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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.

430     C H A P T E R

8     HYPOTHESIS TE STING WITH TWO SAMPLES

EXAMPLE

See Minitab steps
on page 464.

1

A Two-Sample t-Test for the Difference Between Means
The results of a state mathematics test for random samples of students taught
by two different teachers at the same school are shown at the left. Can you
conclude that there is a difference in the mean mathematics test scores for
the students of the two teachers? Use a = 0.10. Assume the populations are
normally distributed and the population variances are not equal.

Sample Statistics for
State Mathematics Test Scores
Teacher 1

Teacher 2

x1 = 473
s1 = 39.7
n1 = 8

x2 = 459
s2 = 24.5
n2 = 18

Solution
Note that s1 and s2 are unknown, the samples are random and independent,
and the populations are normally distributed. So, you can use the t@test. The
claim is “there is a difference in the mean mathematics test scores for the
students of the two teachers.” So, the null and alternative hypotheses are
H0: m1 = m2    and    Ha: m1 ≠ m2.  (Claim)
Because the population variances are not equal and the smaller sample
size is 8, use d.f. = 8 - 1 = 7. Because the test is a two-tailed test with
d.f. = 7 and a = 0.10, the critical values are -t0 = -1.895 and t0 = 1.895.
The rejection regions are t 6 -1.895 and t 7 1.895. The standardized test
statistic is
t =

1 − α = 0.90

=

2

2

= 0.05

−4 − 3 − 2 − 1

1

3

4

− t0 = −1.895 t ≈ 0.922 t0 = 1.895

s21
s22
+
n2
B n1

1473 - 4592 - 0

139.72 2
124.52 2
+
B 8
18

≈ 0.922.

= 0.05
t

0

1x1 - x2 2 - 1m1 - m2 2

Use the t@test (variances are not equal).

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

Round to three decimal places.

The figure at the left shows the location of the rejection regions and the
standardized test statistic t. Because t is not in the rejection region, you fail to
reject the null hypothesis.
Interpretation  There is not enough evidence at the 10% level of significance
to support the claim that the mean mathematics test scores for the students of
the two teachers are different.

Try It Yourself 1
Sample Statistics for
Annual Earnings
High school
diploma

Associate’s
degree

x1 = \$32,493
s1 = \$3118
n1 = 19

x2 = \$40,907
s2 = \$6162
n2 = 16

The annual earnings of 19 people with a high school diploma and 16 people
with an associate’s degree are shown at the left. Can you conclude that there
is a difference in the mean annual earnings based on level of education? Use
a = 0.01. Assume the populations are normally distributed and the population
variances are not equal.  (Adapted from U.S. Census Bureau)
a. Identify the claim and state H0 and Ha.
b. Identify the level of significance a and the degrees of freedom.
c. Find the critical values and identify the rejection regions.
d. Find the standardized test statistic t. Sketch a graph.
e. Decide whether to reject the null hypothesis.
f. Interpret the decision in the context of the original claim.

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

EXAMPLE

431

See TI-84 Plus
steps on page 465.

2

A Two-Sample t-Test for the Difference Between Means
A manufacturer claims that the mean operating cost per mile of its sedans
is less than that of its leading competitor. You conduct a study using
30 randomly selected sedans from the manufacturer and 32 from the leading
competitor. The results are shown at the left. At a = 0.05, can you support the
manufacturer’s claim? Assume the population variances are equal.  (Adapted

Sample Statistics for
Sedan Operating Costs
Manufacturer

Competitor

x1 = \$0.52΋mi
s1 = \$0.05΋mi
n1 = 30

x2 = \$0.55΋mi
s2 = \$0.07΋mi
n2 = 32

from American Automobile Association)

Solution  Note that s1 and s2 are unknown, the samples are random and
independent, and both n1 and n2 are at least 30. So, you can use the t@test. The
claim is “the mean operating cost per mile of the manufacturer’s sedans is less
than that of its leading competitor.” So, the null and alternative hypotheses are

Study Tip

H0: m1 Ú m2    and    Ha: m1 6 m2.  (Claim)
The population variances are equal, so d.f. = n1 + n2 - 2 = 30 + 32 - 2 = 60.
Because the test is a left-tailed test with d.f. = 60 and a = 0.05, the critical
value is t0 = -1.671. The rejection region is t 6 -1.671. To make the
calculation of the standardized test statistic easier, first find the standard error.

