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1 Testing the Difference Between Means (Independent Samples, &#963;[sub(1)] and [sub(2)] Known)

# 1 Testing the Difference Between Means (Independent Samples, &#963;[sub(1)] and [sub(2)] Known)

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S E C T I O N 8 . 1      TESTING THE DIFFERENCE BETWEEN MEANS (INDEPENDENT SAMPLES, S1 AND S2 KNOWN)

419

AN OVERVIEW OF TWO-SAMPLE HYPOTHESIS
TESTING
In this section, you will learn how to test a claim comparing the means of two
different populations using independent samples.
For instance, an Internet service provider is developing a marketing plan and
wants to determine whether there is a difference in the amounts of time male
and female college students spend online each day. The only way to conclude
with certainty that there is a difference is to take a census of all college students,
calculate the mean daily times male students and female students spend online,
and find the difference. Of course, it is not practical to take such a census.
However, it is possible to determine with some degree of certainty whether such
a difference exists.
To determine whether a difference exists, the Internet service provider
begins by assuming that there is no difference in the mean times of the two
populations. That is,
m1 - m2 = 0.

Assume there is no difference.

Then, by taking a random sample from each population, a two-sample hypothesis
test is performed using the test statistic
x1 - x2 = 0.

Test statistic

The Internet service provider obtains the results shown in the next two figures.

Insight
The members in the
two samples, male
college students and
female college
students, are not
matched or paired,
so the samples
are independent.

Sample 1: Male College Students

Sample 2: Female College Students
Population of Female
College Students

Population of Male
College Students
Sample

x1 = 85 min
s1 = 15 min
n1 = 200

x2 = 81 min
s2 = 17 min
n2 = 250

Sample

The figure below shows the sampling distribution of x1 - x2 for many similar
samples taken from two populations for which m1 - m2 = 0. The figure also
shows the test statistic and the standardized test statistic. From the figure, you
can see that it is quite unlikely to obtain sample means that differ by 4 minutes
assuming the actual difference is 0. The difference of the sample means would
be more than 2.5 standard errors from the hypothesized difference of 0!
Performing a two-sample hypothesis test using a level of significance of a = 0.05,
the Internet service provider can conclude that there is a difference in the
amounts of time male college students and female college students spend online
each day.
Sampling Distribution

Test statistic: x1 − x2 = 85 − 81 = 4
−5 −4 −3 −2 −1

0

1

2

3

4

5

Difference in sample means (in minutes)
−3

−2

−1

x1 − x2

Standardized test statistic
z

0

1

2

3

420     C H A P T E R

8     HYPOTHES IS T E STING WITH TWO SAMPLES

It is important to remember that when you perform a two-sample
hypothesis test using independent samples, you are testing a claim concerning
the difference between the parameters in two populations, not the values of the
parameters themselves.

DEFINITION
For a two-sample hypothesis test with independent samples,
1. the null hypothesis H0 is a statistical hypothesis that usually states there
is no difference between the parameters of two populations. The null
hypothesis always contains the symbol … , =, or Ú .
2. the alternative hypothesis Ha is a statistical hypothesis that is true when H0
is false. The alternative hypothesis contains the symbol 7 , ≠, or 6 .

Study Tip
You can also write the null and
alternative hypotheses as
shown below.
b

H0: m1 - m2 = 0
Ha: m1 - m2 ≠ 0

b

H0: m1 - m2 … 0
Ha: m1 - m2 7 0

b

H0: m1 - m2 Ú 0
Ha: m1 - m2 6 0

To write the null and alternative hypotheses for a two-sample hypothesis
parameters from a verbal statement to a mathematical statement. Then,
write its complementary statement. For instance, for a claim about two
population parameters m1 and m2, some possible pairs of null and alternative
hypotheses are
e

H0: m1 = m2
H : m … m2
H : m Ú m2
,     e 0 1
,    and     e 0 1
.
Ha: m1 ≠ m2
Ha: m1 7 m2
Ha: m1 6 m2

Regardless of which hypotheses you use, you always assume there is no
difference between the population means 1m1 = m2 2.

