Ebook Business statistics - A decision - making approach (9th edition): Part 2

www.downloadslide.com

Estimation and Hypothesis
Testing for Two Population
Parameters

From Chapter 10 of Business Statistics, A Decision-Making Approach, Ninth Edition. David F. Groebner,
Patrick W. Shannon and Phillip C. Fry. Copyright © 2014 by Pearson Education, Inc. All rights reserved.




www.downloadslide.com

Quick Prep Links

t Make sure you understand the concepts

t Review material on calculating and
interpreting sample means and standard
deviations.
t Review the normal distribution.

associated with sampling distributions for x
and p.
t Review the steps for developing confidence
interval estimates for a single population
mean and a single population proportion.

t Review the methods for testing hypotheses
about single population means and single

population proportions.

Estimation and Hypothesis
Testing for Two Population
Parameters
1

Estimation for Two Population
Means Using Independent
Samples

Outcome 1. Discuss the logic behind and demonstrate
the techniques for using independent samples to test
hypotheses and develop interval estimates for the
difference between two population means.

2

Hypothesis Tests for Two
Population Means Using Independent Samples

3

Interval Estimation and
Hypothesis Tests for Paired
Samples

Outcome 2. Develop confidence interval estimates and
conduct hypothesis tests for the difference between two
population means for paired samples.

4

Estimation and Hypothesis
Tests for Two Population
Proportions

Outcome 3. Carry out hypothesis tests and establish interval
estimates, using sample data, for the difference between
two population proportions.

Why you need to know
In many business decision-making situations, managers must decide between two or more alternatives.
For example, fleet managers in large companies must decide which model and make of car to purchase
next year. Airlines must decide whether to purchase replacement planes from Boeing or Airbus. When
deciding on a new advertising campaign, a company may need to evaluate proposals from competing advertising agencies. Hiring decisions may require a personnel director to select one employee from a list of applicants. Production managers are often confronted with decisions concerning whether to change a production
process or leave it alone. Each day, consumers purchase a product from among several competing brands.
Fortunately, there are statistical procedures that can help decision makers use sample information to compare different populations. In this text, we introduce these procedures and techniques by discussing methods that
can be used to make statistical comparisons between two populations. Later, we will discuss some methods to
extend this comparison to more than two populations. Whether we are discussing cases involving two populations or those with more than two populations, the techniques we present are all extensions of the statistical tools
involving a single population parameter.

Tatiana Popova/Shutterstock




www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

1

Independent Samples
Samples selected from two or more populations
in such a way that the occurrence of values in
one sample has no influence on the probability
of the occurrence of values in the other
sample(s).

Chapter Outcome 1.

Estimation for Two Population
Means Using Independent Samples

In this section, we examine situations in which we are interested in the difference between
two population means, looking first at the case in which the samples from the two populations
are independent.
We will introduce techniques for estimating the difference between the means of two
populations in the following situations:
1. The population standard deviations are known and the samples are independent.
2. The population standard deviations are unknown and the samples are independent.

Estimating the Difference between Two Population
Means When S1 and S2 Are Known, Using
Independent Samples
Recall that the standard normal distribution z-values were used in establishing the critical value and developing the interval estimate when the population standard deviation was
assumed known and the population distribution is assumed to be normally distributed.1 The
general format for a confidence interval estimate is shown in Equation 1.
Confidence Interval, General Format
Point estimate{ 1Critical value21Standard error2

(1)

In business situations, you will often need to estimate the difference between two population means. For instance, you may wish to estimate the difference in mean starting salaries
between males and females, the difference in mean production output in union and nonunion
factories, or the difference in mean service times at two different fast-food businesses. In
these situations, the best point estimate for m1 -m2 is
Point estimate

x1

x2

In situations in which you know the population standard deviations, s1 and s2, and when
the samples selected from the two populations are independent, an extension of the Central
Limit Theorem tells us that the sampling distribution for all possible differences between x1
and x2 will be approximately normally distributed with a standard error computed as shown
in Equation 2.
Standard Error of x1 − x2 When S1 and S2 Are Known
x1 x2

2
1

2
2

n1

n2

(2)

where:
2
1
2
2

n1 and n2

Variance of population 1
Variance of population 2
Sample sizes from populations 1 and 2

Further, the critical value for determining the confidence interval will be a z-value from
the standard normal distribution. In these circumstances, the confidence interval estimate for
m1 -m2 is found by using Equation 3.

1If

the samples from the two populations are large 1n Ú 302 the normal distribution assumption is not required.




www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

Confidence Interval Estimate for M1 −M2 When S1 and S2 Are Known,

Independent Samples
( x1

x2 )

z

2
1

2
2

n1

n2

(3)

The z-values for several of the most commonly used confidence levels are
Confidence Level

Critical z-value
z
z
z
z

80%
90%

95%
99%

EXAMPLE 1

=
=
=
=

1.28
1.645
1.96
2.575

CONFIDENCE INTERVAL ESTIMATE FOR M1 − M2 WHEN
AND S2 ARE KNOWN, USING INDEPENDENT SAMPLES

S1

Axiom Fitness Axiom Fitness is a small chain of fitness centers located primarily in the
South but with some clubs scattered in other parts of the U.S. and Canada. Recently, the
club in Winston-Salem, North Carolina, worked with a business class from a local university
on a project in which a team of students observed Axiom customers with respect to their
club usage. As part of the study, the students measured the time that customers spent in the
club during a visit. The objective is to estimate the difference in mean time spent per visit for
male and female customers. Previous studies indicate that the standard deviation is 11 minutes
for males and 16 minutes for females. To develop a 95% confidence interval estimate for the
difference in mean times, the following steps are taken:
Step 1 Define the population parameter of interest and select independent

samples from the two populations.
In this case, the company is interested in estimating the difference in mean time
spent in the club between males and females. The measure of interest is m1 -m2.
The student team has selected simple random samples of 100 males and 100
females at different times in the Winston-Salem club.
Step 2 Specify the desired confidence level.
The plan is develop a 95% confidence interval estimate.
Step 3 Compute the point estimate.
The resulting sample means are
Males: x1 = 34.5 minutes

Females: x2 = 42.4 minutes

The point estimate is
x1 -x2 = 34.5-42.4 = -7.9 minutes
Women in the sample spent an average of 7.9 minutes longer in the club.
Step 4 Determine the standard error of the sampling distribution.
The standard error is calculated as
2
1

2
2

n1

n2

112
100

162
100

1.9416

Step 5 Determine the critical value, z, from the standard normal table.
The interval estimate will be developed using a 95% confidence interval.
Because the population standard deviations are known, the critical value is a
z-value from the standard normal table. The critical value is


z = 1.96


www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

Step 6 Develop the confidence interval estimate using Equation 3.
( x1

2
1

2
2

n1

n2

x 2)

z

7.9

1.96

7.9

112
100
3.8056

162
100

The 95% confidence interval estimate for the difference in mean time spent in
the Winston-Salem Axiom Fitness center between men and women is
-11.7056 … 1m1 - m2 2 … - 4.0944

Thus, based on the sample data and the specified confidence level, women spend
on average between 4.09 and 11.71 minutes longer at this Axiom Fitness Center.
>>

END EXAMPLE

TRY PROBLEM 4

Estimating the Difference between Two Means When
S1 and S2 Are Unknown, Using Independent Samples

Chapter Outcome 1.

When estimating a single population mean when the population standard deviation is
unknown, the critical value is a t-value from the t-distribution. This is also the case when you
are interested in estimating the difference between two population means, if the following
assumptions hold:
Assumptions

r The populations are normally distributed.
r The populations have equal variances.
r The samples are independent.
The following application illustrates how a confidence interval estimate is developed
using the t-distribution.

