8.

1 Hypothesis Tests for Two Population Means (Large Samples) 375

E XAMPLE A math instructor wrote two versions of the same test, which he
8.1 believed to be of equal difficulty. He gave the first version (A) to 36
students, and the second version (B) to 41 students. We will con-
sider the two groups to be random samples from the population of
all community college statistics students. The 36 students who took
version A had a mean score of 79.3 with a standard deviation of
11.26. The 41 students who took version B had a mean score of
84.1 with a standard deviation of 9.40. At the 0.05 level of signifi-
cance, test the claim that the two versions were of equal difficulty.

Since one version was not specifically mentioned first in the claim (“the two ver-
sions were of equal difficulty”), we will let version A be population 1 and version B
be population 2.
Step 1
Population 1: version A; population 2: version B
Claim in words: The two versions are of equal difficulty.
Claim: μ1 = μ2
Complement: μ1 ≠ μ2
H0: μ1 = μ2
HA: μ1 ≠ μ2
Step 2
Level of significance: α = 0.05
Step 3
x 1 – x2
Test statistic: z =
s2
1 s 22
+
n1 n2

Step 4
α = 0.05, two-tailed test

0.9500

0.0250 0.0250

–1.96 –1 0 1 1.96

Decision rule: Reject H0 if z < –1.96 or if z > 1.96.

Step 5

x1 – x2
z =
s 12 s 22
+
n1 n2

79.3 – 84.1
=
11.26 2 9.40 2
+
36 41
= –2.01
376 CHAPTER 8 Two-Sample Hypothesis Tests

Decision: Reject H0.

Conclusion: There is sufficient sample evidence to reject the claim that the two
exams were of equal difficulty.

p-value

2 • P(Z < –2.01)

= 2 • 0.0222
0.0222
= 0.0444 ■

–2 –1 0 1 2
– 2.01

You may be wondering why we did not use the one-mean test developed in the
last chapter. For instance, we could have tested the claim that the mean score for ver-
sion A is below 82, and then tested the claim that the mean score for version B is
above 82. The first problem with this approach is that it is very difficult to come up
with that single number to test each mean against ahead of time. The second prob-
lem associated with using two separate tests involves the level of significance. Recall
that using a level of significance of 0.05 means that the probability of rejecting a null
hypothesis that is actually true is 0.05. However, if we perform two separate hypoth-
esis tests at the 0.05 level of significance, then there is a 0.0975 probability of reject-
ing at least one null hypothesis that is actually true. This is calculated as follows.

P(Rejecting at least 1 true null hypothesis) = 1 – P(not rejecting a true null hypothesis)
= 1 – (0.95)2
= 0.0975

Not only is performing two hypothesis tests more tedious and time consuming, it also
increases our chances of making a Type I error.
We continue with two more examples.

E XAMPLE A reading group claims that Americans read more as they grow
8.2 older. A random sample of 115 Americans age 60 or older read for
a mean length of 62.8 minutes per day, with a standard deviation of
18.3 minutes per day. A random sample of 88 Americans between
the ages of 50 and 59 read for a mean length of 54.2 minutes per
day, with a standard deviation of 23.1 minutes per day. At the 0.01
level of significance, test the claim that the mean time spent read-
ing per day by Americans age 60 and older is longer than the mean
time spent reading per day by Americans between the ages of 50
and 59.

We will let Americans age 60 and older be population 1 as they were the first group
mentioned in the claim. We will let Americans between the ages of 50 and 59 be
population 2.

Step 1
Population 1: 60 or older; population 2: 50 to 59 years old
 8.1 Hypothesis Tests for Two Population Means (Large Samples) 377

Claim in words: The mean time spent reading per day by Americans age 60 and
older is longer than the mean time spent reading per day by Americans between
the ages of 50 and 59.
Claim: μ1 > μ2
Complement: μ1 ≤ μ2
H0: μ1 ≤ μ2
HA: μ1 > μ2

Step 2
Level of significance: α = 0.01

Step 3
x 1 – x2
Test statistic: z =
s2
1 s 22
+
n1 n2

Step 4
α = 0.01, right-tailed test

0.9900
Decision rule: Reject H0 if z > 2.33.

0.01
–2 –1 0 1 2
2.33
Step 5

x1 – x2
z =
s 12 s 22
+
n1 n2

62.8 – 54.2
=
18.3 2 23.1 2
+
115 88
= 2.87

Decision: Reject H0.

Conclusion: There is sufficient sample evidence to support the claim that the mean
time spent reading per day by Americans age 60 and older is longer than the
mean time spent reading per day by Americans between the ages of 50 and 59.

p-value

0.9979 P (Z > 2.87) = 1 – 0.9979

= 0.0021 ■

0 2.87
378 CHAPTER 8 Two-Sample Hypothesis Tests

E XAMPLE Many baseball fans prefer the National League to the American
8.3 League because they believe that the National League has better
pitchers, which produces lower scoring games. A random sample of
71 National League baseball games had a mean combined score of
9.7 runs per game, with a standard deviation of 4.94 runs per game.
A random sample of 62 American League baseball games had a
mean combined score of 10.2 runs per game, with a standard devi-
ation of 4.98 runs per game. At the 0.05 level of significance, test
the claim that the mean number of runs scored in a National League
baseball game is less than it is in the American League.

We will let the National League games be population 1 as they were the first group
mentioned in the claim. We will let the American League games be population 2.

Step 1
Population 1: National League games; population 2: American League games
Claim in words: The mean number of runs scored in a National League baseball
game is less than it is in the American League.
Claim: μ1 < μ2
Complement: μ1 ≥ μ2
H0: μ1 ≥ μ2
HA: μ1 < μ2

Step 2
Level of significance: α = 0.05
Step 3
x 1 – x2
Test statistic: z =
s2
1 s 22
+
n1 n2

Step 4
α = 0.05, left-tailed test

0.9500
0.05

–2 –1 0 1 2
–1.645

Decision rule: Reject H0 if z < –1.645.

Step 5

x1 – x2
z =
s 12 s 22
+
n1 n2
 8.1 Hypothesis Tests for Two Population Means (Large Samples) 379

9.7 – 10.2
=
4.94 2 4.98 2
+
71 62
= –0.58

Decision: Fail to reject H0.

