proceed to compute the mean and
standard deviation.
(a) Find the standard deviation from its definition. That
is, find the deviations of each observation from the
mean, square the deviations, then obtain the variance
and the standard deviation.
(b) Interpret the value of sx you obtained in part (a).
98. Feeling sleepy? The first four students to arrive for
a first-period statistics class were asked how much
sleep (to the nearest hour) they got last night. Their
responses were 7, 7, 9, and 9.
(a) Find the standard deviation from its definition. That
is, find the deviations of each observation from the
mean, square the deviations, then obtain the variance
and the standard deviation.
(b) Interpret the value of sx you obtained in part (a).
Starnes-Yates5e_c01_xxiv-081hr3.indd 71
sx
(b)
(c)
10.126
Min
Q1
Med
Q3
Max
IQR
20
27
34
50
86
23
5
10
18.5
29
33
19
Based on the computer output, which distribution
would you guess has a more symmetrical shape?
Explain.
Explain how the IQRs of these two distributions can
be so similar even though the standard deviations
are quite different.
Does it appear that male doctors perform more Csections? Justify your answer.
101. The IQR Is the interquartile range a resistant measure of spread? Give an example of a small data set
that supports your answer.
102. What do they measure? For each of the following
summary statistics, decide (i) whether it could be
used to measure center or spread and (ii) whether it
is resistant.
(a)
Q1 + Q3
Max − Min
(b)
​
2
2
103. SD contest This is a standard deviation contest. You
must choose four numbers from the whole numbers
0 to 10, with repeats allowed.
(a) Choose four numbers that have the smallest possible
standard deviation.
(b) Choose four numbers that have the largest possible
standard deviation.
(c) Is more than one choice possible in either part (a) or
(b)? Explain.
11/13/13 1:08 PM
72
CHAPTER 1
E x p l o r i n g Data
104. Measuring spread Which of the distributions
shown has a larger standard deviation? Justify your
answer.
H. caribaea red
41.90
39.63
38.10
42.01
42.18
37.97
41.93
40.66
38.79
43.09
37.87
38.23
41.47
39.16
38.87
41.69
37.40
37.78
39.78
38.20
38.01
40.57
38.07
36.11
35.68
36.03
36.03
35.45
34.57
38.13
34.63
37.10
H. caribaea yellow
36.78
35.17
37.02
36.82
36.52
36.66
Multiple choice: Select the best answer for Exercises 107
to 110.
105. SSHA scores Here are the scores on the Survey of
Study Habits and Attitudes (SSHA) for 18 first-year
college women:
pg 65
STEP
4
154
103
109
126
137
126
115
137
152
165
140
165
154
129
178
200
101
148
and for 20 first-year college men:
108
92
113
STEP
140
114
169
146
151 70
91
109
115
180
132
187
115
126
75 88
104
Do these data support the belief that men and
women differ in their study habits and attitudes toward learning? (Note that high scores indicate good
study habits and attitudes toward learning.) Follow
the four-step process.
106. Hummingbirds and tropical flower Researchers from Amherst College studied the relationship
between varieties of the tropical flower Heliconia on
the island of Dominica and the different species of
hummingbirds that fertilize the flowers.45 Over time,
the researchers believe, the lengths of the flowers and
the forms of the hummingbirds’ beaks have evolved to
match each other. If that is true, flower varieties fertilized by different hummingbird species should have
distinct distributions of length.
The table below gives length measurements (in
millimeters) for samples of three varieties of
Heliconia, each fertilized by a different species of
hummingbird. Do these data support the researchers’
belief? Follow the four-step process.
4
H. bihai
47.12
48.07
46.75
48.34
46.80
48.15
Starnes-Yates5e_c01_xxiv-081hr3.indd 72
47.12
50.26
46.67
50.12
47.43
46.34
46.44
46.94
46.64
48.36
107.
(a)
(b)
(c)
If a distribution is skewed to the right with no outliers,
mean < median.
(d) mean > median.
mean ≈ median.
(e) We can’t tell without
mean = median. examining the data.
108. The scores on a statistics test had a mean of 81 and
a standard deviation of 9. One student was absent
on the test day, and his score wasn’t included in the
calculation. If his score of 84 was added to the distribution of scores, what would happen to the mean and
standard deviation?
(a) Mean will increase, and standard deviation will
increase.
(b) Mean will increase, and standard deviation will
decrease.
(c) Mean will increase, and standard deviation will stay
the same.
(d) Mean will decrease, and standard deviation will
increase.
(e) Mean will decrease, and standard deviation will
decrease.
109. The stemplot shows the number of home runs hit by
each of the 30 Major League Baseball teams in 2011.
Home run totals above what value should be considered outliers?
09 15
10 3789
11 47
Key: 14|8 is a
12 19
team with 148
13
home runs.
14 89
15 34445
16 239
17 223
18 356
19 1
20 3
21 0
22 2
(a) 173 (b) 210 (c) 222 (d) 229 (e) 257
11/13/13 1:08 PM
73
Section 1.3 Describing Quantitative Data with Numbers
110. Which of the following boxplots best matches the
distribution shown in the histogram?
30
Frequency
25
20
15
10
5
12
rs
on
0
2
4
6 8
Data
10 12
0
(c)
2
4
6 8
Data
to
pa
10 12
Method of communication
(d)
(a)
0
yS
In
te
rn
e
C
el
lp
(b)
rM
SN
ce
/F
ac
Te
eb
le
oo
ph
k
on
e
(la
nd
lin
Te
e)
xt
m
es
sa
gi
ng
0
M
10
pe
(a)
8
tc
ha
6
Data
ne
4
In
2
ho
0
2
4
6 8
Data
10 12
0
2
4
6 8
Data
10 12
(e)
0
2
4
6 8
Data
10 12
Exercises 111 and 112 refer to the following setting.
We used CensusAtSchool’s “Random Data Selector” to
choose a sample of 50 Canadian students who completed a
survey in a recent year.
(b)
Would it be appropriate to make a pie chart for these
data? Why or why not?
Jerry says that he would describe this bar graph as
skewed to the right. Explain why Jerry is wrong.
113. Success in college (1.1) The 2007 Freshman Survey
asked first-year college students about their “habits of
mind”—specific behaviors that college faculty have
identified as being important for student success. One
question asked students, “How often in the past year
did you revise your papers to improve your writing?”
Another asked, “How often in the past year did you
seek feedback on your academic work?” The figure is
a bar graph comparing male and female responses to
these two questions.46
111. How tall are you? (1.2) Here are the students’ heights
(in centimeters).
178
163 150.5 169
173
169
190
178
161 171
170
191
168.5 178.5 173
175160.5
166
164 163
174
160
174
182
167
166
170
170
181 171.5 160
178
157
165
187
168
157.5
145.5 156 182
183
168.5 177
171
162.5 160.5 185.5
Make an appropriate graph to display these data.
Describe the shape, center, and spread of the distribution. Are there any outliers?
112. Let’s chat (1.1) The bar graph displays data on
students’ responses to the question “Which of these
methods do you most often use to communicate with
your friends?”
Starnes-Yates5e_c01_xxiv-081hr3.indd 73
Female
166
80
% “Frequently”
166.5 170
Male
100
54.9%
60
40
49.0%
36.9%
37.6%
Revise papers to
improve writing
Seek feedback
on work
20
0
What does the graph tell us about the habits of mind
of male and female college freshmen?
11/13/13 1:08 PM
74
CHAPTER 1
E x p l o r i n g Data
FRAPPY! Free Response AP® Problem, Yay!
The following problem is modeled after actual AP® Statistics exam
free response questions. Your task is to generate a complete, concise response in 15 minutes.
Directions: Show all your work. Indicate clearly the methods
you use, because you will be scored on the correctness of your
methods as well as on the accuracy and completeness of your
results and explanations.
Using data from the 2010 census, a random sample of 348
U.S. residents aged 18 and older was selected. Among the
variables recorded were gender (male or female), housing status (rent or own), and marital status (married or not married).
The two-way table below summarizes the relationship
between gender and housing status.
Male
Own132
Rent50
Total182
Female
122
44
166
Total
254
94
348
(a) What percent of males in the sample own their
home?
(b) Make a graph to compare the distribution of housing status for males and females.
(c) Using your graph from part (b), describe the relationship between gender and housing status.
(d) The two-way table below summarizes the relationship between marital status and housing status.
Married
Own
172
Rent
40
Total212
Not Married
82
54
136
Total
254
94
348
For the members of the sample, is the relationship between
marital status and housing status stronger or weaker than
the relationship between gender and housing status that
you described in part (c)? Justify your choice using the data
provided in the two-way tables.
After you finish, you can view two example solutions on the book’s
Web site (www.whfreeman.com/tps5e). Determine whether you
think each solution is “complete,” “substantial,” “developing,” or
“minimal.” If the solution is not complete, what improvements would
you suggest to the student who wrote it? Finally, your teacher will
provide you with a scoring rubric. Score your response and note
what, if anything, you would do differently to improve your own
score.
Chapter Review
Introduction: Data Analysis: Making Sense of Data
In this brief section, you learned several fundamental
concepts that will be important throughout the course:
the idea of a distribution and the distinction between
quantitative and categorical variables. You also learned a
strategy for exploring data:
•
Begin by examining each variable by itself. Then move
on to study relationships between variables.
•
Start with a graph or graphs. Then add numerical
summaries.
Section 1.1: Analyzing Categorical Data
In this section, you learned how to display the distribution of
a single categorical variable with pie charts and bar graphs
Starnes-Yates5e_c01_xxiv-081hr3.indd 74
and what to look for when describing these displays. Remember to properly label your graphs! Poor labeling is an
easy way to lose points on the AP® exam. You should also be
able to recognize misleading graphs and be careful to avoid
making misleading graphs yourself.
Next, you learned how to investigate the association
between two categorical variables. Using a two-way table,
you learned how to calculate and display marginal and
conditional distributions. Graphing and comparing conditional distributions allow you to look for an association
between the variables. If there is no association between
the two variables, graphs of the conditional distributions
will look the same. However, if differences in the conditional distributions do exist, there is an association between the variables.
11/13/13 1:08 PM
Section 1.2: Displaying Quantitative Data with Graphs
In this section, you learned how to create three different types of graphs for a quantitative variable: dotplots,
stemplots, and histograms. Each of the graphs has distinct benefits, but all of them are good tools for examining the distribution of a quantitative variable. Dotplots
and stemplots are handy for small sets of data. Histograms
are the best choice when there are a large number of
observations. On the AP® exam, you will be expected to
create each of these types of graphs, label them properly,
and comment on their characteristics.
When you are describing the distribution of a quantitative variable, you should look at its graph for the overall
pattern (shape, center, spread) and striking departures
from that pattern (outliers). Use the acronym SOCS
(shape, outliers, center, spread) to help remember these
four characteristics. Likewise, when comparing distributions, you should include explicit comparison words
such as “is greater than” or “is approximately the same
as.” When asked to compare distributions, a very common mistake on the AP® exam is describing the characteristics of each distribution separately without making
these explicit comparisons.
Section 1.3: Describing Quantitative Data with Numbers
To measure the center of a distribution of quantitative data,
you learned how to calculate the mean and the median of a
distribution. You also learned that the median is a resistant
measure of center but the mean isn’t resistant because it can
be greatly affected by skewness or outliers.
To measure the spread of a distribution of quantitative
data, you learned how to calculate the range, interquartile
range, and standard deviation. The interquartile range (IQR)
is a resistant measure of spread because it ignores the upper
25% and lower 25% of the distribution, but the range isn’t
resistant because it uses only the minimum and maximum
value. The standard deviation is the most commonly used
measure of spread and approximates the typical distance of a
value in the data set from the mean. The standard deviation
is not resistant—it is heavily affected by extreme values.
To identify outliers in a distribution of quantitative data,
you learned the 1.5 × IQR rule. You also learned that boxplots
are a great way to visually summarize the distribution of quantitative data. Boxplots are helpful for comparing distributions
because they make it easy to compare both center (median)
and spread (range, IQR). Yet boxplots aren’t as useful for displaying the shape of a distribution because they do not display
modes, clusters, gaps, and other interesting features.
What Did You Learn?
Learning Objective
Section
Related Example
on Page
Relevant Chapter
Review Exercise(s)
Identify the individuals and variables in a set of data. Intro
3 R1.1
Classify variables as categorical or quantitative. Intro
3
R1.1
Display categorical data with a bar graph. Decide whether it
would be appropriate to make a pie chart. 1.1
9
R1.2, R1.3
Identify what makes some graphs of categorical data deceptive. 1.1
10
R1.3
Calculate and display the marginal distribution of a categorical
variable from a two-way table. 1.1
13
R1.4
Calculate and display the conditional distribution of a categorical
variable for a particular value of the other categorical variable in
a two-way table. 1.1
15
R1.4
Describe the association between two categorical variables by
comparing appropriate conditional distributions. 1.1
17
R1.5
Dotplots: 25
Make and interpret dotplots and stemplots of quantitative data. 1.2Stemplots: 31
R1.6
Describe the overall pattern (shape, center, and spread) of a
distribution and identify any major departures from the pattern
(outliers). 1.2 Dotplots: 26
R1.6, R1.9
Identify the shape of a distribution from a graph as roughly
symmetric or skewed. 1.2
28
R1.6, R1.7, R1.8, R1.9
75
Starnes-Yates5e_c01_xxiv-081hr3.indd 75
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76
CHAPTER 1
E x p l o r i n g Data
What Did You Learn? (continued)
Learning Objective
Section
Related Example
on Page(s)
Relevant Chapter
Review Exercise(s)
Make and interpret histograms of quantitative data. 1.2
33
R1.7, R1.8
Compare distributions of quantitative data using dotplots,
stemplots, or histograms. 1.2
30
R1.8, R1.10
Mean: 49
Calculate measures of center (mean, median). 1.3 Median: 52
R1.6
Calculate and interpret measures of spread (range, IQR, IQR: 55
standard deviation). 1.3 Std. dev.: 60 R1.9
Choose the most appropriate measure of center and spread
in a given setting. 1.3
65
R1.7
Identify outliers using the 1.5 × IQR rule. 1.3
56
R1.6, R1.7, R1.9
Make and interpret boxplots of quantitative data. 1.3
57
R1.7
Use appropriate graphs and numerical summaries to compare
distributions of quantitative variables. 1.3
65
R1.8, R1.10
Chapter 1 Chapter Review Exercises
These exercises are designed to help you review the important
ideas and methods of the chapter.
R1.1 Hit movies According to the Internet Movie Database, Avatar is tops based on box office sales worldwide. The following table displays data on several
popular movies.47
Time Box office
Movie
Year Rating (minutes) Genre (dollars)
Avatar
2009
PG-13
162
Action
2,781,505,847
Titanic
1997
PG-13
194
Drama
1,835,300,000
Harry Potter and
the Deathly
Hallows: Part 2
2011
PG-13
130
Fantasy
1,327,655,619
Transformers:
Dark of the Moon
2011
PG-13
154
Action
1,123,146,996
The Lord of the
Rings: The Return
of the King
R1.2​
Movie ratings The movie rating system we use today
was first established on November 1, 1968. Back
then, the possible ratings were G, PG, R, and X. In
1984, the PG-13 rating was created. And in 1990,
NC-17 replaced the X rating. Here is a summary
of the ratings assigned to movies between 1968 and
2000: 8% rated G, 24% rated PG, 10% rated PG-13,
55% rated R, and 3% rated NC-17.48 Make an appropriate graph for displaying these data.
R1.3 I​ ’d die without my phone! In a survey of over 2000
U.S. teenagers by Harris Interactive, 47% said that
“their social life would end or be worsened without
their cell phone.”49 One survey question asked the
teens how important it is for their phone to have
certain features. The figure below displays data on the
percent who indicated that a particular feature is vital.
60
2003
PG-13
201
Pirates of the
Caribbean: Dead
Man’s Chest
2006
PG-13
151
Toy Story 3
2010
G
103
Action
1,119,929,521
50
46%
39%
40
Action
1,065,896,541
30
Animation 1,062,984,497
19%
20
(a) What individuals does this data set describe?
(b) Clearly identify each of the variables. Which are
quantitative?
(c) Describe the individual in the highlighted row.
Starnes-Yates5e_c01_xxiv-081hr3.indd 76
17%
10
0
Make/receive
calls
Send/receive
text messages
Camera
Send/receive
pictures
11/13/13 1:08 PM
Chapter Review Exercises
Age
Facebook
user?
Younger
(18–22)
Middle
(23–27)
Older
(28 and up)
Yes
78
49
21
No
4
21
46
(a) What percent of the students who responded were
Facebook users? Is this percent part of a marginal
distribution or a conditional distribution? Explain.
(b) What percent of the younger students in the sample
were Facebook users? What percent of the Facebook
users in the sample were younger s­ tudents?
R1.5 Facebook and age Use the data in the previous
­exercise to determine whether there is an association bet­ween Facebook use and age. Give appropriate graphical and numerical evidence to support
your answer.
R1.6​
Density of the earth In 1798, the English scientist
Henry Cavendish measured the density of the earth
several times by careful work with a torsion balance. The variable recorded was the density of the
earth as a multiple of the density of water. Here are
Cavendish’s 29 measurements:51
5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.65
5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.39
5.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85
(a) Present these measurements graphically in a stemplot.
(b) Discuss the shape, center, and spread of the distribution. Are there any outliers?
(c) What is your estimate of the density of the earth
based on these measurements? Explain.
R1.7​
Guinea pig survival times Here are the survival times
in days of 72 guinea pigs after they were injected with
infectious bacteria in a medical experiment.52 Survival
times, whether of machines under stress or cancer
patients after treatment, usually have distributions that
are skewed to the right.
43
45
53
56
56
57
58
66
67
73
74
79
80
80
81
81
81
82
83
83
84
88
89
91
91
92
92
97
99
99 100 100 101 102 102 102
103 104 107 108 109 113 114 118 121 123 126 128
137 138 139 144 145 147 156 162 174 178 179 184
191 198 211 214 243 249 329 380 403 511 522 598
Starnes-Yates5e_c01_xxiv-081hr3.indd 77
R1.8​
Household incomes Rich and poor households differ
in ways that go beyond income. Following are histograms that compare the distributions of household size
(number of people) for low-income and high-income
households.53 Low-income households had annual incomes less than $15,000, and high-income households
had annual incomes of at least $100,000.
60
50
40
Percent
R1.4 F
​ acebook and age Is there a relationship between
Facebook use and age among college students? The
following two-way table displays data for the 219
students who ­responded to the survey.50
(a) M
​ ake a histogram of the data and describe its main
features. Does it show the expected right skew?
(b) Now make a boxplot of the data. Be sure to check
for outliers.
(c) Which measure of center and spread would you
use to summarize the distribution—the mean and
standard deviation or the median and IQR? Justify
your answer.
30
20
10
0
1
2
3
4
5
6
7
Household size, low income
60
50
40
Percent
(a) Explain how the graph gives a misleading i­mpression.
(b) Would it be appropriate to make a pie chart to
­display these data? Why or why not?
(c) Make a graph of the data that isn’t misleading.
77
30
20
10
0
1
2
3
4
5
6
7
Household size, high income
(a) About what percent of each group of households
consisted of two people?
(b) What are the important differences between these
two distributions? What do you think explains these
differences?
Exercises R1.9 and R1.10 refer to the following setting. Do
you like to eat tuna? Many people do. Unfortunately, some
of the tuna that people eat may contain high levels of mercury. Exposure to mercury can be especially hazardous for
pregnant women and small children. How much mercury
is safe to consume? The Food and Drug Administration will
take action (like removing the product from store shelves)
if the mercury concentration in a six-ounce can of tuna is
1.00 ppm (parts per million) or higher.
11/13/13 1:08 PM
78
CHAPTER 1
E x p l o r i n g Data
What is the typical mercury concentration in cans of
tuna sold in stores? A study conducted by Defenders of
Wildlife set out to answer this question. Defenders collected a sample of 164 cans of tuna from stores across the
United States. They sent the selected cans to a laboratory
that is often used by the Environmental Protection Agency
for mercury testing.54
R1.9​
Mercury in tuna A histogram and some computer
output provide information about the mercury concentration in the sampled cans (in parts per million,
ppm).
(a) Interpret the standard deviation in context.
(b) Determine whether there are any outliers.
(c) Describe the shape, center, and spread of the
distribution.
R1.10 Mercury in tuna Is there a difference in the
mercury concentration of light tuna and albacore
tuna? Use the parallel boxplots and the computer
output to write a few sentences comparing the two
distributions.
Descriptive Statistics: Mercury_ppm
Descriptive Statistics: Mercury_ppm
Variable
Mercury
Variable
Mercury
N
164
Q1
0.071
Mean
0.285
Med
0.180
StDev
0.300
Q3
0.380
Min
0.012
Max
1.500
Type
Albacore
Light
Type
Albacore
Light
N
20
144
Q1
0.293
0.059
Mean
0.401
0.269
Med
0.400
0.160
StDev
0.152
0.312
Q3
0.460
0.347
Min
0.170
0.012
Max
0.730
1.500
Chapter 1 AP® Statistics Practice Test
Section I: Multiple Choice Select the best answer for each question.
T1.1 You record the age, marital status, and earned income of a sample of 1463 women. The number and
type of variables you have recorded is
(a)
(b)
(c)
(d)
(e)
3 quantitative, 0 categorical.
4 quantitative, 0 categorical.
3 quantitative, 1 categorical.
2 quantitative, 1 categorical.
2 quantitative, 2 categorical.
T1.2 Consumers Union measured the gas mileage in
miles per gallon of 38 vehicles from the same model
year on a special test track. The pie chart provides
Starnes-Yates5e_c01_xxiv-081hr3.indd 78
information about the country of manufacture of
the model cars tested by Consumers Union. Based
on the pie chart, we conclude that
(a) Japanese cars get significantly lower gas mileage than
cars from other countries.
(b) U.S. cars get significantly higher gas mileage than
cars from other countries.
(c) Swedish cars get gas mileages that are between those
of Japanese and U.S. cars.
(d) cars from France have the lowest gas mileage.
(e) more than half of the cars in the study were from the
United States.
11/13/13 1:08 PM
AP® Statistics Practice Test
79
(c)
20
U.S.
15
10
France
5
Germany
Sweden
Japan
Italy
France
Germany
Italy
Japan
Sweden
U.S.
France
Germany
Italy
Japan
Sweden
U.S.
(d)
20
T1.3 Which of the following bar graphs is equivalent to
the pie chart in Question T1.2?
15
(a)
15
10
5
10
5
(e) None of these.
France
Germany
Italy
Japan
Sweden
U.S.
(b)
T1.4 Earthquake intensities are measured using a device
called a seismograph, which is designed to be most
sensitive to earthquakes with intensities between 4.0
and 9.0 on the Richter scale. Measurements of nine
earthquakes gave the following readings:
25
4.5 L 5.5 H 8.7 8.9 6.0 H 5.2
20
15
10
5
France
Starnes-Yates5e_c01_xxiv-081hr3.indd 79
Germany
Italy
Japan
Sweden
U.S.
(a)
(b)
(c)
(d)
(e)
where L indicates that the earthquake had an intensity b
­ elow 4.0 and an H indicates that the earthquake
had an intensity above 9.0. The median earthquake
intensity of the sample is
5.75. 6.00. 6.47.
8.70.
Cannot be determined.
11/13/13 1:08 PM
80
CHAPTER 1
E x p l o r i n g Data
Questions T1.5 and T1.6 refer to the following setting. In a
statistics class with 136 students, the professor records how
much money (in dollars) each student has in his or her possession during the first class of the semester. The histogram
shows the data that were collected.
60
Frequency
50
40
Business size
30
Response?
Small
Medium
Large
20
Yes
125
81
40
10
No
75
119
160
0
0
10
20
30
40
50
60
70
80
90 100 110
Amount of money
T1.5 The percentage of students with less than $10 in
their possession is closest to
(a) 30%.
(b) 35%. (c) 45%.
(d) 60%.
(e) 70%.
T1.6 Which of the following statements about this distribution is not correct?
(a)
(b)
(c)
(d)
(e)
Questions T1.9 and T1.10 refer to the following setting. A survey
was designed to study how business operations vary according
to their size. Companies were classified as small, medium, or
large. Questionnaires were sent to 200 randomly selected businesses of each size. Because not all questionnaires in a survey
of this type are returned, researchers decided to investigate the
relationship between the response rate and the size of the business. The data are given in the following two-way table:
The histogram is right-skewed.
The median is less than $20.
The IQR is $35.
The mean is greater than the median.
The histogram is unimodal.
T1.7 Forty students took a statistics examination having
a maximum of 50 points. The score distribution is
given in the following stem-and-leaf plot:
0
1
2
3
4
5
28
2245
01333358889
001356679
22444466788
000
The third quartile of the score distribution is equal to
(a) 45. (b) 44. (c) 43. (d) 32. (e) 23.
T1.8 The mean salary of all female workers is $35,000. The
mean salary of all male workers is $41,000. What
must be true about the mean salary of all workers?
(a) It must be $38,000.
(b) It must be larger than the median salary.
(c) It could be any number between $35,000 and
$41,000.
(d) It must be larger than $38,000.
(e) It cannot be larger than $40,000.
Starnes-Yates5e_c01_xxiv-081hr3.indd 80
T1.9
What percent of all small companies receiving
questionnaires responded?
(a) 12.5% (c) 33.3% (e) 62.5%
(b) 20.8% (d) 50.8%
T1.10 Which of the following conclusions seems to be
supported by the data?
(a) There are more small companies than large companies in the survey.
(b) Small companies appear to have a higher response
rate than medium or big companies.
(c) Exactly the same number of companies responded
as didn’t respond.
(d) Overall, more than half of companies responded to
the survey.
(e) If we combined the medium and large companies,
then their response rate would be equal to that of
the small companies.
T1.11 An experiment was conducted to investigate the effect
of a new weed killer to prevent weed growth in onion
crops. Two chemicals were used: the standard weed
killer (C) and the new chemical (W). Both chemicals
were tested at high and low concentrations on a total
of 50 test plots. The percent of weeds that grew in
each plot was recorded. Here are some boxplots of the
results. Which of the following is not a correct statement about the results of this experiment?
W—low conc.
C—low conc.
W—high conc.
*
*
*
C—high conc.
0
10
20
30
40
50
Percent of weeds that grew
11/13/13 1:08 PM
AP® Statistics Practice Test
(a) At both high and low concentrations, the new chemical
(W) gives better weed control than the standard weed
killer (C).
(b) Fewer weeds grew at higher concentrations of both
chemicals.
(c) The results for the standard weed killer (C) are less
variable than those for the new chemical (W).
81
(d) High and low concentrations of either chemical have
approximately the same effects on weed growth.
(e) Some of the results for the low concentration of weed
killer W show fewer weeds growing than some of the
results for the high concentration of W.
Section II: Free Response Show all your work. Indicate clearly the methods you use, because you will be graded on
the correctness of your methods as well as on the accuracy and completeness of your results and explanations.
T1.12 You are interested in how much time students
spend on the Internet each day. Here are data on
the time spent on the Internet (in minutes) for a
particular day r­ eported by a random sample of 30
students at a large high school:
7
20
24
25
25
28
28
30
32
35
42
43
44
45
46
47
48
48
50
51
72
75
77
78
79
83
87
88
135
151
(a) Construct a histogram of these data.
(b) Are there any outliers? Justify your answer.
(c) Would it be better to use the mean and standard
deviation or the median and IQR to describe the
center and spread of this distribution? Why?
T1.13 A study among the Pima Indians of Arizona investigated the relationship between a mother’s diabetic
status and the appearance of birth defects in her children. The results appear in the two-way table below.
Diabetic Status
Birth Defects
Nondiabetic
Prediabetic
Diabetic
754
362
38
31
13
9
None
One or more
Total
Total
(a) Fill in the row and column totals in the margins of
the table.
(b)​Compute (in percents) the conditional distributions
of birth defects for each diabetic status.
(c) Display the conditional distributions in a graph.
Don’t forget to label your graph completely.
(d) Do these data give evidence of an association between
diabetic status and birth defects? Justify your answer.
Starnes-Yates5e_c01_xxiv-081hr3.indd 81
T1.14 The back-to-back stemplot shows the lifetimes of
several Brand X and Brand Y batteries.
Brand X
2110
99775
3221
4
5
Brand Y
1
1
2
2
3
3
4
4
5
5
7
2
6
223334
56889
0
Key: 4|2 represents
420–429 hours.
(a) What is the longest that any battery lasted?
(b) Give a reason someone might prefer a Brand
X battery.
(c) Give a reason someone might prefer a Brand
Y battery.
T1.15 During the early part of the 1994 baseball season,
many fans and players noticed that the number of
home runs being hit seemed unusually large. Here
are the data on the number of home runs hit by
American League and National League teams in
the early part of the 1994 season:
American League: 35 40 43 49 51 54 57 58 58 64 68 68 75 77
National League: 29 31 42 46 47 48 48 53 55 55 55 63 63 67
Compare the distributions of home runs for the two
leagues graphically and numerically. Write a few
sentences summarizing your findings.
11/13/13 1:08 PM
Chapter
2
Introduction
84
Section 2.1
Describing Location in a
Distribution
85
Section 2.2
Density Curves and Normal
Distributions
103
Free Response
AP® Problem, YAY!
134
Chapter 2 Review
134
Chapter 2 Review Exercises
136
Chapter 2 AP® Statistics
Practice Test
137
Starnes-Yates5e_c02_082-139hr2.indd 82
11/13/13 1:32 PM
Modeling Distributions
of Data
case study
Do You Sudoku?
The sudoku craze has officially swept the globe. Here’s what Will Shortz, crossword puzzle editor for the
New York Times, said about sudoku:
As humans we seem to have an innate desire to fill up
empty spaces. This might explain part of the appeal of sudoku, the new international craze, with its empty squares
to be filled with digits. Since April 2005, when sudoku was
introduced to the United States in The New York Post,
more than half the leading American newspapers have
begun printing one or more sudoku a day. No puzzle has
had such a fast introduction in newspapers since the crossword craze of 1924–25.1
Since then, millions of people have made sudoku part of
their daily routines.
Your time: 3 minutes, 19 seconds
0 min
30 mins
Rank: Top 19%
Easy level average time: 5 minutes, 6 seconds.
One of the authors played an online game of sudoku at
www.websudoku.com. The graph provides information
about how well he did. (His time is marked with an arrow.)
In this chapter, you’ll learn more about how to describe the
location of an individual observation—like the author’s
sudoku time—within a distribution.
83
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11/13/13 1:33 PM
84
CHAPTER 2
M o d e l i n g D i s t r i b u t i o n s o f Data
Introduction
Suppose Jenny earns an 86 (out of 100) on her next statistics test. Should she be
satisfied or disappointed with her performance? That depends on how her score
compares with the scores of the other students who took the test. If 86 is the highest score, Jenny might be very pleased. Maybe her teacher will “curve” the grades
so that Jenny’s 86 becomes an “A.” But if Jenny’s 86 falls below the “average” in
the class, she may not be so happy.
Section 2.1 focuses on describing the location of an individual within a distribution. We begin by discussing a familiar measure of position: percentiles. Next,
we introduce a new type of graph that is useful for displaying percentiles. Then
we consider another way to describe an individual’s position that is based on the
mean and standard deviation. In the process, we examine the effects of transforming data on the shape, center, and spread of a distribution.
Sometimes it is helpful to use graphical models called density curves to describe
the location of individuals within a distribution, rather than relying on actual data
values. Such models are especially helpful when data fall in a bell-shaped pattern
called a Normal distribution. Section 2.2 examines the properties of Normal distributions and shows you how to perform useful calculations with them.
Activity
MATERIALS:
Masking tape to mark
number line scale
Where do I stand?
In this Activity, you and your classmates will explore ways to describe where you
stand (literally!) within a distribution.
1. ​Your teacher will mark out a number line on the floor with a scale running
from about 58 to 78 inches.
2. ​Make a human dotplot. Each member of the class should stand at the appropriate location along the number line scale based on height (to the nearest inch).
3. ​Your teacher will make a copy of the dotplot on the board for your reference.
4. ​What percent of the students in the class have heights less than yours? This is
your percentile in the distribution of heights.
5. ​Work with a partner to calculate the mean and standard deviation of the
class’s height distribution from the dotplot. Confirm these values with your
classmates.
6. ​Where does your height fall in relation to the mean: above or below? How
far above or below the mean is it? How many standard deviations above or
below the mean is it? This last number is the z-score corresponding to your
height.
7. ​Class discussion: What would happen to the class’s height distribution if
you converted each data value from inches to centimeters? (There are 2.54
centimeters in 1 inch.) How would this change of units affect the measures
of center, spread, and location (percentile and z-score) that you calculated?
Want to know more about where you stand—in terms of height, weight, or
even body mass index? Do a Web search for “Clinical Growth Charts” at the
National Center for Health Statistics site, www.cdc.gov/nchs.
Starnes-Yates5e_c02_082-139hr2.indd 84
11/13/13 1:33 PM
Section 2.1 Describing Location in a Distribution
2.1
What You Will Learn
•
•
85
Describing Location in a
Distribution
By the end of the section, you should be able to:
Find and interpret the percentile of an individual value
within a distribution of data.
Estimate percentiles and individual values using a
cumulative relative frequency graph.
•
•
Find and interpret the standardized score (z-score) of
an individual value within a distribution of data.
Describe the effect of adding, subtracting, multiplying by, or dividing by a constant on the shape,
center, and spread of a distribution of data.
Here are the scores of all 25 students in Mr. Pryor’s statistics class on their first test:
79
77
81
83
80 77 73 83 74 93 78 80 75 67 73
86 90 79 85 83 89 84 82 77 72
The bold score is Jenny’s 86. How did she perform on this test relative to her
classmates?
The stemplot displays this distribution of test scores. Notice that the distribution is roughly symmetric with no apparent outliers. From the stemplot, we can
see that Jenny did better than all but three students in the class.
6
7
7
8
8
9
7
2334
5777899
00123334
569
03
Key: 7|2 is a student who
scored 72 on the test
Measuring Position: Percentiles
One way to describe Jenny’s location in the distribution of test scores is to tell what
percent of students in the class earned scores that were below Jenny’s score. That is,
we can calculate Jenny’s percentile.
Definition: Percentile
The p th percentile of a distribution is the value with p percent of the observations less
than it.
Using the stemplot, we see that Jenny’s 86 places her fourth from the top of the
class. Because 21 of the 25 observations (84%) are below her score, Jenny is at the
84th percentile in the class’s test score distribution.
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11/13/13 1:33 PM
86
CHAPTER 2
example
M o d e l i n g D i s t r i b u t i o n s o f Data
Mr. Pryor’s First Test
Finding percentiles
Problem: Use the scores on Mr. Pryor’s first statistics test to find the percentiles for the
following students:
(a) ​Norman, who earned a 72.
(b) ​Katie, who scored 93.
(c) ​The two students who earned scores of 80.
Solution:
(a) ​Only 1 of the 25 scores in the class is below Norman’s 72. His percentile is computed as
follows: 1/25 = 0.04, or 4%. So Norman scored at the 4th percentile on this test.
(b) ​Katie’s 93 puts her at the 96th percentile, because 24 out of 25 test scores fall below her
result.
(c) ​Two students scored an 80 on Mr. Pryor’s first test. Because 12 of the 25 scores in the class
were less than 80, these two students are at the 48th percentile.
For Practice Try Exercise
1
Note: Some people define the pth percentile of a distribution as the value with
p percent of observations less than or equal to it. Using this alternative definition
of percentile, it is possible for an individual to fall at the 100th percentile. If we
used this definition, the two students in part (c) of the example would fall at the
56th percentile (14 of 25 scores were less than or equal to 80). Of course, because
80 is the median score, it is also possible to think of it as being the 50th percentile. Calculating percentiles is not an exact science, especially with small data
sets! We’ll stick with the definition of percentile we gave earlier for consistency.
Cumulative Relative Frequency Graphs
Age
Frequency
40–44
2
45–49
7
50–54
13
55–59
12
60–64
7
65–69
3
Starnes-Yates5e_c02_082-139hr2.indd 86
There are some interesting graphs that can be made with percentiles. One of the
most common graphs starts with a frequency table for a quantitative variable. For
instance, the frequency table in the margin summarizes the ages of the first 44
U.S. presidents when they took office.
Let’s expand this table to include columns for relative frequency, cumulative
frequency, and cumulative relative frequency.
• To get the values in the relative frequency column, divide the count in each
class by 44, the total number of presidents. Multiply by 100 to convert to a
percent.
• To fill in the cumulative frequency column, add the counts in the frequency
column for the current class and all classes with smaller values of the variable.
• For the cumulative relative frequency column, divide the entries in the cumulative frequency column by 44, the total number of individuals. Multiply by
100 to convert to a percent.
11/13/13 1:33 PM
Section 2.1 Describing Location in a Distribution
87
Here is the original frequency table with the relative frequency, cumulative frequency, and cumulative relative frequency columns added.
Frequency
Relative
frequency
Cumulative
frequency
40–44
2
2/44 = 0.045, or 4.5%
2
2/44 = 0.045, or 4.5%
45–49
7
7/44 = 0.159, or 15.9%
9
9/44 = 0.205, or 20.5%
50–54
13
13/44 = 0.295, or 29.5%
22
22/44 = 0.500, or 50.0%
55–59
12
12/44 = 0.273, or 27.3%
34
34/44 = 0.773, or 77.3%
60–64
7
7/44 = 0.159, or 15.9%
41
41/44 = 0.932, or 93.2%
65–69
3
3/44 = 0.068, or 6.8%
44
44/44 = 1.000, or 100%
Age
Cumulative relative frequency (%)
Some people refer to cumulative
relative frequency graphs as “ogives”
(pronounced “o-jives”).
Cumulative relative
frequency
To make a cumulative relative frequency graph, we plot a point corresponding to the cumulative relative frequency in each class at the smallest value of the
next class. For example, for the 40 to 44 class, we plot a point at a height of 4.5%
above the age value of 45. This means that 4.5% of presidents were inaugurated
before they were 45 years old. (In other words, age 45 is the 4.5th
percentile of the inauguration age distribution.)
It is customary to start a cumulative relative frequency graph with
a point at a height of 0% at the smallest value of the first class (in this
case, 40). The last point we plot should be at a height of 100%. We
connect consecutive points with a line segment to form the graph.
Figure 2.1 shows the completed cumulative relative frequency graph.
Here’s an example that shows how to interpret a cumulative
relative frequency graph.
100
80
60
40
20
0
40
45
50
55
60
65
Age at inauguration
example
70
fiGURe 2.1 Cumulative relative frequency graph for
the ages of U.S. presidents at inauguration.
Age at Inauguration
Interpreting a cumulative relative frequency graph
What can we learn from Figure 2.1? The graph grows very gradually at first because
few presidents were inaugurated when they were in their 40s. Then the graph gets
very steep beginning at age 50. Why? Because most U.S. presidents were in their 50s
when they were inaugurated. The rapid growth in the graph slows at age 60.
Suppose we had started with only the graph in Figure 2.1, without any of the
information in our original frequency table. Could we figure out what percent of
presidents were between 55 and 59 years old at their inaugurations? Sure. Because
the point at age 60 has a cumulative relative frequency of about 77%, we know
that about 77% of presidents were inaugurated before they were 60 years old.
Similarly, the point at age 55 tells us that about 50% of presidents were younger
than 55 at inauguration. As a result, we’d estimate that about 77% − 50% = 27%
of U.S. presidents were between 55 and 59 when they were inaugurated.
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A cumulative relative frequency graph can be used to describe the position of
an individual within a distribution or to locate a specified percentile of the distribution, as the following example illustrates.
example
Ages of U.S. Presidents
Interpreting cumulative relative frequency graphs
Problem: Use the graph in Figure 2.1 on the previous page to help you answer each question.
​ as Barack Obama, who was first inaugurated at age 47, unusually young?
(a) W
(b) ​Estimate and interpret the 65th percentile of the distribution.
Solution:
Cumulative relative frequency (%)
Cumulative relative frequency (%)
(a) ​To find President Obama’s location in the distribution, we draw a vertical line up from his age (47)
on the horizontal axis until it meets the graphed line. Then we draw a horizontal line from this point
of intersection to the vertical axis. Based on Figure 2.2(a), we would estimate that Barack Obama’s
inauguration age places him at the 11% cumulative relative frequency mark. That is, he’s at the 11th
percentile of the distribution. In other words, about 11% of all U.S. presidents were younger than
Barack Obama when they were inaugurated and about 89% were older.
100
80
60
40
20
11
0
40
45
47
50
55
60
65
70
80
65
60
40
20
0
40
Age at inauguration
(a)
100
45
50
55
58
60
65
70
Age at inauguration
(b)
fiGURe 2.2 The cumulative relative frequency graph of presidents’ ages at inauguration is used to (a) locate
President Obama within the distribution and (b) determine the 65th percentile, which is about 58 years.
(b) ​The 65th percentile of the distribution is the age with cumulative relative frequency 65%. To
find this value, draw a horizontal line across from the vertical axis at a height of 65% until it meets
the graphed line. From the point of intersection, draw a vertical line down to the horizontal axis. In
Figure 2.2(b), the value on the horizontal axis is about 58. So about 65% of all U.S. presidents were
younger than 58 when they took office.
For Practice Try Exercise 9
THINK
ABOUT IT
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Percentiles and quartiles: Have you made the connection between percentiles and the quartiles from Chapter 1? Earlier, we noted that the median (second
quartile) corresponds to the 50th percentile. What about the first quartile, Q1? It’s at
the median of the lower half of the ordered data, which puts it about one-fourth of
the way through the distribution. In other words, Q1 is roughly the 25th percentile.
By similar reasoning, Q3 is approximately the 75th percentile of the distribution.
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Section 2.1 Describing Location in a Distribution
89
CHECK YOUR UNDERSTANDING
1. ​Multiple choice: Select the best answer. Mark receives a score report detailing his performance on a statewide test. On the math section, Mark earned a raw score of 39, which
placed him at the 68th percentile. This means that
(a) ​Mark did better than about 39% of the students who took the test.
(b) ​Mark did worse than about 39% of the students who took the test.
(c) ​Mark did better than about 68% of the students who took the test.
(d) ​Mark did worse than about 68% of the students who took the test.
Cumulative relative frequency (%)
(e) ​Mark got fewer than half of the questions correct on this test.
2. ​Mrs. Munson is concerned about how her daughter’s height
and weight compare with those of other girls of the same age.
She uses an online calculator to determine that her daughter is at
the 87th percentile for weight and the 67th percentile for height.
Explain to Mrs. Munson what this means.
Questions 3 and 4 relate to the following setting. The graph displays the cumulative relative frequency of the lengths of phone calls
made from the mathematics department office at Gabalot High last
month.
100
80
60
40
20
0
0
5
10
15
20
25
30
Call length (minutes)
35
40
45
3. ​About what percent of calls lasted less than 30 minutes? 30
minutes or more?
4. ​Estimate Q1, Q3, and the IQR of the distribution.
Measuring Position: z-Scores
6
7
7
8
8
9
7
2334
5777899
00123334
569
03
Key: 7|2 is a student who
scored 72 on the test
Let’s return to the data from Mr. Pryor’s first statistics test, which are shown in
the stemplot. Figure 2.3 provides numerical summaries from Minitab for these
data. Where does Jenny’s score of 86 fall relative to the mean of this distribution?
Because the mean score for the class is 80, we can see that Jenny’s score is “above
average.” But how much above average is it?
We can describe Jenny’s location in the distribution of her class’s test scores
by telling how many standard deviations above or below the mean her score is.
Because the mean is 80 and the standard deviation is about 6, Jenny’s score of 86
is about one standard deviation above the mean.
Converting observations like this from original values to standard deviation
units is known as standardizing. To standardize a value, subtract the mean of the
distribution and then divide the difference by the standard deviation.
The relationship between the mean and
the median is about what you’d expect
in this fairly symmetric distribution.
Figure 2.3 Minitab output for the scores of Mr. Pryor’s students on their first statistics test.
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Definition: Standardized score (z-score)
If x is an observation from a distribution that has known mean and standard deviation, the standardized score for x is
x − mean
z=
standard deviation
A standardized score is often called a z-score.
