5 Inferences for Two Population Means, Using Paired Samples
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508
CHAPTER 10 Inferences for Two Population Means
EXAMPLE 10.15
Comparing Two Means, Using a Paired Sample
Ages of Married People The U.S. Census Bureau publishes information on the
ages of married people in Current Population Reports. Suppose that we want to
decide whether, in the United States, the mean age of married men differs from the
mean age of married women.
a. Formulate the problem statistically by posing it as a hypothesis test.
b. Explain the basic idea for carrying out the hypothesis test.
c. Suppose that 10 married heterosexual couples in the United States are selected
at random and that the ages, in years, of the people chosen are as shown in the
second and third columns of Table 10.13. Discuss the use of these data to make
a decision concerning the hypothesis test.
TABLE 10.13
Ages, in years, of a random
sample of 10 married couples
Couple
Husband
Wife
Difference, d
1
2
3
4
5
6
7
8
9
10
59
21
33
78
70
33
68
32
54
52
53
22
36
74
64
35
67
28
41
44
6
−1
−3
4
6
−2
1
4
13
8
36
Solution
a. To formulate the problem statistically, we first note that we have one variable—
namely, age—and two populations:
Population 1: All married men
Population 2: All married women.
Let μ1 and μ2 denote the means of the variable “age” for Population 1 and
Population 2, respectively:
μ1 = mean age of all married men
μ2 = mean age of all married women.
We want to perform the hypothesis test
H0: μ1 = μ2 (mean ages of married men and women are the same)
Ha: μ1 = μ2 (mean ages of married men and women differ).
b. Independent samples could be used to carry out the hypothesis test: Take independent simple random samples of, say, 10 married men and 10 married women
and then apply a pooled or nonpooled t-test to the age data obtained. However,
in this case, a paired sample is more appropriate. Here, a pair consists of a married (heterosexual) couple. The variable we analyze is the difference between
the ages of the husband and wife in a couple. By using a paired sample, we can
remove an extraneous source of variation: the variation in the ages among married couples. The sampling error thus made in estimating the difference between
the population means will generally be smaller and, therefore, we are more
likely to detect differences between the population means when such differences
exist.
c. The last column of Table 10.13 contains the difference, d, between the ages
of each of the 10 couples sampled. We refer to each difference as a paired
10.5 Inferences for Two Population Means, Using Paired Samples
509
difference because it is the difference of a pair of observations. For example, in
the first couple, the husband is 59 years old and the wife is 53 years old, giving
a paired difference of 6 years, meaning that the husband is 6 years older than
his wife.
If the null hypothesis of equal mean ages is true, the paired differences of
the ages for the married couples sampled should average about 0; that is, the
sample mean, d, of the paired differences should be roughly 0. If d is too much
different from 0, we would take this as evidence that the null hypothesis is false.
From the last column of Table 10.13, we find that the sample mean of the paired
differences is
di
36
d=
=
= 3.6.
n
10
The question now is, can this difference of 3.6 years be reasonably attributed to
sampling error, or is the difference large enough to indicate that the two populations have different means? To answer that question, we need to know the
distribution of the variable d, which we discuss next.
The Paired t-Statistic
Suppose that x is a variable on each of two populations whose members can be paired.
For each pair, we let d denote the difference between the values of the variable x on
the members of the pair. We call d the paired-difference variable.
It can be shown that the mean of the paired differences equals the difference between the two population means. In symbols,
μd = μ1 − μ2 .
Furthermore, if d is normally distributed, we can apply this equation and our knowledge
of the studentized version of a sample mean (Key Fact 8.5 on page 375) to obtain Key
Fact 10.6.
KEY FACT 10.6
Distribution of the Paired t-Statistic
Suppose that x is a variable on each of two populations whose members can
be paired. Further suppose that the paired-difference variable d is normally
distributed. Then, for paired samples of size n, the variable
t=
d − (μ1 − μ2 )
√
sd/ n
has the t-distribution with df = n − 1.
Note: We use the phrase normal differences as an abbreviation of “the paireddifference variable is normally distributed.”
