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```Copyright © 2005 Pearson Education, Inc.
SEVENTH EDITION and EXPANDED SEVENTH EDITION
Slide 13-1
Chapter 13
Statistics
13.1
Sampling Techniques
Statistics
Statistics is the art and science of gathering,
analyzing, and making inferences from numerical
information (data) obtained in an experiment.
Statistics are divided into two main braches.




Descriptive statistics is concerned with the collection,
organization, and analysis of data.
Inferential statistics is concerned with the making of
generalizations or predictions of the data collected.
Slide 13-4
Statisticians

A statistician’s interest lies in drawing conclusions about
possible outcomes through observations of only a few
particular events.



The population consists of all items or people of interest.
The sample includes some of the items in the population.
When a statistician draws a conclusion from a sample,
there is always the possibility that the conclusion is
incorrect.
Slide 13-5
Types of Sampling



A random sampling occurs if a sample is drawn in such
a way that each time an item is selected, each item has
an equal chance of being drawn.
When a sample is obtained by drawing every nth item
on a list or production line, the sample is a systematic
sample.
A cluster sample is referred to as an area sample
because it is applied on a geographical basis.
Slide 13-6
Types of Sampling continued


Stratified sampling involves dividing the
population by characteristics such as gender,
race, religion, or income.
Convenience sampling uses data that is easily
obtained and can be extremely biased.
Slide 13-7
Example: Identifying Sampling
Techniques



A raffle ticket is drawn by a blindfolded person at a
festival to win a grand prize.
Students at an elementary are classified according to
their present grade level. Then, a random sample of
three students from each grade are chosen to
represent their class.
Every sixth car on highway is stopped for a vehicle
inspection.
Slide 13-8
Example: Identifying Sampling
Techniques continued
Voters are classified based on their polling location. A
random sample of four polling locations are selected.
All the voters from the precinct are included in the
sample.

The first 20 people entering a water park are asked if
they are wearing sunscreen.
Solution:
a) Random
d)
Cluster
b) Stratified
e)
Convenience
c) Systematic

Slide 13-9
13.2
The Misuses of Statistics
Misuses of Statistics
misuse statistics to their own advantage.
When examining statistical information consider the
following:




Was the sample used to gather the statistical data
unbiased and of sufficient size?
Is the statistical statement ambiguous, could it be
interpreted in more than one way?
Slide 13-11

Speedway Airlines and Save
20%”.

Here there is not enough
information given.
 The “Save 20%” could be
off the original ticket price,
the ticket price when you
another airline’s ticket
price.

Salesperson wanted for
Ryan’s Furniture Store.
Average Salary: \$32,000.”
 The word “average” can be
 If most of the salespeople
earn \$20,000 to \$25,000
and the owner earns
\$76,000, this “average
salary” is not a fair
representation.
Slide 13-12
Charts and Graphs

Charts and graphs can also be misleading.


Even though the data is displayed correctly,
adjusting the vertical scale of a graph can give a
different impression.
A circle graph can be misleading if the sum of the
parts of the graphs do not add up to 100%.
Slide 13-13
While each graph presents identical
information, the vertical scales have been
altered.
Sales
Dollars (in thousands)
Dollars (in thousands)
Sales
175
150
125
100
75
50
25
0
99
00
01
02
Years
03
04
500
400
300
200
100
99
00
01
02
03
04
Years
Slide 13-14
13.3
Frequency Distributions
Example

The number of pets per family is recorded for 30
families surveyed. Construct a frequency
distribution of the following data:
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
3
3
3
3
4
4
Slide 13-16
Solution
Number of
Pets
Frequency
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
2
2
0
6
1
10
2
2
2
2
2
2
2
8
3
3
3
3
4
4
3
4
4
2
Slide 13-17
Rules for Data Grouped by Classes



The classes should be of the same “width.”
The classes should not overlap.
Each piece of data should belong to only one
class.
Slide 13-18
Definitions
Classes
 04 
 59 


10  14 
Lower class limits 
 Upper class limits
15  19 
20  24 


25  29 

Midpoint of a class is found by adding the lower and
upper class limits and dividing the sum by 2.
Slide 13-19
Example

The following set of data represents the distance, in
miles, 15 randomly selected second grade students live
from school.
6.8
5.3
9.7
3.8
8.7
0.5
5.9
0.8
5.7
1.3
4.8
9.6
1.5
7.4
0.2
Construct a frequency distribution with the first class 0  2.
Slide 13-20
Solution

First, rearrange the data
from lowest to highest.