A z-score tells us how many standard deviations from the mean an observation falls, and in what direction. Observations larger than the mean have positive
z-scores. Observations smaller than the mean have negative z-scores. For example,
Jenny’s score on the test was x = 86. Her standardized score (z-score) is
z=
x − mean
86 − 80
=
= 0.99
standard deviation
6.07
That is, Jenny’s test score is 0.99 standard deviations above the mean score of the
class.
example
Mr. Pryor’s First Test, Again
Finding and interpreting z-scores
Problem: Use Figure 2.3 on the previous page to find the standardized scores (z-scores) for
each of the following students in Mr. Pryor’s class. Interpret each value in context.
(a) ​Katie, who scored 93.
(b) ​Norman, who earned a 72.
Solution:
(a) ​Katie’s 93 was the highest score in the class. Her corresponding z-score is
z=
93 − 80
= 2.14
6.07
In other words, Katie’s result is 2.14 standard deviations above the mean score for this test.
(b) ​For Norman’s 72, his standardized score is
z=
72 − 80
= −1.32
6.07
Norman’s score is 1.32 standard deviations below the class mean of 80.
For Practice Try Exercise
15(b)
We can also use z-scores to compare the position of individuals in different
distributions, as the following example illustrates.
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Section 2.1 Describing Location in a Distribution
example
91
Jenny Takes Another Test
Using z-scores for comparisons
The day after receiving her statistics test result of 86 from Mr. Pryor, Jenny earned
an 82 on Mr. Goldstone’s chemistry test. At first, she was disappointed. Then Mr.
Goldstone told the class that the distribution of scores was fairly symmetric with a
mean of 76 and a standard deviation of 4.
Problem: On which test did Jenny perform better relative to the class? Justify your answer.
Solution: Jenny’s z-score for her chemistry test result is
z=
82 − 76
= 1.50
4
Her 82 in chemistry was 1.5 standard deviations above the mean score for the class. Because she
scored only 0.99 standard deviations above the mean on the statistics test, Jenny did better relative to the class in chemistry.
For Practice Try Exercise
11
We often standardize observations to express them on a common scale. We
might, for example, compare the heights of two children of different ages by calculating their z-scores. At age 2, Jordan is 89 centimeters (cm) tall. Her height
puts her at a z-score of 0.5; that is, she is one-half standard deviation above the
mean height of 2-year-old girls. Zayne’s height at age 3 is 101 cm, which yields a
z-score of 1. In other words, he is one standard deviation above the mean height
of 3-year-old boys. So Zayne is taller relative to boys his age than Jordan is relative
to girls her age. The standardized heights tell us where each child stands
(pun intended!) in the distribution for his or her age group.
CHECK YOUR UNDERSTANDING
Mrs. Navard’s statistics class has just completed the first three steps of the “Where Do I
Stand?” Activity (page 84). The figure below shows a dotplot of the class’s height distribution, along with summary statistics from computer output.
60
62
64
66
1. ​Lynette, a student in the class, is 65 inches tall. Find and interpret her z-score.
2. ​Another student in the class, Brent, is 74 inches tall. How tall
is Brent compared with the rest of the class? Give appropriate
numerical evidence to support your answer.
68
70
72
74
Height (inches)
Variable
n
x
sx
Min
Q1
Height
25
67
4.29
60
63
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Med Q3
66
69
Max
75
3. ​Brent is a member of the school’s basketball team. The mean
height of the players on the team is 76 inches. Brent’s height
translates to a z-score of −0.85 in the team’s height distribution.
What is the standard deviation of the team members’ heights?
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Transforming Data
To find the standardized score (z-score) for an individual observation, we transform
this data value by subtracting the mean and dividing the difference by the standard
deviation. Transforming converts the observation from the original units of
measurement (inches, for example) to a standardized scale. What effect do these
kinds of transformations—adding or subtracting; multiplying or dividing—have
on the shape, center, and spread of the entire distribution? Let’s investigate using
an interesting data set from “down under.”
Soon after the metric system was introduced in Australia, a group of students
was asked to guess the width of their classroom to the nearest meter. Here are their
guesses in order from lowest to highest:2
8 9
12 13
15 15
25 27
10
13
16
35
10
13
16
38
10
14
16
40
10
14
17
10
14
17
10
15
17
11
15
17
11
15
18
11
15
18
11
15
20
12
15
22
Figure 2.4 includes a dotplot of the data and some numerical summaries.
FIGURE 2.4 ​Fathom dotplot and
summary statistics for Australian
students’ guesses of the classroom
width.
Guess
n
x–
sx Min
44 16.02 7.14 8
Q1
11
Med
15
Q3
17
Max IQR Range
40
6
32
Let’s practice what we learned in Chapter 1 and describe what we see.
Shape: The distribution of guesses appears skewed to the right and bimodal, with
peaks at 10 and 15 meters.
Center: The median guess was 15 meters and the mean guess was about 16 meters.
Due to the clear skewness and potential outliers, the median is a better choice for
summarizing the “typical” guess.
Spread: Because Q1 = 11, about 25% of the students estimated the width of the
room to be fewer than 11 meters. The 75th percentile of the distribution is at about
Q3 = 17. The IQR of 6 meters describes the spread of the middle 50% of students’ guesses. The standard deviation tells us that the typical distance of students’
guesses from the mean was about 7 meters. Because sx is not resistant to extreme
values, we prefer the IQR to describe the variability of this distribution.
Outliers: By the 1.5 × IQR rule, values greater than 17 + 9 = 26 meters or less
than 11 − 9 = 2 meters are identified as outliers. So the four highest guesses—
which are 27, 35, 38, and 40 meters—are outliers.
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Section 2.1 Describing Location in a Distribution
93
Effect of adding or subtracting a constant: By now, you’re probably
wondering what the actual width of the room was. In fact, it was 13 meters wide.
How close were students’ guesses? The student who guessed 8 meters was too low
by 5 meters. The student who guessed 40 meters was too high by 27 meters (and
probably needs to study the metric system more carefully). We can examine the
distribution of students’ guessing errors by defining a new variable as follows:
error = guess − 13
That is, we’ll subtract 13 from each observation in the data set. Try to predict what
the shape, center, and spread of this new distribution will be. Refer to Figure 2.4
as needed.
example
Estimating Room Width
Effect of subtracting a constant
Let’s see how accurate your predictions were (you did make predictions, right?).
Figure 2.5 shows dotplots of students’ original guesses and their errors on the same
scale. We can see that the original distribution of guesses has been shifted to the left.
By how much? Because the peak at 15 meters
in the original graph is located at 2 meters in
the error distribution, the original data values
have been translated 13 units to the left. That
should make sense: we calculated the errors
by subtracting the actual room width, 13 meters, from each student’s guess.
From Figure 2.5, it seems clear that subtracting 13 from each observation did not
affect the shape or spread of the distribution.
But this transformation appears to have decreased the center of the distribution by 13
meters. The summary statistics in the table
below confirm our beliefs.
FIGURE 2.5 ​Fathom dotplots of students’ original guesses of classroom
width and the errors in their guesses.
Guess (m)
Error (m)
n
44
44
–
x
sx
16.02 7.14
3.02 7.14
Min
8
−5
Q1
11
−2
Med
15
2
Q3
17
4
Max
40
27
IQR
6
6
Range
32
32
The error distribution is centered at a value that is clearly positive—the median
error is 2 meters and the mean error is about 3 meters. So the students generally
tended to overestimate the width of the room.
As the example shows, subtracting the same positive number from each value
in a data set shifts the distribution to the left by that number. Adding a positive constant to each data value would shift the distribution to the right by that
constant.
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Let’s summarize what we’ve learned so far about transforming data.
Effect of Adding (or Subtracting) a Constant
Adding the same positive number a to (subtracting a from) each observation
•
•
adds a to (subtracts a from) measures of center and location (mean,
median, quartiles, percentiles), but
does not change the shape of the distribution or measures of spread
(range, IQR, standard deviation).
Effect of multiplying or dividing by a constant: Because our
group of Australian students is having some difficulty with the metric system, it
may not be helpful to tell them that their guesses tended to be about 2 to 3 meters too high. Let’s convert the error data to feet before we report back to them.
There are roughly 3.28 feet in a meter. So for the student whose error was −5
meters, that translates to
−5 meters ×
3.28 feet
= −16.4 feet
1 meter
To change the units of measurement from meters to feet, we multiply each of the
error values by 3.28. What effect will this have on the shape, center, and spread of
the distribution? (Go ahead, make some predictions!)
example
Estimating Room Width
Effect of multiplying by a constant
Figure 2.6 includes dotplots of the students’ guessing errors in meters and feet,
along with summary statistics from computer software. The shape of the two distributions is the same—right-skewed and bimodal. However, the centers and spreads
n
x–
sx
Error (m)
44
3.02
7.14
Error (ft)
44
9.91
23.43
Min
−5
Q1
−2
−16.4 −6.56
Med
2
6.56
Q3
4
Max
27
IQR
6
Range
32
13.12 88.56 19.68 104.96
FIGURE 2.6 ​Fathom dotplots and numerical summaries of students’ errors guessing the width of
their classroom in meters and feet.
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Section 2.1 Describing Location in a Distribution
95
of the two distributions are quite different. The bottom dotplot is centered at a
value that is to the right of the top dotplot’s center. Also, the bottom dotplot shows
much greater spread than the top dotplot.
When the errors were measured in meters, the median was 2 and the mean was
3.02. For the transformed error data in feet, the median is 6.56 and the mean
is 9.91. Can you see that the measures of center were multiplied by 3.28? That
makes sense. If we multiply all the observations by 3.28, then the mean and median should also be multiplied by 3.28.
What about the spread? Multiplying each observation by 3.28 increases the variability of the distribution. By how much? You guessed it—by a factor of 3.28. The
numerical summaries in Figure 2.6 show that the standard deviation, the interquartile range, and the range have been multiplied by 3.28.
We can safely tell our group of Australian students that their estimates of the classroom’s width tended to be too high by about 6.5 feet. (Notice that we choose not
to report the mean error, which is affected by the strong skewness and the three
high outliers.)
As before, let’s recap what we discovered about the effects of transforming
data.
Effect of Multiplying (or Dividing) by a Constant
It is not common to multiply (or divide)
each observation in a data set by a
negative number b. Doing so would
multiply (or divide) the measures of
spread by the absolute value of b.
We can’t have a negative amount of
variability! Multiplying or dividing by a
negative number would also affect the
shape of the distribution as all values
would be reflected over the y axis.
Multiplying (or dividing) each observation by the same positive number b
•
•
•
multiplies (divides) measures of center and location (mean, median,
quartiles, percentiles) by b,
multiplies (divides) measures of spread (range, IQR, standard deviation)
by b, but
does not change the shape of the distribution.
Putting it all together: Adding/subtracting and multiplying/
dividing: What happens if we transform a data set by both adding or sub-
tracting a constant and multiplying or dividing by a constant? For instance, if
we need to convert temperature data from Celsius to Fahrenheit, we have to
use the formula ºF ∙ 9/5(ºC) + 32. That is, we would multiply each of the
observations by 9/5 and then add 32. As the following example shows, we just
use the facts about transforming data that we’ve already established.
example
Too Cool at the Cabin?
Analyzing the effects of transformations
During the winter months, the temperatures at the Starnes’s Colorado cabin
can stay well below freezing (32°F or 0°C) for weeks at a time. To prevent the
pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. She also buys a
digital thermometer that records the indoor temperature each night at midnight.
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Unfortunately, the thermometer is programmed to measure the temperature in
degrees Celsius. A dotplot and numerical summaries of the midnight temperature
readings for a 30-day period are shown below.
Temperature
n
Mean
StDev
Min
Q1
Median
30
8.43
2.27
3.00
7.00
8.50
Q3
Max
10.00 14.00
Problem: Use the fact that °F = (9/5)°C + 32 to help you answer the following questions.
(a) ​Find the mean temperature in degrees Fahrenheit. Does the thermostat setting seem accurate?
(b) ​Calculate the standard deviation of the temperature readings in degrees Fahrenheit. Interpret this
value in context.
(c) ​The 93rd percentile of the temperature readings was 12°C. What is the 93rd percentile temperature in degrees Fahrenheit?
Solution:
(a) ​To convert the temperature measurements from Celsius to Fahrenheit, we multiply each value by
9/5 and then add 32. Multiplying the observations by 9/5 also multiplies the mean by 9/5. Adding
32 to each observation increases the mean by 32. So the mean temperature in degrees Fahrenheit is
(9/5)(8.43) + 32 = 47.17°F. The thermostat doesn’t seem to be very accurate. It is set at 50°F,
but the mean temperature over the 30-day period is about 47°F.
(b) ​Multiplying each observation by 9/5 multiplies the standard deviation by 9/5. However, adding
32 to each observation doesn’t affect the spread. So the standard deviation of the temperature
measurements in degrees Fahrenheit is (9/5)(2.27) = 4.09°F. This means that the typical distance
of the temperature readings from the mean is about 4°F. That’s a lot of variation!
(c) ​Both multiplying by a constant and adding a constant affect the value of the 93rd percentile. To
find the 93rd percentile in degrees Fahrenheit, we multiply the 93rd percentile in degrees Celsius by
9/5 and then add 32: (9/5)(12) + 32 = 53.6°F.
For Practice Try Exercise
19
Let’s look at part (c) of the example more closely. The data value of 12°C is
at the 93rd percentile of the distribution, meaning that 28 of the 30 temperature readings are less than 12°C. When we transform the data, 12°C becomes
53.6°F. The value of 53.6°F is at the 93rd percentile of the transformed distribution because 28 of the 30 temperature readings are less than 53.6°F. What have
we learned? Adding (or subtracting) a constant does not change an individual data
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Section 2.1 Describing Location in a Distribution
97
value’s location within a distribution. Neither does multiplying or dividing by a
positive constant.
THINK
ABOUT IT
This is a result worth noting! If you
start with any set of quantitative data
and convert the values to standardized
scores (z-scores), the transformed
data set will have a mean of 0 and a
standard deviation of 1. The shape of
the two distributions will be the same.
We will use this result to our advantage
in Section 2.2.
Connecting transformations and z-scores: ​
What does all this
transformation business have to do with z-scores? To standardize an observation,
you subtract the mean of the distribution and then divide by the standard deviation. What if we standardized every observation in a distribution?
Returning to Mr. Pryor’s statistics test scores, we recall that the distribution was
roughly symmetric with a mean of 80 and a standard deviation of 6.07. To convert
the entire class’s test results to z-scores, we would subtract 80 from each observation and then divide by 6.07. What effect would these transformations have on the
shape, center, and spread of the distribution?
• Shape: The shape of the distribution of z-scores would be the same as the
shape of the original distribution of test scores. Neither subtracting a constant
nor dividing by a constant would change the shape of the graph. The dotplots
confirm that the combination of these two transformations does not affect the
shape.
• Center: Subtracting 80 from each data value would also reduce the mean
by 80. Because the mean of the original distribution was 80, the mean of the
transformed data would be 0. Dividing each of these new data values by 6.07
would also divide the mean by 6.07. But because the mean is now 0, dividing
by 6.07 would leave the mean at 0. That is, the mean of the z-score distribution would be 0.
• Spread: The spread of the distribution would not be affected by subtracting
80 from each observation. However, dividing each data value by 6.07 would
also divide our common measures of spread by 6.07. The standard deviation
of the distribution of z-scores would therefore be 6.07/6.07 = 1.
The Minitab computer output below confirms the result: If we standardize
every observation in a distribution, the resulting set of z-scores has mean 0 and
standard deviation 1.
Descriptive Statistics: Test scores, z-scores
Variable
n
Mean StDev Minimum
Q1
Median
Q3
Maximum
Test scores 25 80.00 6.07
67.00 76.00 80.00 83.50 93.00
z-scores
25
0.00
1.00
−2.14
−0.66
0.00
0.58
2.14
Many other types of transformations can be very useful in analyzing data. We
have only studied what happens when you transform data through addition, subtraction, multiplication, or division.
Check Your Understanding
The figure on the next page shows a dotplot of the height distribution for Mrs. Navard’s
class, along with summary statistics from computer output.
1. ​Suppose that you convert the class’s heights from inches to centimeters (1 inch =
2.54 cm). Describe the effect this will have on the shape, center, and spread of the
distribution.
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CHAPTER 2
62
64
66
M o d e l i n g D i s t r i b u t i o n s o f Data
68
70
72
2. ​If Mrs. Navard had the entire class stand on a 6-inch-high platform and then had the students measure the distance from the top of
their heads to the ground, how would the shape, center, and spread
of this distribution compare with the original height distribution?
74
Height (inches)
Variable
n
x
sx
Min
Q1
Height
25
67
4.29
60
63
Med Q3
66
69
3. ​Now suppose that you convert the class’s heights to z-scores.
What would be the shape, center, and spread of this distribution?
Explain.
Max
75
DATA EXPLORATION The speed of light
Light travels fast, but it is not transmitted instantly. Light takes over a second to reach
us from the moon and over 10 billion years to reach us from the most distant objects in
the universe. Because radio waves and radar also travel at the speed of light, having an
accurate value for that speed is important in communicating with astronauts and orbiting satellites. An accurate value for the speed of light is also important
to computer designers because electrical signals travel at light speed.
The first reasonably accurate measurements of the speed of light
were made more than a hundred years ago by A. A. Michelson and
Simon Newcomb. The table below contains 66 measurements
made by Newcomb between July and September 1882.3
Newcomb measured the time in seconds that a light signal took to
pass from his laboratory on the Potomac River to a mirror at the base
of the Washington Monument and back, a total distance of about
7400 meters. Newcomb’s first measurement of the passage time of
light was 0.000024828 second, or 24,828 nanoseconds. (There are
109 nanoseconds in a second.) The entries in the table record only
the deviations from 24,800 nanoseconds.
28
25
28
27
24
26
30
29
27
25
33
23
37
28
32
24
29
25
27
25
34
31
28
31
29
−44
19
26
27
27
27
24
30
26
28
16
20
32
33
29
40
36
36
26
16
−2 29 22 24 21
32 36 28 25 21
26 30 22 36 23
32 32 24 39 28
23
The figure provides a histogram and numerical summaries (computed with and without the two outliers) from
Minitab for these data.
1. ​We could convert the passage time measurements to
nanoseconds by adding 24,800 to each of the data values in
the table. What effect would this have on the shape, center,
and spread of the distribution? Be specific.
2. ​After performing the transformation to nanoseconds, we
could convert the measurements from nanoseconds to seconds by dividing each value by 109. What effect would this
have on the shape, center, and spread of the distribution?
Be specific.
Descriptive Statistics: Passage time
3. ​Use the information provided to estiVariable n Mean Stdev Min
Q1
Med Q3 Max
mate the speed of light in meters per secP.Time
66 26.21 10.75 −44 24
27
31 40
ond. Be prepared to justify the method
P.Time*
64 27.75 5.08 16 24.5 27.5 31 40
you used.
Starnes-Yates5e_c02_082-139hr2.indd 98
11/13/13 1:33 PM
Section 2.1 Describing Location in a Distribution
Summary
Section 2.1
•
Two ways of describing an individual’s location within a distribution are
percentiles and z-scores. An observation’s percentile is the percent of the
distribution that is below the value of that observation. To standardize any
observation x, subtract the mean of the distribution and then divide the
difference by the standard deviation. The resulting z-score
z=
•
•
1.
says how many standard deviations x lies above or below the distribution
mean. We can also use percentiles and z-scores to compare the location of
individuals in different distributions.
A cumulative relative frequency graph allows us to examine location within
a distribution. Cumulative relative frequency graphs begin by grouping
the observations into equal-width classes (much like the process of making
a histogram). The completed graph shows the accumulating percent of
observations as you move through the classes in increasing order.
It is common to transform data, especially when changing units of
measurement. When you add a constant a to all the values in a data set,
measures of center (median and mean) and location (quartiles and percentiles)
increase by a. Measures of spread do not change. When you multiply all the
values in a data set by a positive constant b, measures of center, location,
and spread are multiplied by b. Neither of these transformations changes the
shape of the distribution.
Shoes How many pairs of shoes do students have?
Do girls have more shoes than boys? Here are data
from a random sample of 20 female and 20 male
students at a large high school:
Female:
Male:
2.
50
26
26
31
57
19
24
22
23 38
13
50
13
34
23
30
49
13
15 51
14
7
6
5
12
38
8
7
10 10
10
11
4
5
22
7
5
10
35
7
(a)Find and interpret the percentile in the female distribution for the girl with 22 pairs of shoes.
(b) F
ind and interpret the percentile in the male distribution for the boy with 22 pairs of shoes.
(c) Who is more unusual: the girl with 22 pairs of shoes
or the boy with 22 pairs of shoes? Explain.
Starnes-Yates5e_c02_082-139hr2.indd 99
x − mean
standard deviation
Exercises
Section 2.1
pg 86
99
7
8
9
10
11
12
13
14
15
16
Old folks Here is a stemplot of the percents of residents aged 65 and older in the 50 states:
0
8
8
019
16777
01122456778999
0001223344455689
023568
24
9
Key: 15|2 means 15.2% of
this state’s residents are 65
or older
(a) Find and interpret the percentile for Colorado, where
10.1% of the residents are aged 65 and older.
(b) Find and interpret the percentile for Rhode Island,
where 13.9% of the residents are aged 65 and older.
(c) Which of these two states is more unusual? Explain.
11/13/13 1:33 PM
M o d e l i n g D i s t r i b u t i o n s o f Data
3.
Math test Josh just got the results of the statewide
Algebra 2 test: his score is at the 60th percentile.
When Josh gets home, he tells his parents that he got
60 percent of the questions correct on the state test.
Explain what’s wrong with Josh’s interpretation.
4.
Blood pressure Larry came home very excited after a
visit to his doctor. He announced proudly to his wife,
“My doctor says my blood pressure is at the 90th percentile among men like me. That means I’m better off
than about 90% of similar men.” How should his wife,
who is a statistician, respond to Larry’s statement?
5.
Growth charts We used an online growth chart to
find percentiles for the height and weight of a 16-yearold girl who is 66 inches tall and weighs 118 pounds.
According to the chart, this girl is at the 48th percentile for weight and the 78th percentile for height.
Explain what these values mean in plain English.
6.
Run fast Peter is a star runner on the track team. In
the league championship meet, Peter records a time
that would fall at the 80th percentile of all his race
times that season. But his performance places him at
the 50th percentile in the league championship meet.
Explain how this is possible. (Remember that lower
times are better in this case!)
Exercises 7 and 8 involve a new type of graph called a
percentile plot. Each point gives the value of the variable
being measured and the corresponding percentile for one
individual in the data set.
7.
Text me The percentile plot below shows the distribution of text messages sent and received in a two-day
period by a random sample of 16 females from a large
high school.
(a) Describe the student represented by the highlighted
point.
(b) Use the graph to estimate the median number of
texts. Explain your method.
(b) Use the graph to estimate the 30th percentile of the
distribution. Explain your method.
9.
pg 88
Shopping spree The figure below is a cumulative
relative frequency graph of the amount spent by
50 consecutive grocery shoppers at a store.
Cumulative relative frequency (%)
CHAPTER 2
100
90
80
70
60
50
40
30
20
10
0
0
10
20
30
40
50
60
70
80
90
100
Amount spent ($)
(a)​Estimate the interquartile range of this distribution.
Show your method.
(b) What is the percentile for the shopper who spent
$19.50?
(c) Draw the histogram that corresponds to this graph.
10. Light it up! The graph below is a cumulative
relative frequency graph showing the lifetimes
(in hours) of 200 lamps.4
Cumulative relative frequency (%)
100
100
80
60
40
20
0
500
700
900
1100
1300
1500
Lifetimes (hours)
8.
Foreign-born residents The following percentile plot
shows the distribution of the percent of foreign-born
residents in the 50 states.
(a) The highlighted point is for Maryland. Describe what
the graph tells you about this state.
Starnes-Yates5e_c02_082-139hr2.indd 100
(a) Estimate the 60th percentile of this distribution.
Show your method.
(b)​What is the percentile for a lamp that lasted 900 hours?
(c) Draw a histogram that corresponds to this graph.
11/13/13 1:33 PM
101
Section 2.1 Describing Location in a Distribution
11. SAT versus ACT Eleanor scores 680 on the SAT
Mathematics test. The distribution of SAT scores is
symmetric and single-peaked, with mean 500 and
standard deviation 100. Gerald takes the American
College Testing (ACT) Mathematics test and
scores 27. ACT scores also follow a symmetric,
single-peaked distribution—but with mean 18 and
standard deviation 6. Find the standardized scores
for both students. Assuming that both tests measure
the same kind of ability, who has the higher score?
in her hip measured using DEXA. Mary is 35 years
old. Her bone density is also reported as 948 g/cm2,
but her standardized score is z = 0.50. The mean
bone density in the hip for the reference population
of 35-year-old women is 944 grams/cm2.
pg 91
12. Comparing batting averages Three landmarks of
baseball achievement are Ty Cobb’s batting average
of 0.420 in 1911, Ted Williams’s 0.406 in 1941, and
George Brett’s 0.390 in 1980. These batting averages
cannot be compared directly because the distribution
of major league batting averages has changed over the
years. The distributions are quite symmetric, except
for outliers such as Cobb, Williams, and Brett. While
the mean batting average has been held roughly constant by rule changes and the balance between hitting
and pitching, the standard deviation has dropped over
time. Here are the facts:
Decade
Mean
Standard deviation
1910s
0.266
0.0371
1940s
0.267
0.0326
1970s
0.261
0.0317
Find the standardized scores for Cobb, Williams, and
Brett. Who was the best hitter?5
13. Measuring bone density Individuals with low bone
density have a high risk of broken bones (fractures).
Physicians who are concerned about low bone density
(osteoporosis) in patients can refer them for specialized testing. Currently, the most common method for
testing bone density is dual-energy X-ray absorptiometry (DEXA). A patient who undergoes a DEXA test
usually gets bone density results in grams per square
centimeter (g/cm2) and in standardized units.
Judy, who is 25 years old, has her bone density
measured using DEXA. Her results indicate a bone
density in the hip of 948 g/cm2 and a standardized
score of z = −1.45. In the reference population of
25-year-old women like Judy, the mean bone density
in the hip is 956 g/cm2.6
(a) Judy has not taken a statistics class in a few years. Explain to her in simple language what the standardized
score tells her about her bone density.
(b) Use the information provided to calculate the standard
deviation of bone density in the reference population.
14. Comparing bone density Refer to the previous exercise. One of Judy’s friends, Mary, has the bone density
Starnes-Yates5e_c02_082-139hr2.indd 101
(a) Whose bones are healthier—Judy’s or Mary’s? Justify
your answer.
(b) Calculate the standard deviation of the bone density in
Mary’s reference population. How does this compare
with your answer to Exercise 13(b)? Are you surprised?
Exercises 15 and 16 refer to the dotplot and summary statistics of salaries for players on the World Champion 2008
Philadelphia Phillies baseball team.7
0
2
4
6
8
10
12
14
16
Salary (millions)
Variable
n
Mean
Std. dev.
Min
Q1
Med
Q3
Max
Salary
29 3388617 3767484 390000 440000 1400000 6000000 14250000
15. Baseball salaries Brad Lidge played a crucial role as
the Phillies’ “closer,” pitching the end of many games
throughout the season. Lidge’s salary for the 2008
season was $6,350,000.
pg 90
(a) Find the percentile corresponding to Lidge’s salary.
Explain what this value means.
(b)Find the z-score corresponding to Lidge’s salary. Explain what this value means.
16. Baseball salaries Did Ryan Madson, who was paid
$1,400,000, have a high salary or a low salary compared with the rest of the team? Justify your answer
by calculating and interpreting Madson’s percentile
and z-score.
17. The scores on Ms. Martin’s statistics quiz had a
mean of 12 and a standard deviation of 3. Ms. Martin
wants to transform the scores to have a mean of 75
and a standard deviation of 12. What transformations
should she apply to each test score? Explain.
18. Mr. Olsen uses an unusual grading system in his
class. After each test, he transforms the scores to have
a mean of 0 and a standard deviation of 1. Mr. Olsen
then assigns a grade to each student based on the
transformed score. On his most recent test, the class’s
scores had a mean of 68 and a standard deviation of
15. What transformations should he apply to each
test score? Explain.
11/13/13 1:33 PM
102
CHAPTER 2
M o d e l i n g D i s t r i b u t i o n s o f Data
19. Tall or short? Mr. Walker measures the heights (in
inches) of the students in one of his classes. He uses a
computer to calculate the following numerical
summaries:
pg 95
Mean
Std. dev.
Min
Q1
Med
Q3
Max
69.188
3.20
61.5
67.75
69.5
71
74.5
Next, Mr. Walker has his entire class stand on their
chairs, which are 18 inches off the ground. Then he
measures the distance from the top of each student’s
head to the floor.
(a) Find the mean and median of these measurements.
Show your work.
(b) Find the standard deviation and IQR of these measurements. Show your work.
20. Teacher raises A school system employs teachers at
salaries between $28,000 and $60,000. The teachers’
union and the school board are negotiating the form
of next year’s increase in the salary schedule.
(a) If every teacher is given a flat $1000 raise, what will
this do to the mean salary? To the median salary? Explain your answers.
(b) What would a flat $1000 raise do to the extremes and
quartiles of the salary distribution? To the standard
deviation of teachers’ salaries? Explain your answers.
21. Tall or short? Refer to Exercise 19. Mr. Walker converts his students’ original heights from inches to feet.
(a) Find the mean and median of the students’ heights in
feet. Show your work.
(b) Find the standard deviation and IQR of the students’
heights in feet. Show your work.
22. Teacher raises Refer to Exercise 20. If each teacher
receives a 5% raise instead of a flat $1000 raise, the
amount of the raise will vary from $1400 to $3000,
depending on the present salary.
(a) What will this do to the mean salary? To the median
salary? Explain your answers.
(b) Will a 5% raise increase the IQR? Will it increase the
standard deviation? Explain your answers.
23. Cool pool? Coach Ferguson uses a thermometer
to measure the temperature (in degrees Celsius) at
20 different locations in the school swimming pool.
An analysis of the data yields a mean of 25°C and a
standard deviation of 2°C. Find the mean and standard deviation of the temperature readings in degrees
Fahrenheit (recall that °F = (9/5)°C + 32).
24. Measure up Clarence measures the diameter of each
tennis ball in a bag with a standard ruler. Unfortunately, he uses the ruler incorrectly so that each of his
Starnes-Yates5e_c02_082-139hr2.indd 102
measurements is 0.2 inches too large. Clarence’s data
had a mean of 3.2 inches and a standard deviation of
0.1 inches. Find the mean and standard deviation of
the corrected measurements in centimeters (recall
that 1 inch = 2.54 cm).
Multiple choice: Select the best answer for Exercises 25
to 30.
25. Jorge’s score on Exam 1 in his statistics class was at
the 64th percentile of the scores for all students. His
score falls
(a)​between the minimum and the first quartile.
(b) between the first quartile and the median.
(c) between the median and the third quartile.
(d) between the third quartile and the maximum.
(e) at the mean score for all students.
26. When Sam goes to a restaurant, he always tips the
server $2 plus 10% of the cost of the meal. If Sam’s
distribution of meal costs has a mean of $9 and a standard deviation of $3, what are the mean and standard
deviation of the distribution of his tips?
(a) $2.90, $0.30
(b)​$2.90, $2.30
(c) $9.00, $3.00
(d) $11.00, $2.00
(e)​$2.00, $0.90
27. Scores on the ACT college entrance exam follow a
bell-shaped distribution with mean 18 and standard
deviation 6. Wayne’s standardized score on the ACT
was −0.5. What was Wayne’s actual ACT score?
(a) ​5.5 (b) ​
12 (c) ​15 (d) ​17.5 (e) ​21 28. George has an average bowling score of 180 and
bowls in a league where the average for all bowlers
is 150 and the standard deviation is 20. Bill has an
average bowling score of 190 and bowls in a league
where the average is 160 and the standard deviation
is 15. Who ranks higher in his own league, George
or Bill?
(a) Bill, because his 190 is higher than George’s 180.
(b) Bill, because his standardized score is higher than
George’s.
(c) Bill and George have the same rank in their leagues,
because both are 30 pins above the mean.
(d) George, because his standardized score is higher than
Bill’s.
(e) George, because the standard deviation of bowling
scores is higher in his league.
11/13/13 1:33 PM
103
Section 2.2 Density Curves and Normal Distributions
Cumulative Relative Frequency
Exercises 29 and 30 refer to the following setting. The number of absences during the fall semester was recorded for
each student in a large elementary school. The distribution of absences is displayed in the following cumulative
relative frequency graph.
1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0
2
4
6
8
10
Number of Absences
12
Exercises 31 and 32 refer to the following setting. We used
CensusAtSchool’s Random Data Selector to choose a
sample of 50 Canadian students who completed a survey in
a recent year.
31. Travel time (1.2) The dotplot below displays data on
students’ responses to the question “How long does
it usually take you to travel to school?” Describe the
shape, center, and spread of the distribution. Are
there any outliers?
0
14
29. What is the interquartile range (IQR) for the distribution of absences?
(a) 1 (c) ​3 (e) ​14
(b) 2 (d) ​5
(e)​Cannot be determined
2.2
What You Will Learn
R
R
R
R
R
R
R
R
R
R
R
L
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
A
R
R
R
R
A
R
R
L
R
R
R
R
L
A
R
R
R
R
R
R
R
R
Density Curves and Normal
Distributions
By the end of the section, you should be able to:
Estimate the relative locations of the median and mean
on a density curve.
Use the 68–95–99.7 rule to estimate areas (proportions of values) in a Normal distribution.
Use Table A or technology to find (i) the proportion of
z-values in a specified interval, or (ii) a z-score from a
percentile in the standard Normal distribution.
Starnes-Yates5e_c02_082-139hr2.indd 103
100
(b)​Over 10,000 Canadian high school students took the
CensusAtSchool survey that year. What percent of
this population would you estimate is left-handed?
Justify your answer.
(d) Skewed right
•
80
(a)​Make an appropriate graph to display these data.
(c) Skewed left
•
60
32. Lefties (1.1) Students were asked, “Are you righthanded, left-handed, or ambidextrous?” The
responses are shown below (R = right-handed;
L = left-handed; A = ambidextrous).
(a)​Symmetric
•
40
Travel time (minutes)
30. If the distribution of absences was displayed in a
histogram, what would be the best description of the
histogram’s shape?
(b)​Uniform
20
•
•
Use Table A or technology to find (i) the proportion of
values in a specified interval, or (ii) the value that corresponds to a given percentile in any Normal
distribution.
Determine whether a distribution of data is approximately Normal from graphical and numerical evidence.
11/13/13 1:33 PM
104
CHAPTER 2
M o d e l i n g D i s t r i b u t i o n s o f Data
In Chapter 1, we developed a kit of graphical and numerical tools for describing
distributions. Our work gave us a clear strategy for exploring data from a single
quantitative variable.
Exploring Quantitative Data
1. ​Always plot your data: make a graph, usually a dotplot, stemplot, or histogram.
2. ​Look for the overall pattern (shape, center, spread) and for striking departures such as outliers.
3. ​Calculate numerical summaries to briefly describe center and spread.
In this section, we add one more step to this strategy.
4. ​Sometimes the overall pattern of a large number of observations is so regular that we can describe it by a smooth curve.
Density Curves
Figure 2.7 is a histogram of the scores of all 947 seventh-grade students in Gary,
Indiana, on the vocabulary part of the Iowa Test of Basic Skills (ITBS).8 Scores on
this national test have a very regular distribution. The histogram is symmetric,
and both tails fall off smoothly from a single center peak. There are no large
gaps or obvious outliers. The smooth curve drawn through the tops of the
histogram bars in Figure 2.7 is a good description of the overall pattern of
the data.
2
4
6
8
ITBS vocabulary score
example
10
12
FIGURE 2.7 ​Histogram of the Iowa Test of Basic Skills (ITBS) vocabulary
scores of all seventh-grade students in Gary, Indiana. The smooth curve shows
the overall shape of the distribution.
Seventh-Grade Vocabulary Scores
From histogram to density curve
Our eyes respond to the areas of the bars in a histogram. The bar areas represent
relative frequencies (proportions) of the observations. Figure 2.8(a) is a copy of
Figure 2.7 with the leftmost bars shaded. The area of the shaded bars in Figure 2.8(a)
represents the proportion of students with vocabulary scores less than 6.0. There are
287 such students, who make up the proportion 287/947 = 0.303 of all Gary seventhgraders. In other words, a score of 6.0 corresponds to about the 30th percentile.
The total area of the bars in the histogram is 100% (a proportion of 1), because all
of the observations are represented. Now look at the curve drawn through the tops
of the bars. In Figure 2.8(b), the area under the curve to the left of 6.0 is shaded.
In moving from histogram bars to a smooth curve, we make a specific choice:
adjust the scale of the graph so that the total area under the curve is exactly 1. Now
Starnes-Yates5e_c02_082-139hr2.indd 104
11/13/13 1:33 PM
105
Section 2.2 Density Curves and Normal Distributions
2
4
6
8
10
12
2
4
ITBS vocabulary score
(a)
6
8
10
12
ITBS vocabulary score
(b)
FIGURE 2.8 ​(a) The proportion of scores less than or equal to 6.0 in the actual data is 0.303. (b) The proportion of
scores less than or equal to 6.0 from the density curve is 0.293.
the total area represents all the observations, just like with the histogram. We can
then interpret areas under the curve as proportions of the observations.
The shaded area under the curve in Figure 2.8(b) represents the proportion of
students with scores lower than 6.0. This area is 0.293, only 0.010 away from the
actual proportion 0.303. So our estimate based on the curve is that a score of 6.0
falls at about the 29th percentile. You can see that areas under the curve give
good approximations to the actual distribution of the 947 test scores. In practice, it
might be easier to use this curve to estimate relative frequencies than to determine
the actual proportion of students by counting data values.
A curve like the one in the previous example is called a density curve.
Definition: Density curve
A density curve is a curve that
• is always on or above the horizontal axis, and
• has area exactly 1 underneath it.
A density curve describes the overall pattern of a distribution. The area under the
curve and above any interval of values on the horizontal axis is the proportion of all
observations that fall in that interval.
c
Starnes-Yates5e_c02_082-139hr2.indd 105
!
n
Density curves, like distributions, come in many shapes. A density curve is
often a good description of the overall pattern of a distribution. Outliers, which
are departures from the overall pattern, are not described by the curve.
ti
No set of real data is exactly described by a density curve. The curve is an au o
approximation that is easy to use and accurate enough for practical use.
11/13/13 1:33 PM
106
CHAPTER 2
M o d e l i n g D i s t r i b u t i o n s o f Data
Describing Density Curves
Our measures of center and spread apply to density curves as well as to actual sets of
observations. Areas under a density curve represent proportions of the total number
of observations. The median of a data set is the point with half the observations on either side. So the median of a density curve is the “equal-areas point,” the point with
half the area under the curve to its left and the remaining half of the area to its right.
Because density curves are idealized patterns, a symmetric density curve is
exactly symmetric. The median of a symmetric density curve is therefore at its
center. Figure 2.9(a) shows a symmetric density curve with the median marked. It
isn’t so easy to spot the equal-areas point on a skewed curve. There are mathematical ways of finding the median for any density curve. That’s how we marked the
median on the skewed curve in Figure 2.9(b).
What about the mean? The mean of a set of observations is their arithmetic
average. As we saw in Chapter 1, the mean is also the “balance point” of a distribution. That is, if we think of the observations as weights strung out along a thin
rod, the mean is the point at which the rod would balance. This fact is also true of
The long right tail pulls
the mean to the right
of the median.
Mean
Median
Median and mean
(a)
(b)
FIGURE 2.9 ​(a) The median and mean of a symmetric density curve both lie at the center of symmetry. (b) The median and mean of a right-skewed density curve. The mean is pulled away from
the median toward the long tail.
Starnes-Yates5e_c02_082-139hr2.indd 106
11/13/13 1:33 PM
Section 2.2 Density Curves and Normal Distributions
107
density curves. The mean of a density curve is the point at which the curve would
balance if made of solid material. Figure 2.10 illustrates this fact about the mean.
FIGURE 2.10 ​The mean is the balance point of a density curve.
A symmetric curve balances at its center because the two sides are identical.
The mean and median of a symmetric density curve are equal, as in Figure 2.9(a).
We know that the mean of a skewed distribution is pulled toward the long tail.
Figure 2.9(b) shows how the mean of a skewed density curve is pulled toward the
long tail more than the median is.
Distinguishing the Median and Mean of a Density Curve
The median of a density curve is the equal-areas point, the point that
divides the area under the curve in half.
The mean of a density curve is the balance point, at which the curve would
balance if made of solid material.
The median and mean are the same for a symmetric density curve. They
both lie at the center of the curve. The mean of a skewed curve is pulled
away from the median in the direction of the long tail.
You probably noticed that we used
the same notation for the mean and
standard deviation of a population in
Chapter 1, m and s, as we do here for
the mean and standard deviation of a
density curve.
Because a density curve is an idealized description of a distribution of data, we
distinguish between the mean and standard deviation of the density curve and the
mean x– and standard deviation sx computed from the actual observations. The usual
notation for the mean of a density curve is m (the Greek letter mu). We write the
standard deviation of a density curve as s (the Greek letter sigma). We can roughly
locate the mean m of any density curve by eye, as the balance point. There is no easy
way to locate the standard deviation s by eye for density curves in general.
Check Your Understanding
Use the figure shown to answer the following questions.
1. ​Explain why this is a legitimate density curve.
2. ​About what proportion of observations lie between 7
and 8?
Total area under
curve = 1
3. ​Trace the density curve onto your paper. Mark the approximate
location of the median.
4. ​Now mark the approximate location of the mean. Explain why
the mean and median have the relationship that they do in this
case.
Area = 0.12
7
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Normal Distributions
One particularly important class of density curves has already appeared in Figures
2.7, 2.8, and 2.9(a). They are called Normal curves. The distributions they describe are called Normal distributions. Normal distributions play a large role in
statistics, but they are rather special and not at all “normal” in the sense of being
usual or typical. We capitalize Normal to remind you that these curves are special.
Look at the two Normal curves in Figure 2.11. They illustrate several important
facts:
•
•
All Normal curves have the same overall shape: symmetric, single-peaked,
and bell-shaped.
Any specific Normal curve is completely described by giving its mean m and
its standard deviation s.
FIGURE 2.11 ​Two Normal curves, showing the mean m and standard deviation s.
•
•
The mean is located at the center of the symmetric curve and is the same as the
median. Changing m without changing s moves the Normal curve along the
horizontal axis without changing its spread.
The standard deviation s controls the spread of a Normal curve. Curves with
larger standard deviations are more spread out.
The standard deviation s is the natural measure of spread for Normal distributions. Not only do m and s completely determine the shape of a Normal curve,
but we can locate s by eye on a Normal curve. Here’s how. Imagine that you are
skiing down a mountain that has the shape of a Normal curve. At first, you descend at an ever-steeper angle as you go out from the peak:
Fortunately, before you find yourself going straight down, the slope begins to grow
flatter rather than steeper as you go out and down:
c
Starnes-Yates5e_c02_082-139hr2.indd 108
!
n
The points at which this change of curvature takes place are located at a distance s on
either side of the mean m. (Advanced math students know these points as “inflection
points.”) You can feel the change as you run a pencil along a Normal curve and so
find the standard deviation. Remember that m and s alone do not specify
autio
the shape of most distributions. The shape of density curves in general does
not reveal s. These are special properties of Normal distributions.
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109
Section 2.2 Density Curves and Normal Distributions
Definition: Normal distribution and Normal curve
A Normal distribution is described by a Normal density curve. Any particular Normal
distribution is completely specified by two numbers: its mean m and standard deviation s. The mean of a Normal distribution is at the center of the symmetric Normal
curve. The standard deviation is the distance from the center to the change-ofcurvature points on either side.
We abbreviate the Normal distribution with mean m and standard deviation s as
N (m, s).
!
n
Why are the Normal distributions important in statistics? Here are three reasons. First, Normal distributions are good descriptions for some distributions of
real data. Distributions that are often close to Normal include
• scores on tests taken by many people (such as SAT exams and IQ tests),
• repeated careful measurements of the same quantity (like the diameter of a
tennis ball), and
• characteristics of biological populations (such as lengths of crickets and yields
of corn).