Conclusion: There is not sufficient sample evidence to support the claim that the
mean number of runs scored in a National League baseball game is less than it is
in the American League.

p-value

0.2810 P (Z < – 0.58) = 0.2810 ■

–2 –1 0 1 2
–0.58

Confidence Intervals
Recall that the idea of hypothesis testing is based on confidence intervals. There is a
confidence interval that can be constructed for the difference between two population
means, when both sample sizes are large. We begin, as we did before, by calculating
the margin of error. Here is the formula.

s 12 s 22
E = z α/2 +
n1 n2
– –
We then subtract this margin of error from x 1 – x 2 to find the left endpoint of the con-
– –
fidence interval, and we add the margin of error to x 1 – x 2 to find the right endpoint
of the confidence interval. (If you want to create a one-sided confidence interval, see
the “Extra” material at the end of the Section 6.2 exercises.)

(x–1 – x–2 ) – E ≤ (μ1 – μ2 ) ≤ (x–1 – x–2 ) + E

E XAMPLE A math instructor wrote two versions of the same test that he
8.4 believed to be of equal difficulty. He gave the first version (A) to
36 students and the second version (B) to 41 students. We will
consider the two groups to be random samples from the popula-
tion of all community college statistics students. The 36 students
who took version A had a mean score of 79.3 with a standard devi-
ation of 11.26. The 41 students who took version B had a mean
score of 84.1 with a standard deviation of 9.40. Construct a 95%
confidence interval for the difference between the mean score of
version A and the mean score of version B.
380 CHAPTER 8 Two-Sample Hypothesis Tests

We will let population 1 be the scores on version A. We begin by finding the value
for z α/2. Recall that for a 95% confidence interval, z 0.025 = 1.96.

0.9500

0.0250 0.0250

–1 0 1
–1.96 1.96

Next we calculate the margin of error E.

s 12 s 22
E = z α/2 +
n1 n2

11.26 2 9.40 2
= 1.96 +
36 41
= 4.67

Now we construct the interval.

(x–1 – x–2) – E ≤ (μ1 – μ2) ≤ (x–1 – x–2) + E

(79.3 – 84.1) – 4.67 ≤ (μ1 – μ2) ≤ (79.3 – 84.1) + 4.67
–4.8 – 4.67 ≤ (μ1 – μ2) ≤ –4.8 + 4.67
–9.47 ≤ (μ1 – μ2) ≤ –0.13

We are 95% confident that the difference between the two means is between
–0.13 points and –9.47 points. Note that this interval does not contain 0; this is
strong evidence that the two population means are not equal. This conclusion can
be affirmed by checking the hypothesis test of the first example of the section. ■

MICROSOFT EXCEL Hypothesis Test for

Two Population Means
(Large Samples)
Microsoft Excel’s Data Analysis ToolPak has a built-in procedure for a hypothesis test
comparing two population means. Before using it, we must calculate the variance for
each sample, because their procedure assumes that both population variances are
known.
We will step through an example that demonstrates how to use Excel for this
hypothesis test.

E XAMPLE A mathematics instructor has changed his instruction style to

8.5 include more group activities, both inside and outside of class. He
wants to know whether the results of such an instruction style have
 8.1 Hypothesis Tests for Two Population Means (Large Samples) 381

any effect on students’ performance on a standard department

final exam. Here are the final exam scores of 30 students from
the semester before the change in styles, and 32 students from
the semester after the change. Scores are out of a maximum of
50 points. At the 0.05 level of significance, test the claim that the
mean score for both styles is the same.

Before
34 21 36 28 46 31 45 40
48 42 27 41 39 29 37 37
29 46 34 38 43 46 35 34
32 45 27 40 47 41

After
35 39 50 25 25 36 34 31
45 35 37 32 36 31 44 38
34 23 31 36 21 30 44 45
30 39 39 42 28 47 40 27

In column A, enter the scores for the semester before the change in cells A1
through A30. In column B, enter the scores for the semester after the change
in cells B1 through B32.

In cell C1, type Variance #1, and type Variance #2 in cell D1. To calculate the vari-
ance of the scores in sample 1, type the following in cell C2.

=VAR(A1:A30)

To calculate the variance of the scores in sample 2, type the following in cell D2.

=VAR(B1:B32)

Excel tells us that the variance of the first sample is 50.48 and the variance of the
second sample is 52.53.

We proceed with the test by selecting Data Analysis from the Tools menu. When
the dialog box appears, select z-Test: Two Sample for Means and click on OK.
When the z-Test dialog box opens, type A1:A30 in the box labeled Variable 1
Range and type B1:B32 in the box labeled Variable 2 Range. In the box labeled
Hypothesized Mean Difference, enter 0. (We would enter 5 if we claimed that
the first mean was 5 points higher than the second mean.) In the box labeled
Variable 1 Variance (known) enter 50.48, which was the variance that we cal-
culated earlier. In the box labeled Variable 2 Variance (known) enter 52.53. In
the box labeled Alpha, enter 0.05, which is the level of significance for this test.
Click on OK.

Here is a copy of the output, which appears on another worksheet.

z-Test: Two Sample for Means

Variable 1 Variable 2

Mean 37.26666667 35.28125

Known Variance 50.48 52.53
Observations 30 32
(continues)
382 CHAPTER 8 Two-Sample Hypothesis Tests

(continued)
Variable 1 Variable 2

Hypothesized Mean Difference 0

z 1.088945603
P(Z<=z) one-tail 0.138088975
z Critical one-tail 1.644853
P(Z<=z) two-tail 0.276177949
z Critical two-tail 1.959961082

Excel lists the mean for each sample. The value labeled as z is the test statistic,
which is 1.09. Then Excel lists the p-values and critical values for a one-tailed test
and a two-tailed test. Since our test is a two-tailed test, the p-value is 0.2762 and
the critical value to be used in the decision rule is 1.96.

With this information we can write the hypothesis test.

Step 1
Population 1: previous teaching style; population 2: new teaching style
Claim in words: The mean score for both styles is the same.
Claim: μ1 = μ2
Complement: μ1 ≠ μ2
H0: μ1 = μ2
HA: μ1 ≠ μ2

Step 2
Level of significance: α = 0.05

Step 3
x 1 – x2
Test statistic: z =
s2
1 s 22
+
n1 n2

Step 4
α = 0.05, two-tailed test

0.9500

0.0250 0.0250

–1 0 1
–1.96 1.96

Decision rule: Reject H0 if z < –1.96 or if z > 1.96.

