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2 Inferences for Two Population Means, Using Independent Samples: Standard Deviations Assumed Equal†

# 2 Inferences for Two Population Means, Using Independent Samples: Standard Deviations Assumed Equal†

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10.2 Inferences for Two Population Means: σ s Assumed Equal

469

has the standard normal distribution. Replacing σ1 and σ2 with their common value σ

and using some algebra, we obtain the variable

z=

(x¯1 − x¯2 ) − (μ1 − μ2 )

.

σ (1/n 1 ) + (1/n 2 )

(10.1)

However, we cannot use this variable as a basis for the required test statistic because

σ is unknown.

Consequently, we need to use sample information to estimate σ , the unknown population standard deviation. We do so by first estimating the unknown population variance, σ 2 . The best way to do that is to regard the sample variances, s12 and s22 , as two

estimates of σ 2 and then pool those estimates by weighting them according to sample

size (actually by degrees of freedom). Thus our estimate of σ 2 is

sp2 =

(n 1 − 1)s12 + (n 2 − 1)s22

,

n1 + n2 − 2

and hence that of σ is

(n 1 − 1)s12 + (n 2 − 1)s22

.

n1 + n2 − 2

sp =

The subscript “p” stands for “pooled,” and the quantity sp is called the pooled sample

standard deviation.

Replacing σ in Equation (10.1) with its estimate, sp , we get the variable

(x¯1 − x¯ 2 ) − (μ1 − μ2 )

,

sp (1/n 1 ) + (1/n 2 )

which we can use as the required test statistic. Although the variable in Equation (10.1)

has the standard normal distribution, this one has a t-distribution, with which you are

KEY FACT 10.2

Distribution of the Pooled t-Statistic

Suppose that x is a normally distributed variable on each of two populations

and that the population standard deviations are equal. Then, for independent

samples of sizes n1 and n2 from the two populations, the variable

t=

(x¯ 1 − x¯ 2 ) − (μ1 − μ2 )

sp (1/n1 ) + (1/n2 )

has the t-distribution with df = n1 + n2 − 2.

In light of Key Fact 10.2, for a hypothesis test that has null hypothesis H0: μ1 = μ2

(population means are equal), we can use the variable

t=

sp

x¯1 − x¯ 2

(1/n 1 ) + (1/n 2 )

as the test statistic and obtain the critical value(s) or P-value from the t-table, Table IV

in Appendix A. We call this hypothesis-testing procedure the pooled t-test.† Procedure 10.1 provides a step-by-step method for performing a pooled t-test by using either

the critical-value approach or the P-value approach.

† The pooled t-test is also known as the two-sample t-test with equal variances assumed, the pooled two-

variable t-test, and the pooled independent samples t-test.

470

CHAPTER 10 Inferences for Two Population Means

PROCEDURE 10.1 Pooled t-Test

Purpose To perform a hypothesis test to compare two population means, μ1 and μ2

Assumptions

1.

2.

3.

4.

Simple random samples

Independent samples

Normal populations or large samples

Equal population standard deviations

Step 1

The null hypothesis is H0: μ1 = μ2 , and the alternative hypothesis is

Ha: μ1 = μ2

(Two tailed)

Ha: μ1 < μ2

(Left tailed)

or

Step 2

Decide on the signiﬁcance level, α.

Step 3

Compute the value of the test statistic

t=

sp

Ha: μ1 > μ2

.

(Right tailed)

or

x¯ 1 − x¯ 2

,

(1/n1 ) + (1/n2 )

where

sp =

(n1 − 1)s12 + (n2 − 1)s22

.

n1 + n 2 − 2

Denote the value of the test statistic t0 .

CRITICAL-VALUE APPROACH

Step 4

OR

P-VALUE APPROACH

Step 4 The t-statistic has df = n1 + n2 − 2. Use

Table IV to estimate the P-value, or obtain it exactly by

using technology.

The critical value(s) are

−tα

±tα/2

or

or

(Two tailed)

(Left tailed)

(Right tailed)

with df = n1 + n2 − 2. Use Table IV to find the critical

value(s).

P -value

P-value

Reject

H0

Do not

reject H0

Reject

H0

Do not reject H0 Reject

H0

Reject Do not reject H0

H0

−|t0| 0 |t0|

Two tailed

␣ /2

␣ /2

−t␣/2

0

t␣/2

Two tailed

t

−t␣

0

t

0

t␣

Right tailed

Left tailed

t

t

P- value

t0

0

Left tailed

t

0

t0

t

Right tailed

Step 5 If P ≤ α, reject H0 ; otherwise, do not

reject H0 .

Step 5 If the value of the test statistic falls in the

rejection region, reject H0 ; otherwise, do not reject H0 .

Step 6

Interpret the results of the hypothesis test.

Note: The hypothesis test is exact for normal populations and is approximately correct for

large samples from nonnormal populations.