# of miles
from school
Frequency
0.2
0.5
0.8
0-2
5
1.3
1.5
3.8
2.1 - 4.1
1
4.8
5.3
5.7
4.2 - 6.2
4
5.9
6.8
7.4
6.3 - 8.3
2
8.4 -10.4
3
8.7
9.6
9.7
15
Slide 13-21
13.4
Statistical Graphs
Circle Graphs

Circle graphs (also known as pie charts) are
often used to compare parts of one or more
components of the whole to the whole.
Slide 13-23
Example

According to a recent hospital survey of 200 patients the
following table indicates how often hospitals used four
different kinds of painkillers. Use the information to
construct a circle graph illustrating the percent each
painkiller was used.
Aspirin
Ibuprofen
56
104
Acetaminophen
16
Other
24
200
Slide 13-24
Solution

Determine the measure of the corresponding
central angle.
Painkiller
Number of
Patients
Percent of Total
Measure of Central
Angle
Aspirin
56
56
200
 100  28%
0.28  360 = 100.8
Ibuprofen
104
104
200
 100  52%
0.52  360 = 187.2
Acetaminophen
16
16
200
 100  8%
0.08  360 = 28.8
Other
24
24
200
 100  12%
0.12  360 = 43.2
Total
200
100%
360
Slide 13-25
Solution continued

Use a protractor to construct a circle graph and label it
properly.
Hospital Painkiller Use
Ibuprofen
52%
Aspirin
28%
Other
12%
Acetaminophe
n
8%
Slide 13-26
Histogram


A histogram is a graph with observed values on
its horizontal scale and frequencies on it vertical
scale.
# of pets Frequency
Example: Construct a
0
6
histogram of the
1
10
frequency distribution.
2
8
3
4
4
2
Slide 13-27
Solution
Number of Pets per Family
12
Frequency
10
8
6
4
2
0
0
1
2
3
4
Number of Pets
# of pets
Frequency
0
6
1
10
2
8
3
4
4
2
Slide 13-28
Frequency Polygon
Number of Pets per Family
12
Frequency
10
8
6
4
2
0
0
1
2
3
4
Number of Pets
Slide 13-29
Stem-and-Leaf Display



A stem-and-leaf display is a tool that organizes
and groups the data while allowing us to see the
actual values that make up the data.
The left group of digits is called the stem.
The right group of digits is called the leaf.
Slide 13-30
Example

The table below indicates the number of miles
20 workers have to drive to work. construct a
stem-and-leaf display.
12
18
3
8
12
25
21
3
15
4
17
27
43
21
16
12
26
35
14
9
Slide 13-31
Solution

Data

12
18
3
8
12
25
21
3
15
4
17
27
43
21
16
12
26
35
14
9
0 33489
1 22245678
2 11567
3 5
4 3
Slide 13-32
13.5
Measures of Central Tendency
Definitions

An average is a number that is representative
of a group of data.

The arithmetic mean, or simply the mean is
symbolized by x or by the Greek letter mu, .
Slide 13-34
Mean

The mean, x is the sum of the data divided by
the number of pieces of data. The formula for
calculating the mean is
x

x
n

where  x represents the sum of all the data and
n represents the number of pieces of data.
Slide 13-35
Example-find the mean

Find the mean amount of money parents spent
on new school supplies and clothes if 5 parents
randomly surveyed replied as follows: \$327
\$465 \$672 \$150 \$230
x \$327  \$465  \$672  \$150  \$230

x

n
\$1844

 \$368.80
5
5
Slide 13-36
Median


The median is the value in the middle of a set
of ranked data.
Example: Determine the mean of \$327 \$465
\$672 \$150 \$230.
Rank the data from smallest to largest.
\$150 \$230 \$327 \$465 \$672
middle value
(median)
Slide 13-37
Example: Median (even data)

Determine the median of the following set of
data: 8, 15, 9, 3, 4, 7, 11, 12, 6, 4.
Rank the data:
3 4 4 6 7 8 9 11 12 15
There are 10 pieces of data so the median will
lie halfway between the two middle pieces the 7
and 8. The median is (7 + 8)/2 = 7.5
3 4 4 6 7 8 9 11 12 15
Slide 13-38
Mode

The mode is the piece of data that occurs most
frequently.