TWO-SAMPLE z@TEST FOR THE DIFFERENCE
BETWEEN MEANS

In the remainder of this section, you will learn how to perform a z@test for the
difference between two population means m1 and m2 when the samples are
independent. These conditions are necessary to perform such a test.
1. The population standard deviations are known.
2. The samples are randomly selected.
3. The samples are independent.
4. The populations are normally distributed or each sample size is at least 30.
When these conditions are met, the sampling distribution for x1 − x2, the
difference of the sample means, is a normal distribution with mean and standard
error as shown in the table below and the figure at the left.
Sampling Distribution
for x1 − x2

In Words

σ x1 − x2

μ1 − μ 2

x1 − x2

In Symbols

The mean of the difference of the sample
means is the assumed difference between
the two population means. When no
difference is assumed, the mean is 0.

Mean = mx1 - x2

The variance of the sampling distribution
is the sum of the variances of the individual
sampling distributions for x1 and x2. The
standard error is the square root of the
sum of the variances.

Standard error = sx1 - x2

= mx1 - mx2
= m1 - m2

= 2s2x1 + s2x2
=

s21
s22
+
B n1
n2

S E C T I O N 8 . 1      TESTING THE DIFFERENCE BETWEEN MEANS (INDEPENDENT SAMPLES, S1 AND S2 KNOWN)

421

When the conditions on the preceding page are met and the sampling
distribution for x1 - x2 is a normal distribution, you can use the z@test to test
the difference between two population means m1 and m2. The standardized test
statistic takes the form of
z =

1Observed difference2 - 1Hypothesized difference2
.
Standard error

As you read the definition and guidelines for a two-sample z@test, note that if
the null hypothesis states m1 = m2, m1 … m2, or m1 Ú m2, then m1 = m2 is assumed
and the expression m1 - m2 is equal to 0.

TWO-SAMPLE z -TEST FOR THE DIFFERENCE
BETWEEN MEANS

Picturing the World
elementary and secondary school
108,400 in Ohio. In a survey,
200 public elementary and
secondary school teachers in each
state were asked to report their
salary. The results are shown
below. It is claimed that the
mean salary in Ohio is greater
than the mean salary in Georgia.
(Source: National Education Association)

Georgia
x1 = \$52,900
n1 = 200

Ohio

x2 = \$56,700
n2 = 200

Determine a null hypothesis
and alternative hypothesis for
this claim.

A two-sample z@test can be used to test the difference between two population
means m1 and m2 when these conditions are met.
1. Both s1 and s2 are known.
2. The samples are random.
3. The samples are independent.
4. The populations are normally distributed or both n1 Ú 30 and n2 Ú 30.
The test statistic is x1 - x2. The standardized test statistic is
z =

1x1 - x2 2 - 1m1 - m2 2
s21
s22
where  sx1 - x2 =
+
.
sx1 - x2
n2
B n1

GUIDELINES
Using a Two-Sample z-Test for the Difference Between Means
(Independent Samples, S1 and S2 Known)
IN WORDSIN SYMBOLS
1. Verify that s1 and s2 are known, the
samples are random and independent,
and either the populations are normally
distributed or both n1 Ú 30 and n2 Ú 30.
2. State the claim mathematically
State H0 and Ha.
and verbally. Identify the null
and alternative hypotheses.
3. Specify the level of significance.
Identify a.
4. Determine the critical value(s).
Use Table 4 in Appendix B.
5. Determine the rejection region(s).
1x1 - x2 2 - 1m1 - m2 2
6. Find the standardized test statistic
z =
sx1 - x2
and sketch the sampling distribution.
7. Make a decision to reject or fail to
If z is in the rejection region,
reject the null hypothesis.

then reject H0. Otherwise,
fail to reject H0.
8. Interpret the decision in the context
of the original claim.
A hypothesis test for the difference between means can also be performed
using P@values. Use the guidelines above, skipping Steps 4 and 5. After finding
the standardized test statistic, use Table 4 in Appendix B to calculate the
P@value. Then make a decision to reject or fail to reject the null hypothesis. If P
is less than or equal to a, then reject H0. Otherwise, fail to reject H0.