It is important to note that
when using a TI-84 Plus
for the two-sample t-test,
select the Pooled: Yes
input option when the
variances are equal.

sx1 - x2 =
=

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

130 - 1210.052 2 + 132 - 1210.072 2 #
1
1
+
B
A
30
32
30 + 32 - 2

≈ 0.0155416

The standardized test statistic is

1 − α = 0.95

t =

α = 0.05

−3

−2

−1

t
0

1

2

3

t ≈ −1.930 t0 = −1.671

1x1 - x2 2 - 1m1 - m2 2

sx1 - x2

Use the t@test (variances are equal).

10.52 - 0.552 - 0
0.0155416

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

Round to three decimal places.

≈ -1.930.

The figure at the left shows the location of the rejection region and the standardized
test statistic t. Because t is in the rejection region, you reject the null hypothesis.
Interpretation  There is enough evidence at the 5% level of significance to
support the manufacturer’s claim that the mean operating cost per mile of its
sedans is less than that of its competitor’s.

Try It Yourself 2
Sample Statistics for
Minivan Operating Costs
Manufacturer

Competitor

x1 = \$0.56΋mi
s1 = \$0.08΋mi
n1 = 34

x2 = \$0.58΋mi
s2 = \$0.07΋mi
n2 = 38

A manufacturer claims that the mean operating cost per mile of its minivans
is less than that of its leading competitor. You conduct a study using
34 randomly selected minivans from the manufacturer and 38 from the leading
competitor. The results are shown at the left. At a = 0.10, can you support the
manufacturer’s claim? Assume the population variances are equal.  (Adapted
from American Automobile Association)

a. Identify the claim and state H0 and Ha.
b. Identify the level of significance a and the degrees of freedom.
c. Find the critical value and identify the rejection region.
d. Find the standardized test statistic t. Sketch a graph.
e. Decide whether to reject the null hypothesis.
f. Interpret the decision in the context of the original claim. Answer: Page A43

432     C H A P T E R

8.2

8     HYPOTHES IS T E STING WITH TWO SAMPLES

Exercises
BUILDING BASIC SKILLS AND VOCABULARY
1. W
 hat conditions are necessary in order to use the t@test to test the difference
between two population means?
2. Explain how to perform a two-sample t@test for the difference between
two population means.
In Exercises 3 – 8, use Table 5 in Appendix B to find the critical value(s) for the
alternative hypothesis, level of significance a, and sample sizes n1 and n2. Assume
that the samples are random and independent, the populations are normally
distributed, and that the population variances are (a) equal and (b) not equal.
3. Ha: m1 ≠ m2, a = 0.10, n1 = 11, n2 = 14
4. Ha: m1 7 m2, a = 0.01, n1 = 12, n2 = 15
5. Ha: m1 6 m2, a = 0.05, n1 = 7, n2 = 11

6. Ha: m1 ≠ m2, a = 0.01, n1 = 19, n2 = 22
7. Ha: m1 7 m2, a = 0.05, n1 = 13, n2 = 8
8. Ha: m1 6 m2, a = 0.10, n1 = 30, n2 = 32
In Exercises 9–12, 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.
9. C
 laim: m1 = m2; a = 0.01. Assume s21 = s22.
Sample statistics: x1 = 33.7, s1 = 3.5, n1 = 12 and
x2 = 35.5, s2 = 2.2, n2 = 17

