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Chapter 10
Statistical Inference About Means and Proportions
With Two Populations

Inferences About the Difference Between
Two Population Means: s 1 and s 2 Known

Inferences About the Difference Between
Two Population Means: s 1 and s 2 Unknown

Inferences About the Difference Between
Two Population Means: Matched Samples

Inferences About the Difference Between
Two Population Proportions
© 2005 Thomson/South-Western
Slide 1
Inferences About the Difference Between
Two Population Means: s 1 and s 2 Known


Interval Estimation of m 1 – m 2
Hypothesis Tests About m 1 – m 2
© 2005 Thomson/South-Western
Slide 2
Estimating the Difference Between
Two Population Means



Let m1 equal the mean of population 1 and m2 equal
the mean of population 2.
The difference between the two population means is
m1 - m2.
To estimate m1 - m2, we will select a simple random
sample of size n1 from population 1 and a simple
random sample of size n2 from population 2.
Let x1 equal the mean of sample 1 and x2 equal the
mean of sample 2.
 The point estimator of the difference between the
means of the populations 1 and 2 is x1  x2.

© 2005 Thomson/South-Western
Slide 3
Sampling Distribution of x1  x2

Expected Value
E ( x1  x2 )  m1  m 2

Standard Deviation (Standard Error)
s x1  x2 
s12
n1

s 22
n2
where: s1 = standard deviation of population 1
s2 = standard deviation of population 2
n1 = sample size from population 1
n2 = sample size from population 2
© 2005 Thomson/South-Western
Slide 4
Interval Estimation of m1 - m2:
s 1 and s 2 Known

Interval Estimate
x1  x2  z / 2
s 12 s 22

n1 n2
where:
1 -  is the confidence coefficient
© 2005 Thomson/South-Western
Slide 5
Interval Estimation of m1 - m2:
s 1 and s 2 Known

Example: Par, Inc.
Par, Inc. is a manufacturer
of golf equipment and has
developed a new golf ball
that has been designed to
provide “extra distance.”
In a test of driving distance using a mechanical
driving device, a sample of Par golf balls was
compared with a sample of golf balls made by Rap,
Ltd., a competitor. The sample statistics appear on the
next slide.
© 2005 Thomson/South-Western
Slide 6
Interval Estimation of m1 - m2:
s 1 and s 2 Known

Example: Par, Inc.
Sample Size
Sample Mean
Sample #1
Par, Inc.
120 balls
275 yards
Sample #2
Rap, Ltd.
80 balls
258 yards
Based on data from previous driving distance
tests, the two population standard deviations are
known with s 1 = 15 yards and s 2 = 20 yards.
© 2005 Thomson/South-Western
Slide 7
Interval Estimation of m1 - m2:
s 1 and s 2 Known

Example: Par, Inc.
Let us develop a 95% confidence interval estimate
of the difference between the mean driving distances of
the two brands of golf ball.
© 2005 Thomson/South-Western
Slide 8
Estimating the Difference Between
Two Population Means
Population 1
Par, Inc. Golf Balls
m1 = mean driving
distance of Par
golf balls
Population 2
Rap, Ltd. Golf Balls
m2 = mean driving
distance of Rap
golf balls
m1 – m2 = difference between
the mean distances
Simple random sample
of n1 Par golf balls
Simple random sample
of n2 Rap golf balls
x1 = sample mean distance
for the Par golf balls
x2 = sample mean distance
for the Rap golf balls
x1 - x2 = Point Estimate of m1 – m2
© 2005 Thomson/South-Western
Slide 9
Point Estimate of m1 - m2
Point estimate of m1  m2 = x1  x2
= 275  258
= 17 yards
where:
m1 = mean distance for the population
of Par, Inc. golf balls
m2 = mean distance for the population
of Rap, Ltd. golf balls
© 2005 Thomson/South-Western
Slide 10
Interval Estimation of m1 - m2:
s 1 and s 2 Known
x1  x2  z / 2
s12
s 22
(15) 2 ( 20) 2