Nick Barounis/Shutterstock

BUSINESS APPLICATION ESTIMATING THE DIFFERENCE BETWEEN TWO
POPULATION MEANS

RETIREMENT INVESTING A major political issue for the past decade
has focused on the long-term future of the U.S. Social Security system.
Many people who have entered the workforce in the past 20 years believe the
system will not be solvent when they retire, so they are actively investing in
their own retirement accounts. One investment alternative is a tax-sheltered
annuity (TSA) marketed by life insurance companies. Certain people,
depending on occupation, qualify to invest part of their paychecks in a TSA
and to pay no federal income tax on this money until it is withdrawn. While

the money is invested, the insurance companies invest it in either stock or
bond portfolios. A second alternative open to many people is a plan known
as a 401(k), in which employees contribute a portion of their paychecks to
purchase stocks, bonds, or mutual funds. In some cases, employers match all or part of the employee
contributions. In many 401(k) systems, the employees can control how their funds are invested.
A recent study was conducted in North Dakota to estimate the difference in mean annual
contributions for individuals covered by the two plans [TSA or 401(k)]. A simple random
sample of 15 people from the population of adults who are eligible for a TSA investment
was selected. A second sample of 15 people was selected from the population of adults in
North Dakota who have 401(k) plans. The variable of interest is the dollar amount of money
invested in the retirement plan during the previous year. Specifically, we are interested in
estimating m1 -m2 using a 95% confidence interval estimate where:
m1 = Mean dollars invested by the TSA -eligible population during the past year
m2 = Mean dollars invested by the 4011k2-eligible population during the past year




www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

TSA–Eligible

401(k)–Eligible

n1 = 15
x1 = +2,119.70
s1 = +709.70

n2 = 15

x2 = +1,777.70
s2 = +593.90

Before applying the t-distribution, we need to determine whether the assumptions are likely
to be satisfied. First, the samples are considered independent because the amount invested by
one group should have no influence on the likelihood that any specific amount will be found
for the second sample.
Next, Figure 1 shows the sample data and the box and whisker plots for the two samples.
These plots exhibit characteristics that are reasonably consistent with those associated with
normal distributions and approximately equal variances. Although using a box and whisker
plot to check the t-distribution assumptions may seem to be imprecise, studies have shown the
t-distribution to be applicable even when there are small violations of the assumptions. This is
particularly the case when the sample sizes are approximately equal.2
Equation 4 can be used to develop the confidence interval estimate for the difference
between two population means when you have small independent samples.
Confidence Interval Estimate for M1 −M2 When S1 and S2
Are Unknown, Independent Samples
( x1

x 2)

ts p

1
n1

1
n2

(4)

where:
sp
t

FIGURE 1

Note: TSA, tax-sheltered annuity.

Pooled standarrd deviation

Critical t -value from the t -distribution, with degrees of freedom
equaal to n1 n2 2

 | 

Box and Whisker Plot

Sample Information for the
Investment Study



(n1 1)s12 (n2 1)s22
n1 n2 2

3,300
TSA
3,122
3,253

2,021
2,479
2,318
1,407
2,641
1,648
2,439
1,059
2,799
1,714
951
2,372
1,572

401(k)
1,781
2,594
1,615
334
2,322
2,234
2,022
1,603
1,395
1,604
2,676
1,773
1,156
2,092
1,465

2,830
2,330
1,830
1,330
830

TSA
401(k)

330
Box and Whisker Plot
Five-Number Summary
TSA 401(k)
951
334
Minimum
1,465
First Quartile 1,572
2,318
1,773
Median
2,234
d Quartile 2,641
3,253
2,676
Maximum


www.downloadslide.com

E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

To use Equation 4, we must compute the pooled standard deviation, sp. If the equalvariance assumption holds, then both s21 and s22 are estimators of the same population variance,
s2. To use only one of these, say s21, to estimate s2 would be disregarding the information
obtained from the other sample. To use the average of s21 and s22, if the sample sizes were different, would ignore the fact that more information about s2 is obtained from the sample having the larger sample size. We therefore use a weighted average of s21 and s22, denoted as s2p, to
estimate s2, where the weights are the degrees of freedom associated with each sample. The
square root of s2p is known as the pooled standard deviation and is computed using
sp

(n1

1)s12 (n2 1)s22
n1 n2 2

Notice that the sample size we have available to estimate s2 is n1 + n2. However, to produce
sp, we must first calculate s21 and s22. This requires that we estimate m1 and m2 using x1 and
x2, respectively. The degrees of freedom are equal to the sample size minus the parameters
estimated before the variance estimate is obtained. Therefore, our degrees of freedom must
equal n1 + n2 - 2.
For the retirement investing example, the pooled standard deviation is
sp

(n1

1)s12 (n2 1)s22
n1 n2 2

(15 1)(709.7)2 (15 1)(593.9)2
15 15 2

654.37

Using the t-distribution table, the critical t-value for
n1 + n2 - 2 = 15 + 15 - 2 = 28
degrees of freedom and 95% confidence is
t = 2.0484
Now we can develop the interval estimate using Equation 4:
( x1
(2,119.70

x2 )

1, 777.70)
342

tsp

1
n1

1
n2

2.0484(654.37)

1
15

1
15

489.45

Thus, the 95% confidence interval estimate for the difference in mean dollars for people
who invest in a TSA versus those who invest in a 401(k) is
- +147.45 … 1m1 - m2 2 … +831.45
This confidence interval estimate crosses zero and therefore indicates there may be no
difference between the mean contributions to TSA accounts and to 401(k) accounts by adults
in North Dakota. The implication of this result is that the average amount invested by those
individuals who invest in pretax TSA programs is no more or no less than that invested by
those participating in after-tax 401(k) programs. Based on this result, there may be an opportunity to encourage the TSA investors to increase deposits.

EXAMPLE 2

CONFIDENCE INTERVAL ESTIMATE FOR M1 − M2 WHEN S1
AND S2 ARE UNKNOWN, USING INDEPENDENT SAMPLES

stocklight/Shutterstock

Andreason Marketing, Inc. Andreason Marketing, Inc. has
been hired by a major newspaper in the U.S. to estimate the difference in mean time that newspaper subscribers spend reading
the Saturday newspaper when subscribers age 50 and under are
compared with those more than 50 years old. A simple random
sample of six people age 50 or younger and eight people over 50
participated in the study. The estimate can be developed using
the following steps:



www.downloadslide.com

E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

Step 1 Define the population parameter of interest and select independent
samples from the two populations.
The objective here is to estimate the difference between the two age groups
with respect to the mean time spent reading the Saturday edition of the newspaper. The parameter of interest is m1 - m2.
The marketing company has selected simple random samples of six
“younger” and eight “older” people. Because the reading time by one person does not influence the reading time for any other person, the samples are
independent.
Step 2 Specify the confidence level.
The marketing firm wishes to have a 95% confidence interval estimate.
Step 3 Compute the point estimate.
The resulting sample means and sample standard deviations for the two
groups are
age … 50: x1 = 13.6 minutes
s1 = 3.1 minutes
n1 = 6

age 7 50: x2 = 11.2 minutes
s2 = 5.0 minutes
n2 = 8

The point estimate is
x1

x2

13.6 11.2

2.4 minutes

Step 4 Determine the standard error of the sampling distribution.
The pooled standard deviation is computed using
sp

(n1 1) s12 (n2 1) s22
n1 n2 2

(6 1)3.12 (8 1)52
6 8 2

4.31

The standard error is then calculated as
sp

1
n1

1
n1

1
6

4.31

1
8

2.3277

Step 5 Determine the critical value, t, from the t-distribution table.
Because the population standard deviations are unknown, the critical value
will be a t-value from the t-distribution as long as the population variances
are equal and the populations are assumed to be normally distributed.
The critical t for 95% confidence and 6 + 8 - 2 = 12 degrees of
freedom is
t = 2.1788
Step 6 Develop a confidence interval using Equation 4.
( x1

x2 )

tsp

1
n1

1
n2

where:
sp

(n1 1) s12 (n2 1) s22
n1 n2 2

(6 1)3.12 (8 1)52
6 8 2

Then the interval estimate is
2.4
2..4
2.6715


2.1788(4.31)
5.0715
( 1
2)

1
6

7.4715

1
8

4.31


www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

Because the interval crosses zero, we cannot conclude that a difference exists
between the age groups with respect to the mean reading time for the Saturday
edition. Thus, with respect to this factor, it does not seem to matter whether the
person is 50 or younger or over 50.