Second, Normal distributions are good approximations to the results of many
kinds of chance outcomes, like the number of heads in many tosses of a fair coin.
Third, and most important, we will see that many statistical inference procedures
are based on Normal distributions.
Even though many sets of data follow a Normal distribution, many do not. Most
income distributions, for example, are skewed to the right and so are not Normal. Some distributions are symmetric but not Normal or even close to Normal.
The uniform distribution of Exercise 35 (page 128) is one such example.
autio
Non-Normal data, like non-normal people, not only are common but are
sometimes more interesting than their Normal counterparts.
c
Normal curves were first applied to
data by the great mathematician Carl
Friedrich Gauss (1777–1855). He used
them to describe the small errors made
by astronomers and surveyors in
repeated careful measurements of the
same quantity. You will sometimes see
Normal distributions labeled “Gaussian”
in honor of Gauss.
The 68–95–99.7 Rule
2.19
Earlier, we saw that the distribution of Iowa Test of Basic Skills (ITBS) vocabulary
scores for seventh-grade students in Gary, Indiana, is symmetric, single-peaked,
and bell-shaped. Suppose that the distribution of scores over time is exactly Normal with mean m = 6.84 and standard deviation s = 1.55. (These are the mean and
standard deviation of the 947 actual scores.)
The figure shows the Normal density curve
for this distribution with the points 1, 2, and
3 standard deviations from the mean labeled
on the horizontal axis.
How unusual is it for a Gary seventhgrader to get an ITBS score above 9.94? As
the following activity shows, the answer to
this question is surprisingly simple.
3.74
5.29
6.84
8.39
9.94
11.49
ITBS score
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CHAPTER 2
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Activity
MATERIALS:
Computer with
Internet access
PLET
AP
The Normal density curve applet
In this Activity, you will use the Normal Density Curve applet at the book’s Web
site (www.whfreeman.com/tps5e) to explore an interesting property of Normal
distributions. A graph similar to what you will see when you launch the applet is
shown below. The applet finds the area under the curve in the region indicated
by the green flags. If you drag one flag past the other, the applet will show the area
under the curve between the two flags. When the “2-Tail” box is checked, the applet calculates symmetric areas around the mean.
Use the applet to help you answer the following questions.
1. ​If you put one flag at the extreme left of the curve and the second flag exactly
in the middle, what proportion is reported by the applet? Why does this value
make sense?
2. ​If you place the two flags exactly one standard deviation on either side of the
mean, what does the applet say is the area between them?
3. ​What percent of the area under the Normal
curve lies within 2 standard deviations of the mean?
4. ​Use the applet to show that about 99.7% of the
area under the Normal density curve lies within
three standard deviations of the mean. Does this
mean that about 99.7%/2 = 49.85% will lie within
one and a half standard deviations? Explain.
5. Change the mean to 100 and the standard deviation to 15. Then click “Update.” What percent
of the area under this Normal density curve lies
within one, two, and three standard deviations of
the mean?
6. ​Change the mean to 6.84 and the standard deviation to 1.55. (These values
are from the ITBS vocabulary scores in Gary, Indiana.) Answer the question
from Step 5.
7. ​Summarize: Complete the following sentence: “For any Normal density
curve, the area under the curve within one, two, and three standard deviations of
the mean is about ___%, ___%, and ___%.”
Although there are many Normal curves, they all have properties in common.
In particular, all Normal distributions obey the following rule.
Definition: The 68–95–99.7 rule
In a Normal distribution with mean m and standard deviation s:
• Approximately 68% of the observations fall within s of the mean m.
• Approximately 95% of the observations fall within 2s of the mean m.
• Approximately 99.7% of the observations fall within 3s of the mean m.
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Section 2.2 Density Curves and Normal Distributions
Figure 2.12 illustrates the 68–95–99.7 rule. (Some people refer
to this result as the “empirical rule.”) By remembering these three
numbers, you can think about Normal distributions without constantly making detailed calculations.
Here’s an example that shows how we can use the 68–95–
99.7 rule to estimate the percent of observations in a specified
interval.
68% of data
95% of data
99.7% of data
–3
–2
–1
0
1
2
3
FIGURE 2.12 ​The 68–95–99.7 rule for Normal distributions.
Standard deviations
EXAMPLE
ITBS Vocabulary Scores
Using the 68–95–99.7 rule
Problem: The distribution of ITBS vocabulary scores for seventh-graders in Gary, Indiana, is
N (6.84, 1.55).
(a) What percent of the ITBS vocabulary scores are less than 3.74? Show your work.
(b) What percent of the scores are between 5.29 and 9.94? Show your work.
SOLUTION:
(a) ​Notice that a score of 3.74 is exactly two standard deviations below the mean. By the
68–95–99.7 rule, about 95% of all scores are between
m − 2s = 6.84 − (2)(1.55) = 6.84 − 3.10 = 3.74
m + 2s = 6.84 + (2)(1.55) = 6.84 + 3.10 = 9.94
The other 5% of scores are outside this range. Because Normal distributions are symmetric,
half of these scores are lower than 3.74 and half are higher than 9.94. That is, about 2.5% of
the ITBS scores are below 3.74. Figure 2.13(a) shows this reasoning in picture form.
(b) ​Let’s start with a picture. Figure 2.13(b) shows the area under the Normal density curve between 5.29
and 9.94. We can see that about 68% + 13.5% = 81.5% of ITBS scores are between 5.29 and 9.94.
and
About 95%
of scores are
within 2
of .
About 2.5%
of scores are
less than
3.74.
2.19
(a)
3.74
5. 29
6.84
ITBS score
8.39
9.94
About 68%
of scores
95% of
scores
within
2 of
Area
95% –68%
2
=13.5%
11.49
2.19
(b)
3.74
5.29
6.84
8.39
9.94
11.49
ITBS score
FIGURE 2.13 (a) ​Finding the percent of Iowa Test scores less than 3.74. (b) Finding the percent of Iowa Test scores between
5.29 and 9.94.
For Practice Try Exercise
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CHAPTER 2
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Chebyshev’s inequality is an interesting
result, but it is not required for the AP®
Statistics exam.
THINK
ABOUT IT
The 68–95–99.7 rule applies only to Normal distributions. Is there a similar
rule that would apply to any distribution? Sort of. A result known as Chebyshev’s
inequality says that in any distribution, the proportion of observations falling with1
in k standard deviations of the mean is at least 1 − 2 . If k = 2, for example, Chebyk
1
shev’s inequality tells us that at least 1 − 2 = 0.75 of the observations in any distri2
bution are within 2 standard deviations of the mean. For Normal distributions, we
know that this proportion is much higher than 0.75. In fact, it’s approximately 0.95.
All models are wrong, but some are useful! ​The 68–95–99.7 rule
describes distributions that are exactly Normal. Real data such as the actual ITBS
scores are never exactly Normal. For one thing, ITBS scores are reported only to
the nearest tenth. A score can be 9.9 or 10.0 but not 9.94. We use a Normal distribution because it’s a good approximation, and because we think of the knowledge
that the test measures as continuous rather than stopping at tenths.
How well does the 68–95–99.7 rule describe the actual ITBS scores? Well, 900
of the 947 scores are between 3.74 and 9.94. That’s 95.04%, very accurate indeed.
Of the remaining 47 scores, 20 are below 3.74 and 27 are above 9.94. The tails of
the actual data are not quite equal, as they would be in an exactly Normal distribution. Normal distributions often describe real data better in the center of the
distribution than in the extreme high and low tails.
As famous statistician George Box once noted, “All models are wrong, but
some are useful!”
Check Your Understanding
The distribution of heights of young women aged 18 to 24 is approximately N(64.5, 2.5).
1. ​Sketch a Normal density curve for the distribution of young women’s heights. Label
the points one, two, and three standard deviations from the mean.
2. ​What percent of young women have heights greater than 67 inches? Show your
work.
3. ​What percent of young women have heights between 62 and 72 inches? Show
your work.
The Standard Normal Distribution
As the 68–95–99.7 rule suggests, all Normal distributions share many properties.
In fact, all Normal distributions are the same if we measure in units of size s from
the mean m as center. Changing to these units requires us to standardize, just as
we did in Section 2.1:
x−m
z=
s
If the variable we standardize has a Normal distribution, then so does the new
variable z. (Recall that subtracting a constant and dividing by a constant don’t
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Section 2.2 Density Curves and Normal Distributions
113
change the shape of a distribution.) This new distribution with mean m = 0 and
standard deviation s = 1 is called the standard Normal distribution.
Definition: Standard Normal distribution
The standard Normal distribution is the Normal distribution with mean 0 and standard deviation 1 (Figure 2.14).
–3
–2
–1
0
1
2
3
FIGURE 2.14 ​The standard Normal distribution.
If a variable x has any Normal distribution N (m, s) with mean m and standard
deviation s, then the standardized variable
z=
x−m
s
has the standard Normal distribution N (0,1).
An area under a density curve is a proportion of the observations in a distribution. Any question about what proportion of observations lies in some range of
values can be answered by finding an area under the curve. In a standard Normal
distribution, the 68–95–99.7 rule tells us that about 68% of the observations fall
between z = −1 and z = 1 (that is, within one standard deviation of the mean).
What if we want to find the percent of observations that fall between z = −1.25
and z = 1.25? The 68–95–99.7 rule can’t help us.
Because all Normal distributions are the same when we standardize, we can find
areas under any Normal curve from a single table, a table that gives areas under the
curve for the standard Normal distribution. Table A, the standard Normal table, gives
areas under the standard Normal curve. You can find Table A in the back of the book.
Table entry is
area to left of z.
Definition: The standard Normal table
Table A is a table of areas under the standard Normal curve. The table entry for each
value z is the area under the curve to the left of z.
z
z
.00
.01
.02
0.7
.7580 .7611 .7642
0.8
.7881 .7910 .7939
0.9
.8159 .8186 .8212
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For instance, suppose we wanted to find the proportion of observations from
the standard Normal distribution that are less than 0.81. To find the area to the
left of z = 0.81, locate 0.8 in the left-hand column of Table A, then locate the
remaining digit 1 as .01 in the top row. The entry opposite 0.8 and under .01 is
.7910. This is the area we seek. A reproduction of the relevant portion of Table A
is shown in the margin.
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CHAPTER 2
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Table entry for z
is always the area
under the curve
to the left of z.
FIGURE 2.15 ​The area under a
standard Normal curve to the left of
the point z = 0.81 is 0.7910.
–3
–2
–1
Table entry = 0.7910
for z = 0.81.
0
1
2
3
z = 0.81
Figure 2.15 illustrates the relationship between the value z = 0.81 and the area
0.7910. Note that we have made a connection between z-scores and percentiles when the
shape of a distribution is Normal.
EXAMPLE
Standard Normal Distribution
Finding area to the right
z
.07
.08
.09
21.8
.0307 .0301 .0294
21.7
.0384 .0375 .0367
21.6
.0475 .0465 .0455
What if we wanted to find the proportion of observations from the standard
Normal distribution that are greater than −1.78? To find the area to the right
of z = −1.78, locate −1.7 in the left-hand column of Table A, then locate the
remaining digit 8 as .08 in the top row. The corresponding entry is .0375. (See
the excerpt from Table A in the margin.)
This is the area to the left of z = −1.78. To find the area to the right of z = −1.78, we
use the fact that the total area under the standard Normal density curve is 1. So the
desired proportion is 1 − 0.0375 = 0.9625.
Figure 2.16 illustrates the relationship between the value z = −1.78 and the area
0.9625.
Table entry = 0.0375
for z = – 1.78. This is
the area to the left
of z = – 1.78.
FIGURE 2.16 ​The area
under a standard Normal
curve to the right of the
point z = –1.78 is 0.9625.
Starnes-Yates5e_c02_082-139hr2.indd 114
–3
–2
–1
The area to the right
of z = –1.78 is
1 – 0.0375 = 0.9625.
0
1
2
3
z = – 1.78
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115
Section 2.2 Density Curves and Normal Distributions
c
!
EXAMPLE
n
A common student mistake is to look up a z-value in Table A and report autio
the entry corresponding to that z-value, regardless of whether the problem
asks for the area to the left or to the right of that z-value. To prevent making
this mistake, always sketch the standard Normal curve, mark the z-value, and shade
the area of interest. And before you finish, make sure your answer is reasonable in
the context of the problem.
Catching Some “z”s
Finding areas under the standard Normal curve
Problem: Find the proportion of observations from the standard Normal distribution that are
between –1.25 and 0.81.
Solution: From Table A, the area to the left of z = 0.81 is 0.7910 and the area to the left of
z = –1.25 is 0.1056. So the area under the standard Normal curve between these two z-scores is
0.7910 – 0.1056 = 0.6854. Figure 2.17 shows why this approach works.
Area to left of
z = 0.81 is 0.7910
Area between z = –1.25 and z = 0.81 is
o.7910 – 0.1056 = 0.6854.
Area to left of
z = –1.25 is 0.1056.
–
–3
–2
–1
0
1
2
3
=
–3 –2
–1
0
1
2
3
–3
–2
–1
0
1
2
3
FIGURE 2.17 ​One way to find the area between z = –1.25 and z = 0.81 under the standard Normal curve.
Here’s another way to find the desired area. The area to the left of z = –1.25 under the standard
Normal curve is 0.1056. The area to the right of z = 0.81 is 1 – 0.7910 = 0.2090. So the area
between these two z-scores is
1 – (0.1056 + 0.2090) = 1 – 0.3146 = 0.6854
Figure 2.18 shows this approach in picture form.
The area to the
left of z = – 1.25
is 0.1056.
FIGURE 2.18 ​The area under the
standard Normal curve between
z = –1.25 and z = 0.81 is 0.6854.
–3
–2
The area to the
right of z = 0.81
is 0.2090.
–1
0
1
2
3
For Practice Try Exercise
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Working backward: From areas to z-scores: So far, we have used
Table A to find areas under the standard Normal curve from z-scores. What if we
want to find the z-score that corresponds to a particular area? For example, let’s
find the 90th percentile of the standard Normal curve. We’re looking for the zscore that has 90% of the area to its left, as shown in Figure 2.19.
The area to the
left of z is 0.90.
What’s z?
FIGURE 2.19 ​The z-score with
area 0.90 to its left under the
standard Normal curve.
z
.07
.08
.09
1.1
.8790
.8810
.8830
1.2
.8980
.8997
.9015
1.3
.9147
.9162
.9177
–3
–2
–1
0
1
2
3
z = ???
Because Table A gives areas to the left of a specified z-score, all we need to do
is find the value closest to 0.90 in the middle of the table. From the reproduced
portion of Table A, you can see that the desired z-score is z = 1.28. That is, the
area to the left of z = 1.28 is approximately 0.90.
Check Your Understanding
Use Table A in the back of the book to find the proportion of observations from a standard
Normal distribution that fall in each of the following regions. In each case, sketch a standard Normal curve and shade the area representing the region.
1. ​z < 1.39
2. ​z > −2.15
3. ​−0.56 < z < 1.81
Use Table A to find the value z from the standard Normal distribution that satisfies each
of the following conditions. In each case, sketch a standard Normal curve with your value
of z marked on the axis.
4. ​The 20th percentile
PLET
AP
5.
T echnology
Corner
5. 45% of all observations are greater than z
You can use the Normal Density Curve applet at www.whfreeman.com/tps5e
to confirm areas under the standard Normal curve. Just enter mean 0 and standard deviation 1, and then drag the flags to the appropriate locations. Of course,
you can also confirm Normal curve areas with your calculator.
From z-scores to areas,
and vice versa
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
Finding areas: The normalcdf command on the TI-83/84 (normCdf on the TI-89) can be used to find areas under
a Normal curve. The syntax is normalcdf(lower bound,upper bound,mean,standard deviation).
Let’s use this command to confirm our answers to the previous two examples.
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Section 2.2 Density Curves and Normal Distributions
117
1. What proportion of observations from the standard Normal distribution are greater than –1.78?
Recall that the standard Normal distribution has mean 0 and standard deviation 1.
TI-83/84TI-89
•
Press 2nd VARS (Distr) and choose nor- •
malcdf(. OS 2.55 or later: In the dialog box, enter
these values: lower:–1.78, upper:100000, •
m:0, s:1, choose Paste, and then press ENTER .
Older OS: Complete the command normalcdf
(–1.78,100000,0,1) and press ENTER .
In the Stats/List Editor, Press F5 (Distr) and
choose Normal Cdf(.
In the dialog box, enter these values: lower:–1.78,
upper:100000, m:0, s:1, and then choose
ENTER .
Note: We chose 100000 as the upper bound because it’s many, many standard deviations above the mean.
These results agree with our previous answer using Table A: 0.9625.
2.What proportion of observations from the standard Normal distribution are between –1.25 and 0.81?
The screen shots below confirm our earlier result of 0.6854 using Table A.
Working backward: The TI-83/84 and TI-89 invNorm function calculates the value corresponding to a given percentile in a
Normal distribution. For this command, the syntax is invNorm(area to the left,mean,standard deviation).
3. What is the 90th percentile of the standard Normal distribution?
TI-83/84TI-89
•
Press 2nd VARS (Distr) and choose invNorm(. OS 2.55 or later: In the dialog box, enter these values: area:.90, m:0, s:1, choose
Paste, and then press ENTER . Older OS: Complete the command invNorm(.90,0,1) and
press ENTER .
•
In the Stats/List Editor, Press F5 (Distr),
choose Inverse, and Inverse Normal….
•
In the dialog box, enter these values: area:.90,
m:0, s:1, and then choose ENTER .
These results match what we got using Table A.
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Normal Distribution Calculations
We can answer a question about areas in any Normal distribution by standardizing and using Table A or by using technology. Here is an outline of the method
for finding the proportion of the distribution in any region.
How to Find Areas in Any Normal Distribution
Step 1: State the distribution and the values of interest. Draw a Normal
curve with the area of interest shaded and the mean, standard deviation, and
boundary value(s) clearly identified.
Step 2: Perform calculations—show your work! Do one of the following:
(i) Compute a z-score for each boundary value and use Table A or technology to find the desired area under the standard Normal curve; or (ii) use the
normalcdf command and label each of the inputs.
Step 3: Answer the question.
Here’s an example of the method at work.
example
Tiger on the Range
Normal calculations
On the driving range, Tiger Woods practices his swing with a particular club by hitting
many, many balls. Suppose that when Tiger hits his driver, the distance the ball travels follows a Normal distribution with mean 304 yards and standard deviation 8 yards.
PROBLEM: What percent of Tiger’s drives travel at least 290 yards?
solution:
Step 1: State the distribution and the values of interest. The
N (304,8)
272
280
288
296
304
x = 290
z = –1.75
FIGURE 2.20 ​Distance traveled by
Tiger Woods’s drives on the range.
Starnes-Yates5e_c02_082-139hr2.indd 118
312
320
328
336
distance that Tiger’s ball travels follows a Normal distribution with
m = 304 and s = 8. We want to find the percent of Tiger’s drives
that travel 290 yards or more. Figure 2.20 shows the distribution
with the area of interest shaded and the mean, standard deviation,
and boundary value labeled.
Step 2: Perform calculations–show your work! For the boundary
value x = 290, we have
z=
x − m 290 − 304
=
= −1.75
s
8
S o drives of 290 yards or more correspond to z ≥−1.75 under the standard Normal curve.
From Table A, we see that the proportion of observations less than −1.75 is 0.0401. The area to
the right of −1.75 is therefore 1 − 0.0401 = 0.9599. This is about 0.96, or 96%.
Using technology: The command normalcdf(lower:290,upper:100000, m:304,
s:8) also gives an area of 0.9599.
Step 3: Answer the question. About 96% of Tiger Woods’s drives on the range travel at least 290
yards.
For Practice Try Exercise 53(a)
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119
Section 2.2 Density Curves and Normal Distributions
What proportion of Tiger Woods’s drives go exactly 290
yards? There is no area under the Normal density curve in Figure 2.20 exactly
THINK
ABOUT IT
over the point 290. So the answer to our question based on the Normal model
is 0. Tiger Woods’s actual data may contain a drive that went exactly 290 yards
(up to the precision of the measuring device). The Normal distribution is just an
easy-to-use approximation, not a description of every detail in the data. One more
thing: the areas under the curve with x ≥ 290 and x > 290 are the same. According to the Normal model, the proportion of Tiger’s drives that go at least 290 yards
is the same as the proportion that go more than 290 yards.
The key to doing a Normal calculation is to sketch the area you want, then match
that area with the area that the table (or technology) gives you. Here’s another example.
example
Tiger on the Range (Continued)
More complicated calculations
PROBLEM: What percent of Tiger’s drives travel between 305 and 325 yards?
solution:
Step 1: State the distribution and the values of interest. As
N(304,8)
in the previous example, the distance that Tiger’s ball travels follows
a Normal distribution with m = 304 and s = 8. We want to find the
percent of Tiger’s drives that travel between 305 and 325 yards.
Figure 2.21 shows the distribution with the area of interest shaded
and the mean, standard deviation, and boundary values labeled.
Step 2: Perform calculations–show your work! For the boundary
272
280
288
296
304
312
320
328
336
305 − 304
x = 305
x = 325
= 0.13. The standardized score
value x = 305, z =
8
325 − 304
FIGURE 2.21 ​Distance traveled by Tiger Woods’s drives on
for x = 325 is z =
= 2.63.
the range.
8
From Table A, we see that the area between z = 0.13 and z = 2.63 under the standard Normal curve
is the area to the left of 2.63 minus the area to the left of 0.13. Look at the picture below to check
this. From Table A, area between 0.13 and 2.63 = area to the left of 2.63 − area to the left of
0.13 = 0.9957 − 0.5517 = 0.4440.
minus
–3
–2
–1
0
1
2
3
z = 2.63
equals
–3
–2
–1
0
z = 0.13
1
2
3
–3
–2
–1
0
z = 0.13
1
2
3
z = 2.63
Using technology: The command normalcdf(lower:305, upper:325, m:304, s:8)
gives an area of 0.4459.
Step 3: Answer the question. About 45% of Tiger’s drives travel between 305 and 325 yards.
For Practice Try Exercise
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53(b)
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CHAPTER 2
Table A sometimes yields a slightly
different answer from technology.
That’s because we have to round
z-scores to two decimal places before
using Table A.
M o d e l i n g D i s t r i b u t i o n s o f Data
Sometimes we encounter a value of z more extreme than those appearing in
Table A. For example, the area to the left of z = −4 is not given directly in the
table. The z-values in Table A leave only area 0.0002 in each tail unaccounted
for. For practical purposes, we can act as if there is approximately zero area outside
the range of Table A.
Working backwards: From areas to values: The previous two ex-
amples illustrated the use of Table A to find what proportion of the observations
satisfies some condition, such as “Tiger’s drive travels between 305 and 325 yards.”
Sometimes, we may want to find the observed value that corresponds to a given
percentile. There are again three steps.
How to Find Values from Areas in Any Normal Distribution
Step 1: State the distribution and the values of interest. Draw a Normal
curve with the area of interest shaded and the mean, standard deviation, and
unknown boundary value clearly identified.
Step 2: Perform calculations—show your work! Do one of the following:
(i) Use Table A or technology to find the value of z with the indicated area
under the standard Normal curve, then “unstandardize” to transform back to
the original distribution; or (ii) Use the invNorm command and label each of
the inputs.
Step 3: Answer the question.
example
Cholesterol in Young Boys
Using Table A in reverse
High levels of cholesterol in the blood increase the risk of heart disease. For
14-year-old boys, the distribution of blood cholesterol is approximately Normal
with mean m = 170 milligrams of cholesterol per deciliter of blood (mg/dl) and
standard deviation s = 30 mg/dl.9
PROBLEM: What is the 1st quartile of the distribution of blood cholesterol?
solution:
Step 1: State the distribution and the values of interest. The cholesterol level of 14-year-old
boys follows a Normal distribution with m = 170 and s = 30. The 1st quartile is the boundary value x
with 25% of the distribution to its left. Figure 2.22 shows a picture of what we are trying to find.
Step 2: Perform calculations–show your work! Look in the body of Table A for the entry closest
to 0.25. It’s 0.2514. This is the entry corresponding to z = −0.67. So z = −0.67 is the standardized score with area 0.25 to its left. Now unstandardize. We know that the standardized score
x − 170
= −0.67.
for the unknown cholesterol level x is z = −0.67. So x satisfies the equation
30
Solving for x gives
x = 170 + (−0.67)(30) = 149.9
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Section 2.2 Density Curves and Normal Distributions
121
N(170,30)
Area = 0.25
FIGURE 2.22 ​Locating
the 1st quartile of the
cholesterol distribution
for 14-year-old boys.
z = –0.67
x=?
Using technology: The command invNorm(area:0.25, m:170, s:30) gives x = 149.8.
Step 3: Answer the question. The 1st quartile of blood cholesterol levels in 14-year-old boys is
about 150 mg/dl.
For Practice Try Exercise
53(c)
Check Your Understanding
Follow the method shown in the examples to answer each of the following questions.
Use your calculator or the Normal Curve applet to check your answers.
1. ​Cholesterol levels above 240 mg/dl may require medical attention. What percent of
14-year-old boys have more than 240 mg/dl of cholesterol?
2. ​People with cholesterol levels between 200 and 240 mg/dl are at considerable risk for
heart disease. What percent of 14-year-old boys have blood cholesterol between 200 and
240 mg/dl?
3. ​What distance would a ball have to travel to be at the 80th percentile of Tiger
Woods’s drive lengths?
Assessing Normality
The Normal distributions provide good models for some distributions of real data.
Examples include SAT and IQ test scores, the highway gas mileage of 2014 Corvette convertibles, state unemployment rates, and weights of 9-ounce bags of potato
chips. The distributions of some other common variables are usually skewed and
therefore distinctly non-Normal. Examples include economic variables such as personal income and total sales of business firms, the survival times of cancer patients
after treatment, and the lifetime of electronic devices. While experience can suggest
whether or not a Normal distribution is a reasonable model in a particular case, it
is risky to assume that a distribution is Normal without actually inspecting the data.
In the latter part of this course, we will use various statistical inference procedures to try to answer questions that are important to us. These tests involve sampling individuals and recording data to gain insights about the populations from
which they come. Many of these procedures are based on the assumption that
the population is approximately Normally distributed. Consequently, we need to
develop a strategy for assessing Normality.
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CHAPTER 2
M o d e l i n g D i s t r i b u t i o n s o f Data
example
Unemployment in the States
Are the data close to Normal?
Let’s start by examining data on unemployment rates in the 50 states. Here are the
data arranged from lowest (North Dakota’s 4.1%) to highest (Michigan’s 14.7%).10
4.1 4.5 5.0 6.3 6.3 6.4 6.4 6.6 6.7 6.7 6.7 6.9 7.0
7.0 7.2 7.4 7.4 7.4 7.8 8.0 8.0 8.2 8.2 8.4 8.5 8.5
8.6 8.7 8.8 8.9 9.1 9.2 9.5 9.6 9.6 9.7 10.2 10.3 10.5
10.6 10.6 10.8 10.9 11.1 11.5 12.3 12.3 12.3 12.7 14.7
• Plot the data. Make a dotplot, stemplot, or histogram. See if the graph is approximately symmetric and bell-shaped.
Figure 2.23 is a histogram of the state unemployment rates. The graph is roughly
symmetric, single-peaked, and somewhat bell-shaped.
12
Number of states
10
8
6
4
2
0
FIGURE 2.23 ​Histogram of state
unemployment rates.
4
6
8
10
12
14
Unemployment rate
• Check whether the data follow the 68–95–99.7 rule.
We entered the unemployment rates into computer software and requested
summary statistics. Here’s what we got:
Mean = 8.682 Standard deviation = 2.225.
Now we can count the number of observations within one, two, and three
standard deviations of the mean.
Mean ± 1 SD:
Mean ± 2 SD:
Mean ± 3 SD:
6.457 to 10.907
4.232 to 13.132
2.007 to 15.357
36 out of 50 = 72%
48 out of 50 = 96%
50 out of 50 = 100%
These percents are quite close to the 68%, 95%, and 99.7% targets for a Normal
distribution.
If a graph of the data is clearly skewed, has multiple peaks, or isn’t bell-shaped, that’s
evidence that the distribution is not Normal. However, just because a plot of the data
looks Normal, we can’t say that the distribution is Normal. The 68–95–99.7 rule can
give additional evidence in favor of or against Normality. A Normal probability plot
also provides a good assessment of whether a data set follows a Normal distribution.
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123
Section 2.2 Density Curves and Normal Distributions
example
Unemployment in the States
Making a Normal probability plot
Most software packages, including Minitab, Fathom, and JMP, can construct Normal probability plots (sometimes called Normal quantile plots) from entered data.
The TI-83/84 and TI-89 will also make these graphs. Here’s how a Normal probability plot is constructed.
The highlighted point
is (4.1, –2.326). North
Dakota’s 4.1%
unemployment rate is
at the 1st percentile,
which is at z = –2.326
in the standard
Normal distribution.
Expected z-score
3
2
1
1. ​Arrange the observed data values from smallest to largest. Record the percentile
corresponding to each observation (but remember that there are several definitions
of “percentile”). For example, the smallest observation in a set of 50 values is at either
the 0th percentile (because 0 out of 50 values are below this observation) or the 2nd
percentile (because 1 out of 50 values are at or below this observation). Technology
usually “splits the difference,” declaring this minimum value to be at the (0 + 2)/2 =
1st percentile. By similar reasoning, the second-smallest
value is at the 3rd percentile, the third-smallest value is at
the 5th percentile, and so on. The maximum value is at
the (98 + 100)/2 = 99th percentile.
2. ​Use the standard Normal distribution (Table A or
invNorm) to find the z-scores at these same percentiles.
For example, the 1st percentile of the standard Normal distribution is z = −2.326. The 3rd percentile is
z = −1.881; the 5th percentile is z = −1.645; . . . ; the
99th percentile is z = 2.326.
0
–1
–2
–3
2
4
6
8
10
12
14
16
Unemployment rate
FIGURE 2.24 ​Normal probability plot of the percent of unemployed individuals in each of the 50 states.
c
As Figure 2.24 indicates, real data almost always show some departure autio
from Normality. When you examine a Normal probability plot, look for
shapes that show clear departures from Normality. Don’t overreact to minor
wiggles in the plot. When we discuss statistical methods that are based on the Normal model, we will pay attention to the sensitivity of each method to departures
from Normality. Many common methods work well as long as the data are approximately Normal.
!
n
Some software plots the data values on
the horizontal axis and the z-scores on
the vertical axis, while other software
does just the reverse. The TI-83/84 and
TI-89 give you both options. We prefer
the data values on the horizontal axis,
which is consistent with other types of
graphs we have made.
3. ​Plot each observation x against its expected z-score from
Step 2. If the data distribution is close to Normal, the
plotted points will lie close to some straight line. Figure
2.24 shows a Normal probability plot for the state unemployment data. There is a strong linear pattern, which
suggests that the distribution of unemployment rates is
close to Normal.
Interpreting Normal Probability Plots
If the points on a Normal probability plot lie close to a straight line, the data
are approximately Normally distributed. Systematic deviations from a straight
line indicate a non-Normal distribution. Outliers appear as points that are far
away from the overall pattern of the plot.
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CHAPTER 2
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AP® EXAM TIP ​Normal probability plots are not included on the AP® Statistics topic outline.
However, these graphs are very useful for assessing Normality. You may use them on the AP®
exam if you wish—just be sure that you know what you’re looking for (a linear pattern).
Let’s look at an example of some data that are not Normally distributed.
example
Guinea Pig Survival
Assessing Normality
30
3
25
2
Expected z-score
Frequency of survival time
In Chapter 1 Review Exercise R1.7 (page 77), we introduced data on the survival
times in days of 72 guinea pigs after they were injected with infectious bacteria in
a medical experiment.
Problem: Determine whether these data are approximately Normally distributed.
Solution: Let’s follow the first step in our strategy for assessing Normality: plot the data!
Figure 2.25(a) shows a histogram of the guinea pig survival times. We can see that the distribution
is heavily right-skewed. Figure 2.25(b) is a Normal probability plot of the data. The clear curvature in
this graph confirms that these data do not follow a Normal distribution.
We won’t bother checking the 68–95–99.7 rule for these data because the graphs in Figure
2.25 indicate serious departures from Normality.
20
15
10
1
0
–1
–2
5
–3
0
100
200
300
400
Survival time (days)
500
0
600
100
200
300
400
Survival time (days)
500
600
(b)
(a)
FIGURE 2.25 ​(a) Histogram and (b) Normal probability plot of the guinea pig survival data.
For Practice Try Exercise
THINK
ABOUT IT
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63
How can we determine shape from a Normal probability
plot? ​Look at the Normal probability plot of the guinea pig survival data in
Figure 2.25(b). Imagine drawing a line through the leftmost points, which correspond to the smaller observations. The larger observations fall systematically
11/13/13 1:33 PM
Section 2.2 Density Curves and Normal Distributions
3
Expected z-score
2
1
0
–1
–2
–3
0
100
200
300
400
Survival time (days)
500
600
125
to the right of this line. That is, the right-of-center observations have much larger values than expected based on their
percentiles and the corresponding z-scores from the standard
Normal distribution.
This Normal probability plot indicates that the guinea
pig survival data are strongly right-skewed. In a right-skewed
distribution, the largest observations fall distinctly to the right
of a line drawn through the main body of points. Similarly,
left skewness is evident when the smallest observations fall
to the left the line.
If you’re wondering how to make a Normal probability plot on your calculator,
the following Technology Corner shows you the process.
6. T echnology
Corner Normal probability plots
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
To make a Normal probability plot for a set of quantitative data:
•
Enter the data values in L1/list1. We’ll use the state unemployment rates data from page 122.
•
Define Plot1 as shown.
TI-83/84TI-89
•
Use ZoomStat (ZoomData on the TI-89) to see the finished graph.
Interpretation: The Normal probability plot is quite linear, so it is reasonable to believe that the data follow a Normal
distribution.
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CHAPTER 2
M o d e l i n g D i s t r i b u t i o n s o f Data
DATA EXPLORATION The vending machine problem
Have you ever purchased a hot drink from a vending machine? The intended
sequence of events runs something like this. You insert your money into the machine and select your preferred beverage. A cup falls out of the machine, landing
upright. Liquid pours out until the cup is nearly full. You reach in, grab the pipinghot cup, and drink happily.
Sometimes, things go wrong. The machine might swipe your money. Or the cup
might fall over. More frequently, everything goes smoothly until the liquid begins to
flow. It might stop flowing when the cup is only half full. Or the liquid might keep coming until your cup overflows. Neither of these results leaves you satisfied.
The vending machine company wants to keep customers happy. So they have
decided to hire you as a statistical consultant. They provide you with the following
summary of important facts about the vending machine:
• Cups will hold 8 ounces.
• The amount of liquid dispensed varies according to a Normal distribution
centered at the mean m that is set in the machine.
• s = 0.2 ounces.
If a cup contains too much liquid, a customer may get burned from a spill. This
could result in an expensive lawsuit for the company. On the other hand, customers may be irritated if they get a cup with too little liquid from the machine. Given
these issues, what mean setting for the machine would you recommend? Write a
brief report to the vending machine company president that explains your answer.
case closed
Do You Sudoku?
In the chapter-opening Case Study (page 83), one of the authors played
an online game of sudoku. At the end of his game, the graph on the next
page was displayed. The density curve shown was constructed from a histogram of times from 4,000,000 games played in one week at this Web
site. You will now use what you have learned in this chapter to analyze
how well the author did.
1.
2.
3.
Starnes-Yates5e_c02_082-139hr2.indd 126
State and interpret the percentile for the author’s time of
3 minutes and 19 seconds. (Remember that smaller times
indicate better performance.)
Explain why you cannot find the z-score corresponding to
the author’s time.
Suppose the author’s time to finish the puzzle had been 5
minutes and 6 seconds instead.
(a) Would his percentile be greater than 50%, equal to
50%, or less than 50%? Justify your answer.
(b) Would his z-score be positive, negative, or zero? Explain.
11/13/13 1:33 PM
Section 2.2 Density Curves and Normal Distributions
4.
5.
Section 2.2
From long experience, the author’s times to finish an easy sudoku puzzle at this Web site follow
a Normal distribution with mean 4.2 minutes and
standard deviation 0.7 minutes. In what percent of
the games that he plays does the author finish an
easy puzzle in less than 3 minutes and 15 seconds?
Show your work. (Note: 3 minutes and 15 seconds
is not the same as 3.15 seconds!)
The author’s wife also enjoys playing sudoku
online. Her times to finish an easy puzzle at this
Web site follow a Normal distribution with mean
3.8 minutes and standard deviation 0.9 minutes. In
her most recent game, she finished in 3 minutes.
Whose performance is better, relatively speaking:
the author’s 3 minutes and 19 seconds or his wife’s
3 minutes? Justify your answer.
Summary
•
•
•
•
•
•
•
We can describe the overall pattern of a distribution by a density curve. A
density curve always remains on or above the horizontal axis and has total area 1
underneath it. An area under a density curve gives the proportion of observations
that fall in an interval of values.
A density curve is an idealized description of the overall pattern of a distribution
that smooths out the irregularities in the actual data. We write the mean of
a density curve as m and the standard deviation of a density curve as s to
distinguish them from the mean x– and the standard deviation sx of the actual data.
The mean and the median of a density curve can be located by eye. The mean m
is the balance point of the curve. The median divides the area under the curve in
half. The standard deviation s cannot be located by eye on most density curves.
The mean and median are equal for symmetric density curves. The mean of a
skewed curve is located farther toward the long tail than the median is.
The Normal distributions are described by a special family of bell-shaped,
symmetric density curves, called Normal curves. The mean m and standard
deviation s completely specify a Normal distribution N(m,s). The mean is the
center of the curve, and s is the distance from m to the change-of-curvature
points on either side.
All Normal distributions obey the 68–95–99.7 rule, which describes what percent
of observations lie within one, two, and three standard deviations of the mean.
All Normal distributions are the same when measurements are standardized. If
x follows a Normal distribution with mean m and standard deviation s, we can
standardize using
z=
Starnes-Yates5e_c02_082-139hr2.indd 127
127
x−m
s
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CHAPTER 2
M o d e l i n g D i s t r i b u t i o n s o f Data
•
•
The variable z has the standard Normal distribution with mean 0 and standard deviation 1.
Table A at the back of the book gives percentiles for the standard Normal
curve. By standardizing, we can use Table A to determine the percentile
for a given z-score or the z-score corresponding to a given percentile in any
Normal distribution. You can use your calculator or the Normal Curve applet
to perform Normal calculations quickly.
To perform certain inference procedures in later chapters, we will need to
know that the data come from populations that are approximately Normally
distributed. To assess Normality for a given set of data, we first observe the
shape of a dotplot, stemplot, or histogram. Then we can check how well the
data fit the 68–95–99.7 rule for Normal distributions. Another good method
for assessing Normality is to construct a Normal probability plot.
2.2 T echnology
Corners
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
5. From z-scores to areas, and vice versa
6. Normal probability plots
Section 2.2
page 116
page 125
Exercises
33. Density curves Sketch a density curve that might describe a distribution that is symmetric but has two peaks.
34. Density curves Sketch a density curve that might
describe a distribution that has a single peak and is
skewed to the left.
Exercises 35 to 38 involve a special type of density curve—
one that takes constant height (looks like a horizontal line)
over some interval of values. This density curve describes a
variable whose values are distributed evenly (uniformly) over
some interval of values. We say that such a variable has a
uniform distribution.
(b)​The proportion of accidents that occur in the first
mile of the path is the area under the density curve
between 0 miles and 1 mile. What is this area?
(c)​Sue’s property adjoins the bike path between the
0.8 mile mark and the 1.1 mile mark. What
proportion of accidents happen in front of Sue’s
property? Explain.
36. Where’s the bus? Sally takes the same bus to work
every morning. The amount of time (in minutes) that
she has to wait for the bus to arrive is described by the
uniform distribution below.
35. Biking accidents Accidents on a level, 3-mile bike
path occur uniformly along the length of the path.
The figure below displays the density curve that
describes the uniform distribution of accidents.
Height =
1
10
Height = 1/3
0
0
1
2
3
Distance along bike path (miles)
(a)​Explain why this curve satisfies the two requirements
for a density curve.
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10
(a)Explain why this curve satisfies the two requirements
for a density curve.
(b)​On what percent of days does Sally have to wait more
than 8 minutes for the bus?
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129
Section 2.2 Density Curves and Normal Distributions
(c)​On what percent of days does Sally wait between 2.5
and 5.3 minutes for the bus?
37. Biking accidents What is the mean m of the density
curve pictured in Exercise 35? (That is, where would
the curve balance?) What is the median? (That is,
where is the point with area 0.5 on either side?)
38. Where’s the bus? What is the mean m of the density
curve pictured in Exercise 36? What is the median?
39. Mean and median The figure below displays two
density curves, each with three points marked. At
which of these points on each curve do the mean and
the median fall?
A BC
A
(a)
B
44. Potato chips Refer to Exercise 42. Use the 68–95–
99.7 rule to answer the following questions. Show
your work!
(a)​Between what weights do the middle 68% of bags fall?
(b)​What percent of bags weigh less than 9.02 ounces?
(c)​What percent of 9-ounce bags of this brand of potato
chips weigh between 8.97 and 9.17 ounces?
(d)​A bag that weighs 9.07 ounces is at what percentile in
this distribution?
45. Estimating SD The figure below shows two Normal curves, both with mean 0. Approximately what
is the standard deviation of each of these curves?
C
(b)
40. Mean and median The figure below displays two
density curves, each with three points marked. At
which of these points on each curve do the mean and
the median fall?
–1.6
A B C
(a)
AB C
(b)
–1.2
–0.8
–0.4
0
0.4
0.8
1.2
1.6
46. A Normal curve Estimate the mean and standard deviation of the Normal density curve in the figure below.
41. Men’s heights The distribution of heights of adult
American men is approximately Normal with mean
69 inches and standard deviation 2.5 inches. Draw an
accurate sketch of the distribution of men’s heights.
Be sure to label the mean, as well as the points 1, 2,
and 3 standard deviations away from the mean on the
horizontal axis.
42. Potato chips The distribution of weights of 9-ounce
bags of a particular brand of potato chips is approximately Normal with mean m = 9.12 ounces
and standard deviation s = 0.05 ounce. Draw an
accurate sketch of the distribution of potato chip bag
weights. Be sure to label the mean, as well as the
points 1, 2, and 3 standard deviations away from the
mean on the horizontal axis.
43. Men’s heights Refer to Exercise 41. Use the 68–95–
pg 111 99.7 rule to answer the following questions. Show your
work!
(a)​Between what heights do the middle 95% of men fall?
(b)​What percent of men are taller than 74 inches?
(c)​What percent of men are between 64 and 66.5 inches
tall?
(d)​A height of 71.5 inches corresponds to what percentile of adult male American heights?
Starnes-Yates5e_c02_082-139hr2.indd 129
3
4
5
6
7
8
9
10
11
12
13 14
15
16
17
For Exercises 47 to 50, use Table A to find the proportion
of observations from the standard Normal distribution that
satisfies each of the following statements. In each case,
sketch a standard Normal curve and shade the area under
the curve that is the answer to the question.
47. Table A practice
(a)​
z < 2.85
(b)​z > 2.85
(c) ​z > −1.66
(d) ​−1.66 < z < 2.85
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M o d e l i n g D i s t r i b u t i o n s o f Data
48. Table A practice
pg 115
(a)​
z < −2.46
(c) ​0.89 < z < 2.46
(b)​
z > 2.46
(d) −2.95 < z < −1.27
49. More Table A practice
(a)​
z is between −1.33 and 1.65
(b)​
z is between 0.50 and 1.79
50. More Table A practice
(a) z is between −2.05 and 0.78
(b)​
z is between −1.11 and −0.32
For Exercises 51 and 52, use Table A to find the value z
from the standard Normal distribution that satisfies each of
the following conditions. In each case, sketch a standard
Normal curve with your value of z marked on the axis.
51. Working backward
(a)​The 10th percentile.
(b)​34% of all observations are greater than z.
52. Working backward
(a)​The 63rd percentile.
(b)​75% of all observations are greater than z.