Step 5
z = 1.09 (calculated by Excel)
Decision: Fail to reject H0.
Conclusion: There is not sufficient sample evidence to reject the claim that the
mean score for both styles is the same.
 8.1 Hypothesis Tests for Two Population Means (Large Samples) 383

p-value

p-value = 0.2762
0.1381
(calculated by Excel) ■

–2 –1 0 1.09 2

TI-83 Hypothesis Test for Two Population Means

(Large Samples)
The TI-83 can help us perform the hypothesis test for two population means whether
we have the two sets of sample data or two sets of sample statistics. We will step
through an example of each type, beginning with an example using the sample data.

E XAMPLE A mathematics instructor has changed his instruction style to

8.6 include more group activities, both inside and outside of class. He
wants to know whether the results of such an instruction style have
any effect on students’ performance on a standard department
final exam. Here are the final exam scores of 30 students from the
semester before the change in styles, and 32 students from the
semester after the change. Scores are out of a maximum of 50
points. At the 0.05 level of significance, test the claim that the
mean score for both styles is the same.

Before
34 21 36 28 46 31 45 40
48 42 27 41 39 29 37 37
29 46 34 38 43 46 35 34
32 45 27 40 47 41

After
35 39 50 25 25 36 34 31
45 35 37 32 36 31 44 38
34 23 31 36 21 30 44 45
30 39 39 42 28 47 40 27

Enter the scores for the semester before the change in list L1. Enter the scores for
the semester after the change in list L2. Before proceeding with the test, we must
find both sample standard deviations, which we will use as estimates of the popula-
tion standard deviations. We access the STAT CALC menu by pressing STAT and
384 CHAPTER 8 Two-Sample Hypothesis Tests

using the → key to move to the right. Highlight option 1:1-Var Stats and press
ENTER . When you are brought to the main screen, enter the list L1 by pressing
2nd 1 , and then press ENTER . Make note of the standard deviation, which
is 7.10. Now repeat the process for list L2. Its standard deviation is 7.25.

Now we access the STAT TESTS menu by pressing STAT and using the → key
to move to the right. Highlight option 3:2-SampZTest and press ENTER .

Since we have the sample data, highlight Data next to Inpt:. Enter the first sample
standard deviation (7.10) next to μ1. Enter the second sample standard deviation
(7.25) next to μ2. Next to List1: enter L1 by pressing 2nd 1 . Next to List2:
enter L2 by pressing 2nd 2 . Leave Freq1: and Freq2: as 1. Scroll down to
see the alternate hypothesis. This test is a two-tailed test, so highlight ≠μ2 and
press ENTER . Finally, scroll down one more line to highlight Calculate and press
ENTER . Here are the screens that you should see.

The first line of output gives us our test statistic, which rounds to 1.09. The next line
gives us our p-value, which is 0.2761. The remaining information includes the mean,
standard deviation, and size of each sample. If we repeated the exact same proce-
dure but chose Draw instead of Calculate, we would see the following screen.

This screen also includes the value of the test statistic and p-value in addition to
the graph. ■

We can also use the TI-83 to help when we are given only the sample statistics
instead of all of the data.

E XAMPLE A mathematics instructor has changed his instruction style to

8.7 include more group activities, both inside and outside of class. He
wants to know whether the results of such an instruction style have
any effect on students’ performance on a standard department
final exam. Thirty students from the semester before the change in
styles had a mean score of 37.27 points, with a standard deviation
 8.1 Hypothesis Tests for Two Population Means (Large Samples) 385

of 7.10 points. Thirty-two students from the semester after the

change had a mean score of 35.28 points, with a standard devia-
tion of 7.25 points. Scores are out of a maximum of 50 points. At
the 0.05 level of significance, test the claim that the mean score for
both styles is the same.

Access the STAT TESTS menu by pressing STAT and using the → key to move
to the right. Highlight option 3:2-SampZTest and press ENTER .

Since we only have the sample statistics, highlight Stats next to Inpt:. Enter the
first sample standard deviation (7.10) next to μ1. Enter the second sample standard
–
deviation (7.25) next to μ2. Enter the first sample mean (37.27) next to x1:, and
the first sample size (30) next to n1:. Enter the second sample mean (35.28) next
–
to x2:, and the second sample size (32) next to n2:. Scroll down to see the alter-
nate hypothesis. This test is a two-tailed test, so highlight ≠μ2 and press ENTER .
Finally, scroll down one more line to highlight Calculate and press ENTER . Here
are the screens that you should see.

Note that the results are slightly off when we compare them to the previous results.
This is due to the rounding of the sample means and sample standard deviations.
The first line of output gives us our test statistic, which rounds to 1.09. The next line
gives us our p-value, which is 0.2750. The remaining information includes the mean
and size of each sample.

Here is the hypothesis test for the first set of calculations.

Step 1
Population 1: previous teaching style; population 2: new teaching style
Claim in words: The mean score for both styles is the same.
Claim: μ1 = μ2
Complement: μ1 ≠ μ2
H0: μ1 = μ2
HA: μ1 ≠ μ2

Step 2
Level of significance: α = 0.05

Step 3
x 1 – x2
Test statistic: z =
s2
1 s 22
+
n1 n2

Step 4
α = 0.05, two-tailed test
386 CHAPTER 8 Two-Sample Hypothesis Tests

0.9500

0.0250 0.0250

–1 0 1
–1.96 1.96

Decision rule: Reject H0 if z < –1.96 or if z > 1.96.