10. C
 laim: m1 6 m2; a = 0.10. Assume s21 = s22.
Sample statistics: x1 = 0.345, s1 = 0.305, n1 = 11 and
x2 = 0.515, s2 = 0.215, n2 = 9

11. C
 laim: m1 … m2; a = 0.05. Assume s21 ≠ s22.
Sample statistics: x1 = 2410, s1 = 175, n1 = 13 and
x2 = 2305, s2 = 52, n2 = 10

12. C
 laim: m1 7 m2; a = 0.01. Assume s21 ≠ s22.
Sample statistics: x1 = 52, s1 = 4.8, n1 = 32 and
x2 = 50, s2 = 1.2, n2 = 40

USING AND INTERPRETING CONCEPTS
Testing the Difference Between Two Means  In Exercises 13–22,
Sample Statistics for
Annual Food Costs
Dogs

Cats

x1 = \$239
s1 = \$32
n1 = 16

x2 = \$203
s2 = \$21
n2 = 18

TABLE FOR EXERCISE 13

(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 t, (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.
13. Pet Food  A pet association claims that the mean annual costs of food for
dogs and cats are the same. The results for samples of the two types of pets
are shown at the left. At a = 0.10, can you reject the pet association’s claim?
Assume the population variances are not equal.  (Adapted from American Pet
Products Association)

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

Sample Statistics for Amount
Spent by Customers
Burger Stop

Fry World

x1 = \$5.46
s1 = \$0.89
n1 = 22

x2 = \$5.12
s2 = \$0.79
n2 = 30

TABLE FOR EXERCISE 14

433

14. T
 ransactions  A magazine claims that the mean amount spent by a customer
at Burger Stop is greater than the mean amount spent by a customer at
Fry World. The results for samples of customer transactions for the two fast
food restaurants are shown at the left. At a = 0.05, can you support the
magazine’s claim? Assume the population variances are equal.
15. P
 ink Seaperch  A marine biologist claims that the mean length of mature
female pink seaperch is different in fall and winter. A sample of 26 mature
female pink seaperch collected in fall has a mean length of 127 millimeters
and a standard deviation of 14 millimeters. A sample of 31 mature female
pink seaperch collected in winter has a mean length of 117 millimeters
and a standard deviation of 9 millimeters. At a = 0.01, can you support
the marine biologist’s claim? Assume the population variances are equal.
(Source: Fishery Bulletin)

16. B
 lue Crabs  A researcher claims that the stomachs of blue crabs from
Location A contain more fish than the stomachs of blue crabs from Location B.
The stomach contents of a sample of 25 blue crabs from Location A contain
a mean of 320 milligrams of fish and a standard deviation of 60 milligrams.
The stomach contents of a sample of 15 blue crabs from Location B contain
a mean of 280 milligrams of fish and a standard deviation of 80 milligrams.
At a = 0.01, can you support the researcher’s claim? Assume the population
variances are equal.
17. A
 nnual Income  A personnel director from Pennsylvania claims that the
mean household income is greater in Allegheny County than it is in Erie
County. In Allegheny County, a sample of 19 residents has a mean household
income of \$49,700 and a standard deviation of \$8800. In Erie County, a
sample of 15 residents has a mean household income of \$42,000 and a
standard deviation of \$5100. At a = 0.05, can you support the personnel
director’s claim? Assume the population variances are not equal.  (Adapted
from U.S. Census Bureau)

18. A
 nnual Income  A personnel director from Hawaii claims that the mean
household income is the same in Kauai County and Maui County. In
Kauai County, a sample of 18 residents has a mean household income of
\$56,900 and a standard deviation of \$12,100. In Maui County, a sample of
20 residents has a mean household income of \$57,800 and a standard
deviation of \$8000. At a = 0.10, can you reject the personnel director’s
claim? Assume the population variances are not equal.  (Adapted from
U.S. Census Bureau)

19. Tensile Strength  The tensile strength of a metal is a measure of its
ability to resist tearing when it is pulled lengthwise. A new experimental
type of treatment produced steel bars with the tensile strengths (in
newtons per square millimeter) listed below.
Experimental Method:
391  383  333  378  368
401  339  376  366  348
The old method produced steel bars with the tensile strengths (in newtons
per square millimeter) listed below.
Old Method:
362  382  368  398  381  391  400
410  396  411  385  385  395
At a = 0.01, can you support the claim that the new treatment makes a
difference in the tensile strength of steel bars? Assume the population
variances are equal.