53. Length of pregnancies The length of human pregnancies from conception to birth varies according
to a distribution that is approximately Normal with
mean 266 days and standard deviation 16 days.
pg 118
(a)​At what percentile is a pregnancy that lasts 240 days
(that’s about 8 months)?
pg 119
(b)​What percent of pregnancies last between 240
and 270 days (roughly between 8 months and
9 months)?
pg 120
(c)​How long do the longest 20% of pregnancies last?
54. IQ test scores Scores on the Wechsler Adult Intelligence Scale (a standard IQ test) for the 20 to 34 age
group are approximately Normally distributed with
m = 110 and s = 25.
(a)​At what percentile is an IQ score of 150?
(b)​What percent of people aged 20 to 34 have IQs
between 125 and 150?
(c)MENSA is an elite organization that admits as members people who score in the top 2% on IQ tests.
What score on the Wechsler Adult Intelligence Scale
would an individual aged 20 to 34 have to earn to
qualify for MENSA membership?
55. Put a lid on it! At some fast-food restaurants, customers who want a lid for their drinks get them from a
large stack left near straws, napkins, and condiments.
Starnes-Yates5e_c02_082-139hr2.indd 130
The lids are made with a small amount of flexibility
so they can be stretched across the mouth of the
cup and then snugly secured. When lids are too
small or too large, customers can get very frustrated,
especially if they end up spilling their drinks. At one
particular restaurant, large drink cups require lids
with a “diameter” of between 3.95 and 4.05 inches.
The restaurant’s lid supplier claims that the diameter
of their large lids follows a Normal distribution with
mean 3.98 inches and standard deviation 0.02 inches.
Assume that the supplier’s claim is true.
(a)​What percent of large lids are too small to fit? Show
your method.
(b)​What percent of large lids are too big to fit? Show
your method.
(c)​Compare your answers to parts (a) and (b). Does it
make sense for the lid manufacturer to try to make
one of these values larger than the other? Why or why
not?
56. I think I can! An important measure of the performance of a locomotive is its “adhesion,” which is the
locomotive’s pulling force as a multiple of its weight.
The adhesion of one 4400-horsepower diesel locomotive varies in actual use according to a Normal distribution with mean m = 0.37 and standard deviation
s = 0.04.
(a) For a certain small train’s daily route, the locomotive
needs to have an adhesion of at least 0.30 for the train
to arrive at its destination on time. On what proportion of days will this happen? Show your method.
(b)​An adhesion greater than 0.50 for the locomotive will
result in a problem because the train will arrive too
early at a switch point along the route. On what proportion of days will this happen? Show your method.
(c)​Compare your answers to parts (a) and (b). Does it
make sense to try to make one of these values larger
than the other? Why or why not?
57. Put a lid on it! Refer to Exercise 55. The supplier is
considering two changes to reduce the percent of its
large-cup lids that are too small to 1%. One strategy is
to adjust the mean diameter of its lids. Another option
is to alter the production process, thereby decreasing
the standard deviation of the lid diameters.
(a)If the standard deviation remains at s = 0.02 inches,
at what value should the supplier set the mean diameter of its large-cup lids so that only 1% are too small
to fit? Show your method.
(b) If the mean diameter stays at m = 3.98 inches, what
value of the standard deviation will result in only 1%
of lids that are too small to fit? Show your method.
11/13/13 1:33 PM
131
Section 2.2 Density Curves and Normal Distributions
(c)​Which of the two options in parts (a) and (b) do you
think is preferable? Justify your answer. (Be sure to
consider the effect of these changes on the percent of
lids that are too large to fit.)
58. I think I can! Refer to Exercise 56. The locomotive’s
manufacturer is considering two changes that could
reduce the percent of times that the train arrives late.
One option is to increase the mean adhesion of the
locomotive. The other possibility is to decrease the
variability in adhesion from trip to trip, that is, to
reduce the standard deviation.
(a) If
​ the standard deviation remains at s = 0.04, to what
value must the manufacturer change the mean adhesion of the locomotive to reduce its proportion of late
arrivals to only 2% of days? Show your work.
(b)​If the mean adhesion stays at m = 0.37, how much
must the standard deviation be decreased to ensure
that the train will arrive late only 2% of the time?
Show your work.
(c) ​Which of the two options in parts (a) and (b) do you
think is preferable? Justify your answer. (Be sure to
consider the effect of these changes on the percent of
days that the train arrives early to the switch point.)
59. Deciles The deciles of any distribution are the values
at the 10th, 20th, . . . , 90th percentiles. The first and
last deciles are the 10th and the 90th percentiles,
respectively.
(a) ​What are the first and last deciles of the standard Normal distribution?
(b)​The heights of young women are approximately Normal with mean 64.5 inches and standard deviation 2.5
inches. What are the first and last deciles of this distribution? Show your work.
60. Outliers The percent of the observations that are
classified as outliers by the 1.5 × IQR rule is the same
in any Normal distribution. What is this percent?
Show your method clearly.
61. Flight times An airline flies the same route at the
same time each day. The flight time varies according to
a Normal distribution with unknown mean and standard deviation. On 15% of days, the flight takes more
than an hour. On 3% of days, the flight lasts 75 minutes
or more. Use this information to determine the mean
and standard deviation of the flight time distribution.
62. Brush your teeth The amount of time Ricardo
spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation.
Ricardo spends less than one minute brushing his
teeth about 40% of the time. He spends more than
Starnes-Yates5e_c02_082-139hr2.indd 131
two minutes brushing his teeth 2% of the time. Use
this information to determine the mean and standard
deviation of this distribution.
63. Sharks Here are the lengths in feet of 44 great white
sharks:11
pg 124
18.7
16.4
13.2
19.1
12.3
16.7
15.8
16.2
18.6
17.8
14.3
22.8
16.4 15.7
16.2 12.6
16.6 9.4
16.8 13.6
18.3
17.8
18.2
13.2
14.6
13.8
13.2
15.7
15.8
12.2
13.6
19.7
14.9
15.2
15.3
18.7
17.6
14.7
16.1
13.2
12.1
12.4
13.5
16.8
(a)​Enter these data into your calculator and make a histogram. Include a sketch of the graph on your paper. Then
calculate one-variable statistics. Describe the shape, center, and spread of the distribution of shark lengths.
(b) Calculate the percent of observations that fall within 1,
2, and 3 standard deviations of the mean. How do these
results compare with the 68–95–99.7 rule?
(c) Use your calculator to construct a Normal probability
plot. Include a sketch of the graph on your paper. Interpret this plot.
(d)​Having inspected the data from several different perspectives, do you think these data are approximately
Normal? Write a brief summary of your assessment
that combines your findings from parts (a) through (c).
64. Density of the earth In 1798, the English scientist
Henry Cavendish measured the density of the earth
several times by careful work with a torsion balance.
The variable recorded was the density of the earth as a
multiple of the density of water. Here are Cavendish’s
29 measurements:12
5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.65
5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.39
5.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85
(a) Enter
​
these data into your calculator and make a histogram. Include a sketch of the graph on your paper.
Then calculate one-variable statistics. Describe the
shape, center, and spread of the distribution of density
measurements.
(b)​Calculate the percent of observations that fall within
1, 2, and 3 standard deviations of the mean. How do
these results compare with the 68–95–99.7 rule?
(c) Use your calculator to construct a Normal probability
plot. Include a sketch of the graph on your paper. Interpret this plot.
(d)​Having inspected the data from several different
perspectives, do you think these data are approximately Normal? Write a brief summary of your assessment that combines your findings from parts (a)
through (c).
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M o d e l i n g D i s t r i b u t i o n s o f Data
65. Runners’ heart rates The figure below is a Normal
probability plot of the heart rates of 200 male runners
after six minutes of exercise on a treadmill.13 The
distribution is close to Normal. How can you see
this? Describe the nature of the small deviations from
Normality that are visible in the plot.
Mean
Std. Dev.
Min
Max
10614
8049
1873
30823
Based on the relationship between the mean, standard deviation, minimum, and maximum, is it reasonable to believe that the distribution of Michigan
tuitions is approximately Normal? Explain.
3
2
Expected z-score
67. Is Michigan Normal? We collected data on the tuition
charged by colleges and universities in Michigan.
Here are some numerical summaries for the data:
68. Weights aren’t Normal The heights of people of the
same gender and similar ages follow Normal distributions reasonably closely. Weights, on the other hand,
are not Normally distributed. The weights of women
aged 20 to 29 have mean 141.7 pounds and median
133.2 pounds. The first and third quartiles are 118.3
pounds and 157.3 pounds. What can you say about
the shape of the weight distribution? Why?
1
0
–1
–2
Multiple choice: Select the best answer for Exercises 69 to 74.
69. Two measures of center are marked on the density
curve shown. Which of the following is correct?
–3
70
80
90
100
110
120
130
140
150
Heart rate (beats per minute)
66. Carbon dioxide emissions The figure below is a
Normal probability plot of the emissions of carbon
dioxide per person in 48 countries.14 In what ways is
this distribution non-Normal?
Expected z-score
3
2
(a)​The median is at the yellow line and the mean is at
the red line.
1
(b)​The median is at the red line and the mean is at the
yellow line.
0
(c)​The mode is at the red line and the median is at the
yellow line.
–1
(d)​The mode is at the yellow line and the median is at
the red line.
(e)​The mode is at the red line and the mean is at the
yellow line.
–2
–3
0
2
4
6
8
10
12
14
16
18
CO2 emissions (metric tons per person)
Starnes-Yates5e_c02_082-139hr2.indd 132
20
Exercises 70 to 72 refer to the following setting. The
weights of laboratory cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2
grams. The following figure is the Normal curve for this
distribution of weights.
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133
Section 2.2 Density Curves and Normal Distributions
If the distribution of points was displayed in a histogram, what would be the best description of the
histogram’s shape?
(a)​Approximately Normal
(b)​Symmetric but not approximately Normal
(c)​Skewed left
(d)​Skewed right
(e)​Cannot be determined
A
B
C
D
E
F
G
70. Point C on this Normal curve corresponds to
(a) 84 grams. (c) ​78 grams. (e) ​74 grams.
(b)​82 grams. (d) ​76 grams.
71. About what percent of the cockroaches have weights
between 76 and 84 grams?
40
(c) ​
68% (e) ​34%
(d) 47.5%
35
72. About what proportion of the cockroaches will have
weights greater than 83 grams?
(a) 0.0228
(b)​
0.0668
(c) ​0.1587
(d) ​0.9332
(e) ​0.0772
73. A different species of cockroach has weights that
follow a Normal distribution with a mean of 50
grams. After measuring the weights of many of these
cockroaches, a lab assistant reports that 14% of the
cockroaches weigh more than 55 grams. Based on
this report, what is the approximate standard deviation
of weights for this species of cockroaches?
(a) 4.6 (d) ​14.0
(b) 5.0 (e) Cannot determine without more information.
(c)​6.2
74. The following Normal probability plot shows the
distribution of points scored for the 551 players in the
2011–2012 NBA season.
Miles per gallon
(a) 99.7%
(b)​95%
75. Gas it up! (1.3) Interested in a sporty car? Worried
that it might use too much gas? The Environmental Protection Agency lists most such vehicles in
its “two-seater” or “minicompact” categories. The
figure shows boxplots for both city and highway gas
mileages for our two groups of cars. Write a few
sentences comparing these distributions.
30
25
20
15
10
5
0
Two city
Two hwy
Mini city
Mini hwy
76. Python eggs (1.1) How is the hatching of water
python eggs influenced by the temperature of the
snake’s nest? Researchers assigned newly laid eggs
to one of three temperatures: hot, neutral, or cold.
Hot duplicates the extra warmth provided by the
mother python, and cold duplicates the absence
of the mother. Here are the data on the number of
eggs and the number that hatched:15
Probability Plot of Points
Normal
3
Expected z-score
2
1
Cold
Neutral
Hot
Number of eggs
27
56
104
Number hatched
16
38
75
(a)​Make a two-way table of temperature by outcome
(hatched or not).
0
-1
-2
-3
0
Starnes-Yates5e_c02_082-139hr2.indd 133
500
1000
Points
1500
2000
(b)​Calculate the percent of eggs in each group that
hatched. The researchers believed that eggs would be
less likely to hatch in cold water. Do the data support
that belief?
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M o d e l i n g D i s t r i b u t i o n s o f Data
FRAPPY! Free Response AP® Problem, Yay!
The following problem is modeled after actual AP® Statistics exam
free response questions. Your task is to generate a complete, concise response in 15 minutes.
Directions: Show all your work. Indicate clearly the methods
you use, because you will be scored on the correctness of your
methods as well as on the accuracy and completeness of your
results and explanations.
The distribution of scores on a recent test closely followed a Normal distribution with a mean of 22 points and
a standard deviation of 4 points.
(a) What proportion of the students scored at least 25
points on this test?
(b) What is the 31st percentile of the distribution of
test scores?
(c) The teacher wants to transform the test scores so that
they have an approximately Normal distribution with
a mean of 80 points and a standard deviation of 10
points. To do this, she will use a formula in the form:
new score = a + b (old score)
Find the values of a and b that the teacher should
use to transform the distribution of test scores.
(d) Before the test, the teacher gave a review assignment for homework. The maximum score on
the assignment was 10 points. The distribution
of scores on this assignment had a mean of 9.2
points and a standard deviation of 2.1 points.
Would it be appropriate to use a Normal distribution to calculate the proportion of students
who scored below 7 points on this assignment?
Explain.
After you finish, you can view two example solutions on the book’s
Web site (www.whfreeman.com/tps5e). Determine whether you
think each solution is “complete,” “substantial,” “developing,” or
“minimal.” If the solution is not complete, what improvements
would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and
note what, if anything, you would do differently to improve your
own score.
Chapter Review
Section 2.1: Describing Location in a Distribution
In this section, you learned two different ways to describe
the location of individuals in a distribution, percentiles
and standardized scores (z-scores). Percentiles describe
the location of an individual by measuring what percent
of the observations in the distribution have a value less
than the individual’s value. A cumulative relative frequency graph is a handy tool for identifying percentiles
in a distribution. You can use it to estimate the percentile
for a particular value of a variable or estimate the value of
the variable at a particular percentile.
Standardized scores (z-scores) describe the location of
an individual in a distribution by measuring how many
standard deviations the individual is above or below the
mean. To find the standardized score for a particular ob-
Starnes-Yates5e_c02_082-139hr2.indd 134
servation, transform the value by subtracting the mean
and dividing the difference by the standard deviation. Besides describing the location of an individual in a distribution, you can also use z-scores to compare observations
from different distributions—standardizing the values
puts them on a standard scale.
You also learned to describe the effects on the shape,
center, and spread of a distribution when transforming
data from one scale to another. Adding a positive constant
to (or subtracting it from) each value in a data set changes
the measures of location but not the shape or spread of
the distribution. Multiplying or dividing each value in a
data set by a positive constant changes the measures of
location and measures of spread but not the shape of the
distribution.
11/13/13 1:33 PM
Section 2.2: Density Curves and Normal Distributions
In this section, you learned how density curves are used to
model distributions of data. An area under a density curve
gives the proportion of observations that fall in a specified
interval of values. The total area under a density curve is
1, or 100%.
The most commonly used density curve is called a
Normal curve. The Normal curve is symmetric, singlepeaked, and bell-shaped with mean m and standard deviation s. For any distribution of data that is approximately
Normal in shape, about 68% of the observations will be
within 1 standard deviation of the mean, about 95% of
the observations will be within 2 standard deviations of
the mean, and about 99.7% of the observations will be
within 3 standard deviations of the mean. Conveniently,
this relationship is called the 68–95–99.7 rule.
When observations do not fall exactly 1, 2, or 3 standard
deviations from the mean, you learned how to use Table A
(or technology) to identify the proportion of values in any
specified interval under a Normal curve. You also learned
how to use Table A (or technology) to determine the value of an individual that falls at a specified percentile in
a Normal distribution. On the AP® exam, it is extremely
important that you clearly communicate your methods
when answering questions that involve the Normal distribution. You must specify the shape (Normal), center (m),
and spread (s) of the distribution; identify the region under
the Normal curve that you are working with; and correctly
calculate the answer with work shown. Shading a Normal
curve with the mean, standard deviation, and boundaries
clearly identified is a great start.
Finally, you learned how to determine whether a distribution of data is approximately Normal using graphs (dotplots, stemplots, histograms) and the 68–95–99.7 rule. You
also learned that a Normal probability plot is a great way to
determine whether the shape of a distribution is approximately Normal. The more linear the Normal probability
plot, the more Normal the distribution of the data.
What Did You Learn?
Learning Objective
Section
Related Example
on Page(s)
Relevant Chapter
Review Exercise(s)
Find and interpret the percentile of an individual
value within a distribution of data.
2.1
86
R2.1
Estimate percentiles and individual values using a
cumulative relative frequency graph.
2.1
87, 88
R2.2
Find and interpret the standardized score (z-score) of
an individual value within a distribution of data.
2.1
90, 91
R2.1
Describe the effect of adding, subtracting, multiplying by, or dividing by a constant on the shape, center, and spread of a distribution of data.
2.1
93, 94, 95
R2.3
Estimate the relative locations of the median and mean on a density curve.
2.2
Use the 68–95–99.7 rule to estimate areas
(proportions of values) in a Normal distribution.
2.2
Discussion on
106–107
R2.4
111
R2.5
Use Table A or technology to find (i) the proportion of z-values in a specified interval, or (ii) a z-score from a 114, 115
percentile in the standard Normal distribution.
2.2
Discussion on 116
R2.6
Use Table A or technology to find (i) the proportion of
values in a specified interval, or (ii) the value that corresponds to a given percentile in any Normal distribution.
2.2
118, 119, 120
R2.7, R2.8, R2.9
Determine whether a distribution of data is approximately Normal
from graphical and numerical evidence.
122, 123, 124
R2.10, R2.11
2.2
135
Starnes-Yates5e_c02_082-139hr2.indd 135
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CHAPTER 2
M o d e l i n g D i s t r i b u t i o n s o f Data
Chapter 2 Chapter Review Exercises
These exercises are designed to help you review the important ideas and methods of the chapter.
R2.1Is Paul tall? According to the National Center for
Health Statistics, the distribution of heights for
15-year-old males has a mean of 170 centimeters
(cm) and a standard deviation of 7.5 cm. Paul is 15
years old and 179 cm tall.
(a) Find the z-score corresponding to Paul’s height. Explain what this value means.
(b) Paul’s height puts him at the 85th percentile among
15-year-old males. Explain what this means to someone who knows no statistics.
R2.2 Computer use Mrs. Causey asked her students how
much time they had spent using a computer during the
previous week. The following figure shows a cumulative relative frequency graph of her students’ responses.
(a) Suppose we converted each student’s guess from feet
to meters (3.28 ft = 1 m). How would the shape of
the distribution be affected? Find the mean, median,
standard deviation, and IQR for the transformed data.
(b) The actual width of the room was 42.6 feet. Suppose
we calculated the error in each student’s guess as follows: guess − 42.6. Find the mean and standard deviation of the errors. Justify your answers.
R2.4 What the mean means The figure below is a
density curve. Trace the curve onto your paper.
(a) Mark the approximate location of the median. Explain your choice of location.
(b) Mark the approximate location of the mean. Explain
your choice of location.
(a) At what percentile does a student who used her computer for 7 hours last week fall?
(b) Estimate the interquartile range (IQR) from the
graph. Show your work.
Cumulative relative frequency (%)
100
90
80
70
60
50
R2.5 H
​ orse pregnancies Bigger animals tend to carry
their young longer before birth. The length of horse
pregnancies from conception to birth varies according to a roughly Normal distribution with mean 336
days and standard deviation 3 days. Use the 68–95–
99.7 rule to answer the following questions.
40
30
20
10
0
0
3
6
9
12
15
18
21
24
27
30
Hours per week
R2.3 Aussie, Aussie, Aussie A group of Australian students
were asked to estimate the width of their classroom in
feet. Use the dotplot and summary statistics below to
answer the following questions.
(a) Almost all (99.7%) horse pregnancies fall in what interval of lengths?
(b) What percent of horse pregnancies are longer than
339 days? Show your work.
R2.6 S
​ tandard Normal distribution Use Table A (or
technology) to find each of the following for a standard Normal distribution. In each case, sketch a standard Normal curve and shade the area of interest.
(a) The proportion of observations with −2.25 < z < 1.77
Variable
n
Mean
Guess ft 66 43.70
Stdev Minimum
12.50
Starnes-Yates5e_c02_082-139hr2.indd 136
24.00
Q1
Median
Q3
Maximum
35.50
42.00
48.00
94.00
(b) The number z such that 35% of all observations are
greater than z
11/13/13 1:33 PM
AP® Statistics Practice Test
R2.7 L
​ ow-birth-weight babies Researchers in Norway analyzed data on the birth weights of 400,000 newborns
over a 6-year period. The distribution of birth weights
is Normal with a mean of 3668 grams and a standard
deviation of 511 grams.16 Babies that weigh less than
2500 grams at birth are classified as “low birth weight.”
(a) What percent of babies will be identified as low birth
weight? Show your work.
(b) Find the quartiles of the birth weight distribution.
Show your work.
R2.8 K
​ etchup A fast-food restaurant has just installed a
new automatic ketchup dispenser for use in preparing its burgers. The amount of ketchup dispensed by
the machine follows a Normal distribution with mean
1.05 ounces and standard deviation 0.08 ounce.
(a) If the restaurant’s goal is to put between 1 and 1.2
ounces of ketchup on each burger, what percent of
the time will this happen? Show your work.
137
received A’s. What are the mean and standard deviation of the scores? Show your work.
R2.10 F
​ ruit fly thorax lengths Here are the lengths in
millimeters of the thorax for 49 male fruit flies:17
0.64 0.64 0.64 0.68 0.68 0.68 0.72 0.72 0.72 0.72 0.74 0.76 0.76
0.76 0.76 0.76 0.76 0.76 0.76 0.78 0.80 0.80 0.80 0.80 0.80 0.82
0.82 0.84 0.84 0.84 0.84 0.84 0.84 0.84 0.84 0.84 0.84 0.88 0.88
0.88 0.88 0.88 0.88 0.88 0.88 0.92 0.92 0.92 0.94
Are these data approximately Normally distributed?
Give appropriate graphical and numerical evidence
to support your answer.
R2.11 Assessing Normality A Normal probability plot of a
set of data is shown here. Would you say that these
measurements are approximately Normally distributed? Why or why not?
3
(b) Suppose that the manager adjusts the machine’s settings so that the mean amount of ketchup dispensed
is 1.1 ounces. How much does the machine’s standard deviation have to be reduced to ensure that at
least 99% of the restaurant’s burgers have between 1
and 1.2 ounces of ketchup on them?
2
Expected z-score
1
R2.9 G
​ rading managers Many companies “grade on a
bell curve” to compare the performance of their managers and professional workers. This forces the use of
some low performance ratings, so that not all workers
are listed as “above average.” Ford Motor Company’s “performance management process” for a time
assigned 10% A grades, 80% B grades, and 10% C
grades to the company’s 18,000 managers. Suppose
that Ford’s performance scores really are Normally
distributed. This year, managers with scores less than
25 received C’s, and those with scores above 475
0
–1
–2
–3
Data values
Chapter 2 AP® Statistics Practice Test
Section I: Multiple Choice ​Select the best answer for each question.
T2.1 M
​ any professional schools require applicants to take
a standardized test. Suppose that 1000 students take
such a test. Several weeks after the test, Pete receives
his score report: he got a 63, which placed him at the
73rd percentile. This means that
(a)​Pete’s score was below the median.
Starnes-Yates5e_c02_082-139hr2.indd 137
(b)​Pete did worse than about 63% of the test takers.
(c) Pete did worse than about 73% of the test takers.
(d)​Pete did better than about 63% of the test takers.
(e)​Pete did better than about 73% of the test takers.
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M o d e l i n g D i s t r i b u t i o n s o f Data
T2.2 F
​ or the Normal distribution shown, the standard deviation is closest to
(a) 4.83 inches
(d) ​8.93 inches
(b)​5.18 inches
(e) The standard deviation
cannot be computed from
the given information.
(c)​6.04 inches
T2.6 T
​ he figure shown is the density curve of a distribution.
Seven values are marked on the density curve. Which
of the following statements is true?
–8
(a)​0 –6
–4
–2
0
2
4
6
(b) ​1 (c) ​2 (d) ​3 8
10
12
(e) ​5
T2.3 R
​ ainwater was collected in water collectors at 30 different sites near an industrial complex, and the amount
of acidity (pH level) was measured. The mean and
standard deviation of the values are 4.60 and 1.10,
respectively. When the pH meter was recalibrated
back at the laboratory, it was found to be in error. The
error can be corrected by adding 0.1 pH units to all
of the values and then multiplying the result by 1.2.
The mean and standard deviation of the corrected pH
measurements are
(a)​
5.64, 1.44 (c) ​
5.40, 1.44 (e) ​5.64, 1.20
(b)​
5.64, 1.32 (d) ​5.40, 1.32
100
C
D
E F G
(a)​The mean of the distribution is E.
(b)​The area between B and F is 0.50.
(c)​The median of the distribution is C.
(d)​The 3rd quartile of the distribution is D.
T2.7 I​ f the heights of a population of men follow a Normal
distribution, and 99.7% have heights between 5′0″
and 7′0″, what is your estimate of the standard deviation of the heights in this population?
(a) 1″ (b) ​3″ (c) ​4″ (d) ​6″ (e) ​12″
T2.8 W
​ hich of the following is not correct about a standard
Normal distribution?
80
Percent
B
(e)​The area between A and G is 1.
T2.4 T
​ he figure shows a cumulative relative frequency
graph of the number of ounces of alcohol consumed
per week in a sample of 150 adults who report drinking alcohol occasionally. About what percent of these
adults consume between 4 and 8 ounces per week?
40
(a)​The proportion of scores that satisfy 0 < z < 1.5 is
0.4332.
20
(b)​The proportion of scores that satisfy z < −1.0 is 0.1587.
60
(c)​The proportion of scores that satisfy z > 2.0 is 0.0228.
0
0
2
4
6
8
10
12
14
16
18
Consumption (oz)
(a)​20%
A
(b) ​40% (c) ​50% (d) ​60%
(e) ​80%
T2.5​
The average yearly snowfall in Chillyville is Normally
distributed with a mean of 55 inches. If the snowfall
in Chillyville exceeds 60 inches in 15% of the years,
what is the standard deviation?
Starnes-Yates5e_c02_082-139hr2.indd 138
(d)​The proportion of scores that satisfy z < 1.5 is 0.9332.
(e)​The proportion of scores that satisfy z > −3.0 is 0.9938.
Questions T2.9 and T2.10 refer to the following setting. Until the
scale was changed in 1995, SAT scores were based on a scale
set many years ago. For Math scores, the mean under the old
scale in the 1990s was 470 and the standard deviation was 110.
In 2009, the mean was 515 and the standard deviation was 116.
11/13/13 1:33 PM
AP® Statistics Practice Test
T2.9
​ hat is the standardized score (z-score) for a student
W
who scored 500 on the old SAT scale?
(a)​−30 (b) −0.27 (c) −0.13 (d) ​0.13 (e) ​0.27
T2.10 G
​ ina took the SAT in 1994 and scored 500. Her
cousin Colleen took the SAT in 2013 and scored
530. Who did better on the exam, and how can you
tell?
139
(a)​Colleen—she scored 30 points higher than Gina.
(b)​Colleen—her standardized score is higher than Gina’s.
(c)​Gina—her standardized score is higher than Colleen’s.
(d)​Gina—the standard deviation was bigger in 2013.
(e) The two cousins did equally well—their z-scores are
the same.
Section II: Free Response ​Show all your work. Indicate clearly the methods you use, because you will be graded on
the correctness of your methods as well as on the accuracy and completeness of your results and explanations.
T2.11 A
​ s part of the President’s Challenge, students can
attempt to earn the Presidential Physical Fitness
Award or the National Physical Fitness Award by
meeting qualifying standards in five events: curl-ups,
shuttle run, sit and reach, one-mile run, and pullups. The qualifying standards are based on the 1985
School Population Fitness Survey. For the Presidential Award, the standard for each event is the 85th
percentile of the results for a specific age group and
gender among students who participated in the 1985
survey. For the National Award, the standard is the
50th percentile. To win either award, a student must
meet the qualifying standard for all five events.
Jane, who is 9 years old, did 40 curl-ups in one
minute. Matt, who is 12 years old, also did 40 curlups in one minute. The qualifying standard for the
Presidential Award is 39 curl-ups for Jane and 50
curl-ups for Matt. For the National Award, the standards are 30 and 40, respectively.
(a)​
Compare Jane’s and Matt’s performances using
percentiles. Explain in language simple enough for
someone who knows little statistics to understand.
(b) Who has the higher standardized score (z-score),
Jane or Matt? Justify your answer.
T2.12 T
​ he army reports that the distribution of head circumference among male soldiers is approximately
Starnes-Yates5e_c02_082-139hr2.indd 139
Normal with mean 22.8 inches and standard deviation 1.1 inches.
(a) A male soldier whose head circumference is 23.9
inches would be at what percentile? Show your
method clearly.
(b) The army’s helmet supplier regularly stocks helmets that fit male soldiers with head circumferences between 20 and 26 inches. Anyone with a
head circumference outside that interval requires
a customized helmet order. What percent of male
soldiers require custom helmets?
(c) Find the interquartile range for the distribution of
head circumference among male soldiers.
T2.13 A
​ study recorded the amount of oil recovered from
the 64 wells in an oil field. Here are descriptive statistics for that set of data from Minitab.
Descriptive Statistics: Oilprod
Variable
n
Oilprod
64 48.25 37.80 40.24 2.00 204.90 21.40 60.75
Mean Median StDev
Min
Max
Q1
Q3
Does the amount of oil recovered from all wells in this
field seem to follow a Normal distribution? Give appropriate statistical evidence to support your answer.
11/13/13 1:33 PM
Chapter
3
Introduction
142
Section 3.1
Scatterplots and Correlation
143
Section 3.2
Least-Squares Regression
164
Free Response
AP® Problem, Yay!
199
Chapter 3 Review
200
Chapter 3 Review Exercises
202
Chapter 3 AP® Statistics
Practice Test
203
Starnes-Yates5e_c03_140-205hr3.indd 140
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Describing
Relationships
case study
How Faithful Is Old Faithful?
Frequency
The Starnes family visited Yellowstone National Park in hopes of seeing the Old Faithful geyser erupt.
They had only about four hours to spend in the park. When they pulled into the parking lot near Old
Faithful, a large crowd of people was headed back to their cars from the geyser. Old Faithful had just
­finished erupting. How long would the Starnes family have to wait until the next eruption?
Let’s look at some data. Figure 3.1 shows a histogram of times (in minutes) between c­ onsecutive
50
­eruptions of Old Faithful in the month before the
40
Starnes family’s visit. The shortest interval was
47 ­
minutes, and the longest was 113 minutes.
30
That’s a lot of variability! The distribution has two
clear peaks—one at about 60 minutes and the o­ ther
20
at about 90 minutes.
10
If the Starnes family hopes for a 60-minute
gap between eruptions, but the actual interval is
0
closer to 90 minutes, the kids will get impatient.
40
50
60
70
80
90
100
110
120
Interval (minutes)
If they plan for a 90-minute interval and go somewhere else in the park, they won’t get back in time FIGURE 3.1 ​Histogram of the interval (in minutes) between
to see the next eruption if the gap is only about eruptions of the Old Faithful geyser in the month prior to the
Starnes family’s visit.
60 ­minutes. What should the Starnes family do?
Later in the chapter, you’ll answer this ­question. For now,
keep this in mind: to understand one variable (like eruption
interval), you often have to look at how it is related to other
variables.
141
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CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
Introduction
Investigating relationships between variables is central to what we do in statistics.
When we understand the relationship between two variables, we can use the value
of one variable to help us make predictions about the other variable. In Section 1.1,
we explored relationships between categorical variables, such as the gender of a
young person and his or her opinion about future income. The association between
these two variables suggests that males are generally more optimistic about their
future income than females.
In this chapter, we investigate relationships between two quantitative variables.
Does knowing the number of points a football team scores per game tell us anything about how many wins it will have? What can we learn about the price of a
used car from the number of miles it has been driven? Are there any variables that
might help the Starnes family predict how long it will be until the next eruption
of Old Faithful?
Activity
Materials:
Meterstick, handprint, and
math department roster
(from Teacher’s Resource
Materials) for each group of
three to four students; one
sheet of graph paper per
student
​CSI Stats:
The case of the missing cookies
Mrs. Hagen keeps a large jar full of cookies on her desk for her students. Over
the past few days, a few cookies have disappeared. The only people with access
to Mrs. Hagen’s desk are the other math teachers at her school. She asks her colleagues whether they have been making withdrawals from the cookie jar. No one
confesses to the crime.
But the next day, Mrs. Hagen catches a break—she finds a clear handprint on
the cookie jar. The careless culprit has left behind crucial evidence! At this point,
Mrs. Hagen calls in the CSI Stats team (your class) to help her identify the prime
suspect in “The Case of the Missing Cookies.”
1. Measure the height and hand span of each member of your group to the
nearest centimeter (cm). (Hand span is the maximum distance from the tip of
the thumb to the tip of the pinkie finger on a person’s fully stretched-out hand.)
2. ​Your teacher will make a data table on the board with two columns, labeled
as follows:
Hand span (cm) Height (cm)
Send a representative to record the data for each member of your group in the table.
3. ​Copy the data table onto your graph paper very near the left margin of the
page. Next, you will make a graph of these data. Begin by constructing a set of
coordinate axes. Allow plenty of space on the page for your graph. Label the
horizontal axis “Hand span (cm)” and the vertical axis “Height (cm).”
4. ​Since neither hand span nor height can be close to 0 cm, we want to start
our horizontal and vertical scales at larger numbers. Scale the horizontal
axis in 0.5-cm increments starting with 15 cm. Scale the vertical axis in 5-cm
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Section 3.1 Scatterplots and Correlation
i­ ncrements starting with 135 cm. Refer to the sketch in the margin
for comparison.
Height (cm)
160
5. ​Plot each point from your class data table as accurately as you can
on the graph. Compare your graph with those of your group members.
155
150
6. ​As a group, discuss what the graph tells you about the relationship
between hand span and height. Summarize your observations in a
­sentence or two.
145
140
135
15
15.5
16
16.5
17
Hand span (cm)
3.1
What You Will Learn
•
•
•
143
17.5
7. ​Ask your teacher for a copy of the handprint found at the scene
and the math department roster. Which math teacher does your group
believe is the “prime suspect”? Justify your answer with ­appropriate
statistical evidence.
Scatterplots and Correlation
By the end of the section, you should be able to:
Identify explanatory and response variables in situations
where one variable helps to explain or influences the other.
Make a scatterplot to display the relationship between
two quantitative variables.
Describe the direction, form, and strength of a
­relationship displayed in a scatterplot and identify outliers in a scatterplot.
•
•
•
•
Interpret the correlation.
Understand the basic properties of correlation,
­including how the correlation is influenced by outliers.
Use technology to calculate ­correlation.
Explain why association does not imply causation.
Most statistical studies examine data on more than one variable. Fortunately,
analysis of several-variable data builds on the tools we used to examine individual
variables. The principles that guide our work also remain the same:
• Plot the data, then add numerical summaries.
• Look for overall patterns and departures from those patterns.
• When there’s a regular overall pattern, use a simplified model to describe it.
Explanatory and Response Variables
We think that car weight helps explain accident deaths and that smoking influences life expectancy. In these relationships, the two variables play different roles.
Accident death rate and life expectancy are the response variables of interest. Car
weight and number of cigarettes smoked are the explanatory variables.
Definition: Response variable, explanatory variable
A response variable measures an outcome of a study. An explanatory variable
may help explain or predict changes in a response variable.
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144
CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
You will often see explanatory variables
called independent variables and
response variables called dependent
variables. Because the words
“independent” and “dependent” have
other meanings in statistics, we won’t
use them here.
EXAMPLE
It is easiest to identify explanatory and response variables when we actually
specify values of one variable to see how it affects another variable. For instance,
to study the effect of alcohol on body temperature, researchers gave several different amounts of alcohol to mice. Then they measured the change in each
mouse’s body temperature 15 minutes later. In this
case, amount of alcohol is the explanatory variable,
and change in body temperature is the response
variable. When we don’t specify the values of either
variable but just observe both variables, there may
or may not be explanatory and response variables. Whether there are depends on how
you plan to use the data.
Linking SAT Math and Critical
Reading Scores
Explanatory or response?
Julie asks, “Can I predict a state’s mean SAT Math score if I know its mean SAT
Critical Reading score?” Jim wants to know how the mean SAT Math and Critical
Reading scores this year in the 50 states are related to each other.
Problem: For each student, identify the explanatory variable and the response variable if p­ ossible.
solution: Julie is treating the mean SAT Critical Reading score as the explanatory variable and
the mean SAT Math score as the response variable. Jim is simply interested in exploring the relationship between the two variables. For him, there is no clear explanatory or response variable.
For Practice Try Exercise
1
In many studies, the goal is to show that changes in one or more explanatory
variables actually cause changes in a response variable. However, other explanatory-­
response relationships don’t involve direct causation. In the alcohol and mice study,
alcohol actually causes a change in body temperature. But there is no cause-and-effect
relationship between SAT Math and Critical Reading scores. Because the scores are
closely related, we can still use a state’s mean SAT Critical Reading score to predict its
mean Math score. We will learn how to make such predictions in Section 3.2.
Check Your Understanding
Identify the explanatory and response variables in each setting.
1.​How does drinking beer affect the level of alcohol in people’s blood? The legal limit
for driving in all states is 0.08%. In a study, adult volunteers drank different numbers of
cans of beer. Thirty minutes later, a police officer measured their blood alcohol levels.
2. The National Student Loan Survey provides data on the amount of debt for recent
college graduates, their current income, and how stressed they feel about college debt. A
sociologist looks at the data with the goal of using amount of debt and income to explain
the stress caused by college debt.
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Section 3.1 Scatterplots and Correlation
145
Displaying Relationships: Scatterplots
625
Mean Math score
600
575
550
525
500
475
450
0
10
20
30
40
50
60
Percent taking SAT
FIGURE 3.2 ​Scatterplot of the
mean SAT Math score in each state
against the percent of that state’s
high school graduates who took the
SAT. The dotted lines intersect at
the point (21, 570), the values for
Colorado.
Here’s a helpful way to remember: the
eXplanatory variable goes on the x
axis.
70
80
90
The most useful graph for displaying the relationship between
two quantitative variables is a scatterplot. Figure 3.2 shows a
scatterplot of the percent of high school graduates in each state
who took the SAT and the state’s mean SAT Math score in a
recent year. We think that “percent taking” will help explain
“mean score.” So “percent taking” is the explanatory variable
and “mean score” is the response variable. We want to see
how mean score changes when percent taking changes, so
we put percent taking (the explanatory variable) on the horizontal axis. Each point represents a single state. In Colorado,
for example, 21% took the SAT, and their mean SAT Math
score was 570. Find 21 on the x (horizontal) axis and 570 on
the y (vertical) axis. Colorado appears as the point (21, 570).
Definition: Scatterplot
A scatterplot shows the relationship between two quantitative variables measured
on the same individuals. The values of one variable appear on the horizontal axis, and
the values of the other variable appear on the vertical axis. Each individual in the data
appears as a point in the graph.
Always plot the explanatory variable, if there is one, on the horizontal axis (the
x axis) of a scatterplot. As a reminder, we usually call the explanatory variable x
and the response variable y. If there is no explanatory-response distinction, either
variable can go on the horizontal axis.
We used computer software to produce Figure 3.2. For some problems, you’ll
be expected to make scatterplots by hand. Here’s how to do it.
How to Make a Scatterplot
1. ​Decide which variable should go on each axis.
2. ​Label and scale your axes.
3. ​Plot individual data values.
The following example illustrates the process of constructing a scatterplot.
Example
SEC Football
Making a scatterplot
At the end of the 2011 college football season, the University of ­Alabama
defeated Louisiana State University for the national championship. Interestingly, both of these teams were from the Southeastern Conference
(SEC). Here are the average number of points scored per game and number of wins for each of the twelve teams in the SEC that season.1
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146
CHAPTER 3
Team
D e s c r i b i n g R e l at i o n s h i p s
Alabama
Arkansas
Auburn
Florida
Georgia
Kentucky
Points per game
34.8
36.8
25.7
25.5
32.0
15.8
Wins
12
11
Team
8
7
10
5
Louisiana State
Mississippi
Mississippi State
South Carolina
Tennessee
Vanderbilt
Points per game
35.7
16.1
25.3
30.1
20.3
26.7
Wins
13
2
7
11
5
6
PROBLEM: Make a scatterplot of the relationship between points per game and wins.
SOLUTION: We follow the steps described earlier to make the scatterplot.
14
12
Wins
10
8
6
4
2
15
20
25
30
35
40
Points per game
FIGURE 3.3 ​Completed scatterplot of points per game
and wins for the teams in the SEC. The dotted lines
intersect at the point (34.8, 12), the values for Alabama.
1. Decide which variable should go on which axis. The number of wins a football
team has depends on the number of points they score. So we’ll use points
per game as the explanatory variable (x axis) and wins as the response
­variable (y axis).
2. Label and scale your axes. We labeled the x axis “Points per game” and the
y axis “Wins.” Because the teams’ points per game vary from 15.8 to 36.8,
we chose a horizontal scale starting at 15 points, with tick marks every
5 points. The teams’ wins vary from 2 to 13, so we chose a vertical scale
starting at 0 with tick marks every 2 wins.
3. Plot individual data values. The first team in the table, Alabama, scored
34.8 points per game and had 12 wins. We plot this point directly above
34.8 on the horizontal axis and to the right of 12 on the vertical axis, as
shown in Figure 3.3. For the second team in the list, Arkansas, we add the
point (36.8, 11) to the graph. By adding the points for the remaining ten
teams, we get the completed scatterplot in Figure 3.3.
For Practice Try Exercise
5
Describing Scatterplots
625
Mean Math score
600
575
550
525
500
475
450
0
10
20
Starnes-Yates5e_c03_140-205hr3.indd 146
To describe a scatterplot, follow the basic strategy of data analysis from Chapters
1 and 2: look for patterns and important departures from those patterns. Let’s
take a closer look at the scatterplot from Figure 3.2. What do we see?
• The graph shows a clear direction: the overall pattern moves from upper left
to lower right. That is, states in which higher percents of high school graduates take the SAT tend to have lower mean SAT Math
scores. We call this a negative association between the two
­variables.
•The form of the relationship is slightly curved. More important, most states fall into one of two distinct clusters.
In about half of the states, 25% or fewer graduates took
the SAT. In the other half, more than 40% took the SAT.
•The strength of a relationship in a scatterplot is determined by how closely the points follow a clear form.
The overall relationship in Figure 3.2 is moderately
30
40
50
60
70
80
90
strong: states with similar percents taking the SAT tend
Percent taking SAT
to have roughly similar mean SAT Math scores.
11/13/13 1:20 PM
Section 3.1 Scatterplots and Correlation
•
THINK
ABOUT IT
147
Two states stand out in the scatterplot: West Virginia at (19, 501) and Maine at
(87, 466). These points can be ­described as outliers because they fall outside
the overall ­pattern.
What explains the clusters? ​There are two widely used college entrance
exams, the SAT and the American College Testing (ACT) exam. Each state usually
favors one or the other. The ACT states cluster at the left of Figure 3.2 and the SAT
states at the right. In ACT states, most students who take the SAT are applying to a
selective college that prefers SAT scores. This select group of students has a higher
mean score than the much larger group of students who take the SAT in SAT states.
How to Examine a Scatterplot
As in any graph of data, look for the overall pattern and for striking departures from that pattern.
•
•
You can describe the overall pattern of a scatterplot by the direction,
form, and strength of the relationship.
An important kind of departure is an outlier, an individual value that
falls outside the overall pattern of the relationship.
Let’s practice examining scatterplots using the SEC football data from the
­previous example.
Example
SEC Football
Describing a scatterplot
In the last example, we constructed the scatterplot shown below that displays the
average number of points scored per game and the number of wins for college
football teams in the Southeastern Conference.
14
12
Wins
10
8
6
4
2
15
20
25
30
Points per game
35
40
PROBLEM: Describe what the scatterplot reveals about the relationship between points per game and wins.