 17  1. 96

n1 n2
120
80
17 + 5.14 or 11.86 yards to 22.14 yards
We are 95% confident that the difference between
the mean driving distances of Par, Inc. balls and Rap,
Ltd. balls is 11.86 to 22.14 yards.
© 2005 Thomson/South-Western
Slide 11
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Known
 Hypotheses
H0 : m1  m2  D0 H0 : m1  m2  D0 H0 : m1  m2  D0
H a : m1  m2  D0 H a : m1  m2  D0 H a : m1  m2  D0
Left-tailed
Right-tailed
Two-tailed
 Test Statistic
z
( x1  x2 )  D0
s 12
n1
© 2005 Thomson/South-Western

s 22
n2
Slide 12
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Known

Example: Par, Inc.
Can we conclude, using
 = .01, that the mean driving
distance of Par, Inc. golf balls
is greater than the mean driving
distance of Rap, Ltd. golf balls?
© 2005 Thomson/South-Western
Slide 13
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Known
 p –Value and Critical Value Approaches
1. Develop the hypotheses.
H0: m1 - m2 < 0
Ha: m1 - m2 > 0
where:
m1 = mean distance for the population
of Par, Inc. golf balls
m2 = mean distance for the population
of Rap, Ltd. golf balls
2. Specify the level of significance.
© 2005 Thomson/South-Western
 = .01
Slide 14
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Known
 p –Value and Critical Value Approaches
3. Compute the value of the test statistic.
z
( x1  x2 )  D0
s 12
n1
z

s 22
n2
(235  218)  0
(15)2 (20)2

120
80
© 2005 Thomson/South-Western

17
 6.49
2.62
Slide 15
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Known
 p –Value Approach
4. Compute the p–value.
For z = 6.49, the p –value < .0001.
5. Determine whether to reject H0.
Because p–value <  = .01, we reject H0.
At the .01 level of significance, the sample evidence
indicates the mean driving distance of Par, Inc. golf
balls is greater than the mean driving distance of Rap,
Ltd. golf balls.
© 2005 Thomson/South-Western
Slide 16
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Known
 Critical Value Approach
4. Determine the critical value and rejection rule.
For  = .01, z.01 = 2.33
Reject H0 if z > 2.33
5. Determine whether to reject H0.
Because z = 6.49 > 2.33, we reject H0.
The sample evidence indicates the mean driving
distance of Par, Inc. golf balls is greater than the mean
driving distance of Rap, Ltd. golf balls.
© 2005 Thomson/South-Western
Slide 17
Inferences About the Difference Between
Two Population Means: s 1 and s 2 Unknown


Interval Estimation of m 1 – m 2
Hypothesis Tests About m 1 – m 2
© 2005 Thomson/South-Western
Slide 18
Interval Estimation of m1 - m2:
s 1 and s 2 Unknown
When s 1 and s 2 are unknown, we will:
• use the sample standard deviations s1 and s2
as estimates of s 1 and s 2 , and
• replace z/2 with t/2.
© 2005 Thomson/South-Western
Slide 19
Interval Estimation of m1 - m2:
s 1 and s 2 Unknown

Interval Estimate
x1  x2  t / 2
s12 s22

n1 n2
Where the degrees of freedom for t/2 are:
2
s s 
  
n1 n2 

df 
2 2
2 2
1  s1 
1  s2 
  
 
n1  1  n1  n2  1  n2 
2
1
© 2005 Thomson/South-Western
2
2
Slide 20
Difference Between Two Population Means:
s 1 and s 2 Unknown

Example: Specific Motors
Specific Motors of Detroit
has developed a new automobile
known as the M car. 24 M cars
and 28 J cars (from Japan) were road
tested to compare miles-per-gallon (mpg) performance.
The sample statistics are shown on the next slide.
© 2005 Thomson/South-Western
Slide 21
Difference Between Two Population Means:
s 1 and s 2 Unknown

Example: Specific Motors
Sample #1
M Cars
24 cars
29.8 mpg
2.56 mpg
Sample #2
J Cars
28 cars
27.3 mpg
1.81 mpg
© 2005 Thomson/South-Western
Sample Size
Sample Mean
Sample Std. Dev.
Slide 22
Difference Between Two Population Means:
s 1 and s 2 Unknown