>>

END EXAMPLE

TRY PROBLEM 1

What If the Population Variances Are Not Equal? If you have reason to believe that
the population variances are substantially different, Equation 4 is not appropriate for computing the confidence interval. Instead of computing the pooled standard deviation as part of the
confidence interval formula, we use Equations 5 and 6.
Confidence Interval for M1 − M2 When S1 and S2 Are Unknown and Not
Equal, Independent Samples
( x1 x 2 )

t

s12
n1

s22
n2

(5)

where:
t is from the t-distribution with degrees of freedom computed using
Degrees of Freedom for Estimating Difference between
Population Means When S1 and S2 Are Not Equal
df

Andre Blais/Shutterstock

EXAMPLE 3

(s12 /n1
(s12 /n1 )2
n1 1

s22 /n2 )2
(s22 /n2 )2
n2 1

(6)

ESTIMATING M1 − M2 WHEN THE POPULATION VARIANCES
ARE NOT EQUAL

Citibank The marketing managers at Citibank are planning to roll out a
new marketing campaign addressed at increasing bank card use. As one
part of the campaign, the company will be offering a low interest rate
incentive to induce people to spend more money using its charge cards.
However, the company is concerned whether this plan will have a different impact on married card holders than on unmarried card holders. So,
prior to starting the marketing campaign nationwide, the company tests
it on a random sample of 30 unmarried and 25 married customers. The managers wish to estimate the difference in mean credit card spending for unmarried versus married for a two-week
period immediately after being exposed to the marketing campaign. Based on past data, the
managers have reason to believe the spending distributions for unmarried and married will be
approximately normally distributed, but they are unwilling to conclude the population variances for spending are equal for the two populations.
A 95% confidence interval estimate for the difference in population means can be developed using the following steps:
Step 1 Define the population parameter of interest.
The parameter of interest is the difference between the mean dollars spent on
credit cards by unmarried versus married customers in the two-week period

after being exposed to Citi’s new marketing program.
Step 2 Specify the confidence level.
The research manager wishes to have a 95% confidence interval estimate.
Step 3 Compute the point estimate.
Independent samples of 30 unmarried and 25 married customers were
taken, and the credit card spending for each sampled customer during



www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

the two-week period was recorded. The following sample results were
observed:

Mean
St. Dev.

Unmarried

Married

$455.10
$102.40

$268.90
$ 77.25

The point estimate is the difference between the two sample means:
x1

Point estimate

x2

455.10

268.90

186.20

Step 4 Determine the standard error of the sampling distribution.
The standard error is calculated as
s12
n1

s22
n2

102.40 2
30

77.252
25

24.25

Step 5 Determine the critical value, t, from the t-distribution table.
Because we are unable to assume the population variances are equal, we must
first use Equation 6 to calculate the degrees of freedom for the t-distribution.

This is done as follows:
df

(s12 /n1
(s12 /n1 )2
n1 1

s22 /n2 )2
(s22 /n2 )2
n2 1

(102.40 2 /30 77.252 /25)2
(102.40 2 /330)2 (77.252 / 25)2
29
24

346, 011.98
6, 586.81

52.53

Thus, the degrees of freedom (rounded down) will be 52. At the 95% confidence
level, using the t-distribution table, the approximate t-value is 2.0086. Note,
since there is no entry for 52 degrees of freedom in the table, we have selected
the t-value associated with 95% confidence and 50 degrees of freedom, which
provides a slightly larger t-value than would have been the case for 52 degrees
of freedom. Thus, the interval estimate will be generously wide.
Step 6 Develop the confidence interval estimate using Equation 5.
The confidence interval estimate is computed using
( x1

x2 )

t

s12
n1

s22
n2

Then the interval estimate is
$455.10

102.40 2
30

$268.90

2.0086

$186.20
$137.48

$ 48.72
( 1
2)

77.252
25

$234.992

$137.48 ————— $234.92
The test provides evidence to conclude unmarried customers, after being
introduced to the marketing program, spend more than married customers, on
average, by anywhere from $137.48 to $234.92 in the two weeks following the
marketing campaign. But before concluding that the campaign is more effective
for unmarried than married customers, the managers would want to compare
these results with data from customer accounts prior to the marketing campaign.
>>

END EXAMPLE

TRY PROBLEM 6




www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

MyStatLab
10-1: Exercises
Skill Development
10-1. The following information is based on independent
random samples taken from two normally distributed
populations having equal variances:
n1 = 15
x1 = 50

s1 = 5

n2 = 13
x2 = 53
s2 = 6

Based on the sample information, determine the 90%
confidence interval estimate for the difference between
the two population means.
10-2. The following information is based on independent
random samples taken from two normally distributed
populations having equal variances:
n1 = 24
x1 = 130
s1 = 19

n2 = 28
x2 = 125
s2 = 17.5

29
35
21

25
35
29

Sample 2
31

37
34

42
42
46

39
40
39

38
43
35

10-4. Construct a 95% confidence interval estimate for the
difference between two population means based on the
following information:
Population 1

Population 2

x1 = 355
s1 = 34
n1 = 50

x2 = 320
s2 = 40
n2 = 80

10-5. Construct a 95% percent confidence interval for
the difference between two population means using
the following sample data that have been selected
from normally distributed populations with different
population variances:
Sample 1
473
346
388

386
438
388

406
391
456

Sample 2
379
328
429

349
398
363

x1 = 42.3 x2 = 32.4

a. If s1 = 3 and s2 = 2 and the sample sizes are

n1 = 50 and n2 = 50, construct a 95% confidence
interval for the difference between the two
population means.
b. If s1 = s2, s1 = 3, and s2 = 2, and the sample
sizes are n1 = 10 and n2 = 10, construct a 95%
confidence interval for the difference between the
two population means.
c. If s1 ≠ s2, s1 = 3, and s2 = 2, and the sample
sizes are n1 = 10 and n2 = 10, construct a 95%
confidence interval for the difference between the
two population means.

Business Applications

Based on the sample information, determine the 95%
confidence interval estimate for the difference between
the two population means.
10-3. Construct a 90% confidence interval estimate for the
difference between two population means given the
following sample data selected from two normally
distributed populations with equal variances:
Sample 1

10-6. Two random samples were selected independently
from populations having normal distributions. The
following statistics were extracted from the samples:

359
401
437

346
411
388

395
384
273

10-7. Amax Industries operates two manufacturing facilities
that specialize in doing custom manufacturing
work for the semiconductor industry. The facility
in Denton, Texas, is highly automated, whereas
the facility in Lincoln, Nebraska, has more manual
functions. For the past few months, both facilities
have been working on a large order for a specialized
product. The vice president of operations is interested
in estimating the difference in mean time it takes to
complete a part on the two lines. To do this, he has
requested that a random sample of 15 parts at each
facility be tracked from start to finish and the time
required be recorded. The following sample data were
recorded:
Denton, Texas