Hypothesis Tests for the Means of Two Populations,
Using a Paired Sample
We now present a hypothesis-testing procedure based on a paired sample for comparing the means of two populations when the paired-difference variable is normally
distributed. In light of Key Fact 10.6, for a hypothesis test that has null hypothesis
H0: μ1 = μ2 , we can use the variable
t=
d
√
sd / n
as the test statistic and obtain the critical value(s) or P-value from the t-table, Table IV.
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CHAPTER 10 Inferences for Two Population Means
We call this hypothesis-testing procedure the paired t-test. Note that the paired
t-test is simply the one-mean t-test applied to the paired-difference variable with null
hypothesis H0: μd = 0. Procedure 10.6 provides a step-by-step method for performing
a paired t-test by using either the critical-value approach or the P-value approach.
PROCEDURE 10.6 Paired t-Test
Purpose To perform a hypothesis test to compare two population means, μ1 and μ2
Assumptions
1.
2.
Simple random paired sample
Normal differences or large sample
Step 1
The null hypothesis is H0: μ1 = μ2 , and the alternative hypothesis is
Ha: μ1 = μ2
(Two tailed)
or
Ha: μ1 < μ2
(Left tailed)
Step 2
Decide on the signiﬁcance level, α.
Step 3
Compute the value of the test statistic
t=
Ha: μ1 > μ2
.
(Right tailed)
or
d
√
sd / n
and denote that value t0 .
CRITICAL-VALUE APPROACH
Step 4
OR
P-VALUE APPROACH
Step 4 The t-statistic has df = n − 1. Use Table IV
to estimate the P-value, or obtain it exactly by using
technology.
The critical value(s) are
−tα
tα
±tα/2
or
or
(Two tailed)
(Left tailed)
(Right tailed)
with df = n − 1. 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
−t␣/2
0
t␣/2
Two tailed
t
t0
0
Left tailed
t
0
t0
t
Right tailed
␣
␣
␣ /2
t
P- value
−t␣
0
t
0
t␣
Right tailed
Left tailed
t
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 differences and is approximately correct
for large samples and nonnormal differences.
Properties and guidelines for use of the paired t-test are the same as those given for
the one-mean z-test in Key Fact 9.7 on page 411 when applied to paired differences. In
particular, the paired t-test is robust to moderate violations of the normality assumption
but, even for large samples, can sometimes be unduly affected by outliers because the
10.5 Inferences for Two Population Means, Using Paired Samples
511
sample mean and sample standard deviation are not resistant to outliers. Here are two
other important points:
r Do not apply the paired t-test to independent samples, and, likewise, do not apply a
pooled or nonpooled t-test to a paired sample.
r The normality assumption for a paired t-test refers to the distribution of the paireddifference variable, not to the two distributions of the variable under consideration.
EXAMPLE 10.16
The Paired t-Test
Ages of Married People We now return to the hypothesis test posed in Example 10.15. A random sample of 10 married couples gave the data on ages, in years,
shown in the second and third columns of Table 10.13 on page 508. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean age
of married men differs from the mean age of married women?
Solution First, we check the two conditions required for using the paired t-test,
as listed in Procedure 10.6.
FIGURE 10.15
Normal score
Normal probability plot of the paired
differences in Table 10.13
r Assumption 1 is satisfied because we have a simple random paired sample. Each
pair consists of a married couple.
r Because the sample size, n = 10, is small, we need to examine issues of normality
and outliers. (See the first bulleted item in Key Fact 9.7 on page 411.) To do so,
we construct in Fig. 10.15 a normal probability plot for the sample of paired
differences in the last column of Table 10.13. This plot reveals no outliers and is
roughly linear. So we can consider Assumption 2 satisfied.
From the preceding items, we see that the paired t-test can be used to conduct
the required hypothesis test. We apply Procedure 10.6.
3
2
1
0
–1
–2
–3
Step 1 State the null and alternative hypotheses.
Let μ1 denote the mean age of all married men, and let μ2 denote the mean age of
all married women. Then the null and alternative hypotheses are, respectively,
–3
0
3
6
9
H0: μ1 = μ2 (mean ages are equal)
Ha: μ1 = μ2 (mean ages differ).
12 15
Paired difference (yr)
Note that the hypothesis test is two tailed.
Step 2 Decide on the signiﬁcance level, α.
We are to perform the test at the 5% significance level, so α = 0.05.