422     C H A P T E R

8     HYPOTHESIS TE STING WITH TWO SAMPLES

See TI-84 Plus
steps on page 465.

2

EXAMPLE

A Two-Sample z-Test for the Difference Between Means

Sample Statistics for
Credit Card Debt
California

Illinois

x1 = \$4777
n1 = 250

x2 = \$4866
n2 = 250

A credit card watchdog group claims that there is a difference in the mean
credit card debts of households in California and Illinois. The results of
a random survey of 250 households from each state are shown at the left.
The two samples are independent. Assume that s1 = \$1045 for California
and s2 = \$1350 for Illinois. Do the results support the group’s claim? Use
a = 0.05.  (Source: PlasticEconomy.com)

Solution
Note that s1 and s2 are known, the samples are random and independent, and
both n1 and n2 are at least 30. So, you can use the z@test. The claim is “there
is a difference in the mean credit card debts of households in California and
Illinois.” So, the null and alternative hypotheses are
H0: m1 = m2    and    Ha: m1 ≠ m2.  (Claim)

Study Tip
In Example 2, you can also use a
P@value to perform the hypothesis
test. For instance, the test is a
two-tailed test, so the P@value is
equal to twice the area to the left
of z = -0.82, or
210.20612 = 0.4122.

Because the test is a two-tailed test and the level of significance is a = 0.05,
the critical values are -z0 = -1.96 and z0 = 1.96. The rejection regions are
z 6 -1.96 and z 7 1.96. The standardized test statistic is
z =

=

Because 0.4122 7 0.05,
you fail to reject H0.

1x1 - x2 2 - 1m1 - m2 2

Use the z@test.

14777 - 48662 - 0

Assume m1 = m2, so m1 - m2 = 0.

Round to two decimal places.

s21
s22
+
n2
B n1

10452
13502
+
B 250
250

≈ -0.82.
1 − α = 0.95

2

2

= 0.025

−3

−2

−1

= 0.025
z

0

1

2

3

− z0 = −1.96 z ≈ − 0.82 z0 = 1.96

The figure at the left shows the location of the rejection regions and the
standardized test statistic z. Because z is not in the rejection region, you fail to
reject the null hypothesis.
Interpretation  There is not enough evidence at the 5% level of significance
to support the group’s claim that there is a difference in the mean credit card
debts of households in California and Illinois.

Try It Yourself 2
A survey indicates that the mean annual wages for forensic science
technicians working for local and state governments are \$55,950 and \$51,100,
respectively. The survey includes a randomly selected sample of size 100
from each government branch. Assume that the population standard
deviations are \$6200 (local) and \$5575 (state). The two samples are
independent. At a = 0.10, is there enough evidence to conclude that
there is a difference in the mean annual wages?  (Source: U.S. Bureau of
Labor Statistics)

a. Identify the claim and state H0 and Ha.
b. Identify the level of significance a.
c. Find the critical values and identify the rejection regions.
d. Find the standardized test statistic z. Sketch a graph.
e. Decide whether to reject the null hypothesis.
f. Interpret the decision in the context of the original claim.