Lincoln, Nebraska

x1 = 56.7 hours
s1 = 7.1 hours

x2 = 70.4 hours
s1 = 8.3 hours

Assuming that the populations are normally distributed
with equal population variances, construct and interpret
a 95% confidence interval estimate.
10-8. A credit card company operates two customer service
centers: one in Boise and one in Richmond. Callers
to the service centers dial a single number, and a
computer program routs callers to the center having
the fewest calls waiting. As part of a customer service
review program, the credit card center would like to
determine whether the average length of a call (not
including hold time) is different for the two centers.
The managers of the customer service centers are
willing to assume that the populations of interest are
normally distributed with equal variances. Suppose



www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

a random sample of phone calls to the two centers is
selected and the following results are reported:

Sample Size
Sample Mean (seconds)
Sample St. Dev. (seconds)

Boise

Richmond

120
195
35.10

135
216
37.80

a. Using the sample results, develop a 90% confidence
interval estimate for the difference between the two
population means.
b. Based on the confidence interval constructed in part
a, what can be said about the difference between the
average call times at the two centers?
10-9. A pet food producer manufactures and then fills
25-pound bags of dog food on two different production
lines located in separate cities. In an effort to determine
whether differences exist between the average fill rates
for the two lines, a random sample of 19 bags from line
1 and a random sample of 23 bags from line 2 were
recently selected. Each bag’s weight was measured and
the following summary measures from the samples were
reported:

Sample Size, n
Sample Mean, x

Sample Standard Deviation, s

Production
Line 1

Production
Line 2

19
24.96
0.07

23
25.01
0.08

Management believes that the fill rates of the two lines
are normally distributed with equal variances.
a. Calculate the point estimate for the difference
between the population means of the two lines.
b. Develop a 95% confidence interval estimate of the
true mean difference between the two lines.
c. Based on the 95% confidence interval estimate
calculated in part b, what can the managers of the
production lines conclude about the differences
between the average fill rates for the two lines?
10-10. Two companies that manufacture batteries for electronics
products have submitted their products to an independent
testing agency. The agency tested 200 of each company’s
batteries and recorded the length of time the batteries

lasted before failure. The following results were
determined:
Company A

Company B

x = 41.5 hours
s = 3.6

x = 39.0 hours
s = 5.0

a. Based on these data, determine the 95% confidence
interval to estimate the difference in average life of
the batteries for the two companies. Do these data
indicate that one company’s batteries will outlast the
other company’s batteries on average? Explain.


b. Suppose the manufacturers of each of these batteries
wished to warranty their batteries. One small
company to which they both ship batteries receives
shipments of 200 batteries weekly. If the average
length of time to failure of the batteries is less than a
specified number, the manufacturer will refund the
company’s purchase price of that set of batteries.
What value should each manufacturer set if they wish
to refund money on at most 5% of the shipments?
10-11. Wilson Construction and Concrete Company is known
as a very progressive company that is willing to try

new ideas to improve its products and service. One
of the key factors of importance in concrete work
is the time it takes for the concrete to “set up.” The
company is considering a new additive that can be
put in the concrete mix to help reduce the setup time.
Before going ahead with the additive, the company
plans to test it against the current additive. To do this,
14 batches of concrete are mixed using each of the
additives. The following results were observed:
Old Additive

New Additive

x = 17.2 hours
s = 2.5 hours

x = 15.9 hours
s = 1.8 hours

a. Use these sample data to construct a 90%
confidence interval estimate for the difference in
mean setup time for the two concrete additives. On
the basis of the confidence interval produced, do
you agree that the new additive helps reduce the
setup time for cement? (Assume the populations are
normally distributed.) Explain your answer.
b. Assuming that the new additive is slightly more
expensive than the old additive, do the data support
switching to the new additive if the managers of
the company are primarily interested in reducing

average setup time?
10-12. A working paper (Mark Aguiar and Erik Hurst,
“Measuring Trends in Leisure: The Allocation of Time
over Five Decades,” 2006) for the Federal Reserve
Bank of Boston concluded that average leisure time
spent per week by women in 2003 was 33.80 hours and
37.56 hours for men. The sample standard deviations
were 40 and 70, respectively. These results were
obtained from samples of women and men of size 8,492
and 6,752, respectively. In this study, leisure refers to
the time individuals spent socializing, in passive leisure,
in active leisure, volunteering, in pet care, gardening,
and recreational child care. Assume that the amount
of leisure time spent by men and women have normal
distributions with equal population variances.
a. Determine the pooled estimate of the common
populations’ standard deviation.
b. Produce the margin of error to estimate the
difference of the two population means with a
confidence level of 95%.


www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

c. Calculate a 95% confidence interval for the
difference in the average leisure time between
women and men.
d. Do your results in part c indicate that the average
amount of men’s leisure time was larger than that of

women in 2003? Support your assertions.
e. Would your conclusion in part d change if you did
not assume the population variances were equal?
10-13. The Graduate Management Admission Council
reported a shift in the job-hunting strategies among
second-year masters of business administration (MBA)
candidates. Even though their prospective base salary
has increased from $81,900 to $93,770 from 2002 to
2005, it appears that MBA candidates are submitting
fewer job applications. Data obtained from online
surveys of 1,442 MBA candidates at 30 business
school programs indicate that in 2002 the average
number of job applications per candidate was 38.9 and
2.0 in 2005. The sample variances were 64 and 0.32,
respectively.
a. Examine the sample variances. Conjecture
whether this sample evidence indicates that the
two population variances are equal to each other.
Support your assertion.
b. On the basis of your answer in part a, construct a
99% confidence interval for the difference in the
average number of job applications submitted by
MBA candidates between 2002 and 2005.
c. Using your result in part b, is it plausible that the
difference in the average number of job applications
submitted is 36.5? Is it plausible that the difference
in the average number of job applications submitted
is 37? Are your answers to these two questions
contradictory? Explain.

Computer Database Exercises
10-14. Logston Enterprises operates a variety of businesses
in and around the St. Paul, Minnesota, area.
Recently, the company was notified by the law firm
representing several female employees that a lawsuit
was going to be filed claiming that males were given
preferential treatment when it came to pay raises by
the company. The Logston human resources manager
has requested that an estimate be made of the
difference between mean percentage raises granted
to males versus females. Sample data are contained
in the file Logston Enterprises. She wants you to
develop and interpret a 95% confidence interval
estimate. She further states that the distribution of
percentage raises can be assumed approximately
normal, and she expects the population variances to
be about equal.

10-15. The owner of the A.J. Fitness Center is interested in
estimating the difference in mean years that female
members have been with the club compared with male
members. He wishes to develop a 95% confidence
interval estimate. Sample data are in the file called
AJ Fitness. Assuming that the sample data are
approximately normal and that the two populations
have equal variances, develop and interpret the
confidence interval estimate. Discuss the result.
10-16. Platinum Billiards, Inc., based in Jacksonville, Florida,
is a retailer of billiard supplies. It stands out among
billiard suppliers because of the research it does to

assure its products are top notch. One experiment was
conducted to measure the speed attained by a cue ball
struck by various weighted pool cues. The conjecture
is that a light cue generates faster speeds while
breaking the balls at the beginning of a game of pool.
Anecdotal experience has indicated that a billiard cue
weighing less than 19 ounces generates faster speeds.
Platinum used a robotic arm to investigate this claim.
The research generated the data given in the file titled
Breakcue.
a. Calculate the sample standard deviation and
mean speed produced by cues in the two weight
categories: (1) under 19 ounces and (2) at or above
19 ounces.
b. Calculate a 95% confidence interval for the
difference in the average speed of a cue ball
generated by each of the weight categories.
c. Is the anecdotal evidence correct? Support your
assertion.
d. What assumptions are required so that your results
in part b would be valid?
10-17. The Federal Reserve reported in its comprehensive
Survey of Consumer Finances, released every three
years, that the average income of families in the United
States declined from 2001 to 2004. This was the first
decline since 1989–1992. A sample of incomes was
taken in 2001 and repeated in 2004. After adjusting
for inflation, the data that arise from these samples are
given in a file titled Federal Reserve.
a. Determine the percentage decline indicated by the

two samples.
b. Using these samples, produce a 90% confidence
interval for the difference in the average family
income between 2001 and 2004.
c. Is it plausible that there has been no decline in the
average income of U.S. families? Support your
assertion.
d. How large an error could you have made by using
the difference in the sample means to estimate the
difference in the population means?
END EXERCISES 10-1




www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

Chapter Outcome 1.