Regarding Assumptions 1 and 2, we note that the pooled t-test can also be used

as a method for comparing two means with a designed experiment. Additionally, the

pooled t-test is robust to moderate violations of Assumption 3 (normal populations)

but, even for large samples, can sometimes be unduly affected by outliers because the

sample mean and sample standard deviation are not resistant to outliers. The pooled

t-test is also robust to moderate violations of Assumption 4 (equal population standard

10.2 Inferences for Two Population Means: σ s Assumed Equal

471

deviations) provided the sample sizes are roughly equal. We will say more about the

robustness of the pooled t-test at the end of Section 10.3.

How can the conditions of normality and equal population standard deviations (Assumptions 3 and 4, respectively) be checked? As before, normality can be checked by

using normal probability plots.

Checking equal population standard deviations can be difficult, especially when

the sample sizes are small. As a rough rule of thumb, you can consider the condition of

equal population standard deviations met if the ratio of the larger to the smaller sample

standard deviation is less than 2. Comparing stem-and-leaf diagrams, histograms, or

boxplots of the two samples is also helpful; be sure to use the same scales for each pair

of graphs.†

EXAMPLE 10.3

The Pooled t-Test

Faculty Salaries Let’s return to the salary problem of Example 10.2, in which we

want to perform a hypothesis test to decide whether the mean salaries of faculty in

private institutions and public institutions are different.

Independent simple random samples of 35 faculty members in private institutions and 30 faculty members in public institutions yielded the data in Table 10.5.

At the 5% significance level, do the data provide sufficient evidence to conclude that

mean salaries for faculty in private and public institutions differ?

TABLE 10.5

Annual salaries (\$1000s) for 35 faculty

members in private institutions and

30 faculty members in public institutions

TABLE 10.6

Summary statistics for

the samples in Table 10.5

Private

institutions

Public

institutions

x¯1 = 98.19

s1 = 26.21

n 1 = 35

x¯2 = 83.18

s2 = 23.95

n 2 = 30

Sample 1 (private institutions)

Sample 2 (public institutions)

97.3 85.9 118.8 93.9 66.6 109.2 64.9 59.9 115.7 126.1 50.3 133.1 89.3

83.1 100.6 99.3 94.9 94.4 139.3 108.8 82.5 67.1 60.7 79.9 50.1 81.7

158.1 142.4 85.0 108.2 116.3 141.5 51.4 83.9 102.5 109.9 105.1 67.9 107.5

125.6 70.6 74.6 69.9 115.4 84.6 92.0 54.9 41.5 59.5 65.9 76.9 66.9

97.2 55.1 126.6 116.7 76.0 109.6 63.0 85.9 113.9 70.3 90.1 99.7 96.7

Solution First, we find the required summary statistics for the two samples, as

shown in Table 10.6. Next, we check the four conditions required for using the

pooled t-test, as listed in Procedure 10.1.

r The samples are given as simple random samples and, therefore, Assumption 1

is satisfied.

r The samples are given as independent samples and, therefore, Assumption 2 is

satisfied.

r The sample sizes are 35 and 30, both of which are large; furthermore, Figs. 10.2

and 10.3, both on the next page, suggest no outliers for either sample. So, we can

consider Assumption 3 satisfied.

r According to Table 10.6, the sample standard deviations are 26.21 and 23.95.

These statistics are certainly close enough for us to consider Assumption 4 satisfied, as we also see from the boxplots in Fig. 10.3.

† The assumption of equal population standard deviations is sometimes checked by performing a formal hypothesis

test, called the two-standard-deviations F-test. We don’t recommend that strategy because, although the pooled

t-test is robust to moderate violations of normality, the two-standard-deviations F-test is extremely nonrobust to

such violations. As the noted statistician George E. P. Box remarked, “To make a preliminary test on variances

[standard deviations] is rather like putting to sea in a rowing boat to find out whether conditions are sufficiently

calm for an ocean liner to leave port!”

CHAPTER 10 Inferences for Two Population Means

FIGURE 10.3

Boxplots of the salary data for faculty in

private institutions and public

institutions

Normal score

FIGURE 10.2

Normal probability plots of the sample

data for faculty in (a) private institutions

and (b) public institutions

3

2

1

0

–1

–2

–3

Normal score

472

3

2

1

0

–1

–2

–3

50 70 90 110 130 150 170

30 50 70 90 110 130 150

Salary (\$1000s)

Salary (\$1000s)

(a) Private institutions

(b) Public institutions

Public

Private

30

50

70

90

110

130

150

170

Salary (\$1000s)

The preceding items suggest that the pooled t-test can be used to carry out the

hypothesis test. We apply Procedure 10.1.

Step 1 State the null and alternative hypotheses.