S E C T I O N 8 . 1      TESTING THE DIFFERENCE BETWEEN MEANS (INDEPENDENT SAMPLES, S1 AND S2 KNOWN)

EXAMPLE
Sample Statistics for
Daily Cost of Meals and
Texas

Virginia

x1 = \$234
n1 = 25

x2 = \$240
n2 = 20

423

3

Using Technology to Perform a Two-Sample z-Test
A travel agency claims that the average daily cost of meals and lodging for
vacationing in Texas is less than the average daily cost in Virginia. The table at
the left shows the results of a random survey of vacationers in each state. The
two samples are independent. Assume that s1 = \$19 for Texas and s2 = \$24
for Virginia, and that both populations are normally distributed. At a = 0.01,
is there enough evidence to support the claim? [H0: m1 Ú m2 and Ha: m1 6 m2
(claim)]  (Source: American Automobile Association)

Solution  Note that s1 and s2 are known, the samples are random and

independent, and the populations are normally distributed. So, you can use the
z@test. The top two displays show how to set up the hypothesis test using a TI-84
Plus. The remaining displays show the results of selecting Calculate or Draw.

Study Tip
Note that the TI-84 Plus
displays P ≈ 0.1808.
Because P 7 a, you
fail to reject the
null hypothesis.

T I - 8 4 PLUS

T I - 8 4 PLUS

2-SampZTest

Inpt:Data Stats
s1:19

s2:24

x1:234

n1:25

x2:240
â n2:20

2-SampZTest
s2:24
á

x1:234

n1:25

x2:240

n2:20

µ1:≠µ2 <µ2 >µ2

Calculate Draw

T I - 8 4 PLUS

T I - 8 4 PLUS

2-SampZTest
µ1<µ2

z= - .912448597

p=.1807662795

x1=234

x2=240
â n1=25

z=-.9124

p=.1808

Because the test is a left-tailed test and a = 0.01, the rejection region is
z 6 -2.33. The standardized test statistic z ≈ -0.91 is not in the rejection
region, so you fail to reject the null hypothesis.
Interpretation  There is not enough evidence at the 1% level of significance
to support the travel agency’s claim.

Try It Yourself 3
Sample Statistics for
Daily Cost of Meals and

x1 = \$296
n1 = 15

x2 = \$293
n2 = 20

A travel agency claims that the average daily cost of meals and lodging for
vacationing in Alaska is greater than the average daily cost in Colorado. The
table at the left shows the results of a random survey of vacationers in each
state. The two samples are independent. Assume that s1 = \$24 for Alaska and
s2 = \$19 for Colorado, and that both populations are normally distributed.
At a = 0.05, is there enough evidence to support the claim? [H0: m1 … m2 and
Ha: m1 7 m2 (claim)]  (Source: American Automobile Association)
a. Use technology to find the test statistic or the P@value.
b. Decide whether to reject the null hypothesis.
c. Interpret the decision in the context of the original claim. Answer: Page A43

424     C H A P T E R

8.1

8     HYPOTHES IS T E STING WITH TWO SAMPLES

Exercises
BUILDING BASIC SKILLS AND VOCABULARY
1. W
 hat is the difference between two samples that are dependent and two
samples that are independent? Give an example of each.
2. E
 xplain how to perform a two-sample z@test for the difference between two
population means using independent samples with s1 and s2 known.
3. Describe another way you can perform a hypothesis test for the difference
between the means of two populations using independent samples with s1
and s2 known that does not use rejection regions.
4. W
 hat conditions are necessary in order to use the z@test to test the difference
between two population means?
In Exercises 5– 8, classify the two samples as independent or dependent. Explain

5. Sample 1: The maximum bench press weights for 53 football players

Sample 2: The maximum bench press weights for the same 53 football
players after completing a weight lifting program
6. Sample 1: The IQ scores of 60 females

Sample 2: The IQ scores of 60 males
7. Sample 1: The average speed of 23 powerboats using an old hull design

Sample 2: The average speed of 14 powerboats using a new hull design
8. Sample 1: The commute times of 10 workers when they use their own
vehicles

Sample 2: The commute times of the same 10 workers when they use
public transportation
In Exercises 9 and 10, use the TI-84 Plus display to make a decision to reject or
fail to reject the null hypothesis at the level of significance. Make your decision
using the standardized test statistic and using the P-value. Assume the sample sizes
are equal.
9. a = 0.0510.  a = 0.01

In Exercises 11 –14, test the claim about the difference between two population
means m1 and m2 at the level of significance a. Assume the samples are random
and independent, and the populations are normally distributed. If convenient,
use technology.