434     C H A P T E R

8     HYPOTHES IS T E STING WITH TWO SAMPLES

20. Tensile Strength  An engineer wants to compare the tensile strengths of
steel bars that are produced using a conventional method and an experimental
method. (The tensile strength of a metal is a measure of its ability to resist
tearing when pulled lengthwise.) To do so, the engineer randomly selects
steel bars that are manufactured using each method and records the tensile
strengths (in newtons per square millimeter) listed below.
Experimental Method:
395  389  421  394  407  411  389  402  422
416  402  408  400  386  411  405  389
Conventional Method:
362  352  380  382  413  384  400
378  419  379  384  388  372  383
At a = 0.10, can the engineer support the claim that the experimental
method produces steel with a greater mean tensile strength? Assume the
population variances are not equal.
21. Teaching Methods  A new method of teaching reading is being tested on
third grade students. A group of third grade students is taught using the
new curriculum. A control group of third grade students is taught using
the old curriculum. The reading test scores for the two groups are shown
in the back-to-back stem-and-leaf plot.
Old Curriculum

9
9 9
9 8 8 4 3 3 2 1
7 6 4 2 2 1 0 0

New Curriculum

3
4
5
6
7
8

3
2 4
0 1 1 4 7 7 7 7 7 8 9 9
0 1 1 2 3 3 4 9
2 4

Key: 9 0 4 0 3 = 49 for old curriculum and 43 for new curriculum

At a = 0.10, is there enough evidence to support the claim that the new
old method does? Assume the population variances are equal.
22. Teaching Methods  Two teaching methods and their effects on science
test scores are being reviewed. A group of students is taught in traditional
lab sessions. A second group of students is taught using interactive
simulation software. The science test scores for the two groups are shown
in the back-to-back stem-and-leaf plot.

4
9 9 8 8 7 6 6 3 2 1 0
9 8 5 1 1 1 0 0
2 0

Interactive Simulation Software

6
7
8
9

0 4 5 5 7 7 8
0 0 3 4 7 8 8 9 9
1 3 9

Key: 0 0 9 0 1 = 90 for traditional and 91 for interactive

At a = 0.05, can you support the claim that the mean science test score
is lower for students taught using the traditional lab method than it is for
students taught using the interactive simulation software? Assume the
population variances are equal.

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

435

EXTENDING CONCEPTS
Constructing Confidence Intervals for M1 − M2  When the sampling
distribution for x1 - x2 is approximated by a t-distribution and the population
variances are not equal, you can construct a confidence interval for m1 - m2, as
shown below.