2

Hypothesis Tests for Two Population
Means Using Independent Samples

You are going to encounter situations that will require you to test whether two populations
have equal means or whether one population mean is larger (or smaller) than another. These
hypothesis-testing applications are just an extension of the hypothesis-testing process for
a single population mean. They also build directly on the estimation process introduced in
Section 1.

In this section, we will introduce hypothesis-testing techniques for the difference between
the means of two populations in the following situations:
1. The population standard deviations are known and the samples are independent.
2. The population standard deviations are unknown and the samples are independent.
The remainder of this section presents examples of hypothesis tests for these different
situations.

Testing for M1 − M2 When S1 and S2 Are Known,
Using Independent Samples
Samples are considered to be independent when the samples from the two populations are
taken in such a way that the occurrence of values in one sample has no influence on the probability of occurrence of the values in the second sample. In special cases in which the population standard deviations are known and the samples are independent, the test statistic is a
z-value computed using Equation 7.
z-Test Statistic for M1 − M2 When S1 and S2 Are Known,
Independent Samples
z

( x1

x2) (
2
1

1
2
2

n1

n2

2)

(7)

where:
1m1 -m22 = Hypothesized difference in population means

If the calculated z-value using Equation 7 exceeds the critical z-value from the standard normal distribution, the null hypothesis is rejected. Example 4 illustrates the use of this test
statistic.
EXAMPLE 4

HYPOTHESIS TEST FOR M1 − M2 WHEN
KNOWN, INDEPENDENT SAMPLES

S1

AND

S2

ARE

Brooklyn Brick, Inc. Brooklyn Brick, Inc. is a Pennsylvania-based company that makes
bricks and concrete blocks for the building industry. One product is a brick facing material
that looks like a real brick but is much thinner. The ideal thickness is 0.50 inches. The bricks
that the company makes must be very uniform in their dimension so brickmasons can build
straight walls. The company has two plants that produce brick facing products, and the technology used at the two plants is slightly different. At plant 1, the standard deviation in the
thickness of brick facing products is known to be 0.025 inches, and the standard deviation at
plant 2 is 0.034 inches. These are known values. However, the company is interested in determining whether there is a difference in the average thickness of brick facing products made at
the two plants. Specifically, the company wishes to know whether plant 2 also provides brick

facing products that have a greater mean thickness than the products produced at plant 1. If
the test determines that plant 2 does provided thicker materials than plant 1, the managers will



www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

How to do it

(Example 4)

The Hypothesis-Testing Process
for Two Population Means
The hypothesis-testing process
for tests involving two population
means introduced in this section is
essentially the same as for a single
population mean. The process is
composed of the following steps:

have the maintenance department attempt to adjust the process to reduce the mean thickness.
To test this, you can use the following steps:
Step 1 Specify the population parameter of interest.
This is m1 - m2, the difference in the two population means.
Step 2 Formulate the appropriate null and alternative hypotheses.
We are interested in determining whether the mean thickness for plant 2
exceeds that for plant 1. The following null and alternative hypotheses are
specified:
H0: m1 - m2 Ú 0.0

1. Specify the population parameter of interest.

H0: m1
HA: m1
H0: m1
HA: m1
H0: m1
HA: m1

-

m2
m2
m2
m2
m2
m2

= c
two-tailed test
≠c
… c
one-tailed test
7c
Ú c
6 c one-tailed test

Step 3 Specify the significance level for the test.
The test will be conducted using a = 0.05.

Step 4 Determine the rejection region and state the decision rule.
Because the population standard deviations are assumed to be known, the
critical value is a z-value from the standard normal distribution. This test
is a one-tailed lower-tail test, with a = 0.05. From the standard normal
distribution, the critical z-value is
-z0.05 = -1.645
The decision rule compares the test statistic found in Step 5 to the critical
z-value.

where c = any specified number.

If z 6 -1.645, reject the null hypothesis;
Otherwise, do not reject the null hypothesis.

3. Specify the significance level
1a2 for testing the hypothesis.
Alpha is the maximum allowable probability of committing a
Type I statistical error.

4. Determine the rejection region
and develop the decision rule.

5. Compute the test statistic or the
p-value. Of course, you must
first select simple random samples from each population and
compute the sample means.

6. Reach a decision. Apply the
decision rule to determine
whether to reject the null

hypothesis.

7. Draw a conclusion.

HA: m1 6 m2

HA: m1 - m2 6 0.0

2. Formulate the appropriate null
and alternative hypotheses. The
null hypothesis should contain
the equality. Possible formats
for hypotheses testing concerning two populations means are

H0: m1 Ú m2
or

Alternatively, you can state the decision rule in terms of a p-value, as follows:
If p@value 6 a = 0.05, reject the null hypothesis;
Otherwise, do not reject the null hypothesis.
Step 5 Compute the test statistic.
Select simple random samples of brick facing pieces from the two populations
and compute the sample means. A simple random sample of 100 brick facing
pieces is selected from plant 1’s production, and another simple random sample
of 100 brick facing pieces is selected from plant 2’s production. The samples
are independent because the thicknesses of the brick pieces made by one plant
can in no way influence the thicknesses of the bricks made by the other plant.
The means computed from the samples are
x1

0.501 inches

and

x2

0.509 inches

The test statistic is obtained using Equation 7.
z

z

( x1

x2 ) (
2
1

1
2
2

n1

n2

0.501 0.509

2)

0

0.0252

0.034 2

100

100

1.90

Step 6 Reach a decision.
The critical -z0.05 = -1.645, and the test statistic value was computed to be
z = -1.90. Applying the decision rule,
Because z = -1.90 6 -1.645, reject the null hypothesis.
Figure 2 illustrates this hypothesis test.



www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

FIGURE 2

|

Example 4 Hypothesis Test

H0:
HA:

1
1

$
,

2
2

H0:
HA:

or

1–
1

–

2$
2,

0
0

= 0.05

Rejection Region
= 0.05

z
0

– z0.05 = –1.645
z = –1.90
Test Statistic:
(x1 – x2) – (
z =
2
1

n1

+

1–

2)

=

2
2

(0.501– 0.509) – 0

= –1.90

0.0252 0.0342
+
100
100

n2

Decision Rule:
Since z = –1.90 < z = –1.645, reject H0.
Conclude that the brick facings made by plant 2 have a larger mean thickness
than those made by plant 1.

Step 7 Draw a conclusion.
There is statistical evidence to conclude that the brick facings made by plant 2 have
a larger mean thickness than those made by plant 1. Thus, the managers of Brooklyn
Brick, Inc. need to take action to reduce the mean thicknesses from plant 2.
>>

END EXAMPLE

TRY PROBLEM 21

Using p-Values The z-test statistic computed in Example 4 indicates that the difference in sample means is 1.90 standard errors below the hypothesized difference of zero.
Because this falls below the z critical level of –1.645, the null hypothesis is rejected. You could
have also tested this hypothesis using the p-value approach. The p-value for this one-tailed
test is the probability of a z-value in a standard normal distribution being less than –1.90. From the
standard normal table, the probability associated with z = -1.90 is 0.4713. Then the p-value is
p@value = 0.5000 - 0.4713 = 0.0287
The decision rule to use with p-values is

If p@value 6 a, reject the null hypothesis;
Otherwise, do not reject the null hypothesis.
Because
p@value = 0.0287 6 a = 0.05
reject the null hypothesis and conclude that the mean brick facing thickness for plant 2 is
greater than the mean thickness for products produced by plant 1.