11.Claim: m1 = m2; a = 0.1.

Population statistics: s1 = 3.4 and s2 = 1.5

Sample statistics: x1 = 16, n1 = 29 and x2 = 14, n2 = 28

S E C T I O N 8 . 1      TESTING THE DIFFERENCE BETWEEN MEANS (INDEPENDENT SAMPLES, S1 AND S2 KNOWN)

425

12.Claim: m1 7 m2; a = 0.10.

Population statistics: s1 = 40 and s2 = 15

Sample statistics: x1 = 500, n1 = 100 and x2 = 495, n2 = 75
13. Claim: m1 6 m2; a = 0.05.

Population statistics: s1 = 75 and s2 = 105

Sample statistics: x1 = 2435, n1 = 35 and x2 = 2432, n2 = 90
14.Claim: m1 … m2; a = 0.03.

Population statistics: s1 = 136 and s2 = 215

Sample statistics: x1 = 5004, n1 = 144 and x2 = 4895, n2 = 156

USING AND INTERPRETING CONCEPTS
Testing the Difference Between Two Means  In Exercises 15–24,

(a) identify the claim and state H0, and Ha, (b) find the critical value(s) and identify
the rejection region(s), (c) find the standardized test statistic z, (d) decide whether
to reject or fail to reject the null hypothesis, and (e) interpret the decision in the
context of the original claim. Assume the samples are random and independent,
and the populations are normally distributed. If convenient, use technology.
15. B
 raking Distances  To compare the braking distances for two types of tires,
a safety engineer conducts 35 braking tests for each type. The mean braking
distance for Type A is 42 feet. Assume the population standard deviation
is 4.7 feet. The mean braking distance for Type B is 45 feet. Assume the
population standard deviation is 4.3 feet. At a = 0.10, can the engineer
support the claim that the mean braking distances are different for the two
types of tires?  (Adapted from Consumer Reports)
16. M
 eal-Replacement Diets  To compare the amounts spent in the first three
months by clients of two meal-replacement diets, a researcher randomly
selects 20 clients of each diet. The mean amount spent for Diet A is \$643.
Assume the population standard deviation is \$89. The mean amount
spent for Diet B is \$588. Assume the population standard deviation is \$75.
At a = 0.01, can the researcher support the claim that the mean amount
spent in the first three months by clients of Diet A is greater than the mean
amount spent in the first three months by clients of Diet B?  (Adapted from
Consumer Reports)

17. W
 ind Energy  An energy company wants to choose between two regions
in a state to install energy-producing wind turbines. A researcher claims
that the wind speed in Region A is less than the wind speed in Region B.
To test the regions, the average wind speed is calculated for 60 days in
each region. The mean wind speed in Region A is 14.0 miles per hour. Assume
the population standard deviation is 2.9 miles per hour. The mean wind
speed in Region B is 15.1 miles per hour. Assume the population standard
deviation is 3.3 miles per hour. At a = 0.05, can the company support the
researcher’s claim?
18. R
 epair Costs: Washing Machines  You want to buy a washing machine,
and a salesperson tells you that the mean repair costs for Model A and
Model B are equal. You research the repair costs. The mean repair cost of
24 Model A washing machines is \$208. Assume the population standard
deviation is \$18. The mean repair cost of 26 Model B washing machines
is \$221. Assume the population standard deviation is \$22. At a = 0.05, can
you reject the salesperson’s claim?