Sample Statistics for
Driving Distances
Golfer 1

Golfer 2

x1 = 267 yd
s1 = 6 yd
n1 = 9

x2 = 244 yd
s2 = 12 yd
n2 = 5

TABLE FOR EXERCISE 23
Sample Statistics for
African Elephant Lifespans
Wild

Zoo

x1 = 56.0 yr
s1 = 8.6 yr
n1 = 20

x2 = 16.9 yr
s2 = 3.8 yr
n2 = 12

TABLE FOR EXERCISE 24

1x1 - x2 2 - tc

s21
s22
s21
s22
+
6 m1 - m2 6 1x1 - x2 2 + tc
+
n2
Bn1 n2
B n1

where d.f. is the smaller of n1 - 1 and n2 - 1

In Exercises 23 and 24, construct the indicated confidence interval for m1 - m2.
Assume the populations are approximately normal with unequal variances.
23. G
 olf  To compare the mean driving distances for two golfers, you randomly
select several drives from each golfer. The results are shown at the left.
Construct a 90% confidence interval for the difference in mean driving
distances for the two golfers.
24. E
 lephants  To compare the mean lifespans of African elephants in the wild
and in a zoo, you randomly select several lifespans from both locations. The
results are shown at the left. Construct a 95% confidence interval for the
difference in mean lifespans of elephants in the wild and in a zoo.  (Adapted
from Science Magazine)

Constructing Confidence Intervals for M1 − M2  When the sampling
distribution for x1 - x2 is approximated by a t-distribution and the populations have
equal variances, you can construct a confidence interval for m1 - m2, as shown below.
1x1 - x2 2 - tc s
n #
where s
n =

1
1
1
1
+
6 m1 - m2 6 1x1 - x2 2 + tc s
n #
+
n2
n2
A n1
A n1

1n1 - 12s21 + 1n2 - 12s22
and d.f. = n1 + n2 - 2
B
n1 + n2 - 2

In Exercises 25 and 26, construct the indicated confidence interval for m1 - m2.
Assume the populations are approximately normal with equal variances.
Sample Statistics for
Kidney Transplants
35 – 49

50 – 64

x1 = 1805 days
s1 = 166 days
n1 = 21

x2 = 1629 days
s2 = 204 days
n2 = 11

TABLE FOR EXERCISE 25

25. K
 idney Transplant Waiting Times  To compare the mean times spent
waiting for a kidney transplant for two age groups, you randomly select
several people in each age group who have had a kidney transplant. The
results are shown at the left. Construct a 95% confidence interval for the
difference in mean times spent waiting for a kidney transplant for the two
age groups.  (Adapted from Organ Procurement and Transplantation Network)
26. C
 omparing Cancer Drugs  In a study, two groups of patients with colorectal
cancer are treated with different drugs. Group A is treated with the drug
Irinotecan and Group B is treated with the drug Fluorouracil. The results
of the study on the number of months in which the groups reported no
cancer-related pain are shown below. Construct a 99% confidence interval
for the difference in mean months with no cancer-related pain for the
two drugs.  (Adapted from The Lancet)
Sample Statistics for Cancer Drugs
Irinotecan
x1 = 10.3 mo
s1 = 1.2 mo
n1 = 52

Fluorouracil
x2 = 8.5 mo
s2 = 1.5 mo
n2 = 50

CASE

STUDY

How Protein Affects Weight Gain in Overeaters

In a study published in the Journal of the American Medical Association, three groups of 18- to 35-year-old
participants overate for an 8-week period. The groups consumed different levels of protein in their diet.
The low protein group’s diet was 5% protein, the normal protein group’s diet was 15% protein, and the high
protein group’s diet was 25% protein. The study found that the low protein group gained considerably less
weight than the normal protein group or the high protein group.
You are a scientist working at a health research firm. The firm wants you to replicate the experiment. You
conduct a similar experiment over an 8-week period. The results of the experiment are shown below.

Weight gain
(after 8 weeks)

Low protein
group

Normal protein
group

High protein
group

x1 = 6.8 lb
s1 = 1.7 lb
n1 = 12

x2 = 13.5 lb
s2 = 2.5 lb
n2 = 16

x3 = 14.2 lb
s3 = 2.1 lb
n3 = 15

EXERCISES
In Exercises 1–3, perform a two-sample t-test to
determine whether the mean weight gains of the two
indicated studies are different. Assume the populations
are normally distributed and the population variances
are equal. For each exercise, write your conclusions as
a sentence. Use a = 0.05.
1. Test the weight gains of the low protein group
against those in the normal protein group.
2. Test the weight gains of the low protein group
against those in the high protein group.