SOLUTION: Direction: In general, it appears that teams that score
more points per game have more wins and teams that score fewer points per
game have fewer wins. We say that there is a positive association between
points per game and wins.
Form: There seems to be a linear pattern in the graph (that is, the overall
pattern follows a straight line).
Strength: Because the points do not vary much from the linear pattern,
the relationship is fairly strong. There do not appear to be any values that
depart from the linear pattern, so there are no outliers.
For Practice Try Exercise
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7
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148
CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
Even when there is a clear association
between two variables in a scatterplot,
the direction of the relationship only
describes the overall trend—not the
relationship for each pair of points. For
example, even though teams that score
more points per game generally have
more wins, Georgia and South Carolina
are exceptions to the overall pattern.
Georgia scored more points per game
than South Carolina (32 versus 30.1)
but had fewer wins (10 versus 11).
So far, we’ve seen relationships with two different directions. The number of
wins generally increases as the points scored per game increases (positive association). The mean SAT score generally goes down as the percent of graduates
taking the test increases (negative association). Let’s give a careful definition for
these terms.
Definition: Positive association, negative association
Two variables have a positive association when above-average values of one tend
to accompany above-average values of the other and when below-average values
also tend to occur together.
Two variables have a negative association when above-average values of one tend
to accompany below-average values of the other.
Of course, not all relationships have a clear direction that we can describe
as a positive association or a negative association. Exercise 9 involves a relationship that doesn’t have a single direction. This next example, however, illustrates a
strong positive association with a simple and important form.
EXAMPLE
The Endangered Manatee
Pulling it all together
Manatees are large, gentle, slow-moving creatures found along the coast of ­Florida.
Many manatees are injured or killed by boats. The table below contains data on
the number of boats registered in Florida (in thousands) and the number of manatees killed by boats for the years 1977 to 2010.2
Florida boat registrations (thousands) and manatees killed by boats
Year
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Boats Manatees
Year
Boats
Manatees
Year
Boats
Manatees
1977
447
13
1989
711
50
2001
944
81
1978
460
21
1990
719
47
2002
962
95
1979
481
24
1991
681
53
2003
978
73
1980
498
16
1992
679
38
2004
983
69
1981
513
24
1993
678
35
2005
1010
79
1982
512
20
1994
696
49
2006
1024
92
1983
526
15
1995
713
42
2007
1027
73
1984
559
34
1996
732
60
2008
1010
90
1985
585
33
1997
755
54
2009
982
97
1986
614
33
1998
809
66
2010
942
83
1987
645
39
1999
830
82
1988
675
43
2000
880
78
11/13/13 1:20 PM
Section 3.1 Scatterplots and Correlation
Problem: Make a scatterplot to show the ­relationship between
100
the number of manatees killed and the number of registered boats.
Describe what you see.
solution: For the scatterplot, we’ll use “boats ­registered”
as the explanatory variable and “manatees killed” as the response
variable. Figure 3.4 is our c­ ompleted scatterplot. There is a positive
­association—more boats registered goes with more manatees
killed. The form of the relationship is linear. That is, the overall
pattern follows a straight line from lower left to upper right. The
relationship is strong because the points don’t deviate greatly from
a line, except for the 4 years that have a high number of boats registered, but fewer deaths than expected based on the linear pattern.
Manatees killed
90
80
70
60
50
40
30
20
10
149
400
500
600
700
800
1000
900
1100
Boats registered in Florida (1000s)
FIGURE 3.4 ​Scatterplot of the number of Florida manatees
killed by boats from 1977 to 2010 against the number of
boats registered in Florida that year.
For Practice Try Exercise
13
Check Your Understanding
In the chapter-opening Case Study (page 141), the Starnes family arrived at Old Faithful
after it had erupted. They wondered how long it would be until the next eruption. Here is
a scatterplot that plots the interval between consecutive eruptions
of Old Faithful against the duration of the previous eruption, for
the month prior to their visit.
120
115
Interval (minutes)
105
1.​Describe the direction of the relationship. Explain why this
makes sense.
95
85
2. What form does the relationship take? Why are there two
clusters of points?
75
65
55
45
1
2
3
Duration (minutes)
Starnes-Yates5e_c03_140-205hr3.indd 149
4
5
3.
How strong is the relationship? Justify your answer.
4.
Are there any outliers?
5. What information does the Starnes family need to predict
when the next eruption will occur?
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CHAPTER 3
7. T ECHNOLOGY
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D e s c r i b i n g R e l at i o n s h i p s
Scatterplots on the calculator
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
Making scatterplots with technology is much easier than constructing them by hand. We’ll use the SEC football data
from page 146 to show how to construct a scatterplot on a TI-83/84 or TI-89.
•
•
Enter the data values into your lists. Put the points per game in L1/list1 and the number of wins in L2/list2.
Define a scatterplot in the statistics plot menu (press F2 on the TI-89). Specify the settings shown below.
•
Use ZoomStat (ZoomData on the TI-89) to obtain a graph. The calculator will set the window dimensions automatically
by looking at the values in L1/list1 and L2/list2.
Notice that there are no scales on the axes and that the axes are not labeled. If you copy a scatterplot from your calculator onto
your paper, make sure that you scale and label the axes.
AP® EXAM TIP ​If you are asked to make a scatterplot on a free-response question, be sure to
label and scale both axes. Don’t just copy an unlabeled calculator graph directly onto your paper.
Measuring Linear Association: Correlation
c
Some people refer to r as the
“correlation coefficient.”
Starnes-Yates5e_c03_140-205hr4.indd 150
!
n
A scatterplot displays the direction, form, and strength of the relationship between
two quantitative variables. Linear relationships are particularly important because a
straight line is a simple pattern that is quite common. A linear relationship is strong
if the points lie close to a straight line and weak if they are widely scattered about a
line. Unfortunately, our eyes are not good judges of how strong a linear ­relationship is.
The two scatterplots in Figure 3.5 (on the facing page) show the same data, autio
but the graph on the right is drawn smaller in a large field. The right-hand
graph seems to show a stronger linear relationship.
Because it’s easy to be fooled by different scales or by the amount of space
around the cloud of points in a scatterplot, we need to use a numerical measure
to supplement the graph. Correlation is the measure we use.
Definition: Correlation r
The correlation r measures the direction and strength of the linear relationship
between two quantitative variables.
11/20/13 6:24 PM
151
Section 3.1 Scatterplots and Correlation
FIGURE 3.5 ​Two Minitab scatterplots of the same data. The straight-line pattern in the graph on
the right appears stronger because of the surrounding space.
How good are you at estimating the
correlation by eye from a scatterplot?
To find out, try an online applet. Just
search for “guess the correlation
applets.”
The correlation r is always a number between −1 and 1. Correlation indicates the
direction of a linear relationship by its sign: r > 0 for a positive association and r < 0
for a negative association. Values of r near 0 indicate a very weak linear relationship.
The strength of the linear relationship increases as r moves away from 0 toward either
−1 or 1. The extreme values r = −1 and r = 1 occur only in the case of a perfect
linear relationship, when the points lie exactly along a straight line.
Figure 3.6 shows scatterplots that correspond to various values of r. To make
the meaning of r clearer, the standard deviations of both variables in these plots
are equal, and the horizontal and vertical scales are the same. The correlation
describes the direction and strength of the linear relationship in each graph.
Correlation r
FIGURE 3.6 ​How correlation measures the strength
of a linear relationship.
Patterns closer to a
straight line have correlations closer to 1 or −1.
Correlation r
0
0.7
Correlation r
Correlation r
0.3
0.9
Correlation r
Correlation r
0.5
0.99
The following Activity lets you explore some important properties of the correlation.
Starnes/Yates/Moore: The Practice of Statistics, 4E
New Fig.: 3.6
Perm. Fig.: 322
First Pass: 2010-03-03
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CHAPTER 3
ACTIVITY
MATERIALS:
Computer with Internet
­connection
PLET
AP
D e s c r i b i n g R e l at i o n s h i p s
​Correlation and Regression applet
Go to the book’s Web site, www.whfreeman.com/tps5e, and launch the Correlation and Regression applet.
1. ​You are going to use the Correlation and Regression applet to make several
scatterplots with 10 points that have correlation close to 0.7.
(a) Start by putting two points on the graph. What’s the value of the correlation?
Why does this make sense?
(b) ​Make a lower-left to upper-right pattern of 10 points with correlation about r = 0.7. (You can drag points up or down to adjust r after
you have 10 points.)
(c) ​Make another scatterplot: this one should have 9 points in a vertical stack at the left of the plot. Add 1 point far to the right and move it
until the correlation is close to 0.7.
(d) ​Make a third scatterplot: make this one with 10 points in a curved
pattern that starts at the lower left, rises to the right, then falls again at
the far right. Adjust the points up or down until you have a very smooth
curve with correlation close to 0.7.
Summarize: If you know that the correlation between two variables is r = 0.7,
what can you say about the form of the relationship?
2. ​Click on the scatterplot to create a group of 10 points in the lower-left corner
of the scatterplot with a strong straight-line pattern (correlation about 0.9).
(a) ​Add 1 point at the upper right that is in line with the first 10. How does the
correlation change?
(b) ​Drag this last point straight down. How small can you make the correlation?
Can you make the correlation negative?
Summarize: What did you learn from Step 2 about the effect of a single point on
the correlation?
Now that you know what information the correlation provides—and doesn’t
provide—let’s look at an example that shows how to interpret it.
Example
SEC Football
Interpreting correlation
PROBLEM: Our earlier scatterplot of the average points per game and number of wins for college
football teams in the SEC is repeated at top right. For these data, r = 0.936.
(a) Interpret the value of r in context.
(b) The point highlighted in red on the scatterplot is Mississippi. What effect does Mississippi have
on the correlation? Justify your answer.
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153
Section 3.1 Scatterplots and Correlation
SOLUTION:
14
(a) The correlation of 0.936 confirms what we see in the scatterplot: there
is a strong, positive linear relationship between points per game and wins in
the SEC.
(b) Mississippi makes the correlation closer to 1 (stronger). If ­Mississippi
were not included, the remaining points wouldn’t be as tightly clustered in a
linear pattern.
12
Wins
10
8
6
4
2
15
25
20
35
30
40
Points per game
For Practice Try Exercise
21
AP® EXAM TIP ​If you’re asked to interpret a correlation, start by looking at a scatterplot of
the data. Then be sure to address direction, form, strength, and outliers (sound familiar?) and
put your answer in context.
Check Your Understanding
The scatterplots below show four sets of real data: (a) repeats the manatee plot in Figure
3.4 (page 149); (b) shows the number of named tropical storms and the number predicted
before the start of hurricane season each year between 1984 and 2007 by William Gray of
Colorado State University; (c) plots the healing rate in micrometers (millionths of a meter) per hour for the two front limbs of several newts in an e­ xperiment; and (d) shows stock
market performance in consecutive years over a ­56-year period. For each graph, estimate
the correlation r. Then interpret the value of r in context.
(a) 100
(b)
30
25
80
Storms observed
Manatees killed
90
70
60
50
40
30
20
15
10
5
20
10
0
400
500
600
700
800
900
1000
1100
0
5
(c)
15
20
25
30
(d)
This year’s percent return
40
Healing rate limb 2
10
Storms predicted
Boats registered in Florida (1000s)
30
20
60
40
20
0
–20
10
10
20
30
Healing rate limb 1
Starnes-Yates5e_c03_140-205hr3.indd 153
40
–20
0
20
40
60
Last year’s percent return
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CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
Calculating Correlation Now that you have some idea of how to interpret
the correlation, let’s look at how it’s calculated.
How to Calculate the Correlation r
Suppose that we have data on variables x and y for n individuals. The values
for the first individual are x1 and y1, the values for the second individual are
x2 and y2, and so on. The means and standard deviations of the two variables
are x– and sx for the x-values, and y– and sy for the y-values. The correlation r
between x and y is
r=
xn − x– yn − y–
x1 − x– y1 − y–
x2 − x– y2 − y–
1
ca
ba
b+a
ba
b + c+ a
ba
bd
sx
sy
sx
sy
sx
sy
n−1
or, more compactly,
r=
xi − x– yi − y–
1
a
ba
b
∙
sx
sy
n−1
The formula for the correlation r is a bit complex. It helps us see what correlation is, but in practice, you should use your calculator or software to find r.
Exercises 19 and 20 ask you to calculate a correlation step-by-step from the definition to solidify its meaning.
The formula for r begins by standardizing the observations. Let’s use the familiar SEC football data to perform the required calculations. The table below shows
the values of points per game x and number of wins y for the SEC college football
teams. For these data, x– = 27.07 and sx = 7.16.
Team
Alabama
Arkansas
Auburn
Florida
Georgia
Kentucky
34.8
36.8
25.7
25.5
32.0
15.8
12
11
Points per game
Wins
Team
8
7
Louisiana State Mississippi Mississippi State South Carolina
Points per game
35.7
Wins
13
16.1
2
25.3
7
30.1
11
10
5
Tennessee
Vanderbilt
20.3
26.7
5
6
The value
xi − x–
sx
in the correlation formula is the standardized points per game (z-score) of the ith
team. For the first team in the table (Alabama), the corresponding z-score is
zx =
34.8 − 27.07
= 1.08
7.16
That is, Alabama’s points per game total (34.8) is a little more than 1
standard deviation above the mean points per game for the SEC teams.
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Section 3.1 Scatterplots and Correlation
Some people like to write the
­correlation formula as
r=
1
∙ zx zy
n−1
to emphasize the product of
standardized scores in the calculation.
THINK
ABOUT IT
155
Standardized values have no units—in this example, they are no longer measured in points.
To standardize the number of wins, we use y– = 8.08 and sy = 3.34. For
12 − 8.08
= 1.17. Alabama’s number of wins (12) is 1.17 stanAlabama, zy =
3.34
dard deviations above the mean number of wins for SEC teams. When we multiply this team’s two z-scores, we get a product of 1.2636. The correlation r is an
“average” of the products of the standardized scores for all the teams. Just
as in the case of the standard deviation sx, the average here divides by one fewer
than the number of individuals. Finishing the calculation reveals that r = 0.936
for the SEC teams.
What does correlation measure? ​The Fathom screen shots below provide more detail. At the left is a scatterplot of the SEC football data with two lines
added—a vertical line at the group’s mean points per game and a horizontal line
at the mean number of wins of the group. Most of the points fall in the upper-right
or lower-left “quadrants” of the graph. That is, teams with above-average points
per game tend to have above-average numbers of wins, and teams with belowaverage points per game tend to have numbers of wins that are below average.
This confirms the positive association between the variables.
Below on the right is a scatterplot of the standardized scores. To get this graph,
we transformed both the x- and the y-values by subtracting their mean and dividing by their standard deviation. As we saw in Chapter 2, standardizing a data set
converts the mean to 0 and the standard deviation to 1. That’s why the vertical and
horizontal lines in the right-hand graph are both at 0.
Notice that all the products of the standardized values will be positive—not
surprising, considering the strong positive association between the variables. What
if there was a negative association between two variables? Most of the points would
be in the upper-left and lower-right “quadrants” and their z-score products would
be negative, resulting in a negative correlation.
Facts about Correlation
How correlation behaves is more important than the details of the formula. Here’s
what you need to know in order to interpret correlation correctly.
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CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
1. ​Correlation makes no distinction between explanatory
and response variables. It makes no difference which variable you call x and which you call y in calculating the correlation. Can you see why from the formula?
xi − x– yi − y–
1
r=
∙ a sx b a sy b
n−1
2. ​Because r uses the standardized values of the observations, r does not change when we change the units of measurement of x, y, or both. Measuring height in centimeters
rather than inches and weight in kilograms rather than
pounds does not change the correlation between height
"He says we've ruined his positive correlation between height and weight." and weight.
3. ​The correlation r itself has no unit of measurement. It is just a number.
c
625
Mean Math score
600
575
550
525
500
475
450
0
10
20
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!
n
Describing the relationship between two variables is more complex autio
than describing the distribution of one variable. Here are some cautions
to keep in mind when you use correlation.
• Correlation does not imply causation. Even when a scatterplot shows a strong
linear relationship between two variables, we can’t conclude that changes in
one variable cause changes in the other. For example, looking at data from
the last 10 years, there is a strong positive relationship between the number of
high school students who own a cell phone and the number of students who
pass the AP® Statistics exam. Does this mean that buying a cell phone will
help you pass the AP® exam? Not likely. Instead, the correlation is positive
because both of these variables are increasing over time.
• Correlation requires that both variables be quantitative, so that it makes sense
to do the arithmetic indicated by the formula for r. We cannot calculate a correlation between the incomes of a group of people and what city they live in
because city is a categorical variable.
• Correlation only measures the strength of a linear relationship between two
variables. Correlation does not describe curved relationships between variables,
no matter how strong the relationship is. A correlation of 0 doesn’t guarantee
that there’s no relationship between two variables, just that there’s no linear
relationship.
• A value of r close to 1 or −1 does not guarantee a linear relationship between
two variables. A scatterplot with a clear curved form can have a correlation
that is close to 1 or −1. For example, the correlation between percent taking the SAT and mean Math score is close
to −1, but the association is clearly curved. Always plot your
data!
• Like the mean and standard deviation, the correlation is
not resistant: r is strongly affected by a few outlying observations. Use r with caution when outliers appear in the
scatterplot.
• Correlation is not a complete summary of two-variable
data, even when the relationship between the variables
30
40
50
60
70
80
90
is linear. You should give the means and standard deviaPercent taking SAT
tions of both x and y along with the correlation.
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157
Section 3.1 Scatterplots and Correlation
Of course, even giving means, standard deviations, and the correlation for “state
SAT Math scores” and “percent taking” will not point out the clusters in Figure 3.2.
Numerical summaries complement plots of data, but they do not replace them.
EXAMPLE
Scoring Figure Skaters
Why correlation doesn’t tell the whole story
Until a scandal at the 2002 Olympics brought change, figure skating was scored
by judges on a scale from 0.0 to 6.0. The scores were often controversial. We have
the scores awarded by two judges, Pierre and Elena, for many skaters. How well do
they agree? We calculate that the correlation between their scores is r = 0.9. But
the mean of Pierre’s scores is 0.8 point lower than Elena’s mean.
These facts don’t contradict each other. They simply give different kinds of information. The mean scores show that Pierre awards lower scores than Elena. But
because Pierre gives every skater a score about 0.8 point lower than Elena does,
the correlation remains high. Adding the same number to all values of either x
or y does not change the correlation. If both judges score the same skaters, the
competition is scored consistently because Pierre and Elena agree on which performances are better than others. The high r shows their agreement. But if Pierre
scores some skaters and Elena others, we should add 0.8 point to Pierre’s scores to
arrive at a fair comparison.
DATA EXPLORATION The SAT essay: Is longer better?
Following the debut of the new SAT Writing test in March 2005, Dr. Les P
­ erelman
from the Massachusetts Institute of Technology stirred controversy by reporting,
“It appeared to me that regardless of what a student wrote, the longer the essay, the
higher the score.” He went on to say, “I have never found a quantifiable predictor
in 25 years of grading that was anywhere as strong as this one. If you just graded
them based on length without ever reading them, you’d be right over 90 percent
of the time.”3 The table below shows the data that Dr. Perelman used to draw his
conclusions.4
Length of essay and score for a sample of SAT essays
Words:
460
422
402
365
357
278
236
201
168
156
133
Score:
6
6
5
5
6
5
4
4
4
3
2
Words:
114
108
100
403
401
388
320
258
236
189
128
3
2
Score:
2
1
1
5
6
6
5
4
4
Words:
67
697
387
355
337
325
272
150
135
Score:
1
6
6
5
5
4
4
2
3
Does this mean that if students write a lot, they are guaranteed high scores?
Carry out your own analysis of the data. How would you respond to each of
Dr. Perelman’s claims?
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CHAPTER 3
Section 3.1
D e s c r i b i n g R e l at i o n s h i p s
Summary
•
•
•
•
•
•
•
•
•
A scatterplot displays the relationship between two quantitative variables
measured on the same individuals. Mark values of one variable on the
horizontal axis (x axis) and values of the other variable on the vertical axis
(y axis). Plot each individual’s data as a point on the graph.
If we think that a variable x may help explain, predict, or even cause changes
in another variable y, we call x an explanatory variable and y a response variable. Always plot the explanatory variable, if there is one, on the x axis of a
scatterplot. Plot the response variable on the y axis.
In examining a scatterplot, look for an overall pattern showing the direction,
form, and strength of the relationship and then look for outliers or other
departures from this pattern.
Direction: If the relationship has a clear direction, we speak of either positive association (above-average values of the two variables tend to occur together) or negative association (above-average values of one variable tend to
occur with below-average values of the other variable).
Form: Linear relationships, where the points show a straight-line pattern, are
an important form of relationship between two variables. Curved relationships and clusters are other forms to watch for.
Strength: The strength of a relationship is determined by how close the
points in the scatterplot lie to a simple form such as a line.
The correlation r measures the strength and direction of the linear association between two quantitative variables x and y. Although you can calculate
a correlation for any scatterplot, r measures strength for only straight-line
relationships.
Correlation indicates the direction of a linear relationship by its sign: r > 0 for
a positive association and r < 0 for a negative association. Correlation ­always
satisfies −1 ≤ r ≤ 1 and indicates the strength of a linear relationship by how
close it is to −1 or 1. Perfect correlation, r = ±1, occurs only when the points
on a scatterplot lie exactly on a straight line.
Remember these important facts about r: Correlation does not imply causation. Correlation ignores the distinction between explanatory and response
variables. The value of r is not affected by changes in the unit of measurement of either variable. Correlation is not resistant, so outliers can greatly
change the value of r.
3.1 T ECHNOLOGY
CORNER
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
7. Scatterplots on the calculator
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page 150
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159
Section 3.1 Scatterplots and Correlation
Exercises
Section 3.1
2.
3.
4.
1000
Treating breast cancer Early on, the most common
treatment for breast cancer was removal of the breast.
It is now usual to remove only the tumor and nearby
lymph nodes, followed by radiation. The change in
­policy was due to a large medical experiment that compared the two treatments. Some breast cancer patients,
chosen at random, were given one or the other treatment.
The patients were closely followed to see how long they
lived following surgery. What are the explanatory and
response variables? Are they categorical or quantitative?
IQ and grades Do students with higher IQ test scores
tend to do better in school? The figure below shows
a scatterplot of IQ and school grade point average
(GPA) for all 78 seventh-grade students in a rural
midwestern school. (GPA was recorded on a 12-point
scale with A+ = 12, A = 11, A− = 10, B+ = 9, . . . ,
D− = 1, and F = 0.)5
12
Grade point average
11
How much gas? Joan is concerned about the
amount of energy she uses to heat her home. The
graph below plots the mean number of cubic feet of
gas per day that Joan used each month against the
average temperature that month (in degrees Fahrenheit) for one heating season.
Gas consumed (cubic feet)
1.
pg 144
growth is response,
water is explanatory,
quantitative
Coral reefs How sensitive to changes in water
temperature are coral reefs? To find out, measure
the growth of corals in aquariums where the water
temperature is controlled at different levels. Growth is
measured by weighing the coral before and after the
experiment. What are the explanatory and response
variables? Are they categorical or quantitative?
800
600
400
200
0
20
25
30
35
40
45
50
55
60
Temperature (degrees Fahrenheit)
(a) D
​ oes the plot show a positive or negative association
between the variables? Why does this make sense?
(b) W
​ hat is the form of the relationship? Is it very strong?
Explain your answers.
(c) E
​ xplain what the point at the bottom right of the plot
represents.
10
5.
9
pg 145
8
7
6
5
4
Heavy backpacks Ninth-grade students at the Webb
Schools go on a backpacking trip each fall. Students are
divided into hiking groups of size 8 by selecting names
from a hat. Before leaving, students and their backpacks
are weighed. The data here are from one hiking group
in a recent year. Make a scatterplot by hand that shows
how backpack weight relates to body weight.
3
2
Body weight (lb):
1
Backpack weight (lb):
0
60
70
80
90
100
110
120
130
140
IQ score
(a) D
​ oes the plot show a positive or negative association
between the variables? Why does this make sense?
(b) W
​ hat is the form of the relationship? Is it very strong?
Explain your answers.
(c) A
​ t the bottom of the plot are several points that we
might call outliers. One student in particular has a
very low GPA despite an average IQ score. What are
the approximate IQ and GPA for this student?
Starnes-Yates5e_c03_140-205hr3.indd 159
6.
120
187
109
103
131
165
158
116
26
30
26
24
29
35
31
28
Bird colonies One of nature’s patterns connects the
percent of adult birds in a colony that return from
the previous year and the number of new adults that
join the colony. Here are data for 13 colonies of
­sparrowhawks:6
Percent return: 74 66 81 52 73 62 52 45 62 46 60 46 38
New adults:
5
6
8 11 12 15 16 17 18 18 19 20 20
Make a scatterplot by hand that shows how the number
of new adults relates to the percent of returning birds.
11/13/13 1:20 PM
pg 147
7.
CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
Heavy backpacks Refer to your graph from ­Exercise 5.
(a) Describe the relationship between body weight and
backpack weight for this group of hikers.
(b) One of the hikers is a possible outlier. Identify the
body weight and backpack weight for this hiker. How
does this hiker affect the form of the association?
8.
Bird colonies Refer to your graph from Exercise 6.
(a) Describe the relationship between number of new sparrowhawks in a colony and percent of returning adults.
(b) For short-lived birds, the association between these
variables is positive: changes in weather and food supply drive the populations of new and returning birds
up or down together. For long-lived territorial birds,
on the other hand, the association is negative because
returning birds claim their territories in the colony
and don’t leave room for new recruits. Which type of
species is the sparrowhawk? Explain.
9. Does fast driving waste fuel? How does the fuel consumption of a car change as its speed increases? Here
are data for a British Ford Escort. Speed is measured in
kilometers per hour, and fuel consumption is measured
in liters of gasoline used per 100 kilometers traveled.7
Speed
(km/h)
Fuel used
(liters/100 km)
Speed
(km/h)
Fuel used
(liters/100 km)
10
21.00
90
7.57
20
13.00
100
8.27
30
10.00
110
9.03
40
8.00
120
9.87
50
7.00
130
10.79
60
5.90
140
11.77
70
6.30
150
12.83
80
6.95
Mass:
36.1 54.6 48.5 42.0 50.6 42.0 40.3 33.1 42.4 34.5 51.1 41.2
Rate:
995 1425 1396 1418 1502 1256 1189 913 1124 1052 1347 1204
(a) Use your calculator to help sketch a scatterplot to
examine the researchers’ belief.
(b) Describe the direction, form, and strength of the
­relationship.
11. Southern education For a long time, the South has
lagged behind the rest of the United States in the performance of its schools. Efforts to improve education
have reduced the gap. We wonder if the South stands
out in our study of state average SAT Math scores. The
figure below enhances the scatterplot in Figure 3.2
(page 145) by plotting 12 southern states in red.
625
Mean SAT Math score
160
600
575
550
525
500
WV
475
450
0
10
20
30
40
50
60
70
80
90
Percent taking
(a) What does the graph suggest about the southern states?
(b) The point for West Virginia is labeled in the graph.
­Explain how this state is an outlier.
12. Do heavier people burn more energy? The study
of dieting described in Exercise 10 collected data
on the lean body mass (in kilograms) and metabolic
rate (in calories) for 12 female and 7 male subjects.
The figure below is a scatterplot of the data for all 19
subjects, with separate symbols for males and females.
(a) Use your calculator to help sketch a scatterplot.
(b) Describe the form of the relationship. Why is it not linear?
Explain why the form of the relationship makes sense.
(c) It does not make sense to describe the variables as either
positively associated or negatively associated. Why?
(d) Is the relationship reasonably strong or quite weak?
Explain your answer.
10. Do heavier people burn more energy? Metabolic
rate, the rate at which the body consumes energy, is
important in studies of weight gain, dieting, and exercise. We have data on the lean body mass and resting
metabolic rate for 12 women who are subjects in a
study of dieting. Lean body mass, given in kilograms,
is a person’s weight leaving out all fat. Metabolic rate
is measured in calories burned per 24 hours. The
­researchers believe that lean body mass is an important influence on metabolic rate.
Starnes-Yates5e_c03_140-205hr3.indd 160
Does the same overall pattern hold for both women
and men? What difference between the sexes do you
see from the graph?
13. Merlins breeding The percent of an animal species
pg 148 in the wild that survives to breed again is often lower
following a successful breeding season. A study of
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Section 3.1 Scatterplots and Correlation
161
merlins (small falcons) in northern Sweden observed
the number of breeding pairs in an isolated area and
the percent of males (banded for identification) that
returned the next breeding season. Here are data for
seven years:8
Breeding pairs:
28
29
29
29
30
32
33
Percent return:
82
83
70
61
69
58
43
(a)
(b)
(c)
(d)
Make a scatterplot to display the relationship between breeding pairs and percent return. Describe
what you see.
14. Does social rejection hurt? We often describe our
emotional reaction to social rejection as “pain.” Does
social rejection cause activity in areas of the brain
that are known to be activated by physical pain? If
it does, we really do experience social and physical
pain in similar ways. Psychologists first included and
then deliberately excluded individuals from a social
activity while they measured changes in brain activity.
After each activity, the subjects filled out questionnaires that assessed how excluded they felt. The table
below shows data for 13 subjects.9 “Social distress” is
measured by each subject’s questionnaire score after
exclusion relative to the score after inclusion. (So values greater than 1 show the degree of distress caused
by exclusion.) “Brain activity” is the change in activity
in a region of the brain that is activated by physical
pain. (So positive values show more pain.)
Subject
Social distress
Brain activity
1
1.26
−0.055
2
1.85
−0.040
3
1.10
−0.026
4
2.50
−0.017
5
2.17
−0.017
6
2.67
0.017
7
2.01
0.021
8
2.18
0.025
9
2.58
0.027
10
2.75
0.033
11
2.75
0.064
12
3.33
0.077
13
3.65
0.124
Make a scatterplot to display the relationship between
social distress and brain activity. Describe what you see.
15. Matching correlations Match each of the following
scatterplots to the r below that best describes it. (Some
r’s will be left over.)
r = −0.9 ​ ​r = −0.7 ​ ​r = −0.3 ​ ​r = 0
r = 0.3 ​ ​
r = 0.7 ​ ​r = 0.9
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(e)
16. Rank the correlations Consider each of the following
relationships: the heights of fathers and the heights
of their adult sons, the heights of husbands and the
heights of their wives, and the heights of women at age
4 and their heights at age 18. Rank the correlations between these pairs of variables from largest to smallest.
Explain your reasoning.
17. Correlation blunders Each of the following statements
contains an error. Explain what’s wrong in each case.
(a) “There is a high correlation between the gender of
­American workers and their income.”
(b) “We found a high correlation (r = 1.09) between
students’ ratings of faculty teaching and ratings made
by other faculty members.”
(c) “The correlation between planting rate and yield of
corn was found to be r = 0.23 bushel.”
18. Teaching and research A college newspaper interviews a psychologist about student ratings of the
teaching of faculty members. The psychologist says,
“The evidence indicates that the correlation between
the research productivity and teaching rating of
faculty members is close to zero.” The paper reports
this as “Professor McDaniel said that good researchers
tend to be poor teachers, and vice versa.” Explain why
the paper’s report is wrong. Write a statement in plain
language (don’t use the word “correlation”) to explain
the psychologist’s meaning.
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19. Dem bones Archaeopteryx is an extinct beast having
feathers like a bird but teeth and a long bony tail
like a reptile. Only six fossil specimens are known.
Because these specimens differ greatly in size, some
scientists think they are different species rather than
individuals from the same species. We will examine
some data. If the specimens belong to the same
species and differ in size because some are younger
than others, there should be a positive linear relationship between the lengths of a pair of bones from all
individuals. An outlier from this relationship would
suggest a different species. Here are data on the
lengths in centimeters of the femur (a leg bone) and
the humerus (a bone in the upper arm) for the five
specimens that preserve both bones:10
Femur (x ):
38
56
59
64
74
Humerus (y ):
41
63
70
72
84
(a) Make a scatterplot. Do you think that all five specimens come from the same species? Explain.
(b) Find the correlation r step by step, using the formula
on page 154. Explain how your value for r matches
your graph in part (a).
20. Data on dating A student wonders if tall women tend
to date taller men than do short women. She measures
herself, her dormitory roommate, and the women in the
adjoining rooms. Then she measures the next man each
woman dates. Here are the data (heights in inches):
Women (x ):
66
64
66
65
70
65
Men (y ):
72
68
70
68
71
65
(a) Make a scatterplot of these data. Based on the scatterplot, do you expect the correlation to be positive or
negative? Near ±1 or not?
(b) Find the correlation r step by step, using the formula on
page 154. Do the data show that taller women tend to
date taller men?
21. Hot dogs Are hot dogs that are high in calories also
high in salt? The figure below is a scatterplot of the
calories and salt content (measured as milligrams of
sodium) in 17 brands of meat hot dogs.11
pg 152
600
550
Sodium (mg)
500
450
(b) What effect does the hot dog brand with the lowest calorie content have on the correlation? Justify your answer.
22. All brawn? The figure below plots the average brain
weight in grams versus average body weight in kilograms for 96 species of mammals.12 There are many
small mammals whose points overlap at the lower left.
(a) The correlation between body weight and brain weight
is r = 0.86. Explain what this value means.
(b) What effect does the elephant have on the correlation? Justify your answer.
4500
Elephant
4000
Brain weight (g)
162
3500
3000
2500
2000
Dolphin
Human
1500
1000
Hippo
500
0
0
400
800
1200 1600 2000 2400 2800
Body weight (kg)
23. Dem bones Refer to Exercise 19.
(a) How would r change if the bones had been measured
in millimeters instead of centimeters? (There are
10 millimeters in a centimeter.)
(b) If the x and y variables are reversed, how would the
correlation change? Explain.
24. Data on dating Refer to Exercise 20.
(a) How would r change if all the men were 6 inches
shorter than the heights given in the table? Does the
correlation tell us if women tend to date men taller
than themselves?
(b) If heights were measured in centimeters rather than
inches, how would the correlation change? (There
are 2.54 centimeters in an inch.)
25. Strong association but no correlation The gas
mileage of an automobile first increases and then
decreases as the speed increases. Suppose that this
relationship is very regular, as shown by the following
data on speed (miles per hour) and mileage (miles
per gallon).
400
Speed:
20
30
40
50
60
350
Mileage:
24
28
30
28
24
300
250
200
150
100
100 110 120 130 140 150 160 170 180 190 200
Calories
(a) The correlation for these data is r = 0.87. Explain
what this value means.
Starnes-Yates5e_c03_140-205hr3.indd 162
(a) Make a scatterplot to show the relationship between
speed and mileage.
(b) Calculate the correlation for these data by hand or
using technology.
(c) Explain why the correlation has the value found in part
(b) even though there is a strong relationship between
speed and mileage.
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Section 3.1 Scatterplots and Correlation
x:
1
2
3
4
10
10
y:
1
3
3
5
1
11
(a) Make a scatterplot to show the relationship between x
and y.
(b) Calculate the correlation for these data by hand or
using technology.
(c) What is responsible for reducing the correlation to the
value in part (b) despite a strong straight-line relationship between x and y in most of the observations?
Multiple choice: Select the best answer for Exercises 27 to 32.
27. You have data for many years on the average price of
a barrel of oil and the average retail price of a gallon
of unleaded regular gasoline. If you want to see how
well the price of oil predicts the price of gas, then you
should make a scatterplot with ______ as the explanatory variable.
(a) the price of oil
(c) the year (e) time
(b) the price of gas
(d) either oil price or gas price
28. In a scatterplot of the average price of a barrel of oil
and the average retail price of a gallon of gas, you
expect to see
(a) very little association.
(b) a weak negative association.
(c) a strong negative association.
(d) a weak positive association.
(e) a strong positive association.
29. The following graph plots the gas mileage (miles per
gallon) of various cars from the same model year versus the weight of these cars in thousands of pounds.
The points marked with red dots correspond to cars
made in Japan. From this plot, we may conclude that
(a) there is a positive association between weight and gas
mileage for Japanese cars.
(b) the correlation between weight and gas mileage for all
the cars is close to 1.
(c) there is little difference between Japanese cars and cars
made in other countries.
(d) Japanese cars tend to be lighter in weight than other cars.
(e) Japanese cars tend to get worse gas mileage than other cars.
Miles per gallon
35
30
25
20
15
2.25
3.00
3.75
Weight (thousands of pounds)
Starnes-Yates5e_c03_140-205hr3.indd 163
4.50
30. If women always married men who were 2 years older
than themselves, what would the correlation between
the ages of husband and wife be?
(a) 2
(b) 1
(c) ​0.5
(d) ​0
(e) Can’t tell without seeing the data
31. The figure below is a scatterplot of reading test scores
against IQ test scores for 14 fifth-grade children.
There is one low outlier in the plot. What effect does
this low outlier have on the correlation?
(a) It makes the correlation closer to 1.
(b) It makes the correlation closer to 0 but still positive.
(c) It makes the correlation equal to 0.
(d) It makes the correlation negative.
(e) It has no effect on the correlation.
Child’s reading test score
26. What affects correlation? Here are some hypothetical data:
120
110
100
90
80
70
60
50
40
30
20
10
90 95 100 105 110 115 120 125 130 135 140 145 150
Child’s IQ test score
32. If we leave out the low outlier, the correlation for
the remaining 13 points in the preceding figure is
closest to
(a) −0.95.
(c) 0.
(e) 0.95.
(b) −0.5.
(d) 0.5.
33. Big diamonds (1.2, 1.3) Here are the weights (in
milligrams) of 58 diamonds from a nodule carried
up to the earth’s surface in surrounding rock. These
data represent a population of diamonds formed in a
single event deep in the earth.13
13.8
3.7 33.8 11.8 27.0 18.9 19.3 20.8 25.4 23.1 7.8
10.9
9.0
9.0 14.4
6.5
7.3
5.6 18.5
1.1 11.2 7.0
7.6
9.0
9.5
7.7
7.6
3.2
6.5
5.4
7.2
7.8 3.5
5.4
5.1
5.3
3.8
2.1
2.1
4.7
3.7
3.8
4.9 2.4
1.4
0.1
4.7
1.5
2.0
0.1
0.1
1.6
3.5
3.7 2.6
4.0
2.3
4.5
Make a graph that shows the distribution of weights
of these diamonds. Describe what you see. Give appropriate numerical measures of center and spread.
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34. College debt (2.2) A report published by the Federal
Reserve Bank of New York in 2012 reported the
results of a nationwide study of college student debt.
Researchers found that the average student loan
balance per borrower is $23,300. They also reported
that about one-quarter of borrowers owe more than
$28,000.14
(a) Assuming that the distribution of student loan
balances is approximately Normal, estimate the
standard deviation of the distribution of student loan
balances.
3.2
What You Will Learn
•
•
•
•
•
•
(b) Assuming that the distribution of student loan balances
is approximately Normal, use your answer to part (a) to
estimate the proportion of borrowers who owe more than
$54,000.
(c) In fact, the report states that about 10% of borrowers
owe more than $54,000. What does this fact indicate
about the shape of the distribution of student loan
balances?
(d) The report also states that the median student loan
balance is $12,800. Does this fact support your conclusion in part (c)? Explain.
Least-Squares Regression
By the end of the section, you should be able to:
Interpret the slope and y intercept of a least-squares
regression line.
Use the least-squares regression line to predict y for a
given x. Explain the dangers of extrapolation.
Calculate and interpret residuals.
Explain the concept of least squares.
Determine the equation of a least-squares regression
line using technology or computer output.
Construct and interpret residual plots to assess whether
a linear model is appropriate.
•
•
•
Interpret the standard deviation of the residuals
and r 2 and use these values to assess how well the
least-squares regression line models the relationship
between two variables.
Describe how the slope, y intercept, standard deviation
of the residuals, and r 2 are influenced by outliers.
Find the slope and y intercept of the least-squares regression line from the means and standard deviations of x and y
and their correlation.
Linear (straight-line) relationships between two quantitative variables are fairly common and easy to understand. In the previous section, we found linear relationships
in settings as varied as sparrowhawk colonies, natural-gas consumption, and Florida
manatee deaths. Correlation measures the direction and strength of these relationships. When a scatterplot shows a linear relationship, we’d like to summarize the
overall pattern by drawing a line on the scatterplot. A regression line summarizes
the relationship between two variables, but only in a specific setting: when one of
the variables helps explain or predict the other. Regression, unlike correlation, requires that we have an explanatory variable and a response variable.
Definition: Regression line
A regression line is a line that describes how a response variable y changes as an
explanatory variable x changes. We often use a regression line to predict the value of
y for a given value of x.
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Section 3.2 Least-Squares Regression
Let’s look at a situation where a regression line provides a useful model.
EXAMPLE
How Much Is That Truck Worth?
Regression lines as models
Everyone knows that cars and trucks lose value the more they are
driven. Can we predict the price of a used Ford F-150 SuperCrew
4 × 4 if we know how many miles it has on the odometer? A random sample of 16 used Ford F-150 SuperCrew 4 × 4s was selected
from among those listed for sale at autotrader.com. The number
of miles driven and price (in dollars) were recorded for each of the
trucks.15 Here are the data:
Miles driven
70,583
129,484
29,932
29,953
24,495
75,678
8359
4447
Price (in dollars)
21,994
9500
29,875
41,995
41,995
28,986
31,891
37,991
Miles driven
34,077
58,023
44,447
68,474
144,162
140,776
29,397
131,385
Price (in dollars)
34,995
29,988
22,896
33,961
16,883
20,897
27,495
13,997
Figure 3.7 is a scatterplot of these data. The plot shows a moderately strong, negative linear association between miles driven and price with no outliers. The correlation is r = −0.815. The line on the plot is a regression line for predicting price
from miles driven.
45,000
Price (in dollars)
40,000
Figure 3.7 Scatterplot
showing the price and
miles driven of used Ford
F-150s, with a regression
line added.
This regression line predicts
price from miles driven.
35,000
30,000
25,000
20,000
15,000
10,000
5000
0
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
Interpreting a Regression Line
A regression line is a model for the data, much like the density curves of
Chapter 2. The equation of a regression line gives a compact mathematical description of what this model tells us about the relationship between the response
variable y and the explanatory variable x.
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Definition: Regression line, predicted value, slope, y intercept
Suppose that y is a response variable (plotted on the vertical axis) and x is an
explanatory variable (plotted on the horizontal axis). A regression line relating y to x
has an equation of the form
y^ = a + bx
In this equation,
• y^ (read “y hat”) is the predicted value of the response variable y for a given
value of the explanatory variable x.
• b is the slope, the amount by which y is predicted to change when x increases
by one unit.
• a is the y intercept, the predicted value of y when x = 0.
Although you are probably accustomed to the form y = mx + b for the equation of a line from algebra, statisticians have adopted a different form for the equation of a regression line. Some use y^ = b0 + b1x. We prefer y^ = a + bx for two
reasons: (1) it’s simpler and (2) your calculator uses this form. Don’t get so caught
up in the symbols that you lose sight of what they mean! The coefficient of x is
always the slope, no matter what symbol is used.
Many calculators and software programs will give you the equation of a regression line from keyed-in data. Understanding and using the line are more important than the details of where the equation comes from.
EXAMPLE
How Much Is That Truck Worth?
Interpreting the slope and y intercept
45,000
The y intercept of the
regression line is a 38,257.
Price (in dollars)
40,000
35,000
The equation of the regression line shown in
Figure 3.7 is
price = 38,257 − 0.1629 (miles driven)
30,000
25,000
20,000
15,000
10,000
5000
0
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
Problem: Identify the slope and y intercept of the
regression line. Interpret each value in context.
solution: The slope b = −0.1629 tells us that the
price of a used Ford F-150 is predicted to go down by 0.1629
­dollars (16.29 cents) for each additional mile that the truck
has been driven. The y intercept a = 38,257 is the predicted
price of a Ford F-150 that has been driven 0 miles.
For Practice Try Exercise
39(a) and (b)
The slope of a regression line is an important numerical description of the relationship between the two variables. Although we need the value of the y intercept
to draw the line, it is statistically meaningful only when the explanatory variable
can actually take values close to zero, as in this setting.