426     C H A P T E R

8     HYPOTHES IS T E STING WITH TWO SAMPLES

19. A
 CT Scores  The mean ACT score for 43 male high school students is
21.1. Assume the population standard deviation is 5.0. The mean ACT score
for 56 female high school students is 20.9. Assume the population standard
deviation is 4.7. At a = 0.01, can you reject the claim that male and female
high school students have equal ACT scores?  (Adapted from ACT, Inc.)
20. A
 CT Scores  A guidance counselor claims that high school students in a
college preparation program have higher ACT scores than those in a general
program. The mean ACT score for 49 high school students who are in a college
preparation program is 22.2. Assume the population standard deviation is 4.8.
The mean ACT score for 44 high school students who are in a general program
is 20.0. Assume the population standard deviation is 5.4. At a = 0.10, can you
support the guidance counselor’s claim?  (Adapted from ACT, Inc.)
21. H
 ome Prices  A real estate agency says that the mean home sales price in
Spring, Texas, is the same as in Austin, Texas. The mean home sales price for
25 homes in Spring is \$127,414. Assume the population standard deviation
is \$25,875. The mean home sales price for 25 homes in Austin is \$112,301.
Assume the population standard deviation is \$27,110. At a = 0.01, is there
enough evidence to reject the agency’s claim?  (Adapted from RealtyTrac)
22. H
 ome Prices  Refer to Exercise 21. Two more samples are taken, one from
Spring and one from Austin. For 50 homes in Spring, x1 = \$124,329. For
50 homes in Austin, x2 = \$110,483. Use a = 0.01. Do the new samples lead
to a different conclusion?
23. Watching More TV?  A sociologist claims that children ages 6 –17 spent
more time watching television in 1981 than children ages 6 –17 do today.
A study was conducted in 1981 to find the time that children ages 6 –17
spent watching television on weekdays. The results (in hours per weekday)
are shown below. Assume the population standard deviation is 0.6 hour.
2.0  2.5  2.1  2.3  2.1  1.6  2.6  2.1  2.1  2.4
2.1  2.1  1.5  1.7  2.1  2.3  2.5  3.3  2.2  2.9
1.5  1.9  2.4  2.2  1.2  3.0  1.0  2.1  1.9  2.2
 ecently, a similar study was conducted. The results are shown below.
R
Assume the population standard deviation is 0.5 hour.
2.9  1.8  0.9  1.6  2.0  1.7  2.5  1.1  1.6  2.0
1.4  1.7  1.7  1.9  1.6  1.7  1.2  2.0  2.6  1.6
1.5  2.5  1.6  2.1  1.7  1.8  1.1  1.4  1.2  2.3
At a = 0.05, can you support the sociologist’s claim?  (Adapted from

University of Michigan’s Institute for Social Research)

24. Spending More Time Sleeping?  A sociologist claims that children
ages 12–14 spent less time sleeping in 1981 than children ages 12–14 do
today. A study was conducted in 1981 to find the time that children ages
12–14 spent sleeping on weekdays. The results (in hours per weekday)
are shown below. Assume the population standard deviation is 0.5 hour.
7.3  7.5  7.7  7.8  6.9  7.9  8.3  7.9  8.0  8.3
7.4  8.5  7.9  6.8  8.2  7.1  7.9  7.6  8.0  8.0
Recently, a similar study was conducted. The results are shown below.
Assume the population standard deviation is 0.4 hour.
9.2  9.1  10.0  9.3  9.6  8.0  9.5  8.2  9.0  8.6
9.2  9.2   8.9  9.1  8.4  9.0  8.8  9.1  8.6  9.0
At a = 0.01, can you support the sociologist’s claim?  (Adapted from

University of Michigan’s Institute for Social Research)

S E C T I O N 8 . 1      TESTING THE DIFFERENCE BETWEEN MEANS (INDEPENDENT SAMPLES, S1 AND S2 KNOWN)

427

25. G
 etting at the Concept  Explain why the null hypothesis H0: m1 = m2 is
equivalent to the null hypothesis H0: m1 - m2 = 0.
26. G
 etting at the Concept  Explain why the null hypothesis H0: m1 Ú m2 is
equivalent to the null hypothesis H0: m1 - m2 Ú 0.