Example: Determine the mode of the data set:
3, 4, 4, 6, 7, 8, 9, 11, 12, 15.
The mode is 4 since is occurs twice and the
other values only occur once.

Slide 13-39
Midrange

The midrange is the value halfway between the lowest
(L) and highest (H) values in a set of data.
lowest value + highest value
Midrange =
2

Example: Find the midrange of the data set \$327, \$465,
\$672, \$150, \$230.
\$150 + \$672
Midrange =
 \$411
2
Slide 13-40
Example

The weights of eight Labrador retrievers
rounded to the nearest pound are 85, 92, 88,
75, 94, 88, 84, and 101. Determine the



a) mean
b) median
c) mode
d) midrange
e) rank the measures of central tendency
from lowest to highest.
Slide 13-41
Example--dog weights 85, 92, 88, 75,
94, 88, 84, 101
85  92  88  75  94  88  84  101
x
8
707

 88.375
8

Mean

Median-rank the data


75, 84, 85, 88, 88, 92, 94, 101
The median is 88.
Slide 13-42
Example--dog weights 85, 92, 88, 75,
94, 88, 84, 101



Mode-the number that occurs most frequently.
The mode is 88.
Midrange = (L + H)/2
= (75 + 101)/2 = 88
Rank the measures
88.375, 88, 88, 88
Slide 13-43
Measures of Position


Measures of position are often used to make
comparisons.
Two measures of position are percentiles and
quartiles.
Slide 13-44
To Find the Quartiles of a Set of Data


Order the data from smallest to largest.
Find the median, or 2nd quartile, of the set of
data. If there are an odd number of pieces of
data, the median is the middle value. If there
are an even number of pieces of data, the
median will be halfway between the two middle
pieces of data.
Slide 13-45
To Find the Quartiles of a Set of Data
continued


The first quartile, Q1, is the median of the lower
half of the data; that is, Q1, is the median of the
data less than Q2.
The third quartile, Q3, is the median of the upper
half of the data; that is, Q3 is the median of the
data greater than Q2.
Slide 13-46
Example: Quartiles

The weekly grocery bills for 23 families are as
follows. Determine Q1, Q2, and Q3.
170
330
225
75
95
210
80
225
160
172
270
170
215
130
190
270
240
310
74
280
270
50
81
Slide 13-47
Example: Quartiles continued

Order the data:
50
75
74
80
81
95
130
160
170
170
172
190
210
215
225
225
240
270
270
270
280
310
330
Q2 is the median of the entire data set which is 190.
Q1 is the median of the numbers from 50 to 172 which is 95.
Q3 is the median of the numbers from 210 to 330 which is 270.
Slide 13-48
13.6
Measures of Dispersion
Measures of Dispersion

Measures of dispersion are used to indicate the

The range is the difference between the highest
and lowest values; it indicates the total spread
of the data.
Slide 13-50
Example: Range


Nine different employees were selected and the
amount of their salary was recorded. Find the
range of the salaries.
\$24,000
\$32,000 \$26,500
\$56,000 \$48,000 \$27,000
\$28,500 \$34,500 \$56,750
Range = \$56,750  \$24,000 = \$32,750
Slide 13-51
Standard Deviation

The standard deviation measures how much
the data differ from the mean.
s
x  x
2
n 1
Slide 13-52
To Find the Standard Deviation of a
Set of Data


1. Find the mean of the set of data.
2. Make a chart having three columns:



Data
Data  Mean
(Data  Mean)2
3. List the data vertically under the
column marked Data.
4. Subtract the mean from each piece
of data and place the difference in
the Data  Mean column.
Slide 13-53
To Find the Standard Deviation of a
Set of Data continued

5.