Testing M1 − M2 When S1 and S2 Are Unknown,
Using Independent Samples



In Section 1 we showed that to develop a confidence interval estimate for the difference between
two population means when the standard deviations are unknown, we used the t-distribution to
obtain the critical value. As you might suspect, this same approach is taken for hypothesis-testing
situations. Equation 8 shows the t-test statistic that will be used when s1 and s2 are unknown.


www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

t-Test Statistic for M1 − M2 When S1 and S2 Are Unknown and Assumed
Equal, Independent Samples
t

( x1

x2) (
sp

1
n1

1

2)

1
n2

df

,

n1

n2

2

(8)

where:
x1 and x2
1

2

n1 and n2
sp

Sample means from populations 1 annd 2
Hypothesized difference between population means
Sample sizes from the two populations
Pooled standard deviation (see Equation 4)

The test statistic in Equation 8 is based on three assumptions:
Assumptions

r Each population has a normal distribution.3
r The two population variances, s21 and s22, are equal.
r The samples are independent.
Notice that in Equation 8, we are using the pooled estimate for the common population
standard deviation that we developed in Section 1.

Tatiana Popova/Shutterstock

BUSINESS APPLICATION HYPOTHESIS TEST FOR THE DIFFERENCE BETWEEN
TWO POPULATION MEANS

RETIREMENT INVESTING (CONTINUED) Recall the earlier
example discussing a study in North Dakota involving retirement
investing. The leaders of the study are interested in determining whether
there is a difference in mean annual contributions for individuals covered
by TSAs and those with 401(k) retirement programs. A simple random
sample of 15 people from the population of adults who are eligible for
a TSA investment was selected. A second sample of 15 people was
selected from the population of adults in North Dakota who have 401(k) plans. The variables of
interest are the dollars invested in the two retirement plans during the previous year.
Specifically, we are interested in testing the following null and alternative hypotheses:

H0: m1 = m2

H0: m1 - m2 = 0.0
HA: m1 - m2 ≠ 0.0

or

HA: m1 ≠ m2

m1 = Mean dollars invested by the TSA - eligible population during the past year
m2 = Mean dollars invested by the 4011k2 - eligible population during the past year
The leaders of the study select a significance level of a = 0.05. The sample results are
TSA–Eligible

401(k)–Eligible

n1 = 15
x1 = +2,119.70
s1 = +709.70

n2 = 15
x2 = +1,777.70
s2 = +593.90

Because the investments by individuals with TSA accounts are in no way influenced by
investments by individuals with 401(k) accounts, the samples are considered independent. The
box and whisker plots shown earlier in Figure 1 are consistent with what might be expected if
the populations have equal variances and are approximately normally distributed.



www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

We are now in a position to complete the hypothesis test to determine whether the mean
dollar amount invested by TSA employees is different from the mean amount invested by
401(k) employees. We first determine the critical values with degrees of freedom equal to
n1 + n2 - 2 = 15 + 15 - 2 = 28
and a = 0.05 for the two-tailed test.4 The appropriate t-values are
t0.025 = {2.0484 = Critical values
To continue the hypothesis test, we compute the pooled standard deviation.
(n1 1)s12 (n2 1)s22
n1 n2 2

sp

(15 1)(709.7)2 (15 1)(593.9)2
15 15 2

654.37

Note that the pooled standard deviation is partway between the two sample standard deviations. Now, keeping in mind that the hypothesized difference between m1 and m2 is zero, we
compute the t-test statistic using Equation 8, as follows:
tϭ

( x1 Ϫ x2 ) Ϫ (
sp

1 Ϫ 2) ϭ

1
1
ϩ
n1 n2

(2,119.70 Ϫ1, 777.70) Ϫ 0.0
ϭ 1.4313
1
1
654.37
ϩ
15 15

This indicates that the difference in sample means is 1.4313 standard errors above the hypothesized difference of zero. Because
t = 1.4313 6 t0.025 = 2.0484
the null hypothesis should not be rejected.
The difference in sample means is attributed to sampling error. Figure 3 summarizes this
hypothesis test. Based on the sample data, there is no statistical justification to believe that
the mean annual investment by individuals eligible for the TSA option is different from those
individuals eligible for the 401(k) plan.

FIGURE 3

|

Hypothesis Test for the
Equality of the Two Population
Means for the North Dakota
Investment Study

Hypothesis:
H0 : 1 – 2 = 0
HA: 1 – 2 = 0

df = n1 + n2 – 2 = 15 + 15 – 2 = 28

Rejection Region
/2 = 0.025

Rejection Region
/2 = 0.025

–t0.025 = –2.0484

0

t0.025 = 2.0484

t

t = 1.4313
Decision Rule:
x1 – x2 = (2,119.70 – 1,777.70) = 342.00
If t > 2.0484, reject H0.
If t < –2.0484, reject H0.
Otherwise, do not reject H0.
Test Statistic:
t=

(x1 – x2) – (
sp

1–

2)

1
1
+
n1 n2

=

(2,119.70 – 1,777.70) – 0.0

= 1.4313

1
1
+
654.37
15 15

where:
sp =

4You



(n1 – 1)s12 + (n2 – 1) s22
=
n1 + n2 – 2

(15 – 1)(709.7)2 + (15 – 1)(593.9)2
= 654.37
15 + 15 – 2

can also use Excel’s T.INV.2T function 1= T.INV.2T10.05,282.


www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

Excel

tutorials

Excel Tutorial

Robert Wilson/Fotolia

BUSINESS APPLICATION USING EXCEL TO TEST FOR THE DIFFERENCE
BETWEEN TWO POPULATION MEANS

SUV VEHICLE MILEAGE Excel has a procedure for
performing the necessary calculations to test hypotheses
involving two population means. Consider a national car
rental company that is interested in testing to determine

whether there is a difference in mean mileage for sport
utility vehicles (SUVs) driven in town versus those driven
on the highway. Based on its experience with regular
automobiles, the company believes the mean highway mileage will exceed the mean city
mileage.
To test this belief, the company has randomly selected 25 SUV rentals driven only on the
highway and another random sample of 25 SUV rentals driven only in the city. The vehicles were
filled with 14 gallons of gasoline. The company then asked each customer to drive the car until
it ran out of gasoline. At that point, the elapsed miles were noted and the miles per gallon (mpg)
were recorded. For their trouble, the customers received free use of the SUV and a coupon valid
for one week’s free rental. The results of the experiment are contained in the file Mileage.
Excel can be used to perform the calculations required to determine whether the
manager’s belief about SUV highway mileage is justified. We first formulate the null and
alternative hypotheses to be tested:
H0: m1 … m2

H0: m1 - m2 … 0.0
or
HA: m1 - m2 7 0.0

HA: m1 7 m2

Population 1 represents highway mileage, and population 2 represents city mileage. The test
is conducted using a significance level of 0.05 = a.
Figure 4 shows the descriptive statistics for the two independent samples.

FIGURE 4

|

Excel 2010 Output—SUV
Mileage Descriptive Statistics

Excel 2010 Instructions:

1. Open file: Mileage.xlsx.
2. Select Data . Data
Analysis.
3. Select Descriptive
Statistics.
4. Define the data range
for all variables to be
analyzed.
5. Select Summary
Statistics.
6. Specify output location.
7. Click OK.




www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

FIGURE 5

|

Excel 2010 Output (PHStat
Add-in) Box and Whisker

Plot—SUV Mileage Test

Excel 2010 Instructions:

1. Open PHStat and
Enable Macros.
2. Open file: Mileage.xlsx.
3. Click Add-Ins.
4. Click PHStat down
arrow.
5. Select Descriptive
Statistics.
6. Select Boxplot.
7. Select data (both
columns including
headings).
8. Click Multiple
Groups–Unstacked.
9. Click OK.
10. Add titles.