Step 5
z = 1.09 (calculated by TI-83)
Decision: Fail to reject H0.
Conclusion: There is not sufficient sample evidence to reject the claim that the
mean score for both styles is the same.

p-value

p-value = 0.2750 (calculated by TI-83) ■

EXERCISES 8.1
Use the standard procedure for all hypothesis tests.
1. A college student randomly surveys 45 female students at her school and finds that
their mean GPA is 3.22, with a standard deviation of 0.31. She also randomly sur-
veys 45 male students and finds that their mean GPA is 3.12 with a standard devi-
ation of 0.31.
(a) At the 0.05 level of significance, test the claim that the mean GPA for female
students at her school is greater than the mean GPA of male students.
(b) Identify potential sources of bias for these samples.
2. A random sample of 88 male deaths in a county over a 1-year period had a mean
age at death of 68.9 years, with a standard deviation of 16.71 years. A random
sample of 84 female deaths in the same county over the same 1-year period had
a mean age at death of 77.1 years, with a standard deviation of 15.65 years. At
the 0.01 level of significance, test the claim that the mean age at death in this
county is lower for males than it is for females.
3. At a college, 60 male students and 63 female students were randomly surveyed.
The males were taking a mean of 12.37 course units, with a standard deviation of
3.740 units. The females were taking a mean of 12.47 units, with a standard devi-
ation of 3.458 units. At the 0.05 level of significance, test the claim that the mean
number of course units is the same for male and female students.
4. A random sample of 45 male college students and 50 female college students was
taken, and each student was asked to write down their height. The male students
had a mean height of 70.4 inches, with a standard deviation of 2.44 inches. The
 8.1 Hypothesis Tests for Two Population Means (Large Samples) 387

female students had a mean height of 64.3 inches, with a standard deviation of
2.86 inches.
(a) At the 0.01 level, test the claim that male college students have a greater
mean height than female college students.
(b) How can we eliminate the bias that may be present with students self-reporting
their heights?
5. At a community college, students seem to perform better during the summer ses-
sion than they do during the fall and spring semesters. Many instructors and
administrators feel that this may be because summer classes have a different stu-
dent makeup than the classes during the fall and spring semesters. A random sam-
ple of 100 students in the spring semester had a mean age of 23.1 years, with a
standard deviation of 7.10 years. A random sample of 30 summer session students
had a mean age of 26.2 years, with a standard deviation of 6.92 years. At the 0.01
level of significance, test the claim that the mean age of spring semester students
is not the same as the mean age of summer session students.
6. A student conducted a study of 103 men and 104 women who had been married
at least one time. The men had a mean age of 25.4 years at their first marriage,
with a standard deviation of 5.19 years. The women had a mean age of 22.6 years
at their first marriage, with a standard deviation of 4.63 years.
(a) At the 0.01 level of significance, test the claim that the mean age of men at their
first marriage is higher than the mean age of women at their first marriage.
(b) If you were going to test the claim, explain how you would gather data to do
so. Be sure to list potential biases that you are seeking to eliminate.
7. The PSAT Mathematics scores of 50 high school juniors and 35 high school soph-
omores were selected at random.
Juniors
24 36 72 48 28 50 44 46 38 72
57 69 48 64 39 48 59 51 59 48
42 37 52 72 47 55 38 58 40 60
60 59 38 42 49 40 31 61 42 47
56 37 62 59 59 44 69 41 60 33
Sophomores
29 52 42 68 38 60 58 49 45 55
30 60 52 37 42 49 43 36 38 52
58 61 31 61 46 58 49 36 59 49
48 47 37 43 31
At the 0.01 level of significance, test the claim that the mean score for juniors on
this test is greater than the mean score of sophomores on this test.
8. Do students do as well on the PSAT verbal test as they do on the PSAT math test?
PSAT verbal scores were obtained for a random sample of 40 high school juniors
and PSAT math scores were obtained for a different sample of 40 high school jun-
iors, as follows.
Verbal
28 41 49 41 40 41 48 47 48 37
56 41 36 49 40 47 44 72 42 50
64 47 56 54 58 71 38 63 40 50
42 53 68 43 38 39 30 70 46 60
Math
34 52 46 56 29 51 49 41 66 42
50 65 50 48 48 45 48 44 67 65
45 44 52 54 46 30 32 58 39 49
52 72 46 59 29 39 35 53 51 41
388 CHAPTER 8 Two-Sample Hypothesis Tests

At the 0.01 level of significance, test the claim that the mean score for the two
exams is the same.
9. It is a widely held belief that males have better mathematical skills than verbal
skills. Here are the SAT math scores for 36 randomly selected male students and
the SAT verbal scores for 30 randomly selected male students.

Math
370 670 460 530 690 640 740 540 610
470 610 660 410 450 720 740 430 480
460 520 580 580 560 510 320 640 650
580 540 580 440 310 260 420 320 410

Verbal
320 280 750 450 520 460 630 580 490
570 450 490 590 530 320 380 520 440
480 610 660 740 580 330 480 590 330
580 500 480
At the 0.05 level of significance, test the claim that the mean SAT math score for
males is greater than the mean SAT verbal score for males.
10. It is a widely held belief that females have better verbal skills than mathematical
skills. Here are the SAT math scores for 42 randomly selected female students and
the SAT verbal scores for 33 randomly selected female students.

Math
320 490 470 280 390 560 520 760 480
460 470 490 460 520 530 580 370 490
550 600 310 480 460 340 400 330 700
570 420 570 750 730 470 640 530 470
310 560 570 330 570 500

Verbal
280 250 510 680 570 360 480 540 600
350 590 390 580 540 510 550 320 510
290 550 550 350 450 530 630 630 630
490 570 690 620 500 380
At the 0.05 level of significance, test the claim that the mean SAT math score for
females is less than the mean SAT verbal score for females.
11. A random sample of 56 departing flights at an airport over a 3-month period had
a mean wait of 14.2 minutes between boarding and takeoff, with a standard devi-
ation of 4.53 minutes. At the same airport, a random sample of 81 incoming
flights over the same 3-month period had a mean wait of 17.5 minutes between
the time that the plane arrived at the gate and the time that the baggage reached
the baggage claim area, with a standard deviation of 9.87 minutes.
(a) At the 0.05 level of significance, test the claim that at this airport the mean
wait for takeoff is less than the mean wait for baggage.
(b) Explain how you would gather data to test the claim that the mean wait for
takeoff is less than the mean wait for baggage at U.S. airports.
12. A survey of 65 beer drinkers ages 21–29 had a mean of 22.7 servings of beer dur-
ing the last month, with a standard deviation of 8.49 servings. A survey of 107 beer
drinkers ages 30–39 had a mean of 19.8 servings of beer during the last month,
with a standard deviation of 6.21 servings. At the 0.05 level of significance, test the
claim that the mean monthly beer consumption of beer drinkers ages 21–29 is
greater than the mean monthly beer consumption of beer drinkers ages 30–39.
(Based on the results of a study by Maritz AmeriPoll.)
 8.1 Hypothesis Tests for Two Population Means (Large Samples) 389