EXTENDING CONCEPTS
Testing a Difference Other Than Zero  Sometimes a researcher is interested

in testing a difference in means other than zero. For instance, you may want to
determine whether the difference between the mean annual salaries for a job differ
by more than a certain amount between two states. In Exercises 27 and 28, you will
test the difference between two means using a null hypothesis of H0: m1 - m2 = k,
H0: m1 - m2 Ú k, or H0: m1 - m2 … k. The standardized test statistic is still
z =

Microbiologists in
Maryland
x1 = \$102,650
n1 = 42
Microbiologists in
California
x2 = \$85,430
n2 = 38

1x1 - x2 2 - 1m1 - m2 2
s21
s22
where    sx1 - x2 =
+
.
sx1 - x2
n2
B n1

27. M
 icrobiologist Salaries  Is the difference between the mean annual salaries
of microbiologists in Maryland and California more than \$10,000? To decide,
you select a random sample of microbiologists from each state. The results
of each survey are shown in the figure. Assume the population standard
deviations are s1 = \$8795 and s2 = \$9250. At a = 0.05, what should you
conclude?  (Adapted from U.S. Bureau of Labor Statistics)
28. R
 egistered Nurse Salaries  Is the difference between the mean annual
salaries of registered nurses in New Jersey and Delaware equal to \$10,000?
To decide, you select a random sample of registered nurses from each state.
The results of each survey are shown in the figure. Assume the population
standard deviations are s1 = \$8345 and s2 = \$7620. At a = 0.01, what
should you conclude?  (Adapted from U.S. Bureau of Labor Statistics)

FIGURE FOR EXERCISE 27
Registered nurses
in New Jersey

Registered nurses
in Delaware

x1 = \$75,820
n1 = 32

x2 = \$70,820
n2 = 30

Constructing Confidence Intervals for M1 − M2  You can construct a
confidence interval for the difference between two population means m1 - m2, as
shown below, when both population standard deviations are known, and either
both populations are normally distributed or both n1 Ú 30 and n2 Ú 30. Also, the
samples must be randomly selected and independent.
1x1 - x2 2 - zc

s21
s22
s21
s22
+
6 m1 - m2 6 1x1 - x2 2 + zc
+
n2
n2
B n1
B n1

In Exercises 29 and 30, construct the indicated confidence interval for m1 - m2.
29. M
 icrobiologist Salaries  Construct a 95% confidence interval for the
difference between the mean annual salaries of microbiologists in Maryland
and California using the data from Exercise 27.
30. R
 egistered Nurse Salaries  Construct a 99% confidence interval for the
difference between the mean annual salaries of registered nurses in
New Jersey and Delaware using the data from Exercise 28.

428     C H A P T E R

8.2

8     HYPOTHESI S T ESTIN G WITH TWO S AM PLES

Testing the Difference Between Means (Independent Samples, s1 and s2 Unknown)

WHAT YOU SHOULD LEARN
• How to perform a two-sample
t-test for the difference
between two means m1 and m2
using independent samples
with s1 and s2 unknown

The Two-Sample t@Test for the Difference Between Means

THE TWO-SAMPLE t@TEST FOR THE DIFFERENCE
BETWEEN MEANS
In Section 8.1, you learned how to test the difference between means when both
population standard deviations are known. In many real-life situations, both
population standard deviations are not known. In this section, you will learn how
to use a t@test to test the difference between two population means m1 and m2
using independent samples from each population when s1 and s2 are unknown.
To use a t@test, these conditions are necessary.
1.  The population standard deviations are unknown.
2.  The samples are randomly selected.
3.  The samples are independent.
4. The populations are normally distributed or each sample size is at least 30.
When these conditions are met, the sampling distribution for the difference
between the sample means x1 - x2 is approximated by a t@distribution with mean
m1 - m2. So, you can use a two-sample t@test to test the difference between the
population means m1 and m2. The standard error and the degrees of freedom of
the sampling distribution depend on whether the population variances s21 and s22
are equal, as shown in the next definition.