The null and alternative hypotheses are, respectively,

H0: μ1 = μ2 (mean salaries are the same)

Ha: μ1 = μ2 (mean salaries are different),

where μ1 and μ2 are the mean salaries of all faculty in private and public institutions,

respectively. Note that the hypothesis test is two tailed.

Step 2 Decide on the signiﬁcance level, α.

The test is to be performed at the 5% significance level, or α = 0.05.

Step 3 Compute the value of the test statistic

t=

sp

x¯ 1 − x¯ 2

,

(1/n1 ) + (1/n2 )

where

sp =

(n1 − 1)s12 + (n2 − 1)s22

.

n1 + n 2 − 2

To find the pooled sample standard deviation, sp , we refer to Table 10.6:

sp =

(35 − 1) · (26.21)2 + (30 − 1) · (23.95)2

= 25.19.

35 + 30 − 2

Referring again to Table 10.6, we calculate the value of the test statistic:

t=

x¯1 − x¯2

98.19 − 83.18

=

= 2.395.

sp (1/n 1 ) + (1/n 2 )

25.19 (1/35) + (1/30)

10.2 Inferences for Two Population Means: σ s Assumed Equal

CRITICAL-VALUE APPROACH

OR

P-VALUE APPROACH

Step 4 The critical values for a two-tailed test

are ±tα/2 with df = n1 + n2 − 2. Use Table IV to

ﬁnd the critical values.

Step 4 The t-statistic has df = n1 + n2 − 2. Use

Table IV to estimate the P-value, or obtain it exactly

by using technology.

From Table 10.6, n 1 = 35 and n 2 = 30 and, therefore,

df = 35 + 30 − 2 = 63. Also, from Step 2, we have

α = 0.05. In Table IV with df = 63, we find that the

critical values are ±tα/2 = ±t0.025 = ±1.998, as shown

in Fig. 10.4A.

From Step 3, the value of the test statistic is t = 2.395.

The test is two tailed, so the P-value is the probability of

observing a value of t of 2.395 or greater in magnitude

if the null hypothesis is true. That probability equals the

FIGURE 10.4A

FIGURE 10.4B

Reject H 0

Do not

reject H 0

473

Reject H 0

P-value

t -curve

df = 63

0.025

0.025

−1.998

0

1.998

Step 5 If the value of the test statistic falls in the

rejection region, reject H0 ; otherwise, do not

reject H0 .

From Step 3, the value of the test statistic is t = 2.395,

which falls in the rejection region (see Fig. 10.4A). Thus

we reject H0 . The test results are statistically significant

at the 5% level.

t

0

t

t = 2.395

From Table 10.6, n 1 = 35 and n 2 = 30 and, therefore,

df = 35 + 30 − 2 = 63. Referring to Fig. 10.4B and to

Table IV with df = 63, we find that 0.01 < P < 0.02.

(Using technology, we obtain P = 0.0196.)

Step 5 If P ≤ α, reject H0 ; otherwise, do not

reject H0 .

From Step 4, 0.01 < P < 0.02. Because the P-value is

less than the specified significance level of 0.05, we reject H0 . The test results are statistically significant at the

5% level and (see Table 9.8 on page 408) provide strong

evidence against the null hypothesis.

Step 6 Interpret the results of the hypothesis test.

Report 10.1

Exercise 10.45

on page 477

Interpretation At the 5% significance level, the data provide sufficient evidence

to conclude that a difference exists between the mean salaries of faculty in private

and public institutions.

Confidence Intervals for the Difference between the Means

of Two Populations with Equal Standard Deviations

We can also use Key Fact 10.2 on page 469 to derive a confidence-interval procedure,

Procedure 10.2, for the difference between two population means, which we call the

pooled t-interval procedure.†

† The pooled t-interval procedure is also known as the two-sample t-interval procedure with equal variances

assumed, the pooled two-variable t-interval procedure, and the pooled independent samples t-interval procedure.

474

CHAPTER 10 Inferences for Two Population Means

PROCEDURE 10.2 Pooled t-Interval Procedure

Purpose To find a confidence interval for the difference between two population means, μ1

and μ2

Assumptions

1.

2.

3.

4.

Simple random samples

Independent samples

Normal populations or large samples

Equal population standard deviations

Step 1 For a conﬁdence level of 1 − α, use Table IV to ﬁnd tα/2 with

df = n1 + n2 − 2.

Step 2

The endpoints of the conﬁdence interval for μ1 − μ2 are

( x¯ 1 − x¯ 2 ) ± tα/2 · sp (1/n1 ) + (1/n2 ),

where sp is the pooled sample standard deviation.

Step 3

Interpret the confidence interval.

Note: The confidence interval is exact for normal populations and is approximately correct

for large samples from nonnormal populations.

EXAMPLE 10.4

The Pooled t-Interval Procedure

Faculty Salaries Obtain a 95% confidence interval for the difference, μ1 − μ2 ,

between the mean salaries of faculty in private and public institutions.