13. A professional bowler claims that his game is better suited to synthetic lanes than
to natural lanes. Here are the scores of 42 randomly selected games on synthetic
lanes and the scores of 42 randomly selected games on natural lanes.
Synthetic
256 269 279 245 290 214 210 235 238
213 279 236 218 222 247 188 249 214
196 222 237 177 246 227 279 267 236
237 244 217 258 264 266 279 238 257
238 224 226 267 256 228

Natural
248 221 246 255 289 244 223 222 255
192 224 203 204 179 244 234 225 218
201 219 214 233 204 212 217 207 203
244 163 215 182 192 203 198 210 225
207 236 224 264 228 164
At the 0.05 level of significance, test the bowler’s claim.
14. A math instructor is not sure that collecting homework from his class on a daily
basis is beneficial to his students. He conducts an experiment by collecting home-
work from one algebra class of 37 students but not collecting homework from
another algebra class of 32 students. Here are the test scores for both classes.
Collected Homework
53 88 68 64 70 67 57 88 83 79
78 55 84 78 78 62 82 76 54 74
85 76 74 71 84 87 75 73 63 64
73 48 82 87 78 78 75

Did Not Collect Homework

60 79 65 74 76 62 71 72 76 77
62 73 66 80 74 74 81 68 61 68
76 76 78 70 79 88 68 82 63 68
76 87
At the 0.05 level of significance, test the claim that collecting homework produces
the same mean score as not collecting homework.
15. Are American League baseball games higher scoring than National League base-
ball games? Here are the number of runs scored in 62 randomly selected American
League games and 71 randomly selected National League games.
American
5 4 12 16 11 10 11 6 12 4
20 11 7 6 8 13 23 16 3 9
4 13 9 17 8 16 4 12 8 9
4 14 7 14 4 9 15 7 14 7
8 9 11 4 8 13 9 11 10 6
6 7 9 17 19 25 17 12 9 6
11 1

National
11 11 6 1 15 19 9 11 12 13
6 27 6 12 11 6 14 13 6 6
(continues)
390 CHAPTER 8 Two-Sample Hypothesis Tests

(continued)
National
10 13 9 16 3 14 9 22 11 9
6 4 6 9 13 13 11 3 8 12
3 13 9 14 5 5 6 13 7 12
3 7 9 22 10 3 10 6 6 6
7 6 11 19 13 10 5 4 10 14
4
At the 0.01 level of significance, test the claim that the mean number of runs
scored in American League games is greater than the mean number of runs scored
in National League games.
16. Do American League baseball games have more hits than National League base-
ball games? Here are the number of hits in 62 randomly selected American League
games and 71 randomly selected National League games.

American
13 16 23 23 23 16 14 14 23 9
32 27 17 12 14 27 31 24 9 17
11 19 18 26 13 28 14 18 18 17
14 21 15 27 16 13 24 19 22 16
20 20 17 14 18 22 20 18 20 18
14 18 21 31 28 33 27 20 16 19
17 7

National
19 20 14 12 25 27 16 17 23 19
20 32 15 23 21 17 23 17 13 20
17 21 20 18 15 20 23 30 19 20
13 9 13 16 19 25 21 14 11 18
8 19 19 24 10 19 12 16 16 22
14 20 16 26 21 9 16 14 16 13
18 12 19 27 24 16 14 18 20 19
12
At the 0.01 level of significance, test the claim that the mean number of hits in
American League games is greater than the mean number of hits in National
League games.
17. Do American League baseball games take longer than National League baseball
games? Here are the lengths, in minutes, of 62 randomly selected American
League games and 71 randomly selected National League games.

American
175 168 185 183 196 151 198 148 189
146 225 199 143 149 156 192 230 178
124 167 156 185 179 185 144 158 157
196 190 166 176 186 153 189 180 205
163 177 154 172 178 186 187 163 181
208 142 200 150 172 166 168 182 219
235 224 202 171 159 153 169 171
 8.1 Hypothesis Tests for Two Population Means (Large Samples) 391

National
170 187 169 129 159 209 181 183 197
170 157 214 154 166 191 178 198 178
161 198 179 150 265 136 143 152 190
213 158 176 146 172 137 200 158 181
202 152 128 161 153 164 169 194 108
146 157 167 169 190 160 177 171 198
159 160 146 188 149 151 172 143 189
204 215 181 141 152 176 180 158
At the 0.05 level of significance, test the claim that the mean time to complete an
American League game is longer than the mean time to complete a National
League game.
18. Here are the prices for a gallon of unleaded regular gas at 48 randomly selected
gas stations in the Los Angeles area and 39 randomly selected gas stations in the
San Francisco Bay area from the week before Christmas in 1999.

Los Angeles
$ 0.949 $ 1.019 $ 0.989 $ 1.039 $ 0.969 $ 0.999
$ 0.979 $ 1.019 $ 1.039 $ 1.009 $ 0.939 $ 0.989
$ 1.059 $ 0.959 $ 1.049 $ 1.029 $ 1.039 $ 0.999
$ 1.099 $ 1.019 $ 1.019 $ 1.029 $ 1.069 $ 0.979
$ 1.039 $ 1.089 $ 1.049 $ 0.989 $ 1.039 $ 1.019
$ 1.019 $ 0.999 $ 1.009 $ 0.949 $ 1.019 $ 1.069
$ 1.069 $ 0.979 $ 0.999 $ 1.029 $ 1.079 $ 0.979
$ 0.999 $ 1.039 $ 0.969 $ 1.059 $ 1.049 $ 0.979