Study Tip
To perform the two-sample t-test
described at the right, you will
need to know whether
the variances of two
populations are equal.
In this chapter, each
example and exercise
will state whether the
variances are equal.
You will learn to test
for differences between
two population variances
in Chapter 10.

TWO-SAMPLE t -TEST FOR THE DIFFERENCE
BETWEEN MEANS
A two-sample t@test is used to test the difference between two population
means m1 and m2 when (1) s1 and s2 are unknown, (2) the samples are random,
(3) the samples are independent, and (4) the populations are normally
distributed or both n1 Ú 30 and n2 Ú 30. The test statistic is x1 - x2, and the
standardized test statistic is
t =

1x1 - x2 2 - 1m1 - m2 2
.
sx1 - x2

Variances are equal: If the population variances are equal, then information
from the two samples is combined to calculate a pooled estimate of the
standard deviation S
n.
s
n =

1n1 - 12s21 + 1n2 - 12s22
B
n1 + n2 - 2

The standard error for the sampling distribution of x1 - x2 is
sx1 - x2 = s
n #

1
1
+
n2
A n1

Variances equal

and d.f. = n1 + n2 - 2.

Variances are not equal: If the population variances are not equal, then the
standard error is
sx1 - x2 =

s21
s22
+
n2
B n1

Variances not equal

and d.f. = smaller of n1 - 1 and n2 - 1.

SECTION 8.2    TESTING THE DIFFERENCE BETWEEN MEANS (INDEPENDENT SAMPLES, S1 AND S2 UNKNOWN)

Picturing the World
American Psychological
Association in the journal
Neuropsychology reported that
children with musical training
showed better verbal memory
than children with no musical
training. The study also showed
that the longer the musical
training, the better the verbal
memory. Suppose you tried
to duplicate the results as
follows. A verbal memory test
with a possible 100 points was
other half had no training and
acted as the control group. The
45 children with training had an
average score of 83.12 with a
standard deviation of 5.7. The
45 students in the control group
had an average score of 79.9
with a standard deviation of 6.2.

429

The requirements for the z@test described in Section 8.1 and the t@test
described in this section are shown in the flowchart below.
Two-Sample Tests for Independent Samples

Are both population
standard deviations
known?

Yes

Are both
populations normal
or are both sample
sizes at least 30?

No
Are both
populations normal
or are both sample
sizes at least 30?

Yes

Use the z-test.

No

No

You cannot use the
z-test or the t-test.

Yes
Are the population
variances equal?

Yes

No

Use the t-test with s x1 − x2 =

Use the t-test with s x1 − x2 = σˆ •

1 1
+
n1 n2

and d.f. = n1 + n2 − 2.

s12 s22
n1 + n2

and d.f. = smaller of n1 − 1 and n2 − 1.

GUIDELINES
At A = 0.05, is there enough
evidence to support the
claim that children with
musical training have better
verbal memory test scores
than those without training?
Assume the population
variances are equal.

Using a Two-Sample t-Test for the Difference Between Means
(Independent Samples, S1 and S2 Unknown)
IN WORDSIN SYMBOLS
1. Verify that s1 and s2 are unknown,
the samples are random and independent,
and either the populations are normally
distributed or both n1 Ú 30 and n2 Ú 30.
2. State the claim mathematically
State H0 and Ha.
and verbally. Identify the null
and alternative hypotheses.
3. Specify the level of significance.
Identify a.
4. Determine the degrees of freedom.d.f. = n1 + n2 - 2 or
d.f. = smaller of n1 - 1
and n2 - 1
5. Determine the critical value(s).
Use Table 5 in Appendix B.
6. Determine the rejection region(s).
1x1 - x2 2 - 1m1 - m2 2
7. Find the standardized test statistic
t =
sx1 - x2
and sketch the sampling distribution.
8. Make a decision to reject or fail to
reject the null hypothesis.

9. Interpret the decision in the context
of the original claim.

If t is in the rejection region,

then reject H0. Otherwise,
fail to reject H0.