Step 3 Compute the value of the test statistic
t=
d
√ .
sd / n
The paired differences (d-values) of the sample pairs are shown in the last column of
Table 10.13. We need to determine the sample mean and sample standard deviation
of those paired differences. We do so in the usual manner:
d=
36
di
=
= 3.6,
n
10
and
sd =
di2 − ( di )2 /n
=
n−1
352 − (36)2 /10
= 4.97.
10 − 1
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CHAPTER 10 Inferences for Two Population Means
Consequently, the value of the test statistic is
t=
CRITICAL-VALUE APPROACH
d
3.6
√ = 2.291.
√ =
sd / n
4.97/ 10
OR
P-VALUE APPROACH
Step 4 The critical values for a two-tailed test
are ±tα/2 with df = n − 1. Use Table IV to ﬁnd the
critical values.
Step 4 The t-statistic has df = n − 1. Use Table IV
to estimate the P-value, or obtain it exactly by using
technology.
We have n = 10 and α = 0.05. Table IV reveals that,
for df = 10 − 1 = 9, ±t0.05/2 = ±t0.025 = ±2.262, as
shown in Fig. 10.16A.
From Step 3, the value of the test statistic is t = 2.291.
The test is two tailed, so the P-value is the probability of
observing a value of t of 2.291 or greater in magnitude
if the null hypothesis is true. That probability equals the
shaded area shown in Fig. 10.16B.
FIGURE 10.16A
Reject H 0
Do not reject H 0
Reject H 0
FIGURE 10.16B
P-value
0.025
−2.262
0.025
0
2.262
t
t
0
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 = 2.291,
which falls in the rejection region depicted in Fig.10.16A.
Thus we reject H0 . The test results are statistically significant at the 5% level.
t = 2.291
Because n = 10, we have df = 10 − 1 = 9. Referring to
Fig. 10.16B and Table IV, we find that 0.02 < P < 0.05.
(Using technology, we found that P = 0.0478.)
Step 5 If P ≤ α, reject H0 ; otherwise, do not
reject H0 .
From Step 4, 0.02 < P < 0.05. 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 10.6
Exercise 10.159
on page 518
Interpretation At the 5% significance level, the data provide sufficient evidence
to conclude that the mean age of married men differs from the mean age of married
women.
Confidence Intervals for the Difference between the Means of
Two Populations, Using a Paired Sample
We can also use Key Fact 10.6 on page 509 to derive a confidence-interval procedure
for the difference between two population means. We call that confidence-interval procedure the paired t-interval procedure.
10.5 Inferences for Two Population Means, Using Paired Samples
513
PROCEDURE 10.7 Paired t-Interval Procedure
Purpose To find a confidence interval for the difference between two population
means, μ1 and μ2
Assumptions
1.
2.
Simple random paired sample
Normal differences or large sample
Step 1
For a conﬁdence level of 1 − α, use Table IV to ﬁnd tα/2 with df = n − 1.
Step 2
The endpoints of the conﬁdence interval for μ1 − μ2 are
sd
d ± tα/2 · √ .
n
Step 3
Interpret the conﬁdence interval.
Note: The confidence interval is exact for normal differences and is approximately
correct for large samples and nonnormal differences.
EXAMPLE 10.17
The Paired t-Interval Procedure
Ages of Married People Use the age data in the second and third columns of
Table 10.13 on page 508 to obtain a 95% confidence interval for the difference,
μ1 − μ2 , between the mean ages of married men and married women.
Solution We apply Procedure 10.7.
Step 1 For a conﬁdence level of 1 − α, use Table IV to ﬁnd tα/2 with
df = n − 1.
For a 95% confidence interval, α = 0.05. From Table IV, we determine that, for
df = n − 1 = 10 − 1 = 9, we have tα/2 = t0.05/2 = t0.025 = 2.262.
Step 2 The endpoints of the conﬁdence interval for μ1 − μ2 are
sd
d ± tα/2 · √ .
n
From Step 1, tα/2 = 2.262 and n = 10. And, from Example 10.16, d = 3.6 and
sd = 4.97. So, the endpoints of the confidence interval for μ1 − μ2 are
4.97
3.6 ± 2.262 · √ ,
10
or 0.04 to 7.16.
Step 3 Interpret the conﬁdence interval.