San Francisco
$ 1.049 $ 1.169 $ 1.029 $ 1.199 $ 1.059 $ 1.099
$ 1.139 $ 1.109 $ 1.129 $ 1.139 $ 1.139 $ 1.189
$ 1.089 $ 1.079 $ 1.149 $ 1.149 $ 1.059 $ 1.179
$ 1.169 $ 1.069 $ 1.139 $ 1.149 $ 1.109 $ 1.099
$ 1.099 $ 1.199 $ 1.129 $ 1.039 $ 1.109 $ 1.089
$ 1.159 $ 1.069 $ 1.089 $ 1.129 $ 1.239 $ 1.149
$ 1.169 $ 1.119 $ 1.169
At the 0.01 level, test the claim that the mean price for a gallon of unleaded reg-
ular gas in the Los Angeles area is lower than the mean price for a gallon of
unleaded regular gas in the San Francisco area.
19. A random sample of 140 women who first married in 1970 had a mean age of 20.8
years, with a standard deviation of 3.1 years. A similar sample of 108 women who
first married in 1998 had a mean age of 25.0 years with a standard deviation of
4.1 years. At the 0.05 level of significance, test the claim that the mean age at
which women get married has increased from 1970 to 1998. (Based on data from
the U.S. Census Bureau.)
20. Do women have more doctor visits per year than men? A random sample of 111
men produced a mean of 3.8 visits per year, with a standard deviation of 2.1 vis-
its. A random sample of 153 women produced a mean of 5.8 visits per year, with
a standard deviation of 1.8 visits. At the 0.05 level of significance, test the claim
that the mean number of doctor visits per year is higher for women than it is for
men. (Based on the results of the March 2000 National Vital Statistics Report.)
21. Do households that have cable TV have higher incomes? A sample of 420 cable
households had a mean annual income of $51,468 with a standard deviation of
392 CHAPTER 8 Two-Sample Hypothesis Tests

$43,329. A sample of 317 households that do not have cable TV had a mean
annual income of $36,735 with a standard deviation of $29,071. At the 0.05 level
of significance, test the claim that the mean annual income of households with
cable TV is higher than the mean annual income of households without cable TV.
(Based on a study by Cablevision Advertising Bureau.)
22. A random sample of 163 men between the ages of 20 and 29 had a mean weight
of 172.1 pounds with a standard deviation of 33.88 pounds. A random sample of
85 men between the ages of 50 and 59 had a mean weight of 189.2 pounds with
a standard deviation of 32.99 pounds. At the 0.05 level of significance, test the
claim that the mean weight of men who are between 20 and 29 years old is less
than the weight of men who are between 50 and 59 years old. (Based on the results
of a study by the Centers for Disease Control and Prevention.)
23. A child development student is interested in the amount of time that children
watch television. A study of 45 children ages 2–5 at the college’s day care center
revealed that they watched 135.4 minutes of television per day, with a standard
deviation of 82.1 minutes. A study of 120 elementary school children ages 6–12
revealed that they watched a mean of 170.6 minutes of television per day, with a
standard deviation of 106.9 minutes.

(a) Construct a 95% confidence interval for the difference between the mean
amount of television watched per day by 2- to 5-year-olds and the mean
amount of television watched per day by 6- to 12-year-olds.
(b) At the 0.05 level of significance, test the claim that the mean amount of tele-
vision watched per day by 2- to 5-year-olds is the same as the mean amount
of television watched per day by 6- to 12-year-olds.
(c) Does your hypothesis test from part (b) reinforce what your confidence inter-
val from part (a) told you? Explain.

24. A plant produces various types of corn chips. Among the varieties of chips that this
plant produces are blue corn tortilla chips and jalapeno tortilla chips, both of
which are sold in 9-ounce bags. A random sample of 50 bags of blue corn tortilla
chips had a mean weight of 9.06 ounces, with a standard deviation of 0.05 ounces.
A random sample of 50 jalapeno tortilla chips had a mean weight of 9.01 ounces,
with a standard deviation of 0.03 ounces.

(a) Construct a 95% confidence interval for the difference between the mean fill
of blue corn tortilla chip bags and the mean fill of jalapeno tortilla chip bags.
(b) At the 0.05 level of significance, test the claim that the mean fill of blue corn
tortilla chip bags is the same as the mean fill of jalapeno tortilla chip bags.
(c) Does your hypothesis test from part (b) reinforce what your confidence inter-
val from part (a) told you? Explain.

? What Is Wrong with This Picture?

The following statistical project contains a major error. Potential errors could be the choice
of the wrong hypothesis test or that the assumptions required for this test are not met. Find
the error, and explain what could be done to correct it.
A student believes that there are more females at private universities than at public uni-
versities. She randomly selects 30 private universities and visits their Web sites, where she
finds the percentage of females at each school. She repeats this for 30 public universities.
Taking these numbers, she tests the claim that the mean number of females at private uni-
versities is higher than the mean number of females at public universities.
What is wrong with this picture?
 8.2 Hypothesis Tests for Two Population Means (Small Samples) 393

M I N I P RO J E C T
Randomly sample at least 30 males and 30 females at your school, and ask
them how many serious relationships they have had. Use these sample data to
test the claim that college males and college females have had the same mean
number of serious relationships at the 0.05 level of significance. In addition to
your complete hypothesis test, include

• Your raw data

• A histogram for the number of serious relationships that the males have
had and a second histogram for the number of serious relationships that
the females have had

Finally, answer the following questions.

• How did you obtain your sample?

• Which type of sampling did you use?
• Was the sample truly random?
• What potential biases may show up in your sample?

SECTION 8.2
Hypothesis Test for Two Population Means
(Small Samples)
In this section we introduce a hypothesis test for comparing the means of two differ-
ent populations, for which at least one of the samples has a size of less than 30. The
two samples must be independent, and both must be drawn from populations that are
normally distributed. If either of the samples are drawn from a population that is not
normally distributed, we cannot use the methods presented in this section. There is a
hypothesis test covered in Section 11.3 that may be used in this case.
Traditionally, there have been two different test statistics that could be used. The
one that we will use does not assume the two samples come from populations that
have equal variances. (There is a second test that does assume the two samples come
from populations with the same variance, but we will leave that test as optional mate-
rial in the “Extra” section that follows Section 9.1.)
The test statistic that we will be using is

x1 – x 2
t =
s 12 s 22
+
n1 n2

where the degrees of freedom are equal to the smaller of n1 – 1 and n2 – 1. There is
actually a formula that we could use to calculate the degrees of freedom, but our
choice of the smaller of these two values is conservative, making it harder to reject a
null hypothesis.
We begin with a few examples of the hypothesis test using this test statistic, and
then we look at how to construct a confidence interval for the difference between two
population means.
394 CHAPTER 8 Two-Sample Hypothesis Tests

E XAMPLE A female college student majoring in math has heard many times
8.8 that males have higher SAT math scores than females. She believes
that this is not true, but instead that the two groups have the same
mean SAT math scores. She randomly samples 20 females at her
school. Their mean score is 509.5, with a standard deviation of
110.62. She also randomly samples 25 males at her school. Their
mean score is 536.4, with a standard deviation of 117.79. Use these
data to test the claim that females and males have the same mean
SAT math score at the 0.05 level of significance.

