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This page intentionally left blank 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 Starnes-Yates5e_FM_endpp.hr.indd 1 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. Starnes-Yates5e_fm_i-xxiii_hr.indd 1 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 2 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 3 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 4 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 5 11/20/13 7:43 PM 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. vi Starnes-Yates5e_fm_i-xxiii_hr2.indd 6 12/2/13 5:46 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 7 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 8 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 9 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 10 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 11 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 12 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 13 11/20/13 7:43 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 21 11/20/13 7:44 PM 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 22 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 Starnes-Yates5e_fm_i-xxiii_hr.indd 23 11/20/13 7:44 PM 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 24 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 1 11/13/13 1:06 PM 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 11/13/13 1:06 PM 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 11/13/13 1:06 PM 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 11/13/13 1:06 PM 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. Starnes-Yates5e_c01_xxiv-081hr3.indd 7 11/13/13 2:47 PM 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 8 11/13/13 1:07 PM 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 9 11/13/13 1:07 PM 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 10 11/13/13 1:07 PM 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. Starnes-Yates5e_c01_xxiv-081hr3.indd 11 11/13/13 1:07 PM 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. Starnes-Yates5e_c01_xxiv-081hr3.indd 12 11/13/13 1:07 PM 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. Starnes-Yates5e_c01_xxiv-081hr3.indd 13 11/13/13 1:07 PM 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. 11/13/13 1:07 PM 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 15 Male 597 = 55.1% 1083 11/13/13 1:07 PM 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. Starnes-Yates5e_c01_xxiv-081hr3.indd 16 11/13/13 1:07 PM 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 17 Male FIGURE 1.7 Segmented bar graph comparing the opinions of males and females. 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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. Starnes-Yates5e_c01_xxiv-081hr3.indd 37 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 55 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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). Starnes-Yates5e_c01_xxiv-081hr3.indd 59 11/13/13 1:07 PM 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 Starnes-Yates5e_c01_xxiv-081hr3.indd 60 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. Starnes-Yates5e_c01_xxiv-081hr3.indd 61 11/13/13 1:07 PM 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. Starnes-Yates5e_c01_xxiv-081hr3.indd 63 11/13/13 1:07 PM 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 11/13/13 1:07 PM 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 11/13/13 2:19 PM 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 11/13/13 1:07 PM 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 11/13/13 1:08 PM 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 Starnes-Yates5e_c02_082-139hr2.indd 83 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. Starnes-Yates5e_c02_082-139hr2.indd 85 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. Starnes-Yates5e_c02_082-139hr2.indd 87 11/13/13 1:33 PM 88 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 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 Starnes-Yates5e_c02_082-139hr2.indd 88 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. 11/13/13 1:33 PM 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. Starnes-Yates5e_c02_082-139hr2.indd 89 11/13/13 1:33 PM 90 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 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. Starnes-Yates5e_c02_082-139hr2.indd 90 11/13/13 1:33 PM 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 Starnes-Yates5e_c02_082-139hr2.indd 91 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? 11/13/13 1:33 PM 92 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 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. Starnes-Yates5e_c02_082-139hr2.indd 92 11/13/13 1:33 PM 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. Starnes-Yates5e_c02_082-139hr2.indd 93 11/13/13 1:33 PM 94 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 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. Starnes-Yates5e_c02_082-139hr2.indd 94 11/13/13 1:33 PM 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. Starnes-Yates5e_c02_082-139hr2.indd 95 11/13/13 1:33 PM 96 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 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 Starnes-Yates5e_c02_082-139hr2.indd 96 11/13/13 1:33 PM 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. Starnes-Yates5e_c02_082-139hr2.indd 97 11/13/13 1:33 PM 98 60 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 Starnes-Yates5e_c02_082-139hr2.indd 107 8 11/13/13 1:33 PM 108 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 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. 11/13/13 1:33 PM 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 Starnes-Yates5e_c02_082-139hr2.indd 109 11/13/13 1:33 PM 110 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 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. Starnes-Yates5e_c02_082-139hr2.indd 110 11/13/13 1:33 PM 111 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 Starnes-Yates5e_c02_082-139hr2.indd 111 43 11/13/13 1:33 PM 112 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 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 Starnes-Yates5e_c02_082-139hr2.indd 112 11/13/13 1:33 PM 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 Starnes-Yates5e_c02_082-139hr2.indd 113 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. 