Minitab Instructions (for similar results):

1.
2.
3.
4.

Open file: Mileage.MTW.
Choose Graph > Boxplot.

Under Multiple Ys, select Simple.
Click OK.

5. In Graph variables,
enter data columns.
6. Click OK.

Figure 5 displays the Excel box and whisker plots for the two samples. Based on these
plots, the normal distribution and equal variance assumptions appear reasonable. We will proceed with the test of means assuming normal distributions and equal variances.
Figure 6 shows the Excel output for the hypothesis test. The mean highway mileage is
19.6468 mpg, whereas the mean for city driving is 16.146. At issue is whether this difference in sample means 119.6468 - 16.146 = 3.5008 mpg2 is sufficient to conclude the mean
highway mileage exceeds the mean city mileage. The one-tail t critical value for a = 0.05 is
shown in Figure 6 to be
t 0.05 = 1.6772
Figure 6 shows that the “t-Stat” value from Excel, which is the calculated test statistic (or
t-value, based on Equation 8), is equal to
t = 2.52
The difference in sample means (3.5008 mpg) is 2.52 standard errors larger than the hypothesized difference of zero. Because the test statistic
t = 2.52 7 t0.05 = 1.6772
we reject the null hypothesis. Thus, the sample data do provide sufficient evidence to conclude
that mean SUV highway mileage exceeds mean SUV city mileage, and this study confirms the
expectations of the rental company managers. This will factor into the company’s fuel pricing.
The output shown in Figures 6 also provides the p-value for the one-tailed test, which can
also be used to test the null hypothesis. Recall, if the calculated p-value is less than alpha, the
null hypothesis should be rejected. The decision rule is
If p@value 6 0.05, reject H0.
Otherwise, do not reject H0.
The p-value for the one-tailed test is 0.0075. Because 0.0075 6 0.05, the null hypothesis is
rejected. This is the same conclusion as the one we reached using the test statistic approach.


www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

FIGURE 6

|

Excel 2010 Output for the
SUV Mileage t-Test for Two
Population Means

Excel 2010 Instructions:

1. Open file: Mileage.xlsx.
2. Select Data . Data
Analysis.
3. Select t-test: Two
Sample Assuming
Equal Variances.
4. Define data ranges for
the two variables of
interest.
5. Set Hypothesized
Difference to 0.0.
6. Set Alpha at 0.05.
7. Specify Output Location.
8. Click OK.
9. Click the Home tab and

adjust decimal points
in output.

Minitab Instructions (for similar results):

1. Open file: Mileage.MTW.
2. Choose Stat > Basic Statistics >
2-Sample t.
3. Choose Samples in different columns.
4. In First, enter the first data column.
5. In Second, enter the other data column.

EXAMPLE 5

6. Check Assume equal variances.
7. Click Options and enter 1 – a
in Confidence level.
8. In Alternative choose
greater than.
9. Click OK. OK.

HYPOTHESIS TEST FOR M1 − M2 WHEN S1 AND
UNKNOWN, USING INDEPENDENT SAMPLES

S2

ARE

Color Printer Ink Cartridges A recent Associated Press news story out of Brussels, Belgium,
indicated the European Union was considering a probe of computer makers after consumers complained that they were being overcharged for ink cartridges. Companies such as Canon, HewlettPackard, and Epson are the printer market leaders and make most of their printer-related profits

by selling replacement ink cartridges. Suppose an independent test agency wishes to conduct a
test to determine whether name-brand ink cartridges generate more color pages on average than
competing generic ink cartridges. The test can be conducted using the following steps:
Step 1 Specify the population parameter of interest.
We are interested in determining whether the mean number of pages printed
by name-brand cartridges (population 1) exceeds the mean pages printed by
generic cartridges (population 2).
Step 2 Formulate the appropriate null and alternative hypotheses.
The following null and alternative hypotheses are specified:
H0: m1 … m2

H0: m1 - m2 … 0.0
HA: m1 - m2 7 0.0

or

HA: m1 7 m2



www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

Step 3 Specify the significance level for the test.
The test will be conducted using a = 0.05.
When the populations have standard deviations that are unknown, the
critical value is a t-value from the t-distribution if the populations are assumed
to be normally distributed and the population variances are assumed to be equal.
A simple random sample of 10 users was selected, and the users were
given a name-brand cartridge. A second sample of 8 users was given generic

cartridges. Both groups used their printers until the ink ran out. The number of
pages printed was recorded. The samples are independent because the pages
printed by users in one group did not in any way influence the pages printed by
users in the second group. The means computed from the samples are
x1 = 322.5 pages and x2 = 298.3 pages
Because we do not know the population standard deviations, these values are
computed from the sample data and are
s1 = 48.3 pages and s2 = 53.3 pages
Suppose previous studies have shown that the number of pages printed by both
types of cartridge tends to be approximately normal with equal variances.
Step 4 Construct the rejection region.
Based on a one-tailed test with a = 0.05, the critical value is a t-value from
the t-distribution with 10 + 8 - 2 = 16 degrees of freedom. From the t-table,
the critical t-value is
t 0.05 = 1.7459 = Critical value
The calculated test statistic from step 5 is compared to the critical t-value to
form the decision rule. The decision rule is
If t 7 1.7459, reject the null hypothesis;
Otherwise, do not reject the null hypothesis.
Step 5 Determine the test statistic using Equation 8.
tϭ

( x1 Ϫ x2 ) Ϫ (
sp

1 Ϫ 2)

1
1
ϩ

n1 n2

The pooled standard deviation is
sp

(n1

1)s12 (n2 1)s22
n1 n2 2

(10 1)48.32 (8 1)53.32
10 8 2

50.55

Then the t-test statistic is
t

(322.5

298.3) 0.0
1 1
50.55
10 8

1.0093

Step 6 Reach a decision.
Because
t = 1.0093 6 t 0.05 = 1.7459

do not reject the null hypothesis.
Figure 7 illustrates the hypothesis test.
Step 7 Draw a conclusion.
Based on these sample data, there is insufficient evidence to conclude that the
mean number of pages produced by name-brand ink cartridges exceeds the
mean for generic cartridges.
>> END EXAMPLE

TRY PROBLEM 20




www.downloadslide.com
E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

FIGURE 7

|

Example 5 Hypothesis Test

or

Rejection Region
t

0
t = 1.0093

where:
sp =

(n1– 1)s12 + (n2 – 1)s22

(10 – 1) 48.32 + (8 – 1) 53.32

=

n1+ n2 – 2

10 + 8 – 2

= 50.55

Test Statistic
t=

(322.5 – 298.3) – 0

=
sp

1
1
+
n1 n2

50.55

= 1.0093

1
1
+
10
8

Decision Rule:
Because t = 1.0093 < t0.05 = 1.7459, do not reject H0.