Since both samples have sizes of less than 30, the test statistic introduced in this
section is the appropriate one. We begin by labeling the populations. We let females
taking the SAT be population 1, and we let males taking the SAT be population 2.

Step 1
Population 1: females; population 2: males
Claim in words: Females and males have the same mean SAT math score.
Claim: μ1 = μ2
Complement: μ1 ≠ μ2
H0: μ1 = μ2
HA: μ1 ≠ μ2

Step 2
Level of significance: α = 0.05
Step 3
x 1 – x2
Test statistic: t = d.f. = smaller of (n 1 – 1) and (n 2 – 1)
s2
1 s 22
+
n1 n2

Step 4
α = 0.05, two-tailed test, with 20 – 1 = 19 degrees of freedom
Decision rule: Reject H0 if t < –2.093 or if t > 2.093.
Step 5

x1 – x2
t =
s 12 s 22
+
n1 n2

509.5 − 536.4
=
110.62 2 117.79 2
+
20 25
= –0.788

Decision: Fail to reject H0.

Conclusion: There is not sufficient sample evidence to reject the claim that the
mean SAT math scores of females and males are equal.
p-value
Using techniques developed in Section 7.2, we place the p-value between 0.20
and 0.50. A calculator or computer would tell us that the actual p-value of this test
is 0.4354. ■
 8.2 Hypothesis Tests for Two Population Means (Small Samples) 395

For the 1997–98 school year, the mean SAT math score for all males was 35 points
higher than the mean SAT math score for all females—so why weren’t we able to reject
the null hypothesis that the two mean scores were equal? One potential problem
could be the small sample sizes. There was a fairly wide gap between the two sample
means (26.9 points), but the small sample sizes kept the value of the test statistic
small. Similar results based on samples of 200 and 250, rather than 20 and 25, yield
a test statistic of approximately –2.5.
Of course, bias could have affected the value of the test statistic as well. One pos-
sible source of bias may come from the students self-reporting their SAT scores. Do we
know that the score that they reported was their actual score? Are there any other
potential biases?

E XAMPLE In tournaments of the Professional Bowler’s Association, there are

8.9 two rounds. The first round is a qualifying round where the bowlers’
total scores determine who advances to the second round. The
second round is a match play round, meaning that two bowlers
compete against each other. In this round, the scores are not
important, only the number of wins. It is believed that there is more
pressure on the individual bowlers in the match play round, and
that causes scores to drop during this round. Scores were random-
ly selected from 18 qualifying games and 24 match play games
from the same tournament. The mean score for the qualifying
games was 230.8, with a standard deviation of 33.83. The mean
score for the match play games was 224.8, with a standard devia-
tion of 27.76. Use these data to test the claim that the mean score
for qualifying games is higher than the mean score for match play
games, at the 0.05 level of significance.

Since both samples have sizes of less than 30, the test statistic introduced in this
section is the appropriate one. We begin by labeling the populations. We let qualify-
ing games be population 1, and we let match play games be population 2.

Step 1
Population 1: Qualifying; population 2: match play
Claim in words: The mean score for qualifying games is higher than the mean
score for match play games.
Claim: μ1 > μ2
Complement: μ1 ≤ μ2
H0: μ1 ≤ μ2
HA: μ1 > μ2

Step 2
Level of significance: α = 0.05

Step 3
x 1 – x2
Test statistic: t = d.f. = smaller of (n 1 – 1) and (n 2 – 1)
s2
1 s 22
+
n1 n2

Step 4
α = 0.05, right-tailed test, with 18 – 1 = 17 degrees of freedom
396 CHAPTER 8 Two-Sample Hypothesis Tests

Decision rule: Reject H0 if t > 1.740.

Step 5

x1 – x2
t =
s 12 s 22
+
n1 n2

230.8 − 224.8
=
33.82 2 27.76 2
+
18 24
= 0.613

Decision: Fail to reject H0.

Conclusion: There is not sufficient sample evidence to support the claim that the mean
score for qualifying games is higher than the mean score for match play games.
p-value
Using techniques developed in Section 7.2, the best that we can say is that the
p-value is higher than 0.25, since the value of our test statistic is lower than the last
entry on our table for 17 degrees of freedom. A calculator or computer would tell us
that the actual p-value of this test is 0.2720. ■

E XAMPLE Many people believe that male racehorses run faster than female
8.10 racehorses. To test this theory, the results of 55 randomly selected
races at Santa Anita Park were selected, all at the distance of 1
mile. Thirty-four of the races were for male racehorses, and 21 of
the races were restricted to fillies and mares (female racehorses).
The time required by the winners of the male races to complete the
1-mile race had a mean of 96.9 seconds, with a standard deviation
of 1.22 seconds. The time required by the winners of the races
restricted to fillies and mares to complete the 1-mile race had a
mean of 97.8 seconds, with a standard deviation of 1.55 seconds.
At the 0.05 level of significance, test the claim that the mean time
for a male racehorse to win a 1-mile race is less than the mean
time for a female racehorse to win a 1-mile race.

Even though one of the samples (male races) has a size that is at least 30, the
small sample test is appropriate for this two-mean test because the other sample
has a size that is less than 30. We begin by labeling the populations. We let male
races be population 1, and we let female races be population 2.
Step 1
Population 1: male races; population 2: female races
Claim in words: The mean time for a male racehorse to win a 1-mile race is less
than the mean time for a female racehorse to win a 1-mile race.
Claim: μ1 < μ2
Complement: μ1 ≥ μ2
H0: μ1 ≥ μ2
HA: μ1 < μ2
Step 2
Level of significance: α = 0.05
 8.2 Hypothesis Tests for Two Population Means (Small Samples) 397

Step 3
x 1 – x2
Test statistic: t = d.f. = smaller of (n 1 – 1) and (n 2 – 1)
s2
1 s 22
+
n1 n2

Step 4
α = 0.05, left-tailed test, with 21 – 1 = 20 degrees of freedom
Decision rule: Reject H0 if t < –1.725.
Step 5

x1 – x2
t =
s 12 s 22
+
n1 n2

96.9 − 97.8
=
1.22 2 1.55 2
+
34 21
= –2.263

Decision: Reject H0.