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Section 3.2 Least-Squares Regression
THINK
ABOUT IT
167
Does a small slope mean that there’s no relationship? ​For the
miles driven and price regression line, the slope b = −0.1629 is a small number.
This does not mean that change in miles driven has little effect on price. The size
of the slope depends on the units in which we measure the two variables. In this setting, the slope is the predicted change in price (in dollars) when the distance driven
increases by 1 mile. There are 100 cents in a dollar. If we measured price in cents
instead of dollars, the slope would be 100 times larger, b = 16.29. You can’t say how
strong a relationship is by looking at the size of the slope of the regression line.
Prediction
We can use a regression line to predict the response y^ for a specific value of the
explanatory variable x. Here’s how we do it.
Example
How Much Is That Truck Worth?
Predicting with a regression line
For the Ford F-150 data, the equation of the regression line is
45,000
Price (in dollars)
40,000
35,000
price = 38,257 − 0.1629 (miles driven)
30,000
25,000
If a used Ford F-150 has 100,000 miles driven, substitute x = 100,000 in the equation. The predicted
price is
20,000
15,000
10,000
price = 38,257 − 0.1629(100,000) = 21,967 dollars
5000
0
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
This prediction is illustrated in Figure 3.8.
FIGURE 3.8 ​Using the regression line to predict price for a Ford
F-150 with 100,000 miles driven.
The accuracy of predictions from a regression line depends on how much the
data scatter about the line. In this case, prices for trucks with similar mileage show
a spread of about $10,000. The regression line summarizes the pattern but gives
only roughly accurate predictions.
Can we predict the price of a Ford F-150 with 300,000 miles driven? We can
certainly substitute 300,000 into the equation of the line. The prediction is
price = 38,257 − 0.1629(300,000) = −10,613 dollars
That is, we predict that we would need to pay someone else $10,613 just to take
the truck off our hands!
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A negative price doesn’t make much sense in this context. Look again at Figure
3.8. A truck with 300,000 miles driven is far outside the set of x values for our data.
We can’t say whether the relationship between miles driven and price remains linear at such extreme values. Predicting price for a truck with 300,000 miles driven
is an extrapolation of the relationship beyond what the data show.
Definition: Extrapolation
Extrapolation is the use of a regression line for prediction far outside the interval
of values of the explanatory variable x used to obtain the line. Such predictions are
often not accurate.
autio
!
n
Few relationships are linear for all values of the explanatory variable.
Don’t make predictions using values of x that are much larger or much
smaller than those that actually appear in your data.
c
Often, using the regression line
to make a prediction for x = 0 is
an extrapolation. That’s why the y
intercept isn’t always statistically
meaningful.
Check Your Understanding
Some data were collected on the weight of a male white laboratory rat for the first 25 weeks
after its birth. A scatterplot of the weight (in grams) and time since birth (in weeks) shows
a fairly strong, positive linear relationship. The linear regression equation weight = 100 +
40(time) models the data fairly well.
1. What is the slope of the regression line? Explain what it
means in context.
2. What’s the y intercept? Explain what it means in context.
3.
Predict the rat’s weight after 16 weeks. Show your work.
4. Should you use this line to predict the rat’s weight at
age 2 years? Use the equation to make the prediction and
think about the reasonableness of the result. (There are 454
grams in a pound.)
Residuals and the Least-Squares
Regression Line
45,000
Price (in dollars)
40,000
Vertical deviation from the line
35,000
Regression line
ŷ = 38,257 2 0.1629x
30,000
25,000
20,000
In most cases, no line will pass exactly through all the
points in a scatterplot. Because we use the line to predict
y from x, the prediction errors we make are errors in y, the
vertical direction in the scatterplot. A good regression line
makes the vertical deviations of the points from the line as
small as possible.
Figure 3.9 shows a scatterplot of the Ford F-150 data
with a regression line added. The prediction errors are
Data point
15,000
10,000
5000
0
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
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FIGURE 3.9 ​Scatterplot of the Ford F-150 data with a regression
line added. A good regression line should make the prediction
errors (shown as bold vertical segments) as small as ­possible.
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Section 3.2 Least-Squares Regression
169
marked as bold segments in the graph. These vertical deviations represent “leftover” variation in the response variable after fitting the regression line. For that
reason, they are called residuals.
Definition: Residual
A residual is the difference between an observed value of the response variable and
the value predicted by the regression line. That is,
residual = observed y − predicted y
= y − y^
The following example shows you how to calculate and interpret a residual.
Example
How Much Is That Truck Worth?
Finding a residual
PROBLEM: Find and interpret the residual for the Ford F-150 that had 70,583 miles driven
and a price of $21,994.
SOLUTION: The regression line predicts a price of
price = 38,257 − 0.1629(70,583) = 26,759 dollars
for this truck, but its actual price was $21,994. This truck’s residual is
residual = observed y − predicted y
= y − y^ = 21,994 − 26,759 = −4765 dollars
That is, the actual price of this truck is $4765 lower than expected, based on its mileage. The actual
price might be lower than predicted as a result of other factors. For example, the truck may have been
in an accident or may need a new paint job.
For Practice Try Exercise
45
The line shown in Figure 3.9 makes the residuals for the 16 trucks “as small
as possible.” But what does that mean? Maybe this line minimizes the sum of
the residuals. Actually, if we add up the prediction errors for all 16 trucks, the
positive and negative residuals cancel out. That’s the same issue we faced when
we tried to measure deviation around the mean in Chapter 1. We’ll solve the
current problem in much the same way: by squaring the residuals. The regression line we want is the one that minimizes the sum of the squared residuals.
That’s what the line shown in Figure 3.9 does for the Ford F-150 data, which is
why we call it the least-squares regression line.
Definition: Least-squares regression line
The least-squares regression line of y on x is the line that makes the sum of the
squared residuals as small as possible.
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45,000
45,000
40,000
40,000
35,000
35,000
30,000
Price (in dollars)
Price (in dollars)
170
Residual = 4765
25,000
20,000
15,000
Squared residual = (4765)2
= 22,705,225
10,000
Sum of squared
residuals 461,300,000
30,000
25,000
20,000
15,000
10,000
5000
5000
0
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
(a)
0
(b)
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
FIGURE 3.10 ​The least-squares idea: make the errors in predicting y as small as possible by
minimizing the sum of the squares of the residuals.
Figure 3.10 gives a geometric interpretation of the least-squares idea for the truck
data. Figure 3.10(a) shows the “squared” residual for the truck with 70,583 miles ­driven
and a price of $21,994. The area of this square is (−4765)(−4765) = 22,705,225.
Figure 3.10(b) shows the squared residuals for all the trucks. The sum of squared residuals is 461,300,000. No other regression line would give a smaller sum of squared
residuals.
Activity
MATERIALS:
Computer with
Internet ­connection
PLET
AP
Investigating properties of the
least-squares regression line
In this Activity, you will use the Correlation and Regression applet at the book’s
Web site, www.whfreeman.com/tps5e, to explore some properties of the leastsquares regression line.
1. ​Click on the scatterplot to create a group of 15 to 20 points from lower left to
upper right with a clear positive straight-line pattern (correlation around 0.7).
2. Click the “Draw your own line” button to select ­starting and
ending points for your own line on the plot. Use the mouse to
adjust the starting and ending points until you have a line that
models the association well.
3. Click the “Show least-squares line” button. How do the two
lines compare? One way to measure this is to compare the “Relative SS,” the ratio of the sum of squared residuals from your
line and the least-squares regression line. If the two lines are exactly the same, the relative sum of squares will be 1. O
­ therwise,
the relative sum of squares will be larger than 1.
4. Press the “CLEAR” button and create another scatterplot as in Step 1. Then
click on “Show least-squares line” and “Show mean X & Y lines.” What do you
notice? Move or add points, one at a time, in your scatterplot to see if this result
continues to hold true.
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Section 3.2 Least-Squares Regression
5. Now click the “Show residuals” ­button. How does an outlier affect the slope
and y intercept of the least-squares ­regression line? Move or add points, one at a
time, to investigate. Does it depend on whether the outlier has an x-value close
to the center of the plot or toward the far edges of the plot?
Your calculator or statistical software will give the equation of the least-squares
line from data that you enter. Then you can concentrate on understanding and
using the regression line.
L​ east-squares regression lines
on the calculator
8. T echnology
Corner
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
Let’s use the Ford F-150 data to show how to find the equation of the least-squares regression line on the TI-83/84
and TI-89. Here are the data again:
Miles driven
70,583
129,484
29,932
29,953
24,495
75,678
8359
4447
Price (in dollars)
21,994
9500
29,875
41,995
41,995
28,986
31,891
37,991
Miles driven
34,077
58,023
44,447
68,474
144,162
140,776
29,397
131,385
Price (in dollars)
34,995
29,988
22,896
33,961
16,883
20,897
27,495
13,997
1. Enter the miles driven data into L1/list1 and the price data into L2/list2. Then make a scatterplot. Refer to the
Technology Corner on page 150.
2. To determine the least-squares regression line:
TI-83/84
•
Press STAT ; choose CALC and then
LinReg(a+bx). OS 2.55 or later: In the dialog
box, enter the following: Xlist:L1, Ylist:L2,
FreqList (leave blank), Store RegEQ:Y1,
and choose Calculate. Older OS: Finish the
­command to read LinReg(a+bx)L1,L2,Y1 and
press ENTER . (Y1 is found under VARS/Y-VARS/
Function.)
TI-89
•
In the Statistics/List Editor, press F4 (CALC);
choose Regressions and then LinReg(a+bx).
•
Enter list1 for the Xlist, list2 for the Ylist; choose to
store the RegEqn to y1(x); and press ENTER
Note: If you do not want to store the equation to Y1, then leave the StoreRegEq prompt blank (OS 2.55 or later) or
use the following command (older OS): LinReg(a+bx) L1,L2.
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3. Graph the regression line. Turn off all other equations in the Y= screen and use ZoomStat/ZoomData to add the
least-squares line to the scatterplot.
4. Save these lists for later use. On the home screen, use the STO▶ key to help execute the command
L1nMILES:L2nPRICE (list1nMILES:list2nPRICE on the TI-89).
Note: If r2 and r do not appear on the TI-83/84 screen, do this one-time series of keystrokes: OS 2.55 or later:
Press MODE and set STAT DIAGNOSTICS to ON. Older OS: Press 2nd 0 (CATALOG), scroll down to
­DiagnosticOn, and press ENTER . Press ENTER again to execute the command. The screen should say “Done.”
Then redo Step 2 to calculate the least-squares line. The r2 and r values should now appear.
AP® EXAM TIP ​When displaying the equation of a least-squares regression line, the
calculator will report the slope and intercept with much more precision than we need.
However, there is no firm rule for how many decimal places to show for answers on the
AP® exam. Our advice: Decide how much to round based on the context of the problem
you are working on.
Check Your Understanding
It’s time to practice your calculator regression skills. Using the familiar SEC football data
in the table below, repeat the steps in the previous Technology Corner. You should get
y^ = −3.7506 + 0.4372x as the equation of the regression line.
Team
Alabama
Arkansas
Auburn
Florida
Georgia
Kentucky
Points per game
34.8
36.8
25.7
25.5
32.0
15.8
Wins
12
11
8
7
Team
Louisiana State
Mississippi
Mississippi State South Carolina
Points per game
35.7
16.1
25.3
Wins
13
2
7
30.1
11
10
5
Tennessee
Vanderbilt
20.3
26.7
5
6
Determining Whether a Linear Model Is
Appropriate: Residual Plots
One of the first principles of data analysis is to look for an overall pattern and for
striking departures from the pattern. A regression line describes the overall pattern
of a linear relationship between an explanatory variable and a response variable.
We see departures from this pattern by looking at the residuals.
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173
Section 3.2 Least-Squares Regression
Example
How Much Is That Truck Worth?
Examining residuals
Let’s return to the Ford F-150 data about the number of miles driven
and price for a random sample of 16 used trucks. In general, trucks
with more miles driven have lower prices. In the Technology Corner,
we confirmed that the equation of the least-squares regression line for
these data is price = 38,257 − 0.1629 (miles driven). The calculator
screen shot in the margin shows a scatterplot of the data with the leastsquares line added.
One truck had 68,474 miles driven and a price of $33,961. This truck
is marked on the scatterplot with an X. Because the point is above the
line on the scatterplot, we know that its actual price is higher than the predicted price. To find out exactly how much higher, we calculate the residual
for this truck. The predicted price for a Ford F-150 with 68,474 miles driven is
y^ = 38,257 − 0.1629(68,474) = $27,103
The residual for this truck is therefore
residual = observed y − predicted y = y − y^ = 33,961 − 27,103 = $6858
This truck costs $6858 more than expected, based on its mileage.
The 16 points used in calculating the equation of the least-squares regression line
produce 16 residuals. Rounded to the nearest dollar, they are
−4765
2289
Most graphing calculators and
statistical software will calculate and
store residuals for you.
Some software packages prefer to plot
the residuals against the predicted
values y^ instead of against the values
of the explanatory variable. The basic
shape of the two plots is the same
because y^ is linearly related to x.
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−7664
1183
−3506
−8121
8617
6858
7728
2110
3057
5572
−5004
−5973
458
−2857
Although residuals can be calculated from any model that is fitted to the data, the
residuals from the least-squares line have a special property: the mean of the leastsquares residuals is always zero. You can check that the sum of the residuals in the
above example is −$18. The sum is not exactly 0 because of rounding errors.
You can see the residuals in the scatterplot of Figure 3.11(a) on the next page
by looking at the vertical deviations of the points from the line. The residual plot
in Figure 3.11(b) makes it easier to study the residuals by plotting them against
the explanatory variable, miles driven. Because the mean of the residuals is always
zero, the horizontal line at zero in Figure 3.11(b) helps orient us. This “residual
= 0” line corresponds to the regression line in Figure 3.11(a).
Definition: Residual plot
A residual plot is a scatterplot of the residuals against the explanatory variable.
Residual plots help us assess whether a linear model is appropriate.
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174
(a)
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D e s c r i b i n g R e l at i o n s h i p s
(b)
45,000
10,000
5000
35,000
30,000
Residual
Price (in dollars)
40,000
25,000
20,000
0
5000
15,000
10,000
10,000
5000
0
0
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
Miles driven
FIGURE 3.11 ​(a) Scatterplot of price versus miles driven, with the least-squares line. (b) Residual
plot for the regression line displayed in Figure 3.11(a). The line at y = 0 marks the sum (and
mean) of the residuals.
Check Your Understanding
Refer to the Ford F-150 miles driven and price data.
1. Find the residual for the truck that had 8359 miles driven and a price of $31,891.
Show your work.
2. Interpret the value of this truck’s residual in context.
3. For which truck did the regression line overpredict price by the most? Justify your
answer.
Examining residual plots ​A residual plot in effect turns the regression
line horizontal. It magnifies the deviations of the points from the line, making it
easier to see unusual observations and patterns. Because it is easier to see an unusual pattern in a residual plot than a scatterplot of the original data, we often use
residual plots to determine if the model we are using is appropriate.
Figure 3.12(a) shows a nonlinear association between two variables and the
least-squares regression line for these data. Figure 3.12(b) shows the residual plot
for these data.
(a)
(b)
FIGURE 3.12 ​(a) A straight line is not a good model for these data. (b) The residual plot has a
curved pattern.
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Section 3.2 Least-Squares Regression
175
Because the form of our model (linear) is not the same as the form of the association (curved), there is an obvious leftover pattern in the residual plot. When
an obvious curved pattern exists in a residual plot, the model we are using is not
appropriate. We’ll look at how to deal with curved relationships in Chapter 12.
When we use a line to model a linear association, there will be no leftover patterns in the residual plot, only random scatter. Figure 3.13 shows the residual plot
for the Ford F-150 data. Because there is only random scatter in the residual plot,
we know the linear model we used is appropriate.
10,000
Residual
5000
FIGURE 3.13 ​The random scatter
of points indicates that the regression line has the same form as
the association, so the line is an
appropriate model.
THINK
ABOUT IT
0
5000
10,000
0
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
Why do we look for patterns in residual plots? The word residual
comes from the Latin word residuum, meaning “left over.” When we calculate a
residual, we are calculating what is left over after subtracting the predicted value
from the observed value:
residual = observed y− predicted y
Likewise, when we look at the form of a residual plot, we are looking at the form that
is left over after subtracting the form of the model from the form of the ­association:
form of residual plot = form of association − form of model
When there is a leftover form in the residual plot, the form of the association and
form of the model are not the same. However, if the form of the association and
form of the model are the same, the residual plot should have no form, other than
random scatter.
9. T echnology
Corner
Residual plots on the calculator
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
Let’s continue the analysis of the Ford F-150 miles driven and price data from the previous Technology Corner
(page 171). You should have already made a scatterplot, calculated the equation of the least-squares regression line, and
graphed the line on your plot. Now, we want to calculate residuals and make a residual plot. Fortunately, your calculator
has already done most of the work. Each time the calculator computes a regression line, it also computes the residuals and stores them in a list named RESID. Make sure to calculate the equation of the regression line before using the
RESID list!
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TI-83/84
TI-89
1. Display the residuals in L3(list3).
•
With L3 highlighted, press 2nd STAT (LIST) and •
­select the RESID list.
With list3 highlighted, press 2nd - (VAR-LINK), arrow down to STATVARS, and select the resid list.
2. Turn off Plot1 and the regression equation. Specify Plot2 with L1/list1 as the x variable and L3/list3 as the y variable.
Use ZoomStat (ZoomData) to see the residual plot.
The x axis in the residual plot serves as a reference line: points above this line correspond to positive residuals and points
below the line correspond to negative residuals.
Note: If you don’t want to see the residuals in L3/list3, you can make a residual plot in one step by using the RESID list
as the y variable in the scatterplot.
Check Your Understanding
In Exercises 5 and 7, we asked you to make and describe a scatterplot for the hiker data
shown in the table below. Here is a residual plot for the least-squares regression of pack
weight on body weight for the 8 hikers.
Body weight (lb):
Backpack weight (lb):
120
187
109
103
131
165
158
116
26
30
26
24
29
35
31
28
1.​One of the hikers had a residual of nearly 4 pounds. Interpret this value.
2.​Based on the residual plot, is a linear model appropriate for these data?
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Section 3.2 Least-Squares Regression
177
How Well the Line Fits the Data:
The Role of s and r 2 in Regression
A residual plot is a graphical tool for determining if a least-squares regression
line is an appropriate model for a relationship between two variables. Once we
determine that a least-squares regression line is appropriate, it makes sense to ask
a follow-up question: How well does the line work? That is, if we use the leastsquares regression line to make predictions, how good will these predictions be?
The Standard Deviation of the Residuals We already know that a
residual measures how far an observed y-value is from its corresponding predicted
value y^. In an earlier example, we calculated the residual for the Ford F-150 with
68,474 miles driven and price $33,961. The residual was $6858, meaning that the
actual price was $6858 higher than we predicted.
To assess how well the line fits all the data, we need to consider
the residuals for each of the 16 trucks, not just one. Using these
residuals, we can estimate the “typical” prediction error when using
the least-squares regression line. To do this, we calculate the standard deviation of the residuals.
10,000
Residual
5000
0
s=
–5000
–10,000
0
20,000
40,000
60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
2
Å
n−2
For the Ford F-150 data, the sum of squared residuals is 461,300,000.
So, the standard deviation of the residuals is
s=
Did you recognize the number
461,300,000? We first encountered this
number on page 170 when illustrating
that the least-squares regression
line minimized the sum of squared
residuals. We’ll see it again shortly.
∙residuals
461,300,000
= 5740 dollars
Å
14
When we use the least-squares regression line to predict the price of a Ford
F-150 using the number of miles it has been driven, our predictions will typically
be off by about $5740. Looking at the residual plot, this seems like a reasonable
value. Although some of the residuals are close to 0, others are close to $10,000
or −$10,000.
Definition: Standard deviation of the residuals (s)
If we use a least-squares line to predict the values of a response variable y from an
explanatory variable x, the standard deviation of the residuals (s) is given by
s=
Å
∙ residuals
n−2
2
=
∙(y − y^ )
i
Å
2
n−2
This value gives the approximate size of a “typical” prediction error (residual).
THINK
ABOUT IT
Does the formula for s look slightly familiar? ​It should. In Chapter 1,
we defined the standard deviation of a set of quantitative data as
sx =
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∙(x − x–)
i
Å
2
n−1
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We interpreted the resulting value as the “typical” distance of the data points from
the mean. In the case of two-variable data, we’re interested in the typical (vertical)
distance of the data points from the regression line. We find this value in much
the same way: by adding up the squared deviations, then averaging (again in a
funny way), and taking the square root to get back to the original units of measurement. Why do we divide by n − 2 this time instead of n − 1? You’ll have to wait
until Chapter 12 to find out.
The Coefficient of Determination There is another numerical quantity
that tells us how well the least-squares line predicts values of the response variable
y. It is r2, the coefficient of determination. Some computer packages call it “R-sq.”
You may have noticed this value in some of the calculator and computer regression output that we showed earlier. Although it’s true that r2 is equal to the square
of r, there is much more to this story.
Example
How Much Is That Truck Worth?
How can we predict y if we don’t know x?
15,000
10,000
60
,0
00
0
20
,0
00
40
,0
00
5000
Miles driven
0,
00
0
20,000
16
25,000
0,
00
0
30,000
14
35,000
0,
00
0
Price (in dollars)
40,000
12
45,000
80
,0
00
10
0,
00
0
Suppose that we randomly selected an additional used Ford F-150 that was on
sale. What should we predict for its price? Figure
3.14 shows a scatterplot of the truck data that we have
studied throughout this section, including the leastsquares regression line. Another horizontal line has
been added at the mean y-value, y– = $27,834. If we
don’t know the number of miles driven for the additional truck, we can’t use the regression line to make
a prediction. What should we do? Our best strategy
is to use the mean price of the other 16 trucks as our
prediction.
FIGURE 3.14 Scatterplot and least-squares regression line for
the Ford F-150 data with a horizontal line added at the mean
price, $27,834.
Figure 3.15(a) on the facing page shows the prediction errors if we use the
average price y– as our prediction for the original group of 16 trucks. We can see
that the sum of the squared residuals for this line is (yi − y–)2 = 1,374,000,000.
This quantity measures the total variation in the y-values from their mean. This is
also the same quantity we use to calculate the standard deviation of the prices, sy.
If we learn the number of miles driven on the additional truck, then we could
use the least-squares line to predict its price. How much better does the regression
line do at predicting prices than simply using the average price y– of all 16 trucks?
Figure 3.15(b) reminds us that the sum of squared residuals for the least-squares
line is residuals2 = 461,300,000. This is the same quantity we used to calculate
the standard deviation of the residuals.
∙
∙
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45
45
40
40
Price (thous ands )
Price (thous ands )
Section 3.2 Least-Squares Regression
35
30
25
20
15
35
30
25
20
15
10
10
-50
(a)
179
0
50
100
150
M ile s Dr iv e n (thous ands )
P rice = 27834
S um of s quares = 1374000000
-50
200
(b)
0
50
100
150
200
M ile s Dr iv e n (thous ands )
P rice = 3.83e+04 - 0.163Miles Driven; r 2 = 0.66
S um of s quares = 461300000
FIGURE 3.15 (a) The sum of squared residuals is 1,374,000,000 if we use the mean price as
our prediction for all 16 trucks. (b) The sum of squares from the least-squares regression line is
461,300,000.
The ratio of these two quantities tells us what proportion of the total variation in
y still remains after using the regression line to predict the values of the response
variable. In this case,
461,300,000
= 0.336
1,374,000,000
This means that 33.6% of the variation in price is unaccounted for by the leastsquares regression line using x = miles driven. This unaccounted-for variation is
likely due to other factors, including the age of the truck or its condition. Taking
this one step further, the proportion of the total variation in y that is accounted for
by the regression line is
1 − 0.336 = 0.664
We interpret this by saying that “66.4% of the variation in price is accounted for by
the linear model relating price to miles driven.”
Definition: The coefficient of determination: r 2
The coefficient of determination r 2 is the fraction of the variation in the values of y
that is accounted for by the least-squares regression line of y on x. We can calculate
r 2 using the following formula:
r2=1−
∙ residuals
∙(y − y–)
i
2
2
If all the points fall directly on the least-squares line, the sum of squared residuals is 0 and r 2 = 1. Then all the variation in y is accounted for by the linear
relationship with x. Because the least-squares line yields the smallest possible sum
of squared prediction errors, the sum of squared residuals can never be more than
the sum of squared deviations from the mean of y. In the worst-case scenario, the
least-squares line does no better at predicting y than y = y– does. Then the two
sums of squares are the same and r2 = 0.
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It seems fairly remarkable that the coefficient of determination is actually the
correlation squared. This fact provides an important connection between correlation and regression. When you see a correlation, square it to get a better feel for
how well the least-squares line fits the data.
THINK
ABOUT IT
What’s the relationship between the standard deviation of
the residuals s and the coefficient of determination r 2? They
are both calculated from the sum of squared residuals. They also both attempt to
answer the question, “How well does the line fit the data?” The standard deviation of the residuals reports the size of a typical prediction error, in the same units
as the response variable. In the truck example, s = 5740 dollars. The value of r2,
however, does not have units and is usually expressed as a percentage between 0%
and 100%, such as r2 = 66.4%. Because these values assess how well the line fits
the data in different ways, we recommend you follow the example of most statistical software and report them both.
Let’s revisit the SEC football data to practice what we have learned.
Example
SEC Football
Residual plots, s, and r2
In Section 3.1, we looked at the relationship between the average number of points scored per game x and the number of wins y for the 12
college football teams in the Southeastern Conference. A scatterplot
with the least-squares regression line and a residual plot are shown. The
equation of the least-squares regression line is y^ = −3.75 + 0.437x.
Also, s = 1.24 and r2 = 0.88.
Problem:
(a) Calculate and interpret the residual for South Carolina, which scored 30.1 points
per game and had 11 wins.
(b) Is a linear model appropriate for these data? Explain.
(c) Interpret the value of s.
(d) Interpret the value of r 2.
2.0
14
1.5
12
1.0
Residual
Wins
10
8
6
0.5
0.0
0.5
1.0
4
1.5
2
2.0
15
20
25
Points per game
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30
35
15
20
25
30
35
Points per game
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181
Section 3.2 Least-Squares Regression
AP® EXAM TIP ​Students
often have a hard time
interpreting the value of r 2
on AP® exam questions.
They frequently leave out key
words in the definition. Our
advice: Treat this as a fillin-the-blank exercise. Write
“____% of the variation
in [response variable
name] is accounted for by
the linear model relating
[response variable name] to
[explanatory variable name].”
Solution:
(a) The predicted amount of wins for South Carolina is
y^ = −3.75 + 0.437(30.1) = 9.40 wins
The residual for South Carolina is
residual = y − y^ = 11 − 9.40 = 1.60 wins
South Carolina won 1.60 more games than expected, based on the number of points they scored per
game.
(b) Because there is no obvious pattern left over in the residual plot, the linear model is appropriate.
(c) When using the least-squares regression line with x = points per game to predict y = the number of wins, we will typically be off by about 1.24 wins.
(d) About 88% of the variation in wins is accounted for by the linear model relating wins to points
per game.
For Practice Try Exercise
55
Interpreting Computer Regression Output
Figure 3.16 displays the basic regression output for the Ford F-150 data from two
statistical software packages: Minitab and JMP. Other software produces very similar output. Each output records the slope and y intercept of the least-squares line.
The software also provides information that we don’t yet need (or understand!),
although we will use much of it later. Be sure that you can locate the slope, the y
intercept, and the values of s and r2 on both computer outputs. Once you understand the statistical ideas, you can read and work with almost any software output.
Minitab
Slope
Predictor
Coef
Constant
Miles Driven
JMP
RSquare
0.664248
RSquare Adj
0.640266
SE Coef
T
P
38257
2446
15.64
0.000
Root Mean Square Error
5740.131
0.16292
0.03096
-5.26
0.000
Mean of Response
27833.69
r2
S = 5740.13
r2
Summary of Fit
y intercept
R-Sq = 66.4%
Observations (or Sum Wgts)
Standard
deviation
of the residuals
16
R-Sq(adj) = 64.0%
Parameter Estimates
Standard deviation of the residuals
Term
Estimate
Std Error
t Ratio
Prob>|t|
Intercept
38257.135
2445.813
15.64
<.0001
Miles Driven
-0.162919
0.030956
-5.26
0.0001
y intercept
Slope
FIGURE 3.16 ​Least-squares regression results for the Ford F-150 data from two statistical software packages. Other software produces similar output.
Example
Using Feet to Predict Height
Interpreting regression output
A random sample of 15 high school students was selected from the U.S.
­CensusAtSchool database. The foot length (in centimeters) and height (in
centimeters) of each student in the sample were recorded. Least-squares
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r­ egression was performed on the data. A scatterplot with the regression line
added, a residual plot, and some computer output from the regression are
shown below.
Predictor
Coef
SE Coef
T
Constant
103.41
19.50
5.30
Foot length
2.7469
0.7833
3.51
S = 7.95126 R-Sq = 48.6% R-Sq(adj) = 44.7%
195
P
0.000
0.004
15
190
10
180
Residual
Height (cm)
185
175
170
165
5
0
5
10
160
15
155
22
24
26
Foot length (cm)
28
22
30
24
26
28
30
Foot length (cm)
Problem:
(a) What is the equation of the least-squares regression line that describes the relationship
between foot length and height? Define any variables that you use.
(b) Interpret the slope of the regression line in context.
(c) Find the correlation.
(d) Is a line an appropriate model to use for these data? Explain how you know.
Solution:
(a) The equation is y^ = 103.41 + 2.7469x, where y^ = predicted height (in centimeters) and x is
foot length (in centimeters). We could also write
predicted height = 103.41 + 2.7469 (foot length)
(b) For each additional centimeter of foot length, the least-squares regression line predicts an
increase of 2.7469 cm in height.
(c) To find the correlation, we take the square root of r 2: r = ±#0.486 = ±0.697. Because
the scatterplot shows a positive association, r = 0.697.
(d) Because the scatterplot shows a linear association and the residual plot has no obvious leftover
patterns, a line is an appropriate model to use for these data.
For Practice Try Exercise
59
Regression to the Mean
Using technology is often the most convenient way to find the equation of a leastsquares regression line. It is also possible to calculate the equation of the leastsquares regression line using only the means and standard deviations of the two
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Section 3.2 Least-Squares Regression
183
variables and their correlation. Exploring this method will highlight an important
relationship between the correlation and the slope of a least-squares regression
line—and reveal why we include the word “regression” in the expression “leastsquares regression line.”
How to Calculate the Least-Squares Regression Line
AP® EXAM TIP ​The formula
sheet for the AP® exam uses
different notation for these
sy
equations: b1 = r and
sx
b0 = y– − b1 x– . That’s because
the least-squares line is written
as y^ = b0 + b1x . We prefer our
simpler versions without the
subscripts!
We have data on an explanatory variable x and a response variable y for n
individuals. From the data, calculate the means x– and y– and the standard deviations sx and sy of the two variables and their correlation r. The least-squares
regression line is the line y^ = a + bx with slope
b=r
and y intercept
sy
sx
a = y– − bx–
The formula for the y intercept comes from the fact that the least-squares regression line always passes through the point (x– , y– ). You discovered this in Step 4 of the
Activity on page 170. Substituting (x– , y– ) into the equation y^ = a + bx produces
the equation y– = a + bx– . Solving this equation for a gives the equation shown in
the definition box, a = y– − bx– .
To see how these formulas work in practice, let’s look at an example.
Example
Using Feet to Predict Height
Calculating the least-squares regression line
In the previous example, we used data from a random sample of 15 high school
students to investigate the relationship between foot length (in centimeters) and
height (in centimeters). The mean and standard deviation of the foot lengths
are x– = 24.76 cm and sx = 2.71 cm. The mean and standard deviation of the
heights are y– = 171.43 cm and sy = 10.69 cm. The correlation between foot
length and height is r = 0.697.
PROBLEM: Find the equation of the least-squares regression line for predicting height from foot
length. Show your work.
SOLUTION: The least-squares regression line of height y on foot length x has slope
b=r
sy
10.69
= 0.697
= 2.75
sx
2.71
The least-squares regression line has y intercept
a = y– − b x– = 171.43 − 2.75(24.76) = 103.34
So, the equation of the least-squares regression line is y^ = 103.34 + 2.75x.
For Practice Try Exercise
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61(a)
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D e s c r i b i n g R e l at i o n s h i p s
There is a close connection between the correlation and the slope of the leastsquares regression line. The slope is
b=r
y = mean height
195
190
Height (cm)
185
180
175
170
165
160
155
22
24
sy r # sy
=
sx
sx
This equation says that along the regression line, a change of 1 standard deviation in x
corresponds to a change of r standard deviations in y. When the variables are perfectly
correlated (r = 1 or r = −1), the change in the predicted response y^ is the same (in
standard deviation units) as the change in x. For example, if r = 1 and x is 2 standard
deviations above its mean, then the corresponding value of y^ will be 2 standard deviations above the mean of y.
However, if the variables are not perfectly correlated (−1 < r < 1), the change
in y^ is less than the change in x, when measured in standard deviation units. To illustrate this property, let’s return to the foot length and height data from the previous
example.
The figure at left shows the regression line y^ = 103.34 + 2.75x. We
ŷ = 103.34 + 2.75x
have added four more lines to the graph: a vertical line at the mean
foot length x– , a vertical line at x– + sx (1 standard deviation above the
y + sy
mean foot length), a horizontal line at the mean height y– , and a horizontal line at y– + sy (1 standard deviation above the mean height).
When a student’s foot length is 1 standard deviation above the
??
mean foot length x– , the predicted height y^ is above the mean height
y– , but not an entire standard deviation above the mean. How far
sx
x + sx
above the mean is the value of y^?
From the graph, we can see that
x = mean foot length
26
28
30
Foot length (cm)
b = slope =
change in y
change in x
=
??
sx
From earlier, we know that
b=
Sir Francis Galton (1822–1911) looked
at data on the heights of children
versus the heights of their parents.
He found that taller-than-average
parents tended to have children who
were taller than average but not quite
as tall as their parents. Likewise,
shorter-than-average parents tended
to have children who were shorter
than average but not quite as short as
their parents. Galton called this fact
“regression to the mean” and used the
symbol r because of the correlation’s
important relationship to regression.
THINK
ABOUT IT
Starnes-Yates5e_c03_140-205hr3.indd 184
r # sy
sx
Setting these two equations equal to each other, we have
?? r # sy
=
sx
sx
Thus, y^ must be r # sy above the mean y– .
In other words, for an increase of 1 standard deviation in the value of the
explanatory variable x, the least-squares regression line predicts an increase of
only r standard deviations in the response variable y. When the correlation isn’t
r = 1 or −1, the predicted value of y is closer to its mean y– than the value of x
is to its mean x– . This is called regression to the mean, because the values of y
“regress” to their mean.
What happens if we standardize both variables? ​Standardizing
a variable converts its mean to 0 and its standard deviation to 1. Doing this to both
x and y will transform the point (x– , y– ) to (0, 0). So the least-squares line for the
standardized values will pass through (0, 0). What about the slope of this line?
From the formula, it’s b = rsy ∕sx. Because we standardized, sx = sy = 1. That
11/13/13 1:21 PM
Section 3.2 Least-Squares Regression
185
means b = r. In other words, the slope is equal to the correlation. The Fathom
screen shot confirms these results. It shows that r2 = 0.49, so r = !0.49 = 0.7,
approximately the same value as the slope of 0.697.
Putting It All Together: Correlation
and Regression
In Chapter 1, we introduced a four-step process for organizing a statistics problem.
Here is another example of the four-step process in action.
Example
STEP
4
Gesell Scores
Putting it all together
Does the age at which a child begins to talk predict a later score on a test of mental
ability? A study of the development of young children recorded the age in months
at which each of 21 children spoke their first word and their Gesell Adaptive Score,
the result of an aptitude test taken much later.16 The data appear in the table below, along with a scatterplot, residual plot, and computer output. Should we use a
linear model to predict a child’s Gesell score from his or her age at first word? If so,
how accurate will our predictions be?
Age (months) at first word and Gesell score
Starnes-Yates5e_c03_140-205hr3.indd 185
Child
Age
Score
Child
Age
Score
Child
Age
Score
1
15
95
8
11
100
15
11
102
2
26
71
9
8
104
16
10
100
3
10
83
10
20
94
17
12
105
4
9
91
11
7
113
18
42
57
5
15
102
12
9
96
19
17
121
6
20
87
13
10
83
20
11
86
7
18
93
14
11
84
21
10
100
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CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
120
30
110
20
100
10
Residual
Score
186
90
80
70
0
10
20
60
30
50
5
10
15
20
25
30
35
Age
40
45
5
10
15
20
25
30
35
40
45
Age
Predictor
Coef
SE Coef
T
P
Constant
109.874
5.068
21.68
0.000
Age
−1.1270
0.3102
−3.63
0.002
S = 11.0229 R-Sq = 41.0% R-Sq(adj) = 37.9%
STATE: Is a linear model appropriate for these data? If so, how well does the least-squares
regression line fit the data?
PLAN: To determine whether a linear model is appropriate, we will look at the scatterplot and
residual plot to see if the association is linear or nonlinear. Then, if a linear model is appropriate,
we will use the standard deviation of the residuals and r 2 to measure how well the least-squares
line fits the data.
DO: The scatterplot shows a moderately strong, negative linear association between age at
first word and Gesell score. There are a couple of outliers in the scatterplot. Child 19 has a very
high Gesell score for his or her age at first word. Also, child 18 didn’t speak his or her first word
until much later than the other children in the study and has a much lower Gesell score. The residual plot does not have any obvious patterns, confirming what we saw in the scatterplot—a linear
model is appropriate for these data.
From the computer output, the equation of the least-squares regression line is y^ = 109.874 −
1.1270x. The standard deviation of the residuals is s = 11.0229. This means that our predictions
will typically be off by 11.0229 points when we use the linear model to predict Gesell scores from
age at first word. Finally, 41% of the variation in Gesell score is accounted for by the linear model
relating Gesell score to age at first word.
CONCLUDE: Although a linear model is appropriate for these data, our predictions might not be
very accurate. Our typical prediction error is about 11 points, and more than half of the variation in
Gesell score is still unaccounted for. Furthermore, we should be hesitant to use this model to make
predictions until we understand the effect of the two outliers on the regression results.
For Practice Try Exercise
67
Correlation and Regression Wisdom
Correlation and regression are powerful tools for describing the relationship
­between two variables. When you use these tools, you should be aware of their
limitations.
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187
Section 3.2 Least-Squares Regression
c
Example
!
n
1.​
The distinction between explanatory and response variables is impor- autio
tant in regression. Least-squares regression makes the distances of
the data points from the line small only in the y direction. If we reverse the roles of the two variables, we get a different least-squares regression
line.This isn’t true for correlation: switching x and y doesn’t affect the value
of r.
Predicting Price, Predicting Miles Driven
Two different regression lines
Figure 3.17(a) repeats the scatterplot of the Ford F-150 data with the least-squares
regression line for predicting price from miles driven. We might also use the data
on these 16 trucks to predict the number of miles driven from the price of the
truck. Now the roles of the variables are reversed: price is the explanatory variable
and miles driven is the response variable. Figure 3.17(b) shows a scatterplot of
these data with the least-squares regression line for predicting miles driven from
price. The two regression lines are very different. The standard deviations of the
residuals are different as well. In (a), the standard deviation is s = 5740 dollars, but
in (b) the standard deviation is s = 28,716 miles. However, no matter which variable we put on the x axis, the value of r2 is 66.4% and the correlation is r = −0.815.
45,000
140,000
35,000
120,000
Miles Driven
Price (in dollars)
160,000
Price 38257 0.16292 (Miles Driven)
s 5740
r2 66.4%
40,000
30,000
25,000
20,000
100,000
Miles Driven 177462 4.0772 (Price)
s 28716
r2 66.4%
80,000
60,000
15,000
40,000
10,000
20,000
0
5000
0
20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000
Miles driven
(a)
5000 10,000 15,000 20,000 25,000 30,000 35,000 40,000 45,000
Price (in dollars)
(b)
autio
!
n
2.​
Correlation and regression lines describe only linear relationships. You
can c­ alculate the correlation and the least-squares line for any relationship between two quantitative variables, but the results are useful
only if the scatterplot shows a linear pattern. Always plot your data!
c
FIGURE 3.17 (a) Scatterplot with least-squares regression line for predicting price from miles
driven. (b) Scatterplot with least-squares regression line for predicting miles driven from price.
The following four scatterplots show very different associations. Which do you
think has the highest correlation?
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CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
10
9
8
7
6
5
4
3
(a)
4
6
8
10
12
14
(b)
16
13
13
11
11
9
9
7
7
5
5
(c)
4
6
8
10
12
14
(d)
16
4
6
8
8
10
12
10
14
12
14
16
16
18
20
c
!
n
Answer: All four have the same correlation, r = 0.816. Furthermore, the leastsquares regression line for each association is exactly the same, y^ = 3 + 0.5x.
These four data sets, developed by statistician Frank Anscombe, illustrate the importance of graphing data before doing calculations.17
3.​
Correlation and least-squares regression lines are not resistant. You al- autio
ready know that the correlation r is not resistant. One unusual point
in a scatterplot can greatly change the value of r. Is the least-squares
line resistant? Not surprisingly, the answer is no.
Let’s revisit the age at first word and Gesell score data to shed some light on this
issue. The scatterplot and residual plot for these data are shown in Figure 3.18.
The two outliers, child 18 and child 19, are indicated on each plot.
30
140
Child 19
20
100
Residual
Gesell score
120
Child 19
80
60
10
0
Child 18
–10
Child 18
–20
40
0
(a)
10
20
30
40
Age at first word (months)
0
50
(b)
10
20
30
40
Age at first word (months)
50
FIGURE 3.18 ​(a) Scatterplot of ­Gesell Adaptive Scores versus the age at first word for 21 children. The line is the
least-squares regression line for predicting Gesell score from age at first word. (b) Residual plot for the regression.
Child 18 and Child 19 are outliers. Each blue point in the graphs stands for two individuals.
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Section 3.2 Least-Squares Regression
140
189
Two children
Gesell adaptive score
19
Child 19 has a very large residual
because Child
this point
lies far from the regres120
sion line. However, Child 18 has a fairly small residual. That’s because Child 18’s
point is close to the line. How do these two outliers affect the regression?
100
Without Child 18
Figure 3.19 shows the results of removing each of these points
on the correlation and the regression line. The graph adds two more regression lines, one calculated after leaving out Child8018 and the other after leaving out Child 19. You
can see that removing the point for Child 18 moves the line quite a bit. (In fact,
60
= 105.630
− 0.779x.) Because of
the equation of the new least-squares
line is y^ Without
Child 19
Child 18
Child 18’s extreme position on the age scale, this point has a strong influence on
40
the position of the regression line.
However, removing Child 19 has little effect
0
10
20
30
40
50
on the regression line.
Age at first word (months)
FIGURE 3.19 ​Three least-squares
regression lines of Gesell score
on age at first word. The green
line is calculated from all the data.
The dark blue line is calculated
leaving out Child 18. Child 18 is
an influential observation because
leaving out this point moves the
regression line quite a bit. The red
line is calculated leaving out only
Child 19.
Gesell adaptive score
140
With all 19 children:
r 0.64
yˆ 109.874 1.127x
Two children
Child 19
120
100
Without Child 19:
r 0.76
yˆ 109.305 1.193x
Without Child 18
Without Child 18:
r 0.33
yˆ 105.630 0.779x
80
60
Without Child 19
Child 18
40
0
10
20
30
40
50
Age at first word (months)
Withofallthe
19 children:
Least-squares lines make the sum
squares of the vertical distances to the
r 0.64
points as small as possible. A point that
is
extreme
in the x direction with no other
yˆ 109.874 1.127x
points near it pulls the line towardWithout
itself. Child
We call
such points influential.
19:
r 0.76
yˆ 109.305 1.193x
Definition: Outliers and influential
in regression
Withoutobservations
Child 18:
r 0.33
An outlier is an observation that lies outside
the overall pattern of the other observayˆ 105.630 0.779x
tions. Points that are outliers in the y direction but not the x direction of a scatterplot
have large residuals. Other outliers may not have large residuals.