Report 10.7
Exercise 10.165
on page 518
Interpretation We can be 95% confident that the difference between the mean
ages of married men and married women is somewhere between 0.04 years and
7.16 years. In other words (see page 465), we can be 95% confident that the mean
age of married men exceeds the mean age of married women by somewhere between 0.04 years and 7.16 years.
Paired versus Independent Samples
Suppose that we want to perform a hypothesis test or obtain a confidence interval in
order to compare two population means. As you know, two sampling designs that are
candidates for the comparison are paired samples and independent samples.
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CHAPTER 10 Inferences for Two Population Means
When a paired sample is possible, it is often preferable to independent samples.
Indeed, as we mentioned in the solution to part (b) of Example 10.15, by using a paired
sample, we can frequently remove extraneous sources of variation. The sampling error
thus made in estimating the difference between the population means will generally be
smaller and, therefore, we are more likely to detect differences between the population
means when such differences exist.
The paired-sample design is preferable when, for the variable under consideration,
there is a positive correlation between the population pairs; that is, when population
pairs tend to both have above-average values of the variable or below-average values
of the variable. This situation is precisely the case for ages of husbands and wives
(Example 10.15)—older husbands tend to be married to older wives.
THE TECHNOLOGY CENTER
Most statistical technologies have programs that automatically perform paired
t-procedures. In this subsection, we present output and step-by-step instructions for
such programs.
EXAMPLE 10.18
Using Technology to Conduct Paired t-Procedures
Ages of Married People The second and third columns of Table 10.13 on page 508
give the ages of 10 randomly selected married couples. Use Minitab, Excel, or the
TI-83/84 Plus to perform the hypothesis test in Example 10.16 and obtain the confidence interval required in Example 10.17.
Solution Let μ1 denote the mean age of all married men, and let μ2 denote the
mean age of all married women. The task in Example 10.16 is to perform the hypothesis test
H0: μ1 = μ2 (mean ages are equal)
Ha: μ1 = μ2 (mean ages differ)
at the 5% significance level; the task in Example 10.17 is to obtain a 95% confidence
interval for μ1 − μ2 .
We applied the paired t-procedures programs to the data, resulting in Output 10.4. Steps for generating that output are presented in Instructions 10.4. Note
to Excel users: For brevity, we have presented only the essential portions of the
actual output.
As shown in Output 10.4, the P-value for the hypothesis test is about 0.048.
Because the P-value is less than the specified significance level of 0.05, we reject H0 . Output 10.4 also shows that a 95% confidence interval for the difference
between the means is from 0.04 to 7.16.
OUTPUT 10.4
MINITAB
Paired t-procedures on the age data
10.5 Inferences for Two Population Means, Using Paired Samples
515
OUTPUT 10.4 (cont.) Paired t-procedures on the age data
EXCEL
TI-83/84 PLUS
Using T-Test
Using TInterval
INSTRUCTIONS 10.4 Steps for generating Output 10.4
MINITAB
1 Store the age data from the second and third columns
of Table 10.13 in columns named HUSBAND and WIFE
2 Choose Stat ➤ Basic Statistics ➤ Paired t. . .
3 Press the F3 key to reset the dialog box
4 Click in the Sample 1 text box and specify HUSBAND
5 Click in the Sample 2 text box and specify WIFE
6 Click the Options. . . button
7 Click in the Confidence level text box and type 95
8 Click the arrow button at the right of the Alternative
hypothesis drop-down list box and select Difference =
hypothesized difference
9 Click OK twice
EXCEL
1 Store the age data from the second and third columns
of Table 10.13 in columns named HUSBAND and WIFE
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
HUSBAND data
5 Click in the Sample 2 selection box and then select the
column of the worksheet that contains the WIFE data
6 In the Data format list, select the Paired samples
option button
7 Click the Options tab
8 Click the arrow button at the right of the Alternative
hypothesis drop-down list box and select Mean 1 –
Mean 2 = D
9 Type 5 in the Significance level (%) text box
10 Click OK
11 Click the Continue button in the XLSTAT – Selections
dialog box
TI-83/84 PLUS
Store the age data from the second and third columns of
Table 10.13 in lists named HUSB and WIFE.