11/13/13 1:33 PM 114 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 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 11/13/13 1:33 PM 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 Starnes-Yates5e_c02_082-139hr2.indd 115 49 11/13/13 1:33 PM 116 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 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. Starnes-Yates5e_c02_082-139hr2.indd 116 11/13/13 1:33 PM 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. Starnes-Yates5e_c02_082-139hr2.indd 117 11/13/13 1:33 PM 118 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 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) 11/13/13 1:33 PM 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 Starnes-Yates5e_c02_082-139hr2.indd 119 53(b) 11/13/13 1:33 PM 120 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 Starnes-Yates5e_c02_082-139hr2.indd 120 11/13/13 1:33 PM 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. Starnes-Yates5e_c02_082-139hr2.indd 121 11/13/13 1:33 PM 122 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. Starnes-Yates5e_c02_082-139hr2.indd 122 12/2/13 4:38 PM 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. Starnes-Yates5e_c02_082-139hr2.indd 123 11/13/13 1:33 PM 124 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 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 Starnes-Yates5e_c02_082-139hr2.indd 124 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. Starnes-Yates5e_c02_082-139hr2.indd 125 11/13/13 1:33 PM 126 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 11/13/13 1:33 PM 128 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. Starnes-Yates5e_c02_082-139hr2.indd 128 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? 11/13/13 1:33 PM 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 11/13/13 1:33 PM 130 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 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). 11/13/13 1:33 PM 132 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 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. 11/13/13 1:33 PM 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? 11/13/13 1:33 PM 134 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 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 11/13/13 1:33 PM 136 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. 11/13/13 1:33 PM 138 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 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 11/13/13 1:20 PM 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 Starnes-Yates5e_c03_140-205hr3.indd 141 11/13/13 1:20 PM 142 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 Starnes-Yates5e_c03_140-205hr3.indd 142 11/13/13 1:20 PM 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. Starnes-Yates5e_c03_140-205hr3.indd 143 11/13/13 1:20 PM 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. Starnes-Yates5e_c03_140-205hr3.indd 144 11/13/13 1:20 PM 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 Starnes-Yates5e_c03_140-205hr3.indd 145 11/13/13 1:20 PM 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 Starnes-Yates5e_c03_140-205hr3.indd 147 7 11/13/13 1:20 PM 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 Starnes-Yates5e_c03_140-205hr3.indd 148 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? 11/13/13 1:20 PM 150 CHAPTER 3 7. T ECHNOLOGY CORNER 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 Starnes-Yates5e_c03_140-205hr3.indd 151 11/13/13 1:20 PM 152 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. Starnes-Yates5e_c03_140-205hr3.indd 152 11/13/13 1:20 PM 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 11/13/13 1:20 PM 154 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. Starnes-Yates5e_c03_140-205hr3.indd 154 11/13/13 1:20 PM 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. Starnes-Yates5e_c03_140-205hr3.indd 155 11/13/13 1:20 PM 156 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 Starnes-Yates5e_c03_140-205hr3.indd 156 ! 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. 11/13/13 1:20 PM 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? Starnes-Yates5e_c03_140-205hr3.indd 157 11/13/13 1:20 PM 158 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 Starnes-Yates5e_c03_140-205hr4.indd 158 page 150 11/20/13 6:25 PM 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 11/13/13 1:20 PM 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 Starnes-Yates5e_c03_140-205hr3.indd 161 (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. 11/13/13 1:20 PM CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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. 11/13/13 1:20 PM 163 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. 11/13/13 1:20 PM 164 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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. Starnes-Yates5e_c03_140-205hr3.indd 164 11/13/13 1:20 PM 165 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. Starnes-Yates5e_c03_140-205hr3.indd 165 11/13/13 1:20 PM 166 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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. Starnes-Yates5e_c03_140-205hr3.indd 166 11/13/13 1:20 PM 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! Starnes-Yates5e_c03_140-205hr3.indd 167 11/13/13 1:20 PM 168 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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 Starnes-Yates5e_c03_140-205hr3.indd 168 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. 11/13/13 1:20 PM 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. Starnes-Yates5e_c03_140-205hr3.indd 169 11/13/13 1:20 PM CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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. Starnes-Yates5e_c03_140-205hr3.indd 170 11/13/13 1:20 PM 171 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. Starnes-Yates5e_c03_140-205hr4.indd 171 11/20/13 6:26 PM 172 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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. Starnes-Yates5e_c03_140-205hr3.indd 172 11/13/13 1:20 PM 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. Starnes-Yates5e_c03_140-205hr3.indd 173 −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. 