An observation is influential for a statistical calculation if removing it would markedly change the result of the calculation. Points that are outliers in the x direction of
a scatterplot are often influential for the least-squares regression line.
We did not need the distinction between outliers and influential observations
in Chapter 1. A single large salary that pulls up the mean salary x– for a group of
workers is an outlier because it lies far above the other salaries. It is also influential, because the mean changes when it is removed. In the regression setting,
however, not all outliers are influential. The least-squares line is most likely to
be heavily influenced by observations that are outliers in the x direction. The
scatterplot will alert you to such observations. Influential points often have small
residuals, because they pull the regression line toward themselves. If you look at
just a residual plot, you may miss influential points.
The best way to verify that a point is influential is to find the regression line
both with and without the unusual point, as in Figure 3.19. If the line moves more
than a small amount when the point is deleted, the point is influential.
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CHAPTER 3
THINK
ABOUT IT
D e s c r i b i n g R e l at i o n s h i p s
How much difference can one point make? ​The strong influence
of Child 18 makes the original regression of Gesell score on age at first word
misleading. The original data have r2 = 0.41. That is, the least-squares line relating age at which a child begins to talk with Gesell score explains 41% of the variation on this later test of mental ability. This relationship is strong enough to be
interesting to parents. If we leave out Child 18, r2 drops to only 11%. The apparent
strength of the association was largely due to a single influential observation.
What should the child development researcher do? She must decide whether
Child 18 is so slow to speak that this individual should not be allowed to influence
the analysis. If she excludes Child 18, much of the evidence for a connection
between the age at which a child begins to talk and later ability score vanishes. If
she keeps Child 18, she needs data on other children who were also slow to begin
talking, so that the analysis no longer depends so heavily on just one child.
We finish with our most important caution about correlation and regression.
c
EXAMPLE
!
n
4.
Association does not imply causation. When we study the relationship autio
between two variables, we often hope to show that changes in the explanatory variable cause changes in the response variable. A strong association between two variables is not enough to draw conclusions about cause
and effect. Sometimes an observed association really does reflect cause and
effect. A household that heats with natural gas uses more gas in colder months
because cold weather requires burning more gas to stay warm. In other cases,
an association is explained by other variables, and the conclusion that x causes
y is not valid.
Does Having More Cars Make You
Live Longer?
Association, not causation
A serious study once found that people with two cars live longer than people
who own only one car.18 Owning three cars is even better, and so on. There
is a substantial positive association between number of cars x and length of
life y.
The basic meaning of causation is that by changing x, we can bring about
a change in y. Could we lengthen our lives by buying more cars? No. The
study used number of cars as a quick indicator of wealth. Well-off people
tend to have more cars. They also tend to live longer, probably because they
are better educated, take better care of themselves, and get better medical
care. The cars have nothing to do with it. There is no cause-and-effect link
between number of cars and length of life.
Associations such as those in the previous example are sometimes called “nonsense associations.” The association is real. What is nonsense is the conclusion
that changing one of the variables causes changes in the other. Another variable—
such as personal wealth in this example—that influences both x and y can create
a strong association even though there is no direct connection between x and y.
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191
Section 3.2 Least-Squares Regression
Remember: It only makes sense to
talk about the correlation between two
quantitative variables. If one or both
variables are categorical, you should
refer to the association between the
two variables. To be safe, you can use
the more general term “association”
when describing the relationship
between any two variables.
Association Does Not Imply Causation
An association between an explanatory variable x and a response variable y,
even if it is very strong, is not by itself good evidence that changes in x actually cause changes in y.
Here is a chance to use the skills you have gained to address the question posed
at the beginning of the chapter.
case closed
How Faithful Is Old Faithful?
In the chapter-opening Case Study (page 141), the Starnes family had
just missed seeing Old Faithful erupt. They wondered how long it would
be until the next eruption. The scatterplot below shows data on the duration (in minutes) and the interval of time until the next eruption (also in
minutes) for each Old Faithful eruption in the month before their visit.
120
Interval (minutes)
110
100
90
80
70
60
50
40
1
1.
2
3
4
Duration (minutes)
5
Describe the nature of the relationship between interval and
duration.
Here is some computer output from a least-squares regression
analysis on these data.
Regression Analysis: Interval versus Duration
30
Predictor
Constant
Duration
Residual
20
10
0
Coef
33.347
13.2854
SE Coef
1.201
0.3404
T
27.76
39.03
P
0.000
0.000
S = 6.49336 R-Sq = 85.4% R-Sq(adj) = 85.3%
-10
-20
1
2
3
4
Duration (minutes)
5
2.
3.
4.
5.
Starnes-Yates5e_c03_140-205hr3.indd 191
Is a linear model appropriate? Justify your answer.
Give the equation of the least-squares regression line. Be sure to
define any variables you use.
Park rangers indicated that the eruption of Old Faithful that just
finished lasted 3.9 minutes. How long do you predict the Starnes
family will have to wait for the next eruption? Show how you arrived at your answer.
The actual time that the Starnes family has to wait is probably
not exactly equal to your prediction in Question 4. Based on the
computer output, about how far off do you expect the prediction
to be? Explain.
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CHAPTER 3
Section 3.2
D e s c r i b i n g R e l at i o n s h i p s
Summary
•
•
•
•
•
•
•
•
•
•
•
A regression line is a straight line that describes how a response variable y ­changes
as an explanatory variable x changes. You can use a regression line to predict the
value of y for any value of x by substituting this x into the equation of the line.
The slope b of a regression line y^ = a + bx is the rate at which the predicted
response y^ changes along the line as the explanatory variable x changes. Specifically, b is the predicted change in y when x increases by 1 unit.
The y intercept a of a regression line y^ = a + bx is the predicted response y^
when the explanatory variable x equals 0. This prediction is of no statistical
use unless x can actually take values near 0.
Avoid extrapolation, the use of a regression line for prediction using values
of the explanatory variable outside the range of the data from which the line
was calculated.
The most common method of fitting a line to a scatterplot is least squares. The
least-squares regression line is the straight line y^ = a + bx that minimizes the
sum of the squares of the vertical distances of the observed points from the line.
You can examine the fit of a regression line by studying the residuals, which
are the differences between the observed and predicted values of y. Be on the
lookout for patterns in the residual plot, which indicate that a linear model
may not be appropriate.
The standard deviation of the residuals s measures the typical size of the
prediction errors (residuals) when using the regression line.
The coefficient of determination r2 is the fraction of the variation in the
response variable that is accounted for by least-squares regression on the explanatory variable.
The least-squares regression line of y on x is the line with slope b = r(sy/sx)
and intercept a = y– − bx–. This line always passes through the point (x–, y–).
Correlation and regression must be interpreted with caution. Plot the data to be
sure that the relationship is roughly linear and to detect outliers. Also look for influential observations, individual points that substantially change the correlation
or the regression line. Outliers in x are often influential for the regression line.
Most of all, be careful not to conclude that there is a cause-and-effect relationship between two variables just because they are strongly associated.
3.2 T echnology
Corners
TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site.
8. Least-squares regression lines on the calculator
9. Residual plots on the calculator
Starnes-Yates5e_c03_140-205hr4.indd 192
page 171
page 175
11/20/13 6:28 PM
193
Section 3.2 Least-Squares Regression
Section 3.2
Exercises
35. What’s my line? You use the same bar of soap to
shower each morning. The bar weighs 80 grams when
it is new. Its weight goes down by 6 grams per day on
average. What is the equation of the regression line
for predicting weight from days of use?
in Joan’s midwestern home. The figure below shows
the original scatterplot with the least-squares line
added. The equation of the least-squares line is
y^ = 1425 − 19.87x.
900
37. Gas mileage We expect a car’s highway gas mileage
to be related to its city gas mileage. Data for all 1198
vehicles in the government’s recent Fuel Economy
Guide give the regression line: predicted highway
mpg = 4.62 + 1.109 (city mpg).
(a) What’s the slope of this line? Interpret this value in ­context.
(b) What’s the y intercept? Explain why the value of the
intercept is not statistically meaningful.
(c) Find the predicted highway mileage for a car that gets
16 miles per gallon in the city.
38. IQ and reading scores Data on the IQ test scores
and reading test scores for a group of fifth-grade
children give the following regression line: predicted
reading score = −33.4 + 0.882(IQ score).
(a) What’s the slope of this line? Interpret this value in
context.
(b) What’s the y intercept? Explain why the value of the
intercept is not statistically meaningful.
(c) Find the predicted reading score for a child with an
IQ score of 90.
39. Acid rain Researchers studying acid rain measured
the acidity of precipitation in a Colorado wilderness
area for 150 consecutive weeks. Acidity is measured
by pH. Lower pH values show higher acidity. The
researchers observed a linear pattern over time.
They reported that the regression line pH = 5.43 −
0.0053(weeks) fit the data well.19
pg 166
(a) Identify the slope of the line and explain what it
means in this setting.
(b) Identify the y intercept of the line and explain what it
means in this setting.
(c) According to the regression line, what was the pH at
the end of this study?
40. How much gas? In Exercise 4 (page 159), we examined the relationship between the average monthly
temperature and the amount of natural gas consumed
Starnes-Yates5e_c03_140-205hr3.indd 193
Gas consumed (cubic feet)
36. What’s my line? An eccentric professor believes that
a child with IQ 100 should have a reading test score
of 50 and predicts that reading score should increase
by 1 point for every additional point of IQ. What
is the equation of the professor’s regression line for
predicting reading score from IQ?
800
700
600
500
400
300
200
30
35
40
45
50
55
Temperature (degrees Fahrenheit)
60
(a) Identify the slope of the line and explain what it
means in this setting.
(b) Identify the y intercept of the line. Explain why it’s
risky to use this value as a prediction.
(c) Use the regression line to predict the amount of
natural gas Joan will use in a month with an average
temperature of 30°F.
41. Acid rain Refer to Exercise 39. Would it be appropriate to use the regression line to predict pH after 1000
months? Justify your answer.
42. How much gas? Refer to Exercise 40. Would it be
appropriate to use the regression line to predict Joan’s
natural-gas consumption in a future month with an
average temperature of 65°F? Justify your answer.
43. Least-squares idea The table below gives a small
set of data. Which of the following two lines fits the
data better: y^ = 1 − x or y^ = 3 − 2x? Use the leastsquares criterion to justify your answer. (Note: Neither
of these two lines is the least-squares regression line
for these data.)
x:
−1
1
1
3
5
y:
2
0
1
−1
−5
44. Least-squares idea In Exercise 40, the line drawn
on the scatterplot is the least-squares regression line.
Explain the meaning of the phrase “least-squares” to
Joan, who knows very little about statistics.
45. Acid rain In the acid rain study of Exercise 39, the
actual pH measurement for Week 50 was 5.08. Find
and interpret the residual for this week.
pg 169
46. How much gas? Refer to Exercise 40. During March,
the average temperature was 46.4°F and Joan used 490
cubic feet of gas per day. Find and interpret the residual
for this month.
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47. Bird colonies Exercise 6 (page 159) examined the
relationship between the number of new birds y and
percent of returning birds x for 13 sparrowhawk colonies. Here are the data once again.
Percent return: 74 66 81 52 73 62 52 45 62 46 60 46 38
New adults:
5 6
8 11 12 15 16 17 18 18 19 20 20
(a) Use your calculator to help make a scatterplot.
(b) Use your calculator’s regression function to find the
equation of the least-squares regression line. Add this
line to your scatterplot from (a).
­ gyptian village of Nahya. Here are the mean weights
E
(in ­kilograms) for 170 infants in Nahya who were
weighed each month during their first year of life:
Age (months):
Weight (kg):
1
2
3
4
5
6
7
8
9 10 11 12
4.3 5.1 5.7 6.3 6.8 7.1 7.2 7.2 7.2 7.2 7.5 7.8
A hasty user of statistics enters the data into software and
computes the least-squares line without plotting the data.
The result is weight = 4.88 + 0.267 (age). A residual
plot is shown below. Would it be appropriate to use this
regression line to predict y from x? Justify your answer.
(c) Explain in words what the slope of the regression line
tells us.
(d) Calculate and interpret the residual for the colony
that had 52% of the sparrowhawks return and 11 new
adults.
48. Do heavier people burn more energy? Exercise
10 (page 160) presented data on the lean body mass
and resting metabolic rate for 12 women who were
subjects in a study of dieting. Lean body mass, given
in kilograms, is a person’s weight leaving out all fat.
Metabolic rate, in calories burned per 24 hours, is the
rate at which the body consumes energy. Here are the
data again.
Mass: 36.1 54.6 48.5 42.0 50.6 42.0 40.3 33.1 42.4 34.5 51.1 41.2
Rate:
995 1425 1396 1418 1502 1256 1189 913 1124 1052 1347 1204
(a) Use your calculator to help make a scatterplot.
(b) Use your calculator’s regression function to find the
equation of the least-squares regression line. Add this
line to your scatterplot from part (a).
(c) Explain in words what the slope of the regression line
tells us.
52. Driving speed and fuel consumption Exercise 9
(page 160) gives data on the fuel consumption y
of a car at various speeds x. Fuel consumption is
measured in liters of gasoline per 100 kilometers
driven and speed is measured in kilometers per hour.
A statistical software package gives the least-squares
regression line and the residual plot shown below.
The regression line is y^ = 11.058 − 0.01466x. Would
it be appropriate to use the regression line to predict y
from x? Justify your answer.
(d) Calculate and interpret the residual for the woman
who had a lean body mass of 50.6 kg and a metabolic
rate of 1502.
49. Bird colonies Refer to Exercise 47.
(a) Use your calculator to make a residual plot. Describe
what this graph tells you about the appropriateness of
using a linear model.
(b) Which point has the largest residual? Explain what
this residual means in context.
50. Do heavier people burn more energy? Refer to
Exercise 48.
(a) Use your calculator to make a residual plot. Describe
what this graph tells you about the appropriateness of
using a linear model.
(b) Which point has the largest residual? Explain what
the value of that residual means in context.
51. Nahya infant weights A study of nutrition in
­developing countries collected data from the
Starnes-Yates5e_c03_140-205hr3.indd 194
53. Oil and residuals The Trans-Alaska Oil Pipeline is
a tube that is formed from 1/2-inch-thick steel and
that carries oil across 800 miles of sensitive arctic and
subarctic terrain. The pipe segments and the welds that
join them were carefully examined before installation.
How accurate are field measurements of the depth of
small defects? The figure below compares the results
of measurements on 100 defects made in the field with
11/13/13 1:21 PM
195
Section 3.2 Least-Squares Regression
measurements of the same defects made in the laboratory.20 The line y = x is drawn on the scatterplot.
of the least-squares regression line is
y^ = −16.2 + 2.07x. Also, s = 10.2 and r2 = 0.736.
160
Free skate score
Field measurement
80
60
40
140
120
100
80
20
45
50
55
0
0
20
40
60
65
70
75
80
Short program score
80
Laboratory measurement
20
(a) Describe the overall pattern you see in the scatterplot,
as well as any deviations from that pattern.
(c) The line drawn on the scatterplot (y = x) is not the
least-squares regression line. How would the slope
and y intercept of the least-squares line compare?
Justify your answer.
54. Oil and residuals Refer to Exercise 53. The following
figure shows a residual plot for the least-squares regression line. Discuss what the residual plot tells you about
the appropriateness of using a linear model.
30
20
10
0
10
20
10
Residual
(b) If field and laboratory measurements all agree, then
the points should fall on the y = x line drawn on the
plot, except for small variations in the measurements.
Is this the case? Explain.
Residual of field measurement
60
0
10
20
30
45
50
55
60
65
70
75
80
Short program score
(a) Calculate and interpret the residual for the gold
medal winner, Yu-Na Kim, who scored 78.50 in the
short program and 150.06 in the free skate.
(b) Is a linear model appropriate for these data? Explain.
(c) Interpret the value of s.
(d) Interpret the value of r2.
56. Age and height A random sample of 195 students
was selected from the United Kingdom using the
CensusAtSchool data selector. The age (in years) x
and height (in centimeters) y was recorded for each
of the students. A regression analysis was performed
using these data. The scatterplot and residual plot
are shown below. The equation of the least-squares
regression line is y^ = 106.1 + 4.21x. Also, s = 8.61
and r2 = 0.274.
190
30
0
20
40
60
80
180
Laboratory measurement
Starnes-Yates5e_c03_140-205hr3.indd 195
Height
55. Olympic figure skating For many people, the women’s figure skating competition is the highlight of the
Olympic Winter Games. Scores in the short program
x and scores in the free skate y were recorded for each
of the 24 skaters who competed in both rounds during
the 2010 Winter Olympics in Vancouver, Canada.21
A regression analysis was performed using these data.
The scatterplot and residual plot follow. The equation
pg 180
170
160
150
140
130
10
11
12
13
14
15
16
17
Age
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D e s c r i b i n g R e l at i o n s h i p s
brain that responds to physical pain goes up as distress
from social exclusion goes up. A scatterplot shows
a moderately strong, linear relationship. The figure
below shows Minitab regression output for these data.
20
Residual
10
0
10
20
10
11
12
13
14
15
16
17
Age
(a) Calculate and interpret the residual for the student
who was 141 cm tall at age 10.
(b) Is a linear model appropriate for these data? Explain.
(c) Interpret the value of s.
(d) Interpret the value of r2.
57. Bird colonies Refer to Exercises 47 and 49. For the
regression you performed earlier, r2 = 0.56 and s = 3.67.
Explain what each of these values means in this setting.
58. Do heavier people burn more energy? Refer to Exercises 48 and 50. For the regression you performed earlier,
r2 = 0.768 and s = 95.08. Explain what each of these
values means in this setting.
59. Merlins breeding Exercise 13 (page 160) gives data on
the number of breeding pairs of merlins in an isolated
area in each of seven years and the percent of males who
returned the next year. The data show that the percent
returning is lower after successful breeding seasons and
that the relationship is roughly linear. The figure below
shows Minitab regression output for these data.
pg 181
Regression Analysis: Percent return versus Breeding pairs
Predictor
Constant
Breeding pairs
S = 7.76227
Coef
266.07
-6.650
SE Coef
52.15
1.736
R-Sq = 74.6%
T
5.10
-3.83
P
0.004
0.012
R-Sq(adj) = 69.5%
(a) What is the equation of the least-squares regression
line for predicting the percent of males that return
from the number of breeding pairs? Use the equation to predict the percent of returning males after a
season with 30 breeding pairs.
(b) What percent of the year-to-year variation in percent
of returning males is accounted for by the straightline relationship with number of breeding pairs the
previous year?
(c) Use the information in the figure to find the correlation r between percent of males that return and
number of breeding pairs. How do you know whether
the sign of r is + or −?
(d) Interpret the value of s in this setting.
60. Does social rejection hurt? Exercise 14 (page 161)
gives data from a study that shows that social exclusion
causes “real pain.” That is, activity in an area of the
Starnes-Yates5e_c03_140-205hr3.indd 196
(a) What is the equation of the least-squares regression
line for predicting brain activity from social distress
score? Use the equation to predict brain activity for
social distress score 2.0.
(b) What percent of the variation in brain activity among
these subjects is accounted for by the straight-line
relationship with social distress score?
(c) Use the information in the figure to find the correlation r between social distress score and brain activity.
How do you know whether the sign of r is + or −?
(d) Interpret the value of s in this setting.
61. Husbands and wives The mean height of married
American women in their early twenties is 64.5 inches
and the standard deviation is 2.5 inches. The mean
height of married men the same age is 68.5 inches, with
standard deviation 2.7 inches. The correlation between
the heights of husbands and wives is about r = 0.5.
pg 183
(a) Find the equation of the least-squares regression line
for predicting a husband’s height from his wife’s height
for married couples in their early 20s. Show your work.
(b) Suppose that the height of a randomly selected wife
was 1 standard deviation below average. Predict the
height of her husband without using the least-squares
line. Show your work.
62. The stock market Some people think that the behavior of the stock market in January predicts its behavior
for the rest of the year. Take the explanatory variable
x to be the percent change in a stock market index in
January and the response variable y to be the change
in the index for the entire year. We expect a positive
correlation between x and y because the change
during January contributes to the full year’s change.
Calculation from data for an 18-year period gives
x– = 1.75% ​ ​sx = 5.36% ​ y– = 9.07%
sy = 15.35% ​ ​r = 0.596
(a) Find the equation of the least-squares line for predicting
full-year change from January change. Show your work.
(b) Suppose that the percent change in a particular January was 2 standard deviations above average. Predict
the percent change for the entire year, without using
the least-squares line. Show your work.
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Section 3.2 Least-Squares Regression
63. Husbands and wives Refer to Exercise 61.
If so, more stumps should produce more beetle larvae.
Here are the data:24
(a) Find r2 and interpret this value in context.
(b) For these data, s = 1.2. Interpret this value.
Stumps:
64. The stock market Refer to Exercise 62.
Beetle larvae:
(a) Find r2 and interpret this value in context.
Stumps:
(b) For these data, s = 8.3. Interpret this value.
Beetle larvae:
65. Will I bomb the final? We expect that students
who do well on the midterm exam in a course will
usually also do well on the final exam. Gary Smith
of Pomona College looked at the exam scores of
all 346 students who took his statistics class over a
10-year period.22 Assume that both the midterm and
final exam were scored out of 100 points.
(a) State the equation of the least-squares regression line
if each student scored the same on the midterm and
the final.
(b) The actual least-squares line for predicting finalexam score y from midterm-exam score x was
y^ = 46.6 + 0.41x. Predict the score of a student who
scored 50 on the midterm and a student who scored
100 on the midterm.
(c) Explain how your answers to part (b) illustrate regression to the mean.
66. It’s still early We expect that a baseball player who
has a high batting average in the first month of the
season will also have a high batting average the rest of
the season. Using 66 Major League Baseball players
from the 2010 season,23 a least-squares regression
line was calculated to predict rest-of-season batting
average y from first-month batting average x. Note: A
player’s batting average is the proportion of times at
bat that he gets a hit. A batting average over 0.300 is
considered very good in Major League Baseball.
(a) State the equation of the least-squares regression line
if each player had the same batting average the rest of
the season as he did in the first month of the season.
(b) The actual equation of the least-squares regression line
is y^ = 0.245 + 0.109x. Predict the rest-of-season batting
average for a player who had a 0.200 batting average
the first month of the season and for a player who had a
0.400 batting average the first month of the season.
2
2
1
3
3
4
3
1
2
5
1
3
10 30 12 24 36 40 43 11 27 56 18 40
2
25
1
2
2
1
8 21 14 16
1
4
6 54
1
2
1
4
9 13 14 50
Can we use a linear model to predict the number of
beetle larvae from the number of stumps? If so, how
accurate will our predictions be? Follow the four-step
process.
STEP
68. Fat and calories The number of calories in a food
item depends on many factors, including the amount
of fat in the item. The data below show the amount
of fat (in grams) and the number of calories in 7 beef
sandwiches at McDonalds.25
4
Sandwich
Fat
Calories
Big Mac®
29
550
26
520
®
Quarter Pounder with Cheese
®
Double Quarter Pounder with Cheese
Hamburger
Cheeseburger
42
750
9
250
12
300
Double Cheeseburger
23
440
McDouble
19
390
Can we use a linear model to predict the number of
calories from the amount of fat? If so, how accurate will
our predictions be? Follow the four-step process.
69. Managing diabetes People with diabetes measure
their fasting plasma glucose (FPG; measured in units
of milligrams per milliliter) after fasting for at least
8 hours. Another measurement, made at regular
medical checkups, is called HbA. This is roughly
the percent of red blood cells that have a glucose
molecule attached. It measures average exposure to
glucose over a period of several months. The table
below gives data on both HbA and FPG for 18 diabetics five months after they had completed a diabetes
education class.27
Subject
HbA
(%)
FPG
(mg/mL)
Subject
HbA
(%)
FPG
(mg/mL)
(c) Explain how your answers to part (b) illustrate regression to the mean.
1
6.1
141
10
8.7
172
2
6.3
158
11
9.4
200
67. Beavers and beetles Do beavers benefit beetles?
Researchers laid out 23 circular plots, each 4 meters
in diameter, in an area where beavers were cutting
down cottonwood trees. In each plot, they counted the
number of stumps from trees cut by beavers and the
number of clusters of beetle larvae. Ecologists think
pg 185 that the new sprouts from stumps are more tender than
other cottonwood growth, so that beetles prefer them.
3
6.4
112
12
10.4
271
4
6.8
153
13
10.6
103
5
7.0
134
14
10.7
172
6
7.1
95
15
10.7
359
7
7.5
96
16
11.2
145
8
7.7
78
17
13.7
147
9
7.9
148
18
19.3
255
STEP
4
Starnes-Yates5e_c03_140-205hr3.indd 197
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D e s c r i b i n g R e l at i o n s h i p s
(a) Make a scatterplot with HbA as the explanatory variable. Describe what you see.
(b) Subject 18 is an outlier in the x direction. What effect
do you think this subject has on the correlation? What
effect do you think this subject has on the equation of
the least-squares regression line? Calculate the correlation and equation of the least-squares regression line
with and without this subject to confirm your answer.
(c) Subject 15 is an outlier in the y direction. What effect
do you think this subject has on the correlation? What
effect do you think this subject has on the equation of
the least-squares regression line? Calculate the correlation and equation of the least-squares regression line
with and without this subject to confirm your answer.
70. Rushing for points What is the relationship between
rushing yards and points scored in the 2011 National
Football League? The table below gives the number
of rushing yards and the number of points scored for
each of the 16 games played by the 2011 Jacksonville
Jaguars.26
Game
Rushing yards
Points scored
1
163
16
2
112
3
3
128
10
4
104
10
5
96
20
6
133
13
7
132
12
8
84
14
Multiple choice: Select the best answer for Exercises 71 to 78.
71. Which of the following is not a characteristic of the
least-squares regression line?
(a) The slope of the least-squares regression line is always
between −1 and 1.
(b) The least-squares regression line always goes through
the point (x– , y– ).
(c) The least-squares regression line minimizes the sum
of squared residuals.
(d) The slope of the least-squares regression line will
always have the same sign as the correlation.
(e) The least-squares regression line is not resistant to
outliers.
72. Each year, students in an elementary school take a
standardized math test at the end of the school year.
For a class of fourth-graders, the average score was 55.1
with a standard deviation of 12.3. In the third grade,
these same students had an average score of 61.7 with
a standard deviation of 14.0. The correlation between
the two sets of scores is r = 0.95. Calculate the equation of the least-squares regression line for predicting a
fourth-grade score from a third-grade score.
(a) y^ = 3.60 + 0.835x
(d) y^ = −11.54 + 1.08x
(b) y^ = 15.69 + 0.835x
(e) Cannot be calculated
without the data.
(c) y^ = 2.19 + 1.08x
73. Using data from the 2009 LPGA tour, a regression
analysis was performed using x = average driving
distance and y = scoring average. Using the output
from the regression analysis shown below, determine
the equation of the least-squares regression line.
9
141
17
10
108
10
11
105
13
12
129
14
Predictor
Constant
Driving Distance
13
116
41
S = 1.01216 R-Sq = 22.1% R-Sq(adj) = 21.6%
14
116
14
15
113
17
16
190
19
(a)
(b)
(c)
(d)
(e)
(a) Make a scatterplot with rushing yards as the explanatory variable. Describe what you see.
(b) The number of rushing yards in Game 16 is an outlier
in the x direction. What effect do you think this game
has on the correlation? On the equation of the leastsquares regression line? Calculate the c­ orrelation and
equation of the least-squares regression line with and
without this game to confirm your answers.
(c) The number of points scored in Game 13 is an outlier in the y direction. What effect do you think this
game has on the correlation? On the equation of the
least-squares regression line? Calculate the correlation and equation of the least-squares regression line
with and without this game to confirm your answers.
Starnes-Yates5e_c03_140-205hr3.indd 198
Coef
87.974
−0.060934
SE Coef
2.391
0.009536
T
36.78
−6.39
P
0.000
0.000
y^ = 87.947 + 2.391x
y^ = 87.947 + 1.01216x
y^ = 87.947 − 0.060934x
y^ = −0.060934 + 1.01216x
y^ = −0.060934 + 87.947x
Exercises 74 to 78 refer to the following setting.
Measurements on young children in Mumbai, India,
found this least-squares line for predicting height y from
arm span x:28
y^ = 6.4 + 0.93x
Measurements are in centimeters (cm).
74. By looking at the equation of the least-squares regression line, you can see that the correlation between
height and arm span is
(a) greater than zero.
(b) less than zero.
11/13/13 1:21 PM
Section 3.2 Least-Squares Regression
(c)
(d)
(e)
75.
(a)
(b)
(c)
(d)
(e)
76.
(a)
(b)
77.
(a)
(b)
(c)
(d)
(e)
78.
(a)
0.93.
6.4.
Can’t tell without seeing the data.
In addition to the regression line, the report on the
Mumbai measurements says that r 2 = 0.95. This
­suggests that
although arm span and height are correlated, arm
span does not predict height very accurately.
height increases by !0.95 = 0.97 cm for each additional centimeter of arm span.
95% of the relationship between height and arm span
is accounted for by the regression line.
95% of the variation in height is accounted for by the
regression line.
95% of the height measurements are accounted for by
the regression line.
One child in the Mumbai study had height 59 cm
and arm span 60 cm. This child’s residual is
​−3.2 cm.
(c) −1.3 cm.
(e) ​62.2 cm.
−2.2 cm.
(d) ​3.2 cm.
Suppose that a tall child with arm span 120 cm and
height 118 cm was added to the sample used in this
study. What effect will adding this child have on the
correlation and the slope of the least-squares regression line?
Correlation will increase, slope will increase.
Correlation will increase, slope will stay the same.
Correlation will increase, slope will decrease.
Correlation will stay the same, slope will stay the same.
Correlation will stay the same, slope will increase.
Suppose that the measurements of arm span and
height were converted from centimeters to meters
by dividing each measurement by 100. How will this
conversion affect the values of r2 and s?
r2 will increase, s will increase.
199
(b) r2 will increase, s will stay the same.
(c) r2 will increase, s will decrease.
(d) r2 will stay the same, s will stay the same.
(e) r2 will stay the same, s will decrease.
Exercises 79 and 80 refer to the following setting.
In its recent Fuel Economy Guide, the Environmental
Protection Agency gives data on 1152 vehicles. There are
a number of outliers, mainly vehicles with very poor gas
mileage. If we ignore the outliers, however, the combined
city and highway gas mileage of the other 1120 or so vehicles is approximately Normal with mean 18.7 miles per
gallon (mpg) and standard deviation 4.3 mpg.
79. In my Chevrolet (2.2) The Chevrolet Malibu with
a four-cylinder engine has a combined gas mileage of
25 mpg. What percent of all vehicles have worse gas
mileage than the Malibu?
80. The top 10% (2.2) How high must a vehicle’s gas
mileage be in order to fall in the top 10% of all
vehicles? (The distribution omits a few high outliers,
mainly hybrid gas-electric vehicles.)
81. Marijuana and traffic accidents (1.1) Researchers
in New Zealand interviewed 907 drivers at age 21.
They had data on traffic accidents and they asked the
drivers about marijuana use. Here are data on the
numbers of accidents caused by these drivers at age
19, broken down by marijuana use at the same age:29
Marijuana use per year
Never 1–10 times 11–50 times 51 ∙ times
Drivers
Accidents caused
452
229
70
156
59
36
15
50
(a) Make a graph that displays the accident rate for each
class. Is there evidence of an association between
marijuana use and traffic accidents?
(b) Explain why we can’t conclude that marijuana use
causes accidents.
FRAPPY! Free Response AP® Problem, Yay!
The following problem is modeled after actual AP® Statistics exam
free response questions. Your task is to generate a complete, concise response in 15 minutes.
Directions: Show all your work. Indicate clearly the methods
you use, because you will be scored on the correctness of your
methods as well as on the accuracy and completeness of your
results and explanations.
Starnes-Yates5e_c03_140-205hr3.indd 199
Two statistics students went to a flower shop and randomly selected 12 carnations. When they got home, the
students prepared 12 identical vases with exactly the same
amount of water in each vase. They put one tablespoon of
sugar in 3 vases, two tablespoons of sugar in 3 vases, and
three tablespoons of sugar in 3 vases. In the remaining
3 vases, they put no sugar. After the vases were prepared,
the students randomly assigned 1 carnation to each vase
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CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
and observed how many hours each flower continued to
look fresh. A scatterplot of the data is shown below.
240
230
Freshness (h)
220
210
200
190
180
170
160
0
1
2
3
Sugar (tbsp)
(a) Briefly describe the association shown in the
­scatterplot.
(b) The equation of the least-squares regression line
for these data is y^ = 180.8 + 15.8x. Interpret the
slope of the line in the context of the study.
(c) Calculate and interpret the residual for the flower
that had 2 tablespoons of sugar and looked fresh for
204 hours.
(d) Suppose that another group of students conducted
a similar experiment using 12 flowers, but included different varieties in addition to carnations.
Would you expect the value of r2 for the second
group’s data to be greater than, less than, or about
the same as the value of r2 for the first group’s data?
Explain.
After you finish, you can view two example solutions on the book’s
Web site (www.whfreeman.com/tps5e). Determine whether you
think each solution is “complete,” “substantial,” “developing,” or
“minimal.” If the solution is not complete, what improvements would
you suggest to the student who wrote it? Finally, your teacher will
provide you with a scoring rubric. Score your response and note
what, if anything, you would do differently to improve your own
score.
Chapter Review
Section 3.1: Scatterplots and Correlation
In this section, you learned how to explore the relationship
between two quantitative variables. As with distributions of
a single variable, the first step is always to make a graph. A
scatterplot is the appropriate type of graph to investigate associations between two quantitative variables. To describe a
scatterplot, be sure to discuss four characteristics: direction,
form, strength, and outliers. The direction of an association might be positive, negative, or neither. The form of
an association can be linear or nonlinear. An association is
strong if it closely follows a specific form. Finally, outliers
are any points that clearly fall outside the pattern of the rest
of the data.
The correlation r is a numerical summary that describes
the direction and strength of a linear association. When
r > 0, the association is positive, and when r < 0, the
association is negative. The correlation will always take
values between −1 and 1, with r = −1 and r = 1 indicating a perfectly linear relationship. Strong linear associations have correlations near 1 or −1, while weak linear
relationships have correlations near 0. However, it isn’t
Starnes-Yates5e_c03_140-205hr3.indd 200
possible to determine the form of an association from
only the correlation. Strong nonlinear relationships can
have a correlation close to 1 or a correlation close to 0,
depending on the association. You also learned that outliers can greatly affect the value of the correlation and
that correlation does not imply causation. That is, we
can’t assume that changes in one variable cause changes
in the other variable, just because they have a correlation close to 1 or –1.
Section 3.2: Least-Squares Regression
In this section, you learned how to use least-squares regression lines as models for relationships between variables that have a linear association. It is important to
understand the difference between the actual data and
the model used to describe the data. For example, when
you are interpreting the slope of a least-squares regression
line, describe the predicted change in the y variable. To
emphasize that the model only provides predicted values, least-squares regression lines are always expressed in
terms of y^ instead of y.
11/13/13 1:21 PM
The difference between the observed value of y and the
predicted value of y is called a residual. Residuals are the key
to understanding almost everything in this section. To find
the equation of the least-squares regression line, find the line
that minimizes the sum of the squared residuals. To see if a
linear model is appropriate, make a residual plot. If there is
no leftover pattern in the residual plot, you know the model
is appropriate. To assess how well a line fits the data, calculate
the standard deviation of the residuals s to estimate the size
of a typical prediction error. You can also calculate r2, which
measures the fraction of the variation in the y variable that is
accounted for by its linear relationship with the x variable.
You also learned how to obtain the equation of a leastsquares regression line from computer output and from
summary statistics (the means and standard deviations of
two variables and their correlation). As with the correlation,
the equation of the least-squares regression line and the values of s and r2 can be greatly influenced by outliers, so be
sure to plot the data and note any unusual values before
making any calculations.
What Did You Learn?
Learning Objective
Section
Related Example
on Page(s)
Relevant Chapter
Review Exercise(s)
Identify explanatory and response variables in situations where one
variable helps to explain or influences the other.
3.1
144
R3.4
Make a scatterplot to display the relationship between two
­quantitative variables.
3.1
145, 148
R3.4
Describe the direction, form, and strength of a relationship
­displayed in a scatterplot and recognize outliers in a scatterplot.
3.1
147, 148
R3.1
Interpret the correlation.
3.1
152
R3.3, R3.4
Understand the basic properties of correlation, including how the
correlation is influenced by outliers.
3.1
152, 156, 157
R3.1, R3.2
Use technology to calculate correlation.
3.1
Activity on 152, 171
R3.4
Explain why association does not imply causation.
3.1
Discussion on 156, 190
R3.6
Interpret the slope and y intercept of a least-squares regression line.
3.2
166
R3.2, R3.4
Use the least-squares regression line to predict y for a given x.
Explain the dangers of extrapolation.
3.2
167, Discussion on 168
(for extrapolation)
R3.2, R3.4, R3.5
Calculate and interpret residuals.
3.2
169
R3.3, R3.4
Explain the concept of least squares.
3.2
Discussion on 169
R3.5
Determine the equation of a least-squares regression line using
technology or computer output.
3.2
Technology Corner on
171, 181
R3.3, R3.4
Construct and interpret residual plots to assess whether a linear
model is appropriate.
3.2
Discussion on 175, 180
R3.3, R3.4
Interpret the standard deviation of the residuals and r and use
these values to assess how well the least-squares regression line
models the relationship between two variables.
3.2
180
R3.3, R3.5
Describe how the slope, y intercept, standard deviation of the
residuals, and r 2 are influenced by outliers.
3.2
Discussion on 188
R3.1
Find the slope and y intercept of the least-squares regression
line from the means and standard deviations of x and y and their
­correlation.
3.2
183
R3.5
2
201
Starnes-Yates5e_c03_140-205hr3.indd 201
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202
CHAPTER 3
D e s c r i b i n g R e l at i o n s h i p s
Chapter 3 Chapter Review Exercises
These exercises are designed to help you review the important
ideas and methods of the chapter.
R3.1 Born to be old? Is there a relationship between the
gestational period (time from conception to birth)
of an animal and its average life span? The figure
shows a scatterplot of the gestational period and average life span for 43 species of animals.30
A
Predictor
Constant
Age
Coef
3704
12188
S = 20870.5
R-Sq = 83.7%
B
SE Coef
8268
1492
T
0.45
8.17
P
0.662
0.000
R-Sq(adj) = 82.4%
160,000
140,000
120,000
30
100,000
Mileage
Life span (years)
40
R3.3 Stats teachers’ cars A random sample of AP® Statistics teachers was asked to report the age (in years)
and mileage of their primary vehicles. A scatterplot
of the data, a least-squares regression printout, and
a residual plot are provided below.
20
80,000
60,000
40,000
20,000
10
0
0
0
100
200
300
400
500
600
(a) Describe the association shown in the scatterplot.
(b) Point A is the hippopotamus. What effect does this
point have on the correlation, the equation of the
least-squares regression line, and the standard deviation of the residuals?
(c) Point B is the Asian elephant. What effect does this
point have on the correlation, the equation of the
least-squares regression line, and the standard deviation of the residuals?
R3.2 Penguins diving A study of king penguins looked
for a relationship between how deep the penguins
dive to seek food and how long they stay under
water.31 For all but the shallowest dives, there is a
linear relationship that is different for different penguins. The study gives a scatterplot for one penguin
titled “The Relation of Dive Duration (y) to Depth
(x).” Duration y is measured in minutes and depth
x is in meters. The report then says, “The regression
equation for this bird is: y^ = 2.69 + 0.0138x.”
(a) What is the slope of the regression line? Interpret
this value.
(b) Does the y intercept of the regression line make any
sense? If so, interpret it. If not, explain why not.
(c) According to the regression line, how long does a
typical dive to a depth of 200 meters last?
(d) Suppose that the researchers reversed the variables,
using x = dive duration and y = depth. What effect
will this have on the correlation? On the equation
of the least-squares regression line?
6
8
10
12
8
10
12
Age
700
Gestation (days)
Starnes-Yates5e_c03_140-205hr3.indd 202
4
60,000
50,000
40,000
30,000
20,000
10,000
0
10,000
20,000
30,000
Residual
0
2
0
2
4
6
Age
(a) Give the equation of the least-squares regression
line for these data. Identify any variables you use.
(b) One teacher reported that her 6-year-old car had
65,000 miles on it. Find and interpret its residual.
(c) What’s the correlation between car age and mileage? Interpret this value in context.
(d) Is a linear model appropriate for these data? Explain how you know.
(e) Interpret the values of s and r2.
R3.4 Late bloomers? Japanese cherry trees tend to blossom early when spring weather is warm and later
when spring weather is cool. Here are some data
on the average March temperature (in °C) and the
day in April when the first cherry blossom appeared
over a 24-year period:32
Temperature (°C):
Days in April
to first bloom:
4.0 5.4 3.2 2.6 4.2 4.7 4.9 4.0 4.9 3.8 4.0 5.1
Temperature (°C):
Days in April
to first bloom:
4.3 1.5 3.7 3.8 4.5 4.1 6.1 6.2 5.1 5.0 4.6 4.0
14
8 11 19 14 14 14 21
13 28 17 19 10 17
3
9 14 13 11
3 11
6
9 11
11/13/13 1:21 PM
AP® Statistics Practice Test
(a) Make a well-labeled scatterplot that’s suitable for predicting when the cherry trees will bloom from the
temperature. Which variable did you choose as the
explanatory variable? Explain.
(b) Use technology to calculate the correlation and the
equation of the least-squares regression line. Interpret the correlation, slope, and y intercept of the line
in this setting.
(c) Suppose that the average March temperature this year
was 8.2°C. Would you be willing to use the equation
in part (b) to predict the date of first bloom? Explain.
(d) Calculate and interpret the residual for the year when the
average March temperature was 4.5°C. Show your work.
(e) Use technology to help construct a residual plot. Describe what you see.
R3.5 What’s my grade? In Professor Friedman’s economics course, the correlation between the students’ total
scores prior to the final examination and their finalexamination scores is r = 0.6. The pre-exam totals for
all students in the course have mean 280 and standard deviation 30. The final-exam scores have mean
75 and standard deviation 8. Professor Friedman has
(a)
(b)
(c)
(d)
R3.6
203
lost Julie’s final exam but knows that her total before
the exam was 300. He decides to predict her finalexam score from her pre-exam total.
Find the equation for the appropriate least-squares
regression line for Professor Friedman’s prediction.
Use the least-squares regression line to predict ­Julie’s
final-exam score.
Explain the meaning of the phrase “least squares” in
the context of this question.
Julie doesn’t think this method accurately predicts
how well she did on the final exam. Determine r2.
Use this result to argue that her actual score could
have been much higher (or much lower) than the
predicted value.
Calculating achievement The principal of a high
school read a study that reported a high correlation
between the number of calculators owned by high
school students and their math achievement. Based
on this study, he decides to buy each student at his
school two calculators, hoping to improve their
math achievement. Explain the flaw in the principal’s reasoning.
Chapter 3 AP® Statistics Practice Test
Section I: Multiple Choice Select the best answer for each question.
T3.2 The British government conducts regular surveys
of household spending. The average weekly household spending (in pounds) on tobacco products and
Starnes-Yates5e_c03_140-205hr3.indd 203
alcoholic beverages for each of 11 regions in Great
Britain was recorded. A scatterplot of spending on
alcohol versus spending on tobacco is shown below.
Which of the following statements is true?
6.5
6.0
Alcohol
T3.1 A school guidance counselor examines the number
of extracurricular activities that students do and their
grade point average. The guidance counselor says,
“The evidence indicates that the correlation between
the number of extracurricular activities a student participates in and his or her grade point average is close
to zero.” A correct interpretation of this statement
would be that
(a) active students tend to be students with poor grades,
and vice versa.
(b) students with good grades tend to be students who
are not involved in many extracurricular activities,
and vice versa.
(c) students involved in many extracurricular activities are
just as likely to get good grades as bad grades; the same is
true for students involved in few extracurricular activities.
(d) there is no linear relationship between number of activities and grade point average for students at this school.
(e) involvement in many extracurricular activities and
good grades go hand in hand.