Solution We apply Procedure 10.2.

Step 1 For a conﬁdence level of 1 − α, use Table IV to ﬁnd tα/2 with

df = n1 + n2 − 2.

For a 95% confidence interval, α = 0.05. From Table 10.6, n 1 = 35 and n 2 = 30,

so df = n 1 + n 2 − 2 = 35 + 30 − 2 = 63. In Table IV, we find that with df = 63,

tα/2 = t0.05/2 = t0.025 = 1.998.

Step 2 The endpoints of the conﬁdence interval for μ1 − μ2 are

( x¯ 1 − x¯ 2 ) ± tα/2 · sp (1/n1 ) + (1/n2 ),

where sp is the pooled sample standard deviation.

From Step 1, tα/2 = 1.998. Also, n 1 = 35, n 2 = 30, and, from Example 10.3, we

know that x¯1 = 98.19, x¯2 = 83.18, and sp = 25.19. Hence the endpoints of the confidence interval for μ1 − μ2 are

(98.19 − 83.18) ± 1.998 · 25.19 (1/35) + (1/30),

or 15.01 ± 12.52. Thus the 95% confidence interval is from 2.49 to 27.53.

Step 3 Interpret the conﬁdence interval.

Report 10.2

Exercise 10.51

on page 478

Interpretation We can be 95% confident that the difference between the mean

salaries of faculty in private institutions and public institutions is somewhere between \$2,490 and \$27,530. In other words (see page 465), we can be 95% confident

that the mean salary of faculty in private institutions exceeds that of faculty in public

institutions by somewhere between \$2,490 and \$27,530.

10.2 Inferences for Two Population Means: σ s Assumed Equal

475

The Relation between Hypothesis Tests

and Confidence Intervals

Hypothesis tests and confidence intervals are closely related. Consider, for example,

a two-tailed hypothesis test for comparing two population means at the significance

level α. In this case, the null hypothesis will be rejected if and only if the (1 − α)level confidence interval for μ1 − μ2 does not contain 0. You are asked to examine

the relation between hypothesis tests and confidence intervals in greater detail in Exercises 10.63–10.65.

THE TECHNOLOGY CENTER

Most statistical technologies have programs that automatically perform pooled tprocedures. In this subsection, we present output and step-by-step instructions for such

programs.

EXAMPLE 10.5

Using Technology to Conduct Pooled t-Procedures

Faculty Salaries Table 10.5 on page 471 shows the annual salaries, in thousands

of dollars, for independent samples of 35 faculty members in private institutions and

30 faculty members in public institutions. Use Minitab, Excel, or the TI-83/84 Plus

to perform the hypothesis test in Example 10.3 and obtain the confidence interval

required in Example 10.4.

Solution Let μ1 and μ2 denote the mean salaries of all faculty in private and public

institutions, respectively. The task in Example 10.3 is to perform the hypothesis test

H0: μ1 = μ2 (mean salaries are the same)

Ha: μ1 = μ2 (mean salaries are different)

at the 5% significance level; the task in Example 10.4 is to obtain a 95% confidence

interval for μ1 − μ2 .

We applied the pooled t-procedures programs to the data, resulting in Output 10.1 on this and the next page. Steps for generating that output are presented

in Instructions 10.1 on the next page. Note to Excel users: For brevity, we have presented only the essential portions of the actual output.

As shown in Output 10.1, the P-value for the hypothesis test is about 0.02. Because the P-value is less than the specified significance level of 0.05, we reject H0 .

Output 10.1 also shows that a 95% confidence interval for the difference between

the means is from 2.49 to 27.54.