What If the Population Variances Are Not Equal? In the previous examples, we
assumed that the population variances were equal, and we carried out the hypothesis test for
two population means using Equation 8. Even in cases in which the population variances
are not equal, the t-test as specified in Equation 8 is generally considered to be appropriate
as long as the sample sizes are equal.5 However, if the sample sizes are not equal and if the
sample data lead us to suspect that the variances are not equal, the t-test statistic must be
approximated using Equation 9.6 In cases in which the variances are not equal, the degrees of
freedom are computed using Equation 10.

t-Test Statistic for M1 − M2 When Population Variances Are
Unknown and Not Assumed Equal
( x1

t

x2) (
s12

1

s22

n1

n2

2)

(9)

Degrees of Freedom for t-Test Statistic When Population
Variances Are Not Equal
df

(s12 /n1
(s12 /n1)2
n1 1

s22 /n2 )2
(s22 /n2 )2
n2 1

(10)

5Studies

show that when the sample sizes are equal or almost equal, the t distribution is appropriate even when one
population variance is twice the size of the other.



www.downloadslide.com
E s t i m a t i on a nd H y po t he s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

MyStatLab
10-2: Exercises
Skill Development
10-18. A decision maker wishes to test the following null and
alternative hypotheses using an alpha level equal to 0.05:
H0: m1 - m2 = 0
HA: m1 - m2 ≠ 0
The population standard deviations are assumed to
be known. After collecting the sample data, the test
statistic is computed to be
z = 1.78
a. Using the test statistic approach, what conclusion
should be reached about the null hypothesis?
b. Using the p-value approach, what decision should
be reached about the null hypothesis?
c. Will the two approaches (test statistic and p-value)
ever provide different conclusions based on the
same sample data? Explain.
10-19. The following null and alternative hypotheses have
been stated:
H0: m1 - m2 = 0
HA: m1 - m2 ≠ 0
To test the null hypothesis, random samples have been
selected from the two normally distributed populations
with equal variances. The following sample data were

observed:
Sample from Population 1
33
39
25

29
39
33

35
41
38

Sample from Population 2
46
46
50

43
44
43

42
47
39

Test the null hypothesis using an alpha level equal
to 0.05.
10-20. Given the following null and alternative hypotheses

H0: m1 Ú m2
HA: m1 6 m2
together with the following sample information
Sample 1

Sample 2

n1 = 14
x1 = 565
s1 = 28.9

n2 = 18
x2 = 578
s2 = 26.3

a. Assuming that the populations are normally
distributed with equal variances, test at the 0.10
level of significance whether you would reject the
null hypothesis based on the sample information.
Use the test statistic approach.
b. Assuming that the populations are normally
distributed with equal variances, test at the 0.05
level of significance whether you would reject the


null hypothesis based on the sample information.
Use the test statistic approach.
10-21. Given the following null and alternative hypotheses,
conduct a hypothesis test using an alpha equal to 0.05.
(Note: The population standard deviations are assumed

to be known.)
H0: m1 … m2
HA: m1 7 m2
The sample means for the two populations are shown
as follows:
x1 = 144
s1 = 11
n1 = 40

x2 = 129
s2 = 16
n2 = 50

10-22. The following statistics were obtained from independent
samples from populations that have normal distributions:

ni
xi
si

1

2

41
25.4
5.6

51
33.2

7.4

a. Use these statistics to conduct a test of hypothesis if
the alternative hypothesis is m1 - m2 6 -4. Use a
significance level of 0.01.
b. Determine the p-value for the test described in
part a.
c. Describe the type of statistical error that could have
been made as a result of your hypothesis test.
10-23. Given the following null and alternative hypotheses
H0: m1 - m2 = 0
HA: m1 - m2 ≠ 0
and the following sample information
Sample 1

Sample 2

n1 = 125
s1 = 31
x1 = 130

n2 = 120
s2 = 38
x2 = 105

a. Develop the appropriate decision rule, assuming a
significance level of 0.05 is to be used.
b. Test the null hypothesis and indicate whether the
sample information leads you to reject or fail to
reject the null hypothesis. Use the test statistic

approach.
10-24. Consider the following two independently chosen
samples whose population variances are not equal to
each other.
Sample 1

12.1

13.4

11.7

10.7

14.0

Sample 2

10.5

9.5

8.2

7.8

11.1


www.downloadslide.com

E s t i m a t i on a nd H y po t h e s i s Te s t i n g f o r Two Po p u l a t i o n Pa ra m e t e rs

a. Using a significance level of 0.025, test the null
hypothesis that m1 - m2 … 0.
b. Calculate the p-value.

Business Applications
10-25. Descent, Inc., produces a variety of climbing and
mountaineering equipment. One of its products is a
traditional three-strand climbing rope. An important
characteristic of any climbing rope is its tensile
strength. Descent produces the three-strand rope on
two separate production lines: one in Bozeman and
the other in Challis. The Bozeman line has recently
installed new production equipment. Descent regularly
tests the tensile strength of its ropes by randomly
selecting ropes from production and subjecting them to
various tests. The most recent random sample of ropes,
taken after the new equipment was installed at the
Bozeman plant, revealed the following:
Bozeman

Challis

x1 = 7,200 lb
s1 = 425
n1 = 25

x2 = 7,087 lb
s2 = 415

n2 = 20

Descent’s production managers are willing to
assume that the population of tensile strengths for
each plant is approximately normally distributed
with equal variances. Based on the sample results,
can Descent’s managers conclude that there is a
difference between the mean tensile strengths of
ropes produced in Bozeman and Challis? Conduct
the appropriate hypothesis test at the 0.05 level of
significance.
10-26. The management of the Seaside Golf Club regularly
monitors the golfers on its course for speed of play.
Suppose a random sample of golfers was taken in
2005 and another random sample of golfers was
selected in 2006. The results of the two samples are as
follows:
2005

2006

x1 = 225
s1 = 20.25
n1 = 36

x2 = 219
s2 = 21.70
n2 = 31

Based on the sample results, can the management

of the Seaside Golf Club conclude that average speed
of play was different in 2006 than in 2005? Conduct
the appropriate hypothesis test at the 0.10 level of
significance. Assume that the management of the club
is willing to accept the assumption that the populations
of playing times for each year are approximately
normally distributed with equal variances.
10-27. The marketing manager for a major retail grocery
chain is wondering about the location of the stores’
dairy products. She believes that the mean amount
spent by customers on dairy products per visit is
higher in stores in which the dairy section is in the

central part of the store compared with stores that have
the dairy section at the rear of the store. To consider
relocating the dairy products, the manager feels that
the increase in the mean amount spent by customers
must be at least 25 cents. To determine whether
relocation is justified, her staff selected a random
sample of 25 customers at stores in which the dairy
section is central in the store. A second sample of 25
customers was selected in stores with the dairy section
at the rear of the store. The following sample results
were observed:
Central Dairy

Rear Dairy

x1 = +3.74
s1 = +0.87

x2 = +3.26
s2 = +0.79

a. Conduct a hypothesis test with a significance level
of 0.05 to determine if the manager should relocate
the dairy products in those stores displaying their
dairy products in the rear of the store.
b. If a statistical error associated with hypothesis
testing was made in this hypothesis test, what error
could it have been? Explain.
10-28. Sherwin-Williams is a major paint manufacturer.
Recently, the research and development (R&D)
department came out with a new paint product
designed to be used in areas that are particularly
prone to periodic moisture and hot sun. They believe
that this new paint will be superior to anything that
Sherwin-Williams or its competitors currently offer.
However, they are concerned about the coverage area
that a gallon of the new paint will provide compared
to their current products. The R&D department set
up a test in which two random samples of paint were
selected. The first sample consisted of 25 one-gallon
containers of the company’s best-selling paint, and the
second sample consisted of 15 one-gallon containers
of the new paint under consideration. The following
statistics were computed from each sample and refer
to the number of square feet that each gallon will
cover:
Best-Selling Paint

New Paint Product

x1 = 423 sq.feet
s1 = 22.4 sq.feet
n1 = 25

x2 = 406 sq.feet
s2 = 16.8 sq.feet
n2 = 15

The R&D managers are concerned the average area
covered per gallon will be less for the new paint than
for the existing product. Based on the sample data,
what should they conclude if they base their conclusion
on a significance level equal to 0.01?
10-29. Albertsons was once one of the largest grocery chains
in the United States, with more than 1,100 grocery
stores, but in the early 2000s, the company began
to feel the competitive pinch from companies like
Wal-Mart and Costco. In January 2006, the company
announced that it would be sold to SuperValu, Inc.,