5.5
5.0
4.5
3.0
3.5
Tobacco
4.0
4.5
(a) The observation (4.5, 6.0) is an outlier.
(b) There is clear evidence of a negative association between spending on alcohol and tobacco.
(c) The equation of the least-squares line for this plot
would be approximately y^ = 10 − 2x.
(d) The correlation for these data is r = 0.99.
(e) The observation in the lower-right corner of the plot is
influential for the least-squares line.
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204
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D e s c r i b i n g R e l at i o n s h i p s
T3.3 The fraction of the variation in the values of y that is
explained by the least-squares regression of y on x is
(a) the correlation.
(b) the slope of the least-squares regression line.
(c) the square of the correlation coefficient.
(d) the intercept of the least-squares regression line.
(e) the residual.
T3.4 An AP® Statistics student designs an experiment to see
whether today’s high school students are becoming too
­calculator-dependent. She prepares two quizzes, both
of which contain 40 questions that are best done using
paper-and-pencil methods. A random sample of 30 students participates in the experiment. Each student takes
both quizzes—one with a calculator and one without—
in a random order. To analyze the data, the student constructs a scatterplot that displays the number of correct
answers with and without a calculator for each of the 30
students. A least-squares regression yields the equation
Calculator = −1.2 + 0.865(Pencil) ​ ​
r = 0.79
Which of the following statements is/are true?
I. If the student had used Calculator as the explanatory
variable, the correlation would remain the same.
II. If the student had used Calculator as the explanatory variable, the slope of the least-squares line would
remain the same.
III. The standard deviation of the number of correct answers on the paper-and-pencil quizzes was larger than
the standard deviation on the calculator quizzes.
(a)​I only
(c) ​III only
(e) ​I, II, and III
(b)​II only
(d) ​I and III only
Questions T3.5 and T3.6 refer to the following setting. Scientists examined the activity level of 7 fish at different temperatures. Fish activity was rated on a scale of 0 (no activity) to 100
(maximal activity). The temperature was measured in degrees
Celsius. A computer regression printout and a residual plot
are given below. Notice that the horizontal axis on the residual
plot is labeled “Fitted value.”
Predictor
Constant
Temperature
Coef
148.62
-3.2167
S = 4.78505
R-Sq = 91.0%
SE Coef
10.71
0.4533
T
13.88
-7.10
P
0.000
0.001
Residual
5.0
2.5
0.0
2.5
5.0
60
65
70
75
80
Fitted value
Starnes-Yates5e_c03_140-205hr3.indd 204
85
90
T3.6 Which of the following gives a correct interpretation of s in this setting?
(a) For every 1°C increase in temperature, fish activity
is predicted to increase by 4.785 units.
(b) The typical distance of the temperature readings
from their mean is about 4.785°C.
(c) The typical distance of the activity level ratings
from the least-squares line is about 4.785 units.
(d) The typical distance of the activity level readings
from their mean is about 4.785.
(e) At a temperature of 0°C, this model predicts an activity level of 4.785.
T3.7 Which of the following statements is not true of
the correlation r between the lengths in inches and
weights in pounds of a sample of brook trout?
(a) r must take a value between −1 and 1.
(b) r is measured in inches.
(c) If longer trout tend to also be heavier, then r > 0.
(d) r would not change if we measured the lengths of
the trout in centimeters instead of inches.
(e) r would not change if we measured the weights of
the trout in kilograms instead of pounds.
T3.8​When we standardize the values of a variable, the
distribution of standardized values has mean 0 and
standard d
­ eviation 1. Suppose we measure two
variables X and Y on each of several subjects. We
standardize both variables and then compute the
least-squares regression line. Suppose the slope of
the least-squares regression line is −0.44. We may
conclude that
(a) the intercept will also be −0.44.
(b) the intercept will be 1.0.
(c) the correlation will be 1/−0.44.
(d) the correlation will be 1.0.
(e) the correlation will also be −0.44.
T3.9 There is a linear relationship between the number of
chirps made by the striped ground cricket and the air
temperature. A least-squares fit of some data collected by a biologist gives the model y^ = 25.2 + 3.3x,
where x is the number of chirps per minute and y^
is the estimated temperature in degrees Fahrenheit.
What is the predicted increase in temperature for an
increase of 5 chirps per minute?
(a) 3.3°F
(c) 25.2°F
(e) 41.7°F
(b) 16.5°F
(d) 28.5°F
R-Sq(adj) = 89.2%
7.5
55
T3.5 What was the activity level rating for the fish at a
temperature of 20°C?
(a) 87 (b) 84 (c) 81 (d) 66 (e) 3
95
T3.10 A data set included the number of people per ­television
set and the number of people per physician for
40 countries. The Fathom screen shot below displays
11/13/13 1:21 PM
AP® Statistics Practice Test
205
a scatterplot of the data with the least-squares regression line added. In Ethiopia, there were 503 people
per TV and 36,660 people per doctor. What effect
would removing this point have on the regression line?
(a)
(b)
(c)
(d)
(e)
Slope would increase; y intercept would increase.
Slope would increase; y intercept would decrease.
Slope would decrease; y intercept would increase.
Slope would decrease; y intercept would decrease.
Slope and y intercept would stay the same.
Section II: Free Response ​Show all your work. Indicate clearly the methods you use, because you will be graded on
the correctness of your methods as well as on the accuracy and completeness of your results and explanations.
Age (months):
36
48
51
54
57
60
Height (cm):
86
90
91
93
94
95
(a) Make a scatterplot of these data.
(b) Using your calculator, find the equation of the leastsquares regression line of height on age.
(c) Use your regression line to predict Sarah’s height at
age 40 years (480 months). Convert your prediction
to inches (2.54 cm = 1 inch).
(d) The prediction is impossibly large. Explain why this
­happened.
T3.12 Drilling down beneath a lake in Alaska yields chemical evidence of past changes in climate. Biological
silicon, left by the skeletons of single-celled creatures
called diatoms, is a measure of the abundance of life
in the lake. A rather complex variable based on the
ratio of certain isotopes relative to ocean water gives
an indirect measure of moisture, mostly from snow.
As we drill down, we look further into the past. Here
is a scatterplot of data from 2300 to 12,000 years ago:
grass more heavily, so there are fewer fires and more
trees grow. Lions feed more successfully when there are
more trees, so the lion population increases. Researchers collected data on one part of this cycle, wildebeest
abundance (in thousands of animals) and the percent
of the grass area burned in the same year. The results of
a least-squares regression on the data are shown here.33
40
30
20
Residual
T3.11 Sarah’s parents are concerned that she seems short
for her age. Their doctor has the following record of
Sarah’s height:
10
0
-10
-20
-30
500
(a) Identify the unusual point in the scatterplot. Explain
what’s unusual about this point.
(b) If this point was removed, describe the effect on
i. the correlation.
ii. the slope and y intercept of the least-squares line.
iii. the standard deviation of the residuals.
T3.13 Long-term records from the Serengeti National Park
in Tanzania show interesting ecological relationships.
When wildebeest are more abundant, they graze the
Starnes-Yates5e_c03_140-205hr3.indd 205
750
1000
Wildebeest (1000s)
1250
1500
Predictor
Coef
SE Coef
T
P
Constant
92.29
10.06
9.17
0.000
Wildebeest (1000s)
−0.05762
0.01035
−5.56
0.000
S = 15.9880 R-Sq = 64.6% R-Sq(adj) = 62.5%
(a) Give the equation of the least-squares regression
line. Be sure to define any variables you use.
(b) Explain what the slope of the regression line means
in this setting.
(c) Find the correlation. Interpret this value in context.
(d) Is a linear model appropriate for describing the relationship between wildebeest abundance and percent
of grass area burned? Support your answer with appropriate evidence.
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206
CHAPTER 4
Designing Studies
Chapter
4
Introduction
208
Section 4.1
Sampling and Surveys
209
Section 4.2
Experiments
234
Section 4.3
Using Studies Wisely
266
Free Response AP®
Problem, YAY!
275
Chapter 4 Review
276
Chapter 4 Review Exercises
278
Chapter 4 AP® Statistics
Practice Test
279
Cumulative AP® Practice Test 1 282
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Section  
207
Designing
Studies
case study
Can Magnets Help Reduce Pain?
Early research showed that magnetic fields affected living tissue in humans. Some doctors have begun to
use magnets to treat patients with chronic pain. Scientists wondered whether this type of therapy really
worked. They designed a study to find out.
Fifty patients with chronic pain were recruited for the study. A doctor identified a painful site on each
patient and asked him or her to rate the pain on a scale from 0 (mild pain) to 10 (severe pain). Then, the
doctor selected a sealed envelope containing a magnet at random from a box with a mixture of active and
inactive magnets. That way, neither the doctor nor the patient knew which type of magnet was being used.
The chosen magnet was applied to the site of the pain for 45 minutes. After “treatment,” each patient was
again asked to rate the level of pain from 0 to 10.
In all, 29 patients were given active magnets and 21 patients received inactive magnets. Scientists decided to focus on the improvement in patients’ pain ratings. Here they are, grouped by the type of magnet
used:1
Active:
10 6 1 10 6 8 5 5 6 8 7 8 7 6 4 4 7 10 6 10 6 5 5 1 0 0 0 0 1
Inactive:
4 3 5
2 1 4 1 0 0 1 0 0 0 0 0 0 0
1 0
0 1
What do the data tell us about whether the active magnets
helped reduce pain? By the end of the chapter, you’ll be
ready to interpret the results of this study.
207
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Introduction
You can hardly go a day without hearing the results of a statistical study. Here are
some examples:
• The National Highway Traffic Safety Administration (NHTSA) reports
that seat belt use in passenger vehicles increased from 84% in 2011 to 86%
in 2012.2
• According to a recent survey, U.S. teens aged 13 to 18 spend
an average of 26.8 hours per week online. Although 59%
of the teens said that posting personal information or photos online is unsafe, 62% said they had posted photos of
­themselves.3
• A recent study suggests that lack of sleep increases the risk of
catching a cold.4
• For their final project, two AP® Statistics students showed
that listening to music while studying decreased subjects’
performance on a memory task.5
Can we trust these results? As you’ll learn in this chapter, that depends on how the
data were produced. Let’s take a closer look at where the data came from in each
of these studies.
Each year, the NHTSA conducts an observational study of seat belt use in
vehicles. The NHTSA sends trained observers to record the actual behavior of
people in vehicles at randomly selected locations across the country. The idea
of an observational study is simple: you can learn a lot just by watching. Or by
asking a few questions, as in the survey of teens’ online habits. Harris Interactive
conducted this survey using a “representative sample” of 655 U.S. 13- to 18-yearolds. Both of these studies use information from a sample to draw conclusions
about some larger population. Section 4.1 examines the issues involved in sampling and surveys.
In the sleep and catching a cold study, 153 volunteers took part. They answered
questions about their sleep habits over a two-week period. Then, researchers gave
them a virus and waited to see who developed a cold. This was a complicated
­observational study. Compare this with the experiment performed by the AP®
Statistics students. They recruited 30 students and divided them into two groups
of 15 by drawing names from a hat. Students in one group tried to memorize a list
of words while listening to music. Students in the other group tried to memorize
the same list of words while sitting in silence. Section 4.2 focuses on designing
experiments.
The goal of many statistical studies is to show that changes in one variable cause changes in another variable. In Section 4.3, we’ll look at why
establishing causation is so difficult, especially in observational studies. We’ll
also consider some of the ethical issues involved in planning and conducting
a study.
Here’s an Activity that gives you a preview of what lies ahead.
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Section 4.1 Sampling and Surveys
ACTIVITY
Materials:
Two index cards, each
with 10 distinct numbers
from 00 to 99 ­written on it
(prepared by your teacher);
clock, watch, or stopwatch
to measure 30 seconds;
and a coin for each pair of
students
209
​See no evil, hear no evil?
Confucius said, “I hear and I forget. I see and I remember. I do and I understand.”
Do people really remember what they see better than what they hear?6 In this
Activity, you will perform an experiment to try to find out.
1. Divide the class into pairs of students by drawing names from a hat.
2. Your teacher will give each pair two index cards with 10 distinct numbers
from 00 to 99 on them. Do not look at the numbers until it is time for you to do
the experiment.
3. Flip a coin to decide which of you is Student 1 and which is Student 2.
Shuffle the index cards and deal one face down to each partner.
4. Student 1 will be the first to attempt a memory task while Student 2 keeps time.
Directions: Study the numbers on the index card for 30 seconds. Then turn the
card over. Recite the alphabet aloud (A, B, C, and so on). Then tell your partner
what you think the numbers on the card are. You may not say more than 10 numbers! Student 2 will record how many numbers you recalled ­correctly.
5. Now it’s Student 2’s turn to do a memory task while Student 1 records the data.
Directions: Your partner will read the numbers on your index card aloud three
times slowly. Next, you will recite the alphabet aloud (A, B, C, and so on) and
then tell your partner what you think the numbers on the card are. You may not
say more than 10 numbers! Student 1 will record how many numbers you recalled
correctly.
6. Your teacher will scale and label axes on the board for parallel dotplots of the results. Plot how many numbers you remembered correctly on the appropriate graph.
7. Did students in your class remember numbers better when they saw them or
when they heard them? Give appropriate evidence to support your answer.
8. Based on the results of this experiment, can we conclude that people in
­general remember better when they see than when they hear? Why or why not?
4.1
What You Will Learn
•
•
•
Sampling and Surveys
By the end of the section, you should be able to:
Identify the population and sample in a statistical study.
Identify voluntary response samples and convenience
­samples. Explain how these sampling methods can lead
to bias.
Describe how to obtain a random sample using slips of
paper, technology, or a table of random digits.
Starnes-Yates5e_c04_206-285hr2.indd 209
•
•
Distinguish a simple random sample from a stratified
random sample or cluster sample. Give the advantages
and disadvantages of each sampling method.
Explain how undercoverage, nonresponse, question
wording, and other aspects of a sample survey can lead
to bias.
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CHAPTER 4
Designing Studies
Suppose we want to find out what percent of young drivers in the United States
text while driving. To answer the question, we will survey 16- to 20-year-olds
who live in the United States and drive. Ideally, we would ask them all (take a
census). But contacting every driver in this age group wouldn’t be practical: it
would take too much time and cost too much money. Instead, we put the question
to a sample chosen to represent the entire population of young drivers.
Definition: Population, census, and sample
The population in a statistical study is the entire group of individuals we want
­information about. A census collects data from every individual in the population.
A sample is a subset of individuals in the population from which we actually
­collect data.
The distinction between population and sample is basic to statistics. To make
sense of any sample result, you must know what population the sample represents.
Here’s an example that illustrates this distinction and also introduces some major
uses of sampling.
EXAMPLE
Sampling Hardwood and Humans
Populations and samples
Problem: Identify the population and the sample in each of the following settings.
(a) A furniture maker buys hardwood in large batches. The supplier is supposed to dry the wood
before shipping (wood that isn’t dry won’t hold its size and shape). The furniture maker chooses five
pieces of wood from each batch and tests their moisture content. If any piece exceeds 12% moisture
content, the entire batch is sent back.
(b) Each week, the Gallup Poll questions a sample of about 1500 adult U.S. residents to determine
national opinion on a wide variety of issues.
Solution:
(a) The population is all the pieces of hardwood in a batch. The sample is the five pieces of wood that
are selected from that batch and tested for moisture content.
(b) Gallup’s population is all adult U.S. residents. Their sample is the 1500 adults who actually
respond to the survey questions.
For Practice Try Exercise
1
The Idea of a Sample Survey
We often draw conclusions about a whole population on the basis of a sample.
Have you ever tasted a sample of ice cream and ordered a cone if the sample tastes
good? Because ice cream is fairly uniform, the single taste represents the whole.
Choosing a representative sample from a large and varied population (like all
young U.S. drivers) is not so easy. The first step in planning a sample survey is to
say exactly what population we want to describe. The second step is to say exactly
what we want to measure, that is, to give exact definitions of our variables.
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Section 4.1 Sampling and Surveys
211
We reserve the term “sample survey” for studies that use an organized plan to
choose a sample that represents some specific population, like the pieces of hardwood and the U.S. adults in the previous example. By our definition, the population in a sample survey can consist of people, animals, or things. Some people use
the terms “survey” or “sample survey” to refer only to studies in which people are
asked one or more questions, like the Gallup Poll of the last example. We’ll avoid
this more restrictive terminology.
EXAMPLE
How Does the Current Population
­Survey Work?
A sample survey
One of the most important government sample surveys in the United States is
the monthly Current Population Survey (CPS). The CPS contacts about 60,000
households each month. It produces the monthly unemployment rate and lots of
other economic and social information. To measure unemployment, we must first
specify the population we want to describe. The CPS defines its population as all
U.S. residents (legal or not) 16 years of age and over who are civilians and are not
in an institution such as a prison. The unemployment rate announced in the news
refers to this specific population.
What does it mean to be “unemployed”? Someone who is not looking for work—
for example, a full-time student—should not be called unemployed just because
she is not working for pay. If you are chosen for the CPS sample, the interviewer
first asks whether you are available to work and whether you actually looked for
work in the past four weeks. If not, you are neither employed nor unemployed—
you are not in the labor force.
If you are in the labor force, the interviewer goes on to ask about employment. If you
did any work for pay or in your own business during the week of the survey, you are
employed. If you worked at least 15 hours in a family business without pay, you are
employed. You are also employed if you have a job but didn’t work because of vacation, being on strike, or other good reason. An unemployment rate of 9.7% means
that 9.7% of the sample was unemployed, using the exact CPS definitions of both
“labor force” and “unemployed.”
The final step in planning a sample survey is to decide how to choose a sample
from the population. Let’s take a closer look at some good and not-so-good sampling methods.
How to Sample Badly
The sampling method that yields
a convenience sample is called
­convenience sampling. Other sampling
methods are named in similarly
obvious ways!
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Suppose we want to know how long students at a large high school spent doing homework last week. We might go to the school library and ask the first
30 students we see about their homework time. The sample we get is known
as a convenience sample.
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CHAPTER 4
Designing Studies
Definition: Convenience sample
Choosing individuals from the population who are easy to reach results in a
­convenience sample.
c
!
n
Convenience sampling often produces unrepresentative data. Consider autio
our sample of 30 students from the school library. It’s unlikely that this
convenience sample accurately represents the homework habits of all students at the high school. In fact, if we were to repeat this sampling process again
and again, we would almost always overestimate the average homework time in
the population. Why? Because students who hang out in the library tend to be
more studious. This is bias: using a method that favors some outcomes over others.
Definition: Bias
The design of a statistical study shows bias if it would consistently underestimate or
consistently overestimate the value you want to know.
c
autio
autio
!
n
Voluntary response samples are also
known as self-selected samples.
Bias is not just bad luck in one sample. It’s the result of a bad study
design that will consistently miss the truth about the population in the
same way. Convenience samples are almost guaranteed to show bias. So
are voluntary r­ esponse samples.
c
AP® EXAM TIP If you’re asked to describe how the design of a study leads to bias, you’re
expected to do two things: (1) identify a problem with the design, and (2) explain how this
problem would lead to an underestimate or overestimate. Suppose you were asked, “Explain
how using your statistics class as a sample to estimate the proportion of all high school
students who own a graphing calculator could result in bias.” You might respond, “This is a
convenience sample. It would probably include a much higher proportion of students with a
graphing calculator than in the population at large because a graphing calculator is required
for the statistics class. So this method would probably lead to an overestimate of the actual
population proportion.”
Definition: Voluntary response sample
Starnes-Yates5e_c04_206-285hr2.indd 212
A voluntary response sample consists of people who choose themselves by
­responding to a general invitation.
Call-in, text-in, write-in, and many Internet polls rely on voluntary
response samples. People who choose to participate in such surveys are
usually not representative of some larger population of interest. Voluntary
response samples attract people who feel strongly about an issue, and who
often share the same opinion. That leads to bias.
!
n
The Internet brings voluntary response
samples to the computer nearest you.
Visit www.misterpoll.com to become
part of the sample in any of dozens of
online polls. As the site says, “None
of these polls are ‘scientific,’ but
do represent the collective opinion
of everyone who participates.”
Unfortunately, such polls don’t tell
you anything about the views of the
population.
11/13/13 2:17 PM
Section 4.1 Sampling and Surveys
EXAMPLE
213
Illegal Immigration
Online polls
Former CNN commentator Lou Dobbs doesn’t like illegal immigration. One
of his shows was largely devoted to attacking a proposal to offer driver’s licenses
to illegal immigrants. During the show, Mr. Dobbs invited his viewers to go
to loudobbs.com to vote on the question “Would you be more or less likely to
vote for a presidential candidate who supports giving driver’s licenses to illegal
aliens? The result: 97% of the 7350 people who voted by the end of the show
said, “Less likely.”
Problem: What type of sample did Mr. Dobbs use in his poll? Explain how this sampling method
could lead to bias in the poll results.
Solution: Mr. Dobbs used a voluntary response sample: people chose to go online and
respond. Those who voted were viewers of Mr. Dobbs’s program, which means that they are likely
to support his views. The 97% poll result is probably an extreme overestimate of the percent of
people in the population who would be less likely to support a presidential candidate with this
position.
For Practice Try Exercise
9
CHECK YOUR UNDERSTANDING
For each of the following situations, identify the sampling method used. Then explain
how the sampling method could lead to bias.
1. A farmer brings a juice company several crates of oranges each week. A company
inspector looks at 10 oranges from the top of each crate before deciding whether to buy
all the oranges.
2. The ABC program Nightline once asked whether the United Nations should
continue to have its headquarters in the United States. Viewers were invited to call one
telephone number to respond “Yes” and another for “No.” There was a charge for calling
either number. More than 186,000 callers responded, and 67% said “No.”
How to Sample Well:
Simple Random Sampling
In convenience sampling, the researcher chooses easy-to-reach members of the
population. In voluntary response sampling, people decide whether to join the
sample. Both sampling methods suffer from bias due to personal choice. The best
way to avoid this problem is to let chance choose the sample. That’s the idea of
random sampling.
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CHAPTER 4
Designing Studies
Definition: Random Sampling
Random sampling involves using a chance process to determine which members
of a population are included in the sample.
In everyday life, some people use the
word “random” to mean haphazard,
as in “that’s so random.” In statistics,
random means “due to chance.”
Don’t say that a sample was chosen
at random if a chance process wasn’t
used to select the individuals.
The easiest way to choose a random sample of n people is to write their names
on identical slips of paper, put the slips in a hat, mix them well, and pull out slips
one at a time until you have n of them. An alternative would be to give each member of the population a distinct number and to use the “hat method” with these
numbers instead of people’s names. Note that this version would work just as well
if the population consisted of animals or things. The resulting sample is called a
simple random sample, or SRS for short.
Definition: Simple Random Sample (SRS)
A simple random sample (SRS) of size n is chosen in such a way that every group
of n individuals in the population has an equal chance to be selected as the sample.
An SRS gives every possible sample of the desired size an equal chance to
be chosen. It also gives each member of the population an equal chance to
be included in the sample. Picture drawing 20 slips (the sample) from a hat
containing 200 identical slips (the population). Any 20 slips have the same
chance as any other 20 to be chosen. Also, each slip has a 1-in-10 chance
(20/200) of being selected.
Some other random sampling methods give each member of the population, but not each sample, an equal chance. We’ll look at some of these
later.
How to Choose a Simple R
­ andom Sample The hat method won’t
work well if the population is large. Imagine trying to take a simple random
sample of 1000 U.S. adults! In practice, most people use random numbers generated by technology to choose samples.
EXAMPLE
Teens on the Internet
Choosing an SRS with technology
The principal at Canyon del Oro High School in Arizona wants student input
about limiting access to certain Internet sites on the school’s computers. He asks
the AP® Statistics teacher, Mr. Tabor, to select a “representative sample” of 10 students. Mr. Tabor decides to take an SRS from the 1750 students enrolled this year.
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Section 4.1 Sampling and Surveys
215
He gets an alphabetical roster from the registrar’s office, and numbers the students from 1 to 1750. Then Mr. Tabor uses the random number generator at
www.randomizer.org to choose 10 distinct numbers between 1 and 1750:
The 10 students on the roster that correspond to the chosen numbers will be on
the principal’s committee.
This example highlights the steps in choosing a simple random sample with
technology.
Choosing an SRS with Technology
Step 1: Label. Give each individual in the population a distinct numerical
label from 1 to N.
It is standard practice to use n for the
sample size and N for the population
size.
Step 2: Randomize. Use a random number generator to obtain n different
integers from 1 to N.
You can also use a graphing calculator to choose an SRS.
10. T echnology
Corner
Choosing an SRS
TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.
Let’s use a graphing calculator to select an SRS of 10 students from the Canyon del Oro High School roster.
1. Check that your calculator’s random number generator is working properly.
TI-83/84
•
Press MATH , then select PRB and randInt(.
Complete the command randInt(1,1750)and
press ENTER .
TI-89
•
Press CATALOG , then F3 (Flash Apps) and choose
randInt(. Complete the command TIStat.
randInt(1,1750)and press ENTER .
Compare your results with those of your classmates. If several students got the same number, you’ll need to seed your
calculator’s random integer generator with different numbers before you proceed. Directions for doing this are given in the
Annotated Teacher’s Edition.
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CHAPTER 4
Designing Studies
2. Randomly generate 10 distinct numbers from 1 to 1750.
Do randInt(1,1750)again. Keep pressing ENTER until you have chosen 10 different labels.
Note: If you have a TI-83/84 with OS 2.55 or later, you can use the command RandIntNoRep(1,1750) to sort the
numbers from 1 to 1750 in random order. The first 10 numbers listed give the labels of the chosen students.
If you don’t have technology handy, you can use a table of random digits to
choose an SRS. We have provided a table of random digits at the back of the book
(Table D). Here is an excerpt.
Table D Random digits
Line
101
19223
95034
05756
28713
96409
12531
42544
82853
102
73676
47150
99400
01927
27754
42648
82425
36290
103
45467
71709
77558
00095
32863
29485
82226
90056
You can think of this table as the result of someone putting the digits 0 to 9 in a
hat, mixing, drawing one, replacing it, mixing again, drawing another, and so on.
The digits have been arranged in groups of five within numbered rows to make
the table easier to read. The groups and rows have no special meaning—Table D
is just a long list of randomly chosen digits. As with technology, there are two steps
in using Table D to choose a random sample.
How to Choose an SRS Using Table D
Step 1: Label. Give each member of the population a numerical label with
the same number of digits. Use as few digits as possible.
Step 2: Randomize. Read consecutive groups of digits of the appropriate
length from left to right across a line in Table D. Ignore any group of digits
that wasn’t used as a label or that duplicates a label already in the sample.
Stop when you have chosen n different labels.
Your sample contains the individuals whose labels you find.
Always use the shortest labels that will cover your population. For instance,
you can label up to 100 individuals with two digits: 01, 02, . . . , 99, 00. As standard practice, we recommend that you begin with label 1 (or 01 or 001 or 0001,
as needed). Reading groups of digits from the table gives all individuals the same
chance to be chosen because all labels of the same length have the same chance
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217
Section 4.1 Sampling and Surveys
to be found in the table. For example, any pair of digits in the table is equally
likely to be any of the 100 possible labels 01, 02, . . . , 99, 00. Here’s an example
that shows how this process works.
EXAMPLE
Spring Break!
Choosing an SRS with Table D
The school newspaper is planning an article on family-friendly
places to stay over spring break at a nearby beach town. The
editors intend to call 4 randomly chosen hotels to ask about their
amenities for families with children. They have an alphabetized
list of all 28 hotels in the town.
Problem: Use Table D at line 130 to choose an SRS of 4 hotels for the
editors to call.
Solution: We’ll use the two-step process for selecting an SRS using
Table D.
Step 1: Label. Two digits are needed to label the 28 hotels. We have added
labels 01 to 28 to the alphabetized list of hotels below.
01
02
03
04
05
06
07
Aloha Kai
Anchor Down
Banana Bay
Banyan Tree
Beach Castle
Best Western
Cabana
08
09
10
11
12
13
14
Captiva
Casa del Mar
Coconuts
Diplomat
Holiday Inn
Lime Tree
Outrigger
15
16
17
18
19
20
21
Palm Tree
Radisson
Ramada
Sandpiper
Sea Castle
Sea Club
Sea Grape
22
23
24
25
26
27
28
Sea Shell
Silver Beach
Sunset Beach
Tradewinds
Tropical Breeze
Tropical Shores
Veranda
Step 2: Randomize. To use Table D, start at the left-hand side of line 130 and read two-digit
groups. Skip any groups that aren’t between 01 and 28, as well as any repeated groups. Continue
until you have chosen four hotels. Here is the beginning of line 130:
69051 64817 87174 09517 84534 06489 87201 97245
The first 10 two-digit groups are
69
Skip
Too big
05
✓
16
✓
48
Skip
Too big
17
✓
87
17
40
Skip
Skip
Skip
Too big Repeat Too big
95
17
Skip
Skip
Too big Repeat
We skip 5 of these 10 groups because they are too high (over 28) and 2 because they are repeats
(both 17s). The hotels labeled 05, 16, and 17 go into the sample. We need one more hotel to complete the sample. Continuing along line 130:
84
Skip
Too big
53
Skip
Too big
40
Skip
Too big
64
Skip
Too big
89
Skip
Too big
87
Skip
Too big
20
✓
Our SRS of 4 hotels for the editors to contact is 05 Beach Castle, 16 Radisson, 17 Ramada, and
20 Sea Club.
For Practice Try Exercise
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CHAPTER 4
Designing Studies
We can trust results from an SRS, as well as from other types of random samples
that we will meet later, because the use of impersonal chance avoids bias. The
following activity shows why random sampling is so important.
ACTIVITY
Who Wrote the Federalist Papers?
The Federalist Papers are a series of 85 essays supporting the ratification of the U.S.
Constitution. At the time they were published, the identity of the authors was a
secret known to just a few people. Over time, however, the authors were identified as Alexander Hamilton, James Madison, and John Jay. The authorship
of 73 of the essays is fairly certain, leaving 12 in dispute. However, thanks
in some part to statistical a­ nalysis,7 most scholars now believe that the 12
disputed essays were written by Madison alone or in collaboration with
Hamilton.8
There are several ways to use statistics to help determine the authorship
of a disputed text. One method is to estimate the average word length in
a disputed text and compare it to the average word lengths of works where
the authorship is not in dispute.
The following passage is the opening paragraph of Federalist Paper #51,9
one of the disputed essays. The theme of this essay is the separation of powers
between the three branches of government.
To what expedient, then, shall we finally resort, for maintaining in practice the
necessary partition of power among the several departments, as laid down in
the Constitution? The only answer that can be given is, that as all these exterior
provisions are found to be inadequate, the defect must be supplied, by so
contriving the interior structure of the government as that its several constituent
parts may, by their mutual relations, be the means of keeping each other in
their proper places. Without presuming to undertake a full development of this
important idea, I will hazard a few general observations, which may perhaps
place it in a clearer light, and enable us to form a more correct judgment of the
principles and structure of the government planned by the convention.
1. Choose 5 words from this passage. Count the number of letters in each of the
words you selected, and find the average word length.
2. Your teacher will draw and label a horizontal axis for a class dotplot. Plot the
average word length you obtained in Step 1 on the graph.
3. Use a table of random digits or a random number generator to select a simple
random sample of 5 words from the 130 words in the opening passage. Count
the number of letters in each of the words you selected, and find the average
word length.
4. Your teacher will draw and label another horizontal axis with the same scale
for a comparative class dotplot. Plot the average word length you obtained in
Step 3 on the graph.
5. How do the dotplots compare? Can you think of any reasons why they might
be different? Discuss with your classmates.
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Other Random Sampling Methods
The basic idea of sampling is straightforward: take an SRS from the population
and use your sample results to gain information about the population. Unfortunately, it’s usually difficult to get an SRS from the population of interest. Imagine
trying to get a simple random sample of all the batteries produced in one day at a
factory. Or an SRS of all U.S. high school students. In either case, it would be difficult to obtain an accurate list of the population from which to draw the sample.
It would also be very time-consuming to collect data from each individual that’s
randomly selected. Sometimes, there are also statistical advantages to using more
complex sampling methods.
One of the most common alternatives to an SRS involves sampling groups
(strata) of similar individuals within the population separately. Then these separate “subsamples” are combined to form one stratified random sample.
Stratum is singular. Strata are plural.
Definition: Stratified random sample and strata
To get a stratified random sample, start by classifying the population into groups
of similar individuals, called strata. Then choose a separate SRS in each stratum
and combine these SRSs to form the sample.
Choose the strata based on facts known before the sample is taken. For example,
in a study of sleep habits on school nights, the population of students in a large high
school might be divided into freshman, sophomore, junior, and senior strata. In a preelection poll, a population of election districts might be divided into urban, suburban,
and rural strata. Stratified random sampling works best when the individuals within
each stratum are similar with respect to what is being measured and when there are
large differences between strata. The following Activity makes this point clear.
ACTIVITY
MATERIALS:
Calculator for each student
Sampling sunflowers
A British farmer grows sunflowers for making sunflower oil. Her field is a­ rranged
in a grid pattern, with 10 rows and 10 columns as shown in the ­figure on the next
page. Irrigation ditches run along the top and bottom of the field. The farmer
would like to estimate the number of healthy plants in the field so she can project
how much money she’ll make from selling them. It would take too much time
to count the plants in all 100 squares, so she’ll accept an estimate based on a
sample of 10 squares.
1. ​Use Table D or technology to take a simple random sample of 10 grid
squares. Record the location (for example, B6) of each square you select.
2. ​This time, you’ll take a stratified random sample using the rows as strata.
Use Table D or technology to randomly select one square from each (horizontal) row. Record the location of each square—for example, Row 1: G,
Row 2: B, and so on.
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3. ​Now, take a stratified random sample using the columns as strata.
Use Table D or technology to randomly select one square from each
(vertical) column. Record the location of each square—for example,
Column A: 4, Column B: 1, and so on.
4. ​The table on page N/DS-5 in the back of the book gives the actual
number of sunflowers in each grid square. Use the information provided to calculate your estimate of the mean number of sunflowers per
square for each of your samples in Steps 1, 2, and 3.
5. ​Make comparative dotplots showing the mean number of sunflowers obtained using the three different sampling methods for all members of the class. Describe any similarities and differences you see.
6. ​Your teacher will provide you with the mean number of sunflowers
in the population of all 100 grid squares in the field. How did the three
sampling methods do?
The dotplots below show the mean number of healthy plants in 100 samples using each of the three sampling methods in the Activity: simple random sampling,
stratified random sampling with rows of the field as strata, and stratified random
sampling with columns of the field as strata. Notice that all three distributions are
centered at about 102.5, the true mean number of healthy plants in all squares of
the field. That makes sense because random sampling yields accurate estimates of
unknown population values.
One other detail stands out in the graphs. There is much less variability in the
estimates using stratified random sampling with the rows as strata. The table on
page N/DS-5 shows the actual number of healthy sunflowers in each grid square.
Notice that the squares within each row contain a similar number of healthy
plants but there are big differences between rows. When we can choose strata that
are “similar within but different between,” stratified random samples give more precise estimates than simple random samples of the same size.
Why didn’t using the columns as strata reduce the variability of the estimates
in a similar way? Because the squares within each column have very different
numbers of healthy plants.
Both simple random sampling and stratified random sampling are hard to use
when populations are large and spread out over a wide area. In that situation, we’d
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prefer a method that selects groups (clusters) of individuals that are “near” one
another. That’s the idea of a cluster sample.
Definition: Cluster sample and clusters
To get a cluster sample, start by classifying the population into groups of individuals
that are located near each other, called clusters. Then choose an SRS of the clusters. All individuals in the chosen clusters are included in the sample.
Remember: strata are ideally “similar
within, but different between,” while
­clusters are ideally “different within,
but similar between.”
EXAMPLE
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Cluster samples are often used for practical reasons, like saving time and
­ oney. Cluster sampling works best when the clusters look just like the populam
tion but on a smaller scale. Imagine a large high school that assigns its students to
homerooms alphabetically by last name. The school administration is considering a new schedule and would like student input. Administrators decide to survey
200 randomly selected students. It would be difficult to track down an SRS of 200
students, so the administration opts for a cluster sample of homerooms. The principal (who knows some statistics) takes a simple random sample of 8 homerooms
and gives the survey to all 25 students in each homeroom.
Cluster samples don’t offer the statistical advantage of better information
about the population that stratified random samples do. That’s because clusters
are often chosen for ease so they may have as much variability as the population
itself.
Be sure you understand the difference between strata and clusters. We autio
want each stratum to contain similar individuals and for there to be large
differences between strata. For a cluster sample, we’d like each cluster to
look just like the population, but on a smaller scale. Here’s an example that compares the random sampling methods we have ­discussed so far.
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In a cluster sample, some people take
an SRS from each cluster rather than
including all members of the cluster.
Sampling at a School Assembly
Strata or clusters?
The student council wants to conduct a survey during the first
five minutes of an all-school assembly in the auditorium about
use of the school library. They would like to announce the results of the survey at the end of the assembly. The student council president asks your statistics class to help carry out the survey.
Problem: There are 800 students present at the assembly. A map of
the auditorium is shown on the next page. Note that students are seated by
grade level and that the seats are numbered from 1 to 800.
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12th grade: Seats 1– 200
Describe how you would use your calculator to select 80 students to complete the survey with each
of the following:
(a) ​Simple random sample
(b) Stratified random sample
(c) Cluster sample
Solution:
Note that cluster sampling is much
more efficient than finding 80 seats
scattered about the auditorium,
as required by both of the other
sampling methods.
(a) To take an SRS, we need to choose 80 of the seat numbers at random. Use randInt(1,800) on your
calculator until 80 different seats are selected. Then give the survey to the students in those seats.
(b) The students in the assembly are seated by grade level. Because students’ library use might be
similar within grade levels but different across grade levels, we’ll use the grade level seating areas as
our strata. Within each grade’s seating area, we’ll select 20 seats at random. For the 9th grade, use
randInt(601,800) to select 20 different seats. Use randInt(401,600) to pick 20 different sophomore seats, randInt(201,400) to get 20 different junior seats, and randInt(1,200) to choose 20
different senior seats. Give the survey to the students in the selected seats.
(c) With the way students are seated, each column of seats from the stage to the back of the auditorium could be used as a cluster. Note that each cluster contains students from all four grade levels, so
each should represent the population well. Because there are 20 clusters, each with 40 seats, we need
to choose 2 clusters at random to get 80 students for the survey. Use randInt(1,20) to select two
clusters, and then give the survey to all 40 students in each column of seats.
For Practice Try Exercise
21
Most large-scale sample surveys use multistage samples that combine two or more
sampling methods. For example, the U.S. Census Bureau carries out a monthly
Current Population Survey (CPS) of about 60,000 households. Researchers start by
choosing a stratified random sample of neighborhoods in 756 of the 2007 geographical areas in the United States. Then they divide each neighborhood into clusters of
four nearby households and select a cluster sample to interview.
Analyzing data from sampling methods more complex than an SRS takes us
beyond basic statistics. But the SRS is the building block of more elaborate methods, and the principles of analysis remain much the same for these other methods.
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CHECK YOUR UNDERSTANDING
The manager of a sports arena wants to learn more about the financial status of the people
who are attending an NBA basketball game. He would like to give a survey to a representative sample of the more than 20,000 fans in attendance. Ticket prices for the game
vary a great deal: seats near the court cost over $100 each, while seats in the top rows
of the arena cost $25 each. The arena is divided into 30 numbered sections, from
101 to 130. Each section has rows of seats labeled with letters from A (nearest the
court) to ZZ (top row of the arena).
1. Explain why it might be difficult to give the survey to an SRS of 200 fans.
2. Which would be a better way to take a stratified random sample of fans: using
the lettered rows or the numbered sections as strata? Explain.
3. Which would be a better way to take a cluster sample of fans: using the
lettered rows or the numbered sections as clusters? Explain.
Inference for Sampling
The purpose of a sample is to give us information about a larger population. The
process of drawing conclusions about a population on the basis of sample data is
called inference because we infer information about the population from what we
know about the sample.
Inference from convenience samples or voluntary response samples would be
misleading because these methods of choosing a sample are biased. We are almost
certain that the sample does not fairly represent the population. The first reason to
rely on random sampling is to avoid bias in choosing a sample.
Still, it is unlikely that results from a random sample are exactly the same as for
the entire population. Sample results, like the unemployment rate obtained from
the monthly Current Population Survey, are only estimates of the truth about the
population. If we select two samples at random from the same population, we will
almost certainly choose different individuals. So the sample results will differ somewhat, just by chance. Properly designed samples avoid systematic bias. But their
results are rarely exactly correct, and we expect them to vary from sample to sample.
EXAMPLE
Going to class
How much do sample results vary?
Suppose that 70% of the students in a large university attended all their classes last
week. Imagine taking a simple random sample of 100 students and recording the
proportion of students in the sample who went to every class last week. Would the
sample proportion be exactly 0.70? Probably not. Would the sample proportion
be close to 0.70? That depends on what we mean by “close.” The following graph
shows the results of taking 500 SRSs, each of size 100, and recording the proportion of students who attended all their classes in each sample.
What do we see? The graph is centered at about 0.70, the population proportion.
All of the sample proportions fall between 0.55 and 0.85. So we shouldn’t be surprised if the difference between the sample proportion and the population proportion is as large as 0.15. The graph also has a very distinctive “bell shape.”
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Dotplot of the sample proportion of students in each of 500 SRSs of size 100
who attended all their classes last week.
The population proportion is 0.70.
Why can we trust random samples? As the previous example illustrates, the results
of random sampling don’t change haphazardly from sample to sample. Because we
deliberately use chance, the results obey the laws of probability that govern chance
behavior. These laws allow us to say how likely it is that sample results are close to
the truth about the population. The second reason to use random sampling is that
the laws of probability allow trustworthy inference about the population. Results
from random samples come with a “margin of error” that sets bounds on the size of
the likely error. We will discuss the details of inference for sampling later.
One point is worth making now: larger random samples give better information
about the population than smaller samples. For instance, let’s look at what happens if we increase the sample size in the example from 100 to 400 students. The
dotplot below shows the results of taking 500 SRSs, each of size 400, and recording
the proportion of students who attended all their classes in each sample. This graph
is also centered at about 0.70. But now all the sample proportions fall between 0.63
and 0.77. So the difference between the sample proportion and the population proportion is at most 0.07. When using SRSs of size 100, this difference could be as
much as 0.15. The moral of the story: by taking a very large random sample, you can
be confident that the sample result is very close to the truth about the population.
Dotplot of the sample proportion of students in each of 500 SRSs of size 400
who attended all their classes last week.
The population proportion is 0.70.
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The Current Population Survey contacts about 60,000 households, so we’d
expect its estimate of the national unemployment rate to be within about 0.1%
of the actual population value. Opinion polls that contact 1000 or 1500 people
give less precise results—we expect the sample result to be within about 3% of
the actual population percent with a given opinion. Of course, only samples
chosen by chance carry this guarantee. Lou Dobbs’s online sample tells us
little about overall American public opinion even though 7350 people clicked
a response.
Sample Surveys: What Can Go Wrong?
The list of individuals from which a
sample will be drawn is called the
sampling frame.
The use of bad sampling methods (convenience or voluntary response) often
leads to bias. Researchers can avoid bad methods by using random sampling to
choose their samples. Other problems in conducting sample surveys are more
difficult to avoid.
Sampling is often done using a list of individuals in the population. Such lists
are seldom accurate or complete. The result is undercoverage.
DEFINITION: Undercoverage
Undercoverage occurs when some members of the population cannot be chosen in
a sample.
Most samples suffer from some degree of undercoverage. A sample survey of
households, for example, will miss not only homeless people but also prison inmates and students in dormitories. An opinion poll conducted by calling landline
telephone numbers will miss households that have only cell phones as well as
households without a phone. The results of national sample surveys therefore
have some bias due to undercoverage if the people not covered differ from the
rest of the population.
Well-designed sample surveys avoid bias in the sampling process. The real
problems start after the sample is chosen.
One of the most serious sources of bias in sample surveys is nonresponse.
Definition: Nonresponse
Nonresponse occurs when an individual chosen for the sample can’t be contacted
or refuses to participate.
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Nonresponse to surveys often exceeds 50%, even with c