6.

7.

8.
Square the values obtained in the Data 
Mean column and record these values in the
(Data  Mean)2 column.
Determine the sum of the values in the
(Data  Mean)2 column.
Divide the sum obtained in step 6 by
n  1, where n is the number of pieces of
data.
Determine the square root of the number
obtained in step 7. This number is the
standard deviation of the set of data.
Slide 13-54
Example

Find the standard deviation of the following
prices of selected washing machines:
\$280, \$217, \$665, \$684, \$939, \$299
Find the mean.
x 665  217  684  280  939  299 3084

x


 514
n
6
6
Slide 13-55
Example continued, mean = 514
Data  Mean
(Data  Mean)2
217
297
(297)2 = 88,209
280
234
54,756
299
215
46,225
665
151
22,801
684
170
28,900
939
425
180,625
0
421,516
Data
Slide 13-56
Example continued, mean = 514


s
421,516
6 1
s
421,516
 290.35
5
The standard deviation is \$290.35.
Slide 13-57
13.7
The Normal Curve
Types of Distributions
Rectangular Distribution

J-shaped distribution
Rectangular Distribution
Frequency

Values
Slide 13-59
Types of Distributions continued

Bimodal

Skewed to right
Slide 13-60
Types of Distributions continued

Skewed to left

Normal
Slide 13-61
Normal Distribution


In a normal distribution, the mean, median, and
mode all have the same value.
Z-scores determine how far, in terms of
standard deviations, a given score is from the
mean of the distribution.
value of piece of data  mean x  
z

standard deviation

Slide 13-62
Example: z-scores



A normal distribution has a mean of 50 and a
standard deviation of 5. Find z-scores for the
following values.
a) 55
b) 60 c) 43
a)
value of piece of data  mean
z
standard deviation
55  50 5
z55 
 1
5
5
A score of 55 is one standard deviation above the mean.
Slide 13-63
Example: z-scores continued

b) z60
60  50 10


2
5
5
A score of 60 is 2 standard deviations above the mean.

c) z43
43  50 7


 1.4
5
5
A score of 43 is 1.4 standard deviations below the
mean.
Slide 13-64
To Find the Percent of Data Between
any Two Values
1.
2.
3.
Draw a diagram of the normal curve,
indicating the area or percent to be
determined.
Use the formula to convert the given
values to z-scores. Indicate these zscores on the diagram.
Look up the percent that corresponds to
each z-score in Table 13.
Slide 13-65
To Find the Percent of Data Between
any Two Values continued
4.


a) When finding the percent of data between two z-scores
on the opposite side of the mean (when one z-score is
positive and the other is negative), you find the sum of the
individual percents.
b) When finding the percent of data between two z-scores
on the same side of the mean (when both z-scores are
positive or both are negative), subtract the smaller percent
from the larger percent.
Slide 13-66
To Find the Percent of Data Between
any Two Values continued


c) When finding the percent of data to the right of
a positive z-score or to the left of a negative zscore, subtract the percent of data between ) and
z from 50%.
d) When finding the percent of data to the left of a
positive z-score or to the right of a negative zscore, add the percent of data between 0 and z to
50%.
Slide 13-67
Example





Assume that the waiting times for customers at a popular restaurant
before being seated for lunch at a popular restaurant before being
seated for lunch are normally distributed with a mean of 12 minutes
and a standard deviation of 3 min.
a) Find the percent of customers who wait for at least 12 minutes
before being seated.
b) Find the percent of customers who wait between 9 and 18
minutes before being seated.
c) Find the percent of customers who wait at least 17 minutes
before being seated.
d) Find the percent of customers who wait less than 8 minutes
before being seated.
Slide 13-68
Solution
wait for at least 12 minutes
Since 12 minutes is the
mean, half, or 50% of
customers wait at least 12
min before being seated.
between 9 and 18 minutes
9  12
z9 
 1.00
3
18  12
z18 
 2.00
3
Use table 13.7 page 801.
34.1% + 47.7%
= 81.8%
Slide 13-69
Solution continued

at least 17 min
17  12
z17 
 1.67
3
Use table 13.7 page 801.
45.3% is between the mean and
1.67.
50%  45.3% = 4.7%
Thus, 4.7% of customers wait at
least 17 minutes.

less than 8 min
8  12
z8 
 1.33
3
Use table 13.7 page 801.
40.8% is between the mean and
1.33.
50%  40.8% = 9.2%
Thus, 9.2% of customers wait
less than 8 minutes.
Slide 13-70
13.8
Linear Correlation and
Regression
Linear Correlation
Linear correlation is used to determine whether there
is a relationship between two quantities and, if so,
how strong the relationship is.