FOR THE PAIRED DIFFERENCES:
1 Press 2nd ➤ LIST, arrow down to HUSB, and press
ENTER
2 Press −
3 Press 2nd ➤ LIST, arrow down to WIFE, and press ENTER
4 Press STO ➧
5 Press 2nd ➤ A-LOCK, type DIFF, and press ENTER
FOR THE HYPOTHESIS TEST:
1 Press STAT, arrow over to TESTS, and press 2
2 Highlight Data and press ENTER
3 Press the down-arrow key, type 0 for μ0 , and press
ENTER
4 Press 2nd ➤ LIST, arrow down to DIFF, and press
ENTER twice
5 Type 1 for Freq and then press ENTER
6 Highlight = μ0 and press ENTER
7 Arrow down to Calculate and press ENTER
FOR THE CI:
1 Press STAT, arrow over to TESTS, and press 8
2 Highlight Data and press ENTER
3 Press the down-arrow key
4 Press 2nd ➤ LIST, arrow down to DIFF, and press
ENTER twice
5 Type 1 for Freq and then press ENTER
6 Type .95 for C-Level and press ENTER twice
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CHAPTER 10 Inferences for Two Population Means
Note to Minitab and Excel users: As we have noted, 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 paired t-procedure twice: once for the
one-tailed hypothesis test and once for the confidence interval specifying a two-tailed
hypothesis test.
Note to TI-83/84 Plus users: The paired t-procedures are just one-mean t-procedures
applied to the paired differences. Since, at the time of this writing, the TI-83/84 Plus
does not have built-in paired t-procedures, we applied its one-mean t-procedures to the
paired differences, as seen in Instructions 10.4.
Exercises 10.5
Understanding the Concepts and Skills
10.139 State one possible advantage of using paired samples instead
of independent samples.
10.140 What constitutes each pair in a paired sample?
10.141 State the two conditions required for performing a paired
t-procedure. How important are those conditions?
10.142 Provide an example (different from the ones considered in
this section) of a procedure based on a paired sample being more appropriate than one based on independent samples.
In Exercises 10.143–10.148, hypothesis tests are proposed. For each
hypothesis test,
a. identify the variable.
b. identify the two populations.
c. identify the pairs.
d. identify the paired-difference variable.
e. determine the null and alternative hypotheses.
f. classify the hypothesis test as two tailed, left tailed, or right tailed.
10.143 TV Viewing. The A. C. Nielsen Company collects data on
the TV viewing habits of Americans and publishes the information in
Nielsen Report on Television. Suppose that you want to use a paired
sample to decide whether the mean viewing times of married men
and married women differ.
10.144 Self-Reported Weight. The article “Accuracy of SelfReported Height and Weight in a Community-Based Sample of
Older African Americans and Whites” (Journal of Gerontology Series A: Biological Sciences and Medical Sciences, Vol. 65A, No. 10,
pp. 1123–1129) by G. Fillenbaum et al. explores the relationship between measured and self-reported height and weight. The
authors sampled African American and White women and men older
than 70 years of age. A hypothesis test is to be performed to decide
whether, on average, self-reported weight is less than measured
weight for the aforementioned age group.
10.145 Hypnosis and Pain. In the paper “An Analysis of Factors
That Contribute to the Efficacy of Hypnotic Analgesia” (Journal
of Abnormal Psychology, Vol. 96, No. 1, pp. 46–51), D. Price and
J. Barber examined the effects of hypnosis on pain. They measured
response to pain using a visual analogue scale (VAS), in centimeters,
where higher VAS indicates greater pain. VAS sensory ratings were
made before and after hypnosis on each of 16 subjects. A hypothesis test is to be performed to decide whether, on average, hypnosis
reduces pain.
10.146 Sports Stadiums and Home Values. In the paper “Housing Values Near New Sporting Stadiums” (Land Economics, Vol. 81,
Issue 3, pp. 379–395), C. Tu examined the effects of construction of
new sports stadiums on home values. Suppose that you want to use a
paired sample to decide whether construction of new sports stadiums
affects the mean price of neighboring homes.