OUTPUT 10.1 Pooled t-procedures on the salary data

MINITAB

476

CHAPTER 10 Inferences for Two Population Means

OUTPUT 10.1 (cont.) Pooled t-procedures on the salary data

TI-83/84 PLUS

EXCEL

Using 2-SampTTest

INSTRUCTIONS 10.1

Using 2-SampTInt

Steps for generating Output 10.1

MINITAB

1 Store the two samples of salary data from Table 10.5 in

columns named PRIVATE and PUBLIC

2 Choose Stat ➤ Basic Statistics ➤ 2-Sample t. . .

3 Press the F3 key to reset the dialog box

4 Select Each sample is in its own column from the

drop-down list box

5 Click in the Sample 1 text box and specify PRIVATE

6 Click in the Sample 2 text box and specify PUBLIC

7 Click the Options. . . button

8 Click in the Confidence level text box and type 95

9 Click the arrow button at the right of the Alternative

hypothesis drop-down list box and select

Difference = hypothesized difference

10 Check the Assume equal variances check box

11 Click OK twice

EXCEL

1 Store the two samples of salary data from Table 10.5 in

columns named PRIVATE and PUBLIC

2 Choose XLSTAT ➤ Parametric tests ➤ Two-sample

t-test and z-test

3 Click the reset button in the lower left corner of the

dialog box

4 Click in the Sample 1 selection box and then select

the column of the worksheet that contains the

PRIVATE data

5 Click in the Sample 2 selection box and then select the

column of the worksheet that contains the PUBLIC data

6 Click the Options tab

7 Click the arrow button at the right of the Alternative

hypothesis drop-down list box and select Mean 1 –

Mean 2 = D

8 Type 5 in the Significance level (%) text box

9 Click OK

10 Click the Continue button in the XLSTAT – Selections

dialog box

TI-83/84 PLUS

Store the two samples of salary data from Table 10.5 in lists

named PRIV and PUBL.

FOR THE HYPOTHESIS TEST:

1 Press STAT, arrow over to TESTS, and press 4

2 Highlight Data and press ENTER

3 Press the down-arrow key

4 Press 2nd ➤ LIST, arrow down to PRIV, and press

ENTER twice

5 Press 2nd ➤ LIST, arrow down to PUBL, and press

ENTER twice

6 Type 1 for Freq1, press ENTER, type 1 for Freq2, and

press ENTER

7 Highlight = μ2 and press ENTER

8 Press the down-arrow key, highlight Yes, and press

ENTER

9 Arrow down to Calculate and press ENTER

FOR THE CI:

1 Press STAT, arrow over to TESTS, and press 0

2 Highlight Data and press ENTER

3 Press the down-arrow key

4 Press 2nd ➤ LIST, arrow down to PRIV, and press

ENTER twice

5 Press 2nd ➤ LIST, arrow down to PUBL, and press

ENTER twice

6 Type 1 for Freq1, press ENTER, type 1 for Freq2, and

press ENTER

7 Type .95 for C-Level and press ENTER

8 Highlight Yes, and press ENTER

9 Press the down-arrow key and press ENTER

10.2 Inferences for Two Population Means: σ s Assumed Equal

477

Note to Minitab and Excel users: Although Minitab and Excel simultaneously perform a hypothesis test and obtain a confidence interval, the type of confidence interval

found depends on the type of hypothesis test. Specifically, Minitab and Excel compute

a two-sided confidence interval for a two-tailed test and a one-sided confidence interval for a one-tailed test. To perform a one-tailed hypothesis test and obtain a two-sided

confidence interval, apply Minitab’s or Excel’s pooled t-procedure twice: once for the

one-tailed hypothesis test and once for the confidence interval specifying a two-tailed

hypothesis test.

Exercises 10.2

Understanding the Concepts and Skills

Applying the Concepts and Skills

10.33 Regarding the four conditions required for using the pooled

t-procedures:

a. what are they?

b. how important is each condition?

10.34 Explain why sp is called the pooled sample standard deviation.

Preliminary data analyses indicate that you can reasonably consider the assumptions for using pooled t-procedures satisfied in Exercises 10.45–10.50. For each exercise, perform the required hypothesis

test by using either the critical-value approach or the P-value

approach.

In each of Exercises 10.35–10.38, we have provided summary statistics for independent simple random samples from two populations.

Preliminary data analyses indicate that the variable under consideration is normally distributed on each population. Decide, in each

case, whether use of the pooled t-test and pooled t-interval procedure

10.45 Doing Time. The Federal Bureau of Prisons publishes data

in Prison Statistics on the times served by prisoners released from

federal institutions for the first time. Independent random samples of

released prisoners in the fraud and firearms offense categories yielded

the following information on time served, in months.

10.35 x¯1 = 468.3, s1 = 38.2, n 1 = 6,

x¯2 = 394.6, s2 = 84.7, n 2 = 14

Fraud

10.36 x¯1 = 115.1, s1 = 79.4, n 1 = 51,

x¯2 = 24.3, s2 = 10.5, n 2 = 19

3.6

5.3

10.7

8.5

11.8

10.37 x¯1 = 118, s1 = 12.04, n 1 = 99,

x¯2 = 110, s2 = 11.25, n 2 = 80

10.38 x¯1 = 39.04, s1 = 18.82, n 1 = 51,

x¯2 = 49.92, s2 = 18.97, n 2 = 53

In each of Exercises 10.39–10.44, we have provided summary statistics for independent simple random samples from two populations.

In each case, use the pooled t-test and the pooled t-interval procedure to conduct the required hypothesis test and obtain the specified

confidence interval.