The linear correlation coefficient, r, is a unitless
measure that describes the strength of the linear
relationship between two variables.



If the value is positive, as one variable increases, the other
increases.
If the value is negative, as one variable increases, the
other decreases.
The variable, r, will always be a value between –1 and 1
inclusive.
Slide 13-72
Scatter Diagrams

A visual aid used with correlation is the scatter diagram,
a plot of points (bivariate data).



The independent variable, x, generally is a quantity that
can be controlled.
The dependant variable, y, is the other variable.
The value of r is a measure of how far a set of points
varies from a straight line.

The greater the spread, the weaker the correlation and the
closer the r value is to 0.
Slide 13-73
Correlation
Slide 13-74
Correlation
Slide 13-75
Linear Correlation Coefficient

The formula to calculate the correlation
coefficient (r) is as follows:
r
n   xy     x   y 

n x
2
   x 
2

n y
2
   y 
2
Slide 13-76
Example: Words Per Minute versus
Mistakes
There are five applicants applying for a job as a medical
transcriptionist. The following shows the results of the
applicants when asked to type a chart. Determine the
correlation coefficient between the words per minute
typed and the number of mistakes.
Applicant
Words per Minute
Mistakes
Ellen
24
8
George
67
11
Phillip
53
12
Kendra
41
10
Nancy
34
9
Slide 13-77
Solution


We will call the words typed per minute, x, and the mistakes, y.
List the values of x and y and calculate the necessary sums.
WPM
Mistakes
x
y
x2
y2
xy
24
8
576
64
192
67
11
4489
121
737
53
12
2809
144
636
41
10
1681
100
410
34
9
1156
81
306
 x = 219
 y = 50
 x2 =10,711
 y2 = 510
 xy = 2,281
Slide 13-78
Solution continued

The n in the formula represents the number of pieces of data. Here
n = 5.
r 
r 

n   xy     x   y 


n  x2   x 
2


n  y 2   y 
2
5  2281   219  50 
5 10,711   219 
2
5  510    50 
2
11,405  10,950
5 10,711  47,961 5  510   2500

455
53,555  47,961 2550  2500
455

 0.86
5594 50
Slide 13-79
Solution continued


Since 0.86 is fairly close to 1, there is a fairly
strong positive correlation.
This result implies that the more words typed
per minute, the more mistakes made.
Slide 13-80
Linear Regression


Linear regression is the process of determining
the linear relationship between two variables.
The line of best fit (line of regression or the least
square line) is the line such that the sum of the
vertical distances from the line to the data
points is a minimum.
Slide 13-81
The Line of Best Fit

Equation:
y  mx  b,
m
where
n   xy     x   y 


n  x2   x 
2
, and b 
 y  m  x 
n
Slide 13-82
Example


Use the data in the previous example to find
the equation of the line that relates the
number of words per minute and the number
of mistakes made while typing a chart.
Graph the equation of the line of best fit on a
scatter diagram that illustrates the set of
bivariate points.
Slide 13-83
Solution

From the previous results, we
know that
m
n   xy     x   y 

n x
2
   x 
2
5(2,281)  (219)(50)
5(10,711)  2192
455
m
5594
m  0.081
m

Now we find the y-intercept, b.
b
b
 y  m  x 
n
50  0.081 219 
5
32.261
b
 6.452
5
Therefore the line of best fit is y = 0.081x + 6.452
Slide 13-84
Solution continued

To graph y = 0.081x + 6.452, plot at least two
points and draw the graph.
x
y
10
7.262
20
8.072
30
8.882
Slide 13-85
Solution continued