4.In which comparisons in Exercises 1–3 did
you find a difference in weight gains? Write a
5. Construct a 95% confidence interval for m1 - m2,
where m1 is the mean weight gain in the normal
protein group and m2 is the mean weight gain in
the high protein group. Assume the populations
are normally distributed and the population
variances are equal. (See Extending Concepts in
Section 8.2 Exercises.)

3.Test the weight gains of the normal protein
group against those in the high protein group.

436     C H A P T E R

8     HYPOTHESIS TE STING WITH TWO SAMPLES

S E C T I O N 8 . 3      TESTING THE DIFFERENCE BETWEEN MEANS (DEPEND ENT SAMPLES)

437

Testing the Difference Between Means (Dependent Samples)

8.3

WHAT YOU SHOULD LEARN
• How to perform a t-test to test
the mean of the differences for
a population of paired data

Study Tip

The t-Test for the Difference Between Means

THE t-TEST FOR THE DIFFERENCE BETWEEN MEANS
In Sections 8.1 and 8.2, you performed two-sample hypothesis tests with
independent samples using the test statistic x1 - x2 (the difference between
the means of the two samples). To perform a two-sample hypothesis test with
dependent samples, you will use a different technique. You will first find the
difference d for each data pair.
d = 1data entry in first sample2 - 1corresponding data entry in second sample2

Recall from Section 8.1
that two samples are
dependent when each
member of one sample
corresponds to a
member of the
other sample.

The test statistic is the mean d of these differences
d =

Σd
.
n

Mean of the differences between paired
data entries in the dependent samples

These conditions are necessary to conduct the test.
1.  The samples are randomly selected.
2.  The samples are dependent (paired).
3. The populations are normally distributed or the number n of pairs of data is
at least 30.
When these conditions are met, the sampling distribution for d, the mean of the
differences of the paired data entries in the dependent samples, is approximated by
a t@distribution with n - 1 degrees of freedom, where n is the number of data pairs.

−t0

μd

t0

d

The symbols listed in the table are used for the t@test for md. Although
formulas are given for the mean and standard deviation of differences, you
should use technology to calculate these statistics.
Symbol

Study Tip
You can also calculate the
standard deviation of the
differences between paired
data entries using the
shortcut formula

sd =

H

Σd 2 - c

1Σd2 2

n - 1

n

d

n

The number of pairs of data

d

The difference between entries in a data pair

md

The hypothesized mean of the differences of paired data in
the population
The mean of the differences between the paired data entries
in the dependent samples

d

d =
sd

.

Description

Σd
n

The standard deviation of the differences between the paired
data entries in the dependent samples
sd =

B

Σ 1d - d2 2
n - 1

438     C H A P T E R

8     HYPOTHES IS T E STING WITH TWO SAMPLES

Picturing the World
The manufacturer of an
appetite suppressant claims
that when its product is taken
while following a low-fat diet
with regular exercise for
4 months, the average weight
loss is 20 pounds. To test this
claim, you studied 12 randomly
selected dieters taking an appetite
suppressant for 4 months. The
dieters followed a low-fat diet with
regular exercise for all 4 months.
The results are shown in the
Weights (in pounds)
of 12 Dieters
Original
weight

Weight after
4th month

1

185

168

2

194

177

3

213

196

4

198

180

5

244

229

6

162

144

7

211

197

8

273

252

9

178

161

10

192

178

11

181

161

12

209

193

At A = 0.10, does your study
provide enough evidence
to reject the manufacturer’s
claim? Assume the weights
are normally distributed.

When you use a t@distribution to approximate the sampling distribution
for d, the mean of the differences between paired data entries, you can use a
t@test to test a claim about the mean of the differences for a population of
paired data.

t -TEST FOR THE DIFFERENCE BETWEEN MEANS
A t@test can be used to test the difference of two population means when these
conditions are met.
1.  The samples are random.
2.  The samples are dependent (paired).
3.  The populations are normally distributed or n Ú 30.
The test statistic is
d =

Σd
n

and the standardized test statistic is
t =

d - md
sd ΋ 1n

.