10.39 x¯1 = 10, s1 = 2.1, n 1 = 15, x¯ 2 = 12, s2 = 2.3, n 2 = 15

a. Two-tailed test, α = 0.05

b. 95% confidence interval

10.40 x¯1 = 10, s1 = 4, n 1 = 15, x¯2 = 12, s2 = 5, n 2 = 15

a. Two-tailed test, α = 0.05

b. 95% confidence interval

10.41 x¯1 = 20, s1 = 4, n 1 = 10, x¯2 = 18, s2 = 5, n 2 = 15

a. Right-tailed test, α = 0.05

b. 90% confidence interval

10.42 x¯1 = 20, s1 = 4, n 1 = 10, x¯2 = 23, s2 = 5, n 2 = 15

a. Left-tailed test, α = 0.05

b. 90% confidence interval

10.43 x¯1 = 20, s1 = 4, n 1 = 20, x¯2 = 24, s2 = 5, n 2 = 15

a. Left-tailed test, α = 0.05

b. 90% confidence interval

10.44 x¯1 = 23, s1 = 10, n 1 = 25, x¯2 = 24, s2 = 12, n 2 = 50

a. Right-tailed test, α = 0.10

b. 95% confidence interval

Firearms

17.9

5.9

7.0

13.9

16.6

25.5

10.4

18.4

19.6

20.9

23.8

17.9

21.9

13.3

16.1

At the 5% significance level, do the data provide sufficient evidence

to conclude that the mean time served for fraud is less than that

for firearms offenses? (Note: x¯1 = 10.12, s1 = 4.90, x¯2 = 18.78, and

s2 = 4.64.)

10.46 Gender and Humor. Sense of humor of 200 male and 200

female students was examined in a paper. The researchers studied

measures of abstract reasoning, verbal intelligence, humor production

ability, and mating success. The students’ humor ability was tested on

the basis of the number of captionswritten for funny images. Following are the number of captions given by a few participants.

Men

10

13

4

14

19

8

12

5

13

5

3

20

15

18

23

27

4

37

Women

9

3

5

22

15

26

10

11

9

33

30

19

14

25

11

9

2

1

8

17

5

22

21

5

20

12

18

27

8

19

3

13

24

14

3

4

7

3

23

13

6

15

At the 1% significance level, do the data provide sufficient evidence

to conclude that, on average, males have a better sense of humor

and, in particular, a better frame of reference than females? (Note:

x¯1 = 11.39, s1 = 4.14, x¯2 = 9.85, and s2 = 3.67.)

478

CHAPTER 10 Inferences for Two Population Means

10.47 Fortified Juice and PTH. V. Tangpricha et al. did a study

to determine whether fortifying orange juice with Vitamin D would

result in changes in the blood levels of five biochemical variables.

One of those variables was the concentration of parathyroid hormone (PTH), measured in picograms/milliliter (pg/mL). The researchers published their results in the paper “Fortification of Orange

Juice with Vitamin D: A Novel Approach for Enhancing Vitamin D

Nutritional Health” (American Journal of Clinical Nutrition, Vol. 77,

pp. 1478–1483). Concentration levels were recorded at the beginning

of the experiment and again at the end of 12 weeks. The following

data, based on the results of the study, provide the decrease (negative

values indicate increase) in PTH levels, in pg/mL, for those drinking

the fortified juice and for those drinking the unfortified juice.

Fortified

−7.7

−4.8

34.4

−20.1

11.2

26.4

−5.0

−40.2

65.8

55.9

−2.2

73.5

65.1

−48.8

13.5

−20.5

0.0

15.0

−6.1

−48.4

40.0

8.8

29.4

−28.7

At the 5% significance level, do the data provide sufficient evidence to

conclude that drinking fortified orange juice reduces PTH level more

than drinking unfortified orange juice? (Note: The mean and standard

deviation for the data on fortified juice are 9.0 pg/mL and 37.4 pg/mL,

respectively, and for the data on unfortified juice, they are 1.6 pg/mL

and 34.6 pg/mL, respectively.)

10.48 Driving Distances. Data on household vehicle miles of travel

(VMT) are compiled annually by the Federal Highway Administration and are published in National Household Travel Survey, Summary of Travel Trends. Independent random samples of 15 midwestern households and 14 southern households provided the following

data on last year’s VMT, in thousands of miles.

Midwest

16.2

14.6

11.2

24.4

9.6

12.9

18.6

16.6

20.3

15.1

South

17.3

10.8

16.6

20.9

18.3

22.2

24.6

18.0

16.0

22.8

19.2

20.2

12.2

17.5

11.5

Lunch Before Recess

Lunch After Recess

x¯ 1 = 3.49

s1 = 0.57

n 1 = 314

x¯2 = 2.73

s2 = 0.54

n 2 = 314

At the 1% significance level, do the data provide sufficient evidence

to conclude that the mean score for food wasted for lunches before

recess exceeds that for lunches after recess?

Unfortified

−45.6

−15.5

online article “Investigation of the School Professionals’perceptions

and Practices Regarding Issues Influencing Recess Placement in Elementary Schools” (National Food Service Management Institute, The

University of Mississippi, 2008). Summary statistics for the score on

the opinion of recess placement issues by randomly selected students

are presented in the following table.