Example: Specific Motors
Let us develop a 90% confidence
interval estimate of the difference
between the mpg performances of
the two models of automobile.
© 2005 Thomson/South-Western
Slide 23
Point Estimate of m 1  m 2
Point estimate of m1  m2 = x1  x2
= 29.8 - 27.3
= 2.5 mpg
where:
m1 = mean miles-per-gallon for the
population of M cars
m2 = mean miles-per-gallon for the
population of J cars
© 2005 Thomson/South-Western
Slide 24
Interval Estimation of m 1  m 2:
s 1 and s 2 Unknown
The degrees of freedom for t/2 are:
2
 (2.56) (1.81) 



24
28


df 
 24.07  24
2
2
1  (2.56) 2 
1  (1.81) 2 

 


24  1  24  28  1  28 
2
2
With /2 = .05 and df = 24, t/2 = 1.711
© 2005 Thomson/South-Western
Slide 25
Interval Estimation of m 1  m 2:
s 1 and s 2 Unknown
x1  x2  t / 2
s12 s22
(2.56)2 (1.81)2
  29.8  27.3  1.711

n1 n2
24
28
2.5 + 1.069 or
1.431 to 3.569 mpg
We are 90% confident that the difference between
the miles-per-gallon performances of M cars and J cars
is 1.431 to 3.569 mpg.
© 2005 Thomson/South-Western
Slide 26
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Unknown

Hypotheses
H0 : m1  m2  D0 H0 : m1  m2  D0 H0 : m1  m2  D0
H a : m1  m2  D0 H a : m1  m2  D0 H a : m1  m2  D0
Left-tailed

Right-tailed
Two-tailed
Test Statistic
t
( x1  x2 )  D0
2
1
2
2
s
s

n1 n2
© 2005 Thomson/South-Western
Slide 27
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Unknown

Example: Specific Motors
Can we conclude, using a
.05 level of significance, that the
miles-per-gallon (mpg) performance
of M cars is greater than the miles-pergallon performance of J cars?
© 2005 Thomson/South-Western
Slide 28
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Unknown
 p –Value and Critical Value Approaches
1. Develop the hypotheses.
H0: m1 - m2 < 0
Ha: m1 - m2 > 0
where:
m1 = mean mpg for the population of M cars
m2 = mean mpg for the population of J cars
© 2005 Thomson/South-Western
Slide 29
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Unknown
 p –Value and Critical Value Approaches
2. Specify the level of significance.
 = .05
3. Compute the value of the test statistic.
t
( x1  x2 )  D0
s12 s22

n1 n2

© 2005 Thomson/South-Western
(29.8  27.3)  0
(2.56)2 (1.81)2

24
28
 4.003
Slide 30
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Unknown
 p –Value Approach
4. Compute the p –value.
The degrees of freedom for t are:
 (2.56) (1.81) 



24
28


2
df 
2
2
2
1  (2.56)2 
1  (1.81)2 

 


24  1  24  28  1  28 
2
 24.07  24
Because t = 4.003 > t.005 = 2.797, the p–value < .005.
© 2005 Thomson/South-Western
Slide 31
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Unknown
 p –Value Approach
5. Determine whether to reject H0.
Because p–value <  = .05, we reject H0.
We are at least 95% confident that the miles-pergallon (mpg) performance of M cars is greater than
the miles-per-gallon performance of J cars?.
© 2005 Thomson/South-Western
Slide 32
Hypothesis Tests About m 1  m 2:
s 1 and s 2 Unknown
 Critical Value Approach
4. Determine the critical value and rejection rule.
For  = .05 and df = 24, t.05 = 1.711
Reject H0 if t > 1.711
5. Determine whether to reject H0.
Because 4.003 > 1.711, we reject H0.
We are at least 95% confident that the miles-pergallon (mpg) performance of M cars is greater than
the miles-per-gallon performance of J cars?.
© 2005 Thomson/South-Western
Slide 33
Inferences About the Difference Between
Two Population Means: Matched Samples
 With a matched-sample design each sampled item
provides a pair of data values.
 This design often leads to a smaller sampling error
than the independent-sample design because
variation between sampled items is eliminated as a
source of sampling error.
© 2005 Thomson/South-Western
Slide 34
Inferences About the Difference Between
Two Population Means: Matched Samples