11/13/13 1:20 PM 174 (a) CHAPTER 3 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. Starnes-Yates5e_c03_140-205hr3.indd 174 11/13/13 1:20 PM 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! Starnes-Yates5e_c03_140-205hr4.indd 175 11/20/13 6:27 PM 176 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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? Starnes-Yates5e_c03_140-205hr3.indd 176 11/13/13 1:20 PM 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 = Starnes-Yates5e_c03_140-205hr3.indd 177 ∙(x − x–) i Å 2 n−1 11/13/13 1:20 PM 178 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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. ∙ ∙ Starnes-Yates5e_c03_140-205hr3.indd 178 11/13/13 1:20 PM 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. Starnes-Yates5e_c03_140-205hr3.indd 179 11/13/13 1:20 PM 180 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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 Starnes-Yates5e_c03_140-205hr3.indd 180 30 35 15 20 25 30 35 Points per game 11/13/13 1:21 PM 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 Starnes-Yates5e_c03_140-205hr3.indd 181 11/13/13 1:21 PM 182 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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 Starnes-Yates5e_c03_140-205hr3.indd 182 11/13/13 1:21 PM 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 Starnes-Yates5e_c03_140-205hr3.indd 183 61(a) 11/13/13 1:21 PM 184 CHAPTER 3 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 11/13/13 1:21 PM 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. Starnes-Yates5e_c03_140-205hr3.indd 186 11/13/13 1:21 PM 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? Starnes-Yates5e_c03_140-205hr3.indd 187 11/13/13 1:21 PM 188 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. Starnes-Yates5e_c03_140-205hr3.indd 188 11/13/13 1:21 PM 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. Starnes-Yates5e_c03_140-205hr3.indd 189 11/13/13 1:21 PM 190 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. Starnes-Yates5e_c03_140-205hr3.indd 190 11/13/13 1:21 PM 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. 11/13/13 1:21 PM 192 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. 11/13/13 1:21 PM 194 CHAPTER 3 D e s c r i b i n g R e l at i o n s h i p s 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 11/13/13 1:21 PM 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 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. 11/13/13 1:21 PM 197 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 11/13/13 1:21 PM 198 CHAPTER 3 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 11/13/13 1:21 PM 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 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 11/13/13 1:21 PM 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. 11/13/13 1:21 PM 204 CHAPTER 3 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. T3.8When 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. 11/13/13 1:21 PM 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 Starnes-Yates5e_c04_206-285hr3.indd 206 11/20/13 6:30 PM 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 Starnes-Yates5e_c04_206-285hr2.indd 207 11/13/13 2:17 PM 208 CHAPTER 4 Designing Studies 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. Starnes-Yates5e_c04_206-285hr2.indd 208 11/13/13 2:17 PM 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. 11/13/13 2:17 PM 210 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. Starnes-Yates5e_c04_206-285hr2.indd 210 11/13/13 2:17 PM 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! Starnes-Yates5e_c04_206-285hr2.indd 211 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. 11/13/13 2:17 PM 212 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. Starnes-Yates5e_c04_206-285hr2.indd 213 11/13/13 2:17 PM 214 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. Starnes-Yates5e_c04_206-285hr2.indd 214 11/13/13 2:17 PM 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. Starnes-Yates5e_c04_206-285hr2.indd 215 11/13/13 2:17 PM 216 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 Starnes-Yates5e_c04_206-285hr2.indd 216 11/13/13 2:17 PM 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 Starnes-Yates5e_c04_206-285hr3.indd 217 11 12/2/13 4:46 PM 218 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. Starnes-Yates5e_c04_206-285hr2.indd 218 11/13/13 2:18 PM Section 4.1 Sampling and Surveys 219 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. Starnes-Yates5e_c04_206-285hr2.indd 219 11/13/13 2:18 PM 220 A CHAPTER 4 B C D E 1 2 3 4 5 6 7 8 9 10 F Designing Studies G H I J 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 Starnes-Yates5e_c04_206-285hr3.indd 220 11/20/13 6:32 PM 221 Section 4.1 Sampling and Surveys 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 ! n 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. c 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. Starnes-Yates5e_c04_206-285hr2.indd 221 11/13/13 2:18 PM 222 CHAPTER 4 Designing Studies STAGE 1 2 21 76 1 78 1 22 762 782 23 763 783 3 4 24 764 784 5 25 765 785 6 26 766 786 7 27 767 787 8 9 28 29 768 769 788 789 10 30 770 790 9th grade: Seats 601– 800 11th grade: Seats 201– 400 11 31 771 791 12 32 33 772 773 792 793 774 794 775 795 776 796 777 20 19 38 37 36 35 34 17 16 15 14 13 18 778 797 40 39 0 779 798 78 799 80 0 10th grade: Seats 401–600 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. Starnes-Yates5e_c04_206-285hr2.indd 222 11/13/13 2:18 PM Section 4.1 Sampling and Surveys 223 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.” Starnes-Yates5e_c04_206-285hr2.indd 223 11/13/13 2:18 PM 224 CHAPTER 4 Designing Studies 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. Starnes-Yates5e_c04_206-285hr2.indd 224 11/13/13 2:18 PM 225 Section 4.1 Sampling and Surveys 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. c Starnes-Yates5e_c04_206-285