9.3

15.8

20.1

18.2

At the 5% significance level, does there appear to be a difference

in last year’s mean VMT for midwestern and southern households?

(Note: x¯1 = 16.23, s1 = 4.06, x¯2 = 17.69, and s2 = 4.42.)

10.49 Indian Spleen Length. Bhavan Dhingra et al. researchers in

India, were interested in ultra-sonographic measurement of liver and

spleen size in healthy Indian children. They published their findings

in the article ”Normal values of liver and spleen size by Ultrasonography in Indian children” (Indian Pediatrics, Vol. 47, pp. 487–492). The

researchers randomly sampled 347 male and 250 female healthy children in India. The mean and standard deviation of the spleen lengths

for the male children were 7.9 cm and 0.94 cm, respectively, and those

for the female children were 7.6 cm and 0.99 cm, respectively. At

the 1% significance level, do the data provide sufficient evidence to

conclude that a difference exists in mean spleen lengths of male and

female Indian children?

10.50 Recess and Wasted Food. Wendy Bounds et al. conducted

a study to determine, among other things, scheduling recess before

lunch is one way to increase children’s food and nutrient consumption at lunch and reduce plate waste. Results were published in the

In Exercises 10.51–10.56, apply Procedure 10.2 on page 474 to obtain

the required confidence interval. Interpret your result in each case.

10.51 Doing Time. Refer to Exercise 10.45 and obtain a 90% confidence interval for the difference between the mean times served by

prisoners in the fraud and firearms offense categories.

10.52 Gender and Humor. Refer to Exercise 10.46 and obtain a

98% confidence interval for the difference between the mean number

of captions for males and females.

10.53 Fortified Juice and PTH. Refer to Exercise 10.47 and find a

90% confidence interval for the difference between the mean reductions in PTH levels for fortified and unfortified orange juice.

10.54 Driving Distances. Refer to Exercise 10.48 and determine a

95% confidence interval for the difference between last year’s mean

VMTs by midwestern and southern households.

10.55 Indian Spleen Length. Refer to Exercise 10.49 and determine a 99% confidence interval for the difference between mean

spleen lengths of Indian male and female children.

10.56 Recess and Wasted Food. Refer to Exercise 10.50 and find a

98% confidence interval for the difference between the mean score for

food wasted for lunches before recess and that for lunches after recess.

Working with Large Data Sets

10.57 Vegetarians and Omnivores. Philosophical and health issues are prompting an increasing number of Taiwanese to switch to

a vegetarian lifestyle. In the paper “LDL of Taiwanese Vegetarians

Are Less Oxidizable than Those of Omnivores” (Journal of Nutrition, Vol. 130, pp. 1591–1596), S. Lu et al. compared the daily intake

of nutrients by vegetarians and omnivores living in Taiwan. Among

the nutrients considered was protein. Too little protein stunts growth

and interferes with all bodily functions; too much protein puts a strain

on the kidneys, can cause diarrhea and dehydration, and can leach

calcium from bones and teeth. Independent random samples of 51 female vegetarians and 53 female omnivores yielded the data, in grams,

on daily protein intake presented on the WeissStats site. Use the technology of your choice to do the following.

a. Obtain normal probability plots, boxplots, and the standard deviations for the two samples.

b. Do the data provide sufficient evidence to conclude that the mean

daily protein intakes of female vegetarians and female omnivores

differ? Perform the required hypothesis test at the 1% significance

level.

10.2 Inferences for Two Population Means: σ s Assumed Equal

c. Find a 99% confidence interval for the difference between the mean

daily protein intakes of female vegetarians and female omnivores.

d. Are your procedures in parts (b) and (c) justified? Explain your

10.58 Children of Diabetic Mothers. The paper “Correlations Between the Intrauterine Metabolic Environment and Blood Pressure in

Adolescent Offspring of Diabetic Mothers” (Journal of Pediatrics,

Vol. 136, Issue 5, pp. 587–592) by N. Cho et al. presented findings

of research on children of diabetic mothers. Past studies have shown

that maternal diabetes results in obesity, blood pressure, and glucosetolerance complications in the offspring. The WeissStats site provides

data on systolic blood pressure, in mm Hg, from independent random

samples of 99 adolescent offspring of diabetic mothers (ODM) and

80 adolescent offspring of nondiabetic mothers (ONM).

a. Obtain normal probability plots, boxplots, and the standard deviations for the two samples.

b. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean systolic blood pressure of ODM

children exceeds that of ONM children?

c. Determine a 95% confidence interval for the difference between

the mean systolic blood pressures of ODM and ONM children.

d. Are your procedures in parts (b) and (c) justified? Explain your

10.59 A Better Golf Tee? An independent golf equipment testing

facility compared the difference in the performance of golf balls hit

off a regular 2-3/4 wooden tee to those hit off a 3 Stinger Competition golf tee. A Callaway Great Big Bertha driver with 10 degrees of

loft was used for the test, and a robot swung the club head at approximately 95 miles per hour. Data on total distance traveled (in yards)

with each type of tee, based on the test results, are provided on the

WeissStats site.

a. Obtain normal probability plots, boxplots, and the standard deviations for the two samples.

b. At the 1% significance level, do the data provide sufficient evidence to conclude that, on average, the Stinger tee improves total

distance traveled?

c. Find a 99% confidence interval for the difference between the

mean total distance traveled with the regular and Stinger tees.

d. Are your procedures in parts (b) and (c) justified? Why or why not?