The degrees of freedom are
d.f. = n - 1.

GUIDELINES
Using the t-Test for the Difference Between Means (Dependent Samples)
IN WORDSIN SYMBOLS
1. Verify that the samples are random
and dependent, and either the
populations are normally
distributed or n 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. Identify the degrees of freedom.
d.f. = n - 1
5. Determine the critical value(s).
Use Table 5 in Appendix B.
6. Determine the rejection region(s).
Σd
7. Calculate d and sd.
d =
n
Σ1d - d2 2

sd =
B n - 1
d - md
8. Find the standardized test statistic
t =
sd ΋ 1n
and sketch the sampling distribution.
9. Make a decision to reject or fail to
reject the null hypothesis.
10. 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.

S E C T I O N 8 . 3      TESTING THE DIFFERENCE BETWEEN MEANS (DEPEND ENT SAMPLES)

EXAMPLE

1

439

See Minitab steps
on page 464.

The t-Test for the Difference Between Means
A shoe manufacturer claims that athletes can increase their vertical jump
heights using the manufacturer’s training shoes. The vertical jump heights
of eight randomly selected athletes are measured. After the athletes have
used the shoes for 8 months, their vertical jump heights are measured again.
The vertical jump heights (in inches) for each athlete are shown in the table.
At a = 0.10, is there enough evidence to support the manufacturer’s claim?
Assume the vertical jump heights are normally distributed.  (Adapted from
Coaches Sports Publishing)
Athlete

1

2

3

4

5

6

7

8

Vertical jump height
(before using shoes)

24

22

25

28

35

32

30

27

Vertical jump height
(after using shoes)

26

25

25

29

33

34

35

30

Solution
Because the samples are random and dependent, and the populations are normally
distributed, you can use the t@test. The claim is that “athletes can increase their
vertical jump heights.” In other words, the manufacturer claims that an athlete’s
vertical jump height before using the shoes will be less than the athlete’s vertical
jump height after using the shoes. Each difference is given by

Before

After

d

d2

24

26

-2

4

22

25

-3

9

25

25

0

0

28

29

-1

1

35

33

2

4

32

34

-2

4

30

35

-5

25

27

30

-3

9

Σ = - 14

Σ = 56

d = 1jump height before shoes2 - 1jump height after shoes2.

The null and alternative hypotheses are

H0: md Ú 0    and    Ha: md 6 0.  (Claim)
Because the test is a left-tailed test, a = 0.10, and d.f. = 8 - 1 = 7, the
critical value is t0 = -1.415. The rejection region is t 6 -1.415. Using the
table at the left, you can calculate d and sd as shown below. Notice that
the shortcut formula is used to calculate the standard deviation.
d =

sd =

Σd
-14
=
= -1.75
n
8

H

1Σd2 2
1 -142 2
d
56 n
8
=
≈ 2.1213
H
8 - 1
n - 1

Σd2 - c

The standardized test statistic is

Study Tip
You can also use technology and
a P@value to perform a hypothesis
test for the difference between
means. For instance, in Example 1,
you can enter the data in Minitab
(as shown on page 464)
and find P = 0.026.
Because P 6 a, you
reject the null
hypothesis.

t =

1 − α = 0.90

d - md
sd ΋ 1n

α = 0.10

2.1213΋ 18

−3

-1.75 - 0

≈ -2.333.

−2

t ≈ − 2.333

−1

t
0

1

2

3

t0 = −1.415

The figure shows the location of the rejection region and the standardized test
statistic t. Because t is in the rejection region, you reject the null hypothesis.
Interpretation  There is enough evidence at the 10% level of significance to
support the shoe manufacturer’s claim that athletes can increase their vertical
jump heights using the manufacturer’s training shoes.