Example: Express Deliveries
A Chicago-based firm has
documents that must be quickly
distributed to district offices
throughout the U.S. The firm
must decide between two delivery
services, UPX (United Parcel Express) and INTEX
(International Express), to transport its documents.
© 2005 Thomson/South-Western
Slide 35
Inferences About the Difference Between
Two Population Means: Matched Samples

Example: Express Deliveries
In testing the delivery times
of the two services, the firm sent
two reports to a random sample
of its district offices with one
report carried by UPX and the
other report carried by INTEX. Do the data on the
next slide indicate a difference in mean delivery
times for the two services? Use a .05 level of
significance.
© 2005 Thomson/South-Western
Slide 36
Inferences About the Difference Between
Two Population Means: Matched Samples
Delivery Time (Hours)
District Office UPX INTEX Difference
Seattle
Los Angeles
Boston
Cleveland
New York
Houston
Atlanta
St. Louis
Milwaukee
Denver
32
30
19
16
15
18
14
10
7
16
© 2005 Thomson/South-Western
25
24
15
15
13
15
15
8
9
11
7
6
4
1
2
3
-1
2
-2
5
Slide 37
Inferences About the Difference Between
Two Population Means: Matched Samples
 p –Value and Critical Value Approaches
1. Develop the hypotheses.
H0: md = 0
Ha: md  
Let md = the mean of the difference values for the
two delivery services for the population
of district offices
© 2005 Thomson/South-Western
Slide 38
Inferences About the Difference Between
Two Population Means: Matched Samples
 p –Value and Critical Value Approaches
2. Specify the level of significance.
 = .05
3. Compute the value of the test statistic.
 di ( 7  6... 5)
d 

 2. 7
n
10
2
76.1
 ( di  d )
sd 

 2. 9
n 1
9
d  md
2.7  0
t

 2.94
sd n 2.9 10
© 2005 Thomson/South-Western
Slide 39
Inferences About the Difference Between
Two Population Means: Matched Samples
 p –Value Approach
4. Compute the p –value.
For t = 2.94 and df = 9, the p–value is between
.02 and .01. (This is a two-tailed test, so we double
the upper-tail areas of .01 and .005.)
5. Determine whether to reject H0.
Because p–value <  = .05, we reject H0.
We are at least 95% confident that there is a
difference in mean delivery times for the two
services?
© 2005 Thomson/South-Western
Slide 40
Inferences About the Difference Between
Two Population Means: Matched Samples
 Critical Value Approach
4. Determine the critical value and rejection rule.
For  = .05 and df = 9, t.025 = 2.262.
Reject H0 if t > 2.262
5. Determine whether to reject H0.
Because t = 2.94 > 2.262, we reject H0.
We are at least 95% confident that there is a
difference in mean delivery times for the two
services?
© 2005 Thomson/South-Western
Slide 41
Inferences About the Difference Between
Two Population Proportions


Interval Estimation of p1 - p2
Hypothesis Tests About p1 - p2
© 2005 Thomson/South-Western
Slide 42
Sampling Distribution of p1  p2

Expected Value
E ( p1  p2 )  p1  p2

Standard Deviation (Standard Error)
s p1  p2 
p1 (1  p1 ) p2 (1  p2 )

n1
n2
where: n1 = size of sample taken from population 1
n2 = size of sample taken from population 2
© 2005 Thomson/South-Western
Slide 43
Sampling Distribution of p1  p2
If the sample sizes are large, the sampling distribution
of p1  p2 can be approximated by a normal probability
distribution.
The sample sizes are sufficiently large if all of these
conditions are met:
n1p1 > 5
n1(1 - p1) > 5
n2p2 > 5
n2(1 - p2) > 5
© 2005 Thomson/South-Western
Slide 44
Sampling Distribution of p1  p2
s p1  p2 
p1 (1  p1 ) p2 (1  p2 )

n1
n2
p1  p2
p1 – p2
© 2005 Thomson/South-Western
Slide 45
Interval Estimation of p1 - p2