Extending the Concepts and Skills

10.60 In this section, we introduced the pooled t-test, which provides a method for comparing two population means. In deriving the

pooled t-test, we stated that the variable

z=

(x¯1 − x¯2 ) − (μ1 − μ2 )

σ (1/n 1 ) + (1/n 2 )

cannot be used as a basis for the required test statistic because σ is

unknown. Why can’t that variable be used as a basis for the required

test statistic?

10.61 The formula for the pooled variance, sp2 , is given on page 469.

Show that, if the sample sizes, n 1 and n 2 , are equal, then sp2 is the

mean of s12 and s22 .

10.62 Simulation. In this exercise, you are to perform a computer

simulation to illustrate the distribution of the pooled t-statistic, given

in Key Fact 10.2 on page 469.

a. Simulate 1000 random samples of size 4 from a normally distributed variable with a mean of 100 and a standard deviation of

16. Then obtain the sample mean and sample standard deviation

of each of the 1000 samples.

479

b. Simulate 1000 random samples of size 3 from a normally distributed variable with a mean of 110 and a standard deviation of

16. Then obtain the sample mean and sample standard deviation

of each of the 1000 samples.

c. Determine the value of the pooled t-statistic for each of the

1000 pairs of samples obtained in parts (a) and (b).

d. Obtain a histogram of the 1000 values found in part (c).

e. Theoretically, what is the distribution of all possible values of the

pooled t-statistic?

f. Compare your results from parts (d) and (e).

10.63 Two-Tailed Hypothesis Tests and CIs. As we mentioned on

page 475, the following relationship holds between hypothesis tests

and confidence intervals: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0: μ1 = μ2 will be rejected

in favor of the alternative hypothesis Ha: μ1 = μ2 if and only if the

(1 − α)-level confidence interval for μ1 − μ2 does not contain 0. In

each case, illustrate the preceding relationship by comparing the results of the hypothesis test and confidence interval in the specified

exercises.

a. Exercises 10.48 and 10.54

b. Exercises 10.49 and 10.55

10.64 Left-Tailed Hypothesis Tests and CIs. If the assumptions

for a pooled t-interval are satisfied, the formula for a (1 − α)-level

upper confidence bound for the difference, μ1 − μ2 , between two

population means is

(x¯1 − x¯2 ) + tα · sp (1/n 1 ) + (1/n 2 ).

For a left-tailed hypothesis test at the significance level α, the null

hypothesis H0: μ1 = μ2 will be rejected in favor of the alternative

hypothesis Ha: μ1 < μ2 if and only if the (1 − α)-level upper confidence bound for μ1 − μ2 is less than or equal to 0. In each case,

illustrate the preceding relationship by obtaining the appropriate upper confidence bound and comparing the result to the conclusion of

the hypothesis test in the specified exercise.

a. Exercise 10.45

b. Exercise 10.46

10.65 Right-Tailed Hypothesis Tests and CIs. If the assumptions

for a pooled t-interval are satisfied, the formula for a (1 − α)-level

lower confidence bound for the difference, μ1 − μ2 , between two

population means is

(x¯ 1 − x¯2 ) − tα · sp (1/n 1 ) + (1/n 2 ).

For a right-tailed hypothesis test at the significance level α, the null

hypothesis H0: μ1 = μ2 will be rejected in favor of the alternative

hypothesis Ha: μ1 > μ2 if and only if the (1 − α)-level lower confidence bound for μ1 − μ2 is greater than or equal to 0. In each case, illustrate the preceding relationship by obtaining the appropriate lower

confidence bound and comparing the result to the conclusion of the

hypothesis test in the specified exercise.

a. Exercise 10.47

b. Exercise 10.50

10.66 Permutation Tests. With the advent of high-speed computing, new procedures have been developed that permit statistical inferences to be performed under less restrictive conditions than those of

classical procedures. Permutation tests constitute one such collection of new procedures. To perform a permutation test to compare two

population means using independent samples, proceed as follows.

1. Combine the two samples.

2. Randomly select n 1 members from the combined sample. Now treat

these n 1 members as the first sample and the remaining n 2 members as the second sample.

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