Conclusion: There is sufficient sample evidence to support the claim that the mean
time for a male racehorse to win a 1-mile race is less than the mean time for a
female racehorse to win a 1-mile race.
p-value

Using techniques developed in Section 7.2, the best that we can say is that the
p-value is between 0.025 and 0.01. A calculator or computer would tell us that the
actual p-value of this test is 0.0150. ■

Confidence Intervals
Recall that the idea of hypothesis testing is based on confidence intervals. There is a
confidence interval that can be constructed for the difference between two population
means, when at least one of the samples is small. We begin, as we did before, by cal-
culating the margin of error. Here is the formula.

s 12 s 22
E = tα/2 + degrees of freedom: the smaller of n 1 – 1 and n 2 – 1
n1 n2

– –
We then subtract this margin of error from x 1 – x 2 to find the left endpoint of the con-
– –
fidence interval, and we add the margin of error to x 1 – x 2 to find the right endpoint
of the confidence interval. (If you want to create a one-sided confidence interval, see
the “Extra” material at the end of the Section 6.2 exercises.)

(x–1 – x–2 ) – E ≤ (μ1 – μ2 ) ≤ (x–1 – x–2 ) + E

E XAMPLE A mathematics instructor holds a supplementary review session the

8.11 day before each exam. Fourteen of the 24 students in his algebra
class attended his review session before the exam. The mean
398 CHAPTER 8 Two-Sample Hypothesis Tests

score on this exam for these 14 students was 80.9 points out of 100,
with a standard deviation of 9.62 points. The mean score for the 10
students who did not attend was 45.4 points out of 100, with a stan-
dard deviation of 16.36 points. Construct a 95% confidence interval
for the difference between the mean scores of algebra students who
attend a review session and the mean scores of algebra students
who do not attend a review session. (Assume that the students in
this class represent a random sample of all algebra students.)

We let population 1 be the algebra students who attend a review session, and pop-
ulation 2 be the algebra students who do not attend a review session. We begin by
finding t α/2. We have 9 degrees of freedom, since the smaller sample had size 10.
For 9 degrees of freedom, t 0.025 = 2.262. Now we calculate the margin of error.

s 12 s 22
E = t α/2 +
n1 n2

9.62 2 16.36 2
= 2.262 +
14 10
= 13.07

Next we construct the confidence interval.

(x–1 – x–2) – E ≤ (μ1 – μ2) ≤ (x–1 – x–2) + E

(80.9 – 45.4) – 13.07 ≤ (μ1 – μ2) ≤ (80.9 – 45.4) + 13.07
35.5 – 13.07 ≤ (μ1 – μ2) ≤ 35.5 + 13.07
22.43 ≤ (μ1 – μ2) ≤ 48.57

We are 95% confident that the mean score of algebra students who attend a
review session will be between 22.43 and 48.57 points higher than the mean score
of algebra students who do not attend a study session. ■

MICROSOFT EXCEL Hypothesis Test for Two

Population Means
(Small Samples)
Microsoft Excel has a built-in procedure for the hypothesis test for two population
means using the t-distribution, assuming unequal variances. It also has a procedure
for when we assume the populations have equal variance, so be careful to select the
right one. We will step through an example showing how to do this using Excel, and
then we will write the hypothesis test based on our results.

E XAMPLE A business student wondered whether the stock market would be

8.12 busier on Mondays or Fridays. She took a random sample of 23
Mondays and 16 Fridays and found the volume (number of shares
sold) for the New York Stock Exchange. Here are the results,
rounded to the nearest million.
 8.2 Hypothesis Tests for Two Population Means (Small Samples) 399

Monday
731 542 559 592 548 629 597 692
542 543 530 564 560 620 564 714
610 690 610 592 774 689 531

Friday
689 669 613 569 579 635 622 557
558 759 725 736 637 785 683 602

At the 0.05 level of significance, test the claim that the mean vol-
ume on Mondays is the same as the mean volume on Fridays.

In a new Excel worksheet, type Monday in cell A1 and Friday in B1. Enter the vol-
umes for Mondays in column A, from cell A2 through A24. Enter the volumes for
Fridays in column B, from cells B2 through B17.

Select Data Analysis from the Tools menu. When the dialog box appears, select
t-Test: Two-Sample Assuming Unequal Variances and click on OK. In the
box labeled Variable 1 Range, enter A1:A24. In the box labeled Variable 2
Range, enter B1:B17. These two cell ranges include the labels Monday and Friday,
so check the box labeled Labels. In the box labeled Hypothesized Mean
Difference, enter 0. Put the level of significance, 0.05, in the box labeled Alpha.
Click on OK to see the results, which will be on a new worksheet. Here is an exam-
ple of what you should see.

t-Test: Two-Sample Assuming Unequal Variances

Monday Friday

Mean 609.6956522 651.125

Variance 5115.857708 5309.583333
Observations 23 16
Hypothesized Mean Difference 0
df 32
t Stat -1.759723569
P(T<=t) one-tail 0.044006409
t Critical one-tail 1.693888407
P(T<=t) two-tail 0.088012817
t Critical two-tail 2.036931619

We see the mean, variance, and size of each sample. The box labeled df contains
the degrees of freedom. Excel uses the actual formula for degrees of freedom,
rather than our conservative estimate. Here it is.

2
⎛s2 s2 ⎞
⎜ 1 + 2 ⎟
⎜n n2 ⎟
⎝ 1 ⎠
d .f . =
2 2
⎛s2 ⎞ ⎛s2 ⎞
1 ⎜ 1 ⎟ 1 ⎜ 2 ⎟
+
n1 – 1⎜ n1 ⎟ n2 – 1⎜n2 ⎟
⎝ ⎠ ⎝ ⎠

(Recall that we have been using 1 less than the smaller sample size as the number
of degrees of freedom, which in this case is 15.) Here is the calculation that pro-
duces 32 degrees of freedom.