Interval Estimate
p1  p2  z / 2
© 2005 Thomson/South-Western
p1 (1  p1 ) p2 (1  p2 )

n1
n2
Slide 46
Interval Estimation of p1 - p2

Example: Market Research Associates
Market Research Associates is
conducting research to evaluate the
effectiveness of a client’s new advertising campaign. Before the new
campaign began, a telephone survey
of 150 households in the test market
area showed 60 households “aware” of
the client’s product.
The new campaign has been initiated with TV and
newspaper advertisements running for three weeks.
© 2005 Thomson/South-Western
Slide 47
Interval Estimation of p1 - p2

Example: Market Research Associates
A survey conducted immediately
after the new campaign showed 120
of 250 households “aware” of the
client’s product.
Does the data support the position
that the advertising campaign has
provided an increased awareness of
the client’s product?
© 2005 Thomson/South-Western
Slide 48
Point Estimator of the Difference Between
Two Population Proportions
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign
p1 = sample proportion of households “aware” of the
product after the new campaign
p2 = sample proportion of households “aware” of the
product before the new campaign
120 60
p1  p2 

 .48  .40  .08
250 150
© 2005 Thomson/South-Western
Slide 49
Interval Estimation of p1 - p2
For  = .05, z.025 = 1.96:
.48(.52) .40(.60)
.48  .40  1.96

250
150
.08 + 1.96(.0510)
.08 + .10
Hence, the 95% confidence interval for the difference
in before and after awareness of the product is
-.02 to +.18.
© 2005 Thomson/South-Western
Slide 50
Hypothesis Tests about p1 - p2

Hypotheses
We focus on tests involving no difference between
the two population proportions (i.e. p1 = p2)
H0 : p1  p2  0
H a : p1  p2  0
Left-tailed
H
H00::
H
Ha::
a
pp1 - pp2 < 00
1
2
pp1 - pp2 > 00
1
2
Right-tailed
© 2005 Thomson/South-Western
H0 : p1  p2  0
H a : p1  p2  0
Two-tailed
Slide 51
Hypothesis Tests about p1 - p2
 Pooled Estimate of Standard Error of p1  p2
s p1  p2
1 1
 p(1  p)   
 n1 n2 
where:
n1 p1  n2 p2
p
n1  n2
© 2005 Thomson/South-Western
Slide 52
Hypothesis Tests about p1 - p2
 Test Statistic
z
( p1  p2 )
 1
1 
p(1  p ) 


n
n
2 
 1
© 2005 Thomson/South-Western
Slide 53
Hypothesis Tests about p1 - p2
Example: Market Research Associates
Can we conclude, using a .05 level
of significance, that the proportion of
households aware of the client’s product
increased after the new advertising
campaign?

© 2005 Thomson/South-Western
Slide 54
Hypothesis Tests about p1 - p2
 p -Value and Critical Value Approaches
1. Develop the hypotheses.
H0: p1 - p2 < 0
Ha: p1 - p2 > 0
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign
© 2005 Thomson/South-Western
Slide 55
Hypothesis Tests about p1 - p2
 p -Value and Critical Value Approaches
2. Specify the level of significance.
 = .05
3. Compute the value of the test statistic.
250(. 48)  150(. 40) 180
p

. 45
250  150
400
s p1  p2  . 45(. 55)( 1
z
 1 ) . 0514
250 150
(.48  .40)  0
.08

 1.56
.0514
.0514
© 2005 Thomson/South-Western
Slide 56
Hypothesis Tests about p1 - p2
 p –Value Approach
4. Compute the p –value.
For z = 1.56, the p–value = .0594
5. Determine whether to reject H0.
Because p–value >  = .05, we cannot reject H0.
We cannot conclude that the proportion of households
aware of the client’s product increased after the new
campaign.
© 2005 Thomson/South-Western
Slide 57
Hypothesis Tests about p1 - p2
 Critical Value Approach
4. Determine the critical value and rejection rule.
For  = .05, z.05 = 1.645
Reject H0 if z > 1.645
5. Determine whether to reject H0.
Because 1.56 < 1.645, we cannot reject H0.
We cannot conclude that the proportion of households
aware of the client’s product increased after the new
campaign.
© 2005 Thomson/South-Western
Slide 58
End of Chapter 10
© 2005 Thomson/South-Western
Slide 59