10.147 Breastmilk and Antioxidants. There is convincing evidence that breastmilk containing antioxidants is important in the prevention of diseases in infants. Researchers A. Xavier et al. studied the
effects of storing breastmilk on antioxidant levels in the article “Total
Antioxidant Concentrations of Breastmilk—An Eye-Opener to the
Negligent” (Journal of Health, Population and Nutrition, Vol. 29,
No. 6, pp. 605–611). Samples of breastmilk were taken from women
and divided into fresh samples that were immediately tested and the
remaining samples that were stored in the refrigerator and tested after
48 hours. A hypothesis test is to be performed to decide whether, on
average, stored breastmilk has a lower total antioxidant capacity.
10.148 Fiber Density. In the article “Comparison of Fiber Counting
by TV Screen and Eyepieces of Phase Contrast Microscopy” (American Industrial Hygiene Association Journal, Vol. 63, pp. 756–761),
I. Moa et al. reported on determining fiber density by two different
methods. The fiber density of 10 samples with varying fiber density
was obtained by using both an eyepiece method and a TV-screen
method. A hypothesis test is to be performed to decide whether, on
average, the eyepiece method gives a greater fiber density reading
than the TV-screen method.
In Exercises 10.149–10.154, the null hypothesis is H0 : μ1 = μ2 and
the alternative hypothesis is as specified. We have provided data from
a simple random paired sample from the two populations under consideration. In each case, use the paired t-test to perform the required
hypothesis test at the 10% significance level.
10.149 Ha : μ1 = μ2
Observation from
Pair
Population 1
Population 2
1
2
3
4
5
6
7
13
16
13
14
12
8
17
11
15
10
8
8
9
14
10.5 Inferences for Two Population Means, Using Paired Samples
10.150 Ha : μ1 < μ2
517
10.154 Ha : μ1 > μ2
Observation from
Observation from
Pair
Population 1
Population 2
Pair
Population 1
Population 2
1
2
3
4
5
6
7
7
4
10
0
20
−1
12
13
9
6
2
19
5
10
1
2
3
4
5
6
7
8
9
10
40
30
34
22
35
26
26
27
11
35
32
29
36
18
31
26
25
25
15
31
10.151 Ha : μ1 > μ2
Applying the Concepts and Skills
Observation from
Pair
Population 1
Population 2
1
2
3
4
5
6
7
8
7
9
12
15
27
16
11
8
6
8
11
14
6
9
5
2
Preliminary data analyses indicate that use of a paired t-test is reasonable in Exercises 10.155–10.160. Perform each hypothesis test by
using either the critical-value approach or the P-value approach.
10.155 Behavioral Genetics. In the article “Growth references for
height, weight and BMI of Twins aged 0–2.5 years” (ACTA Pediatrica, Vol. 97, pp. 1099–1104), the researchers P. Dommelen et al.
determined the size of the growth deficit in Dutch monozygotic and
dizygotic twins aged between 0–2.5 years as compared to the singletons and to construct reference growth charts for twins. The following table shows the difference of the height of the twins at various age
groups from 0 to 2.5 years.
10.152 Ha : μ1 = μ2
5.2
5.0
4.7
4.5
4.3
3.9
3.2
2.7
1.9
4.1
3.4
2.3
4.8
3.8
1.0
Observation from
Pair
Population 1
Population 2
1
2
3
4
5
6
7
8
10
8
13
13
17
12
12
11
12
7
11
16
15
9
12
7
10.153 Ha : μ1 < μ2
Identify the variable under consideration.
Identify the two populations.
Identify the paired-difference variable.
Are the numbers in the table paired differences? Why or why not?
At the 5% significance level, do the data provide sufficient evidence to conclude that the mean heights of twins separately differ?
(Note: d = 3.65 cm and sd = 1.23 cm.)
f. Repeat part (e) at the 1% significance level.
a.
b.
c.
d.
e.
10.156 Sleep. In 1908, W. S. Gosset published “The Probable Error
of a Mean” (Biometrika, Vol. 6, pp. 1–25). In this pioneering paper,
published under the pseudonym “Student,” he introduced what later
became known as Student’s t-distribution. Gosset used the following
data set, which gives the additional sleep in hours obtained by
10 patients who used laevohysocyamine hydrobromide.
Observation from
Pair
Population 1
Population 2
1
2
3
4
5
6
7
8
9
15
22
15
27
24
23
8
20
2
18
25
17
24
30
23
10
27
3
1.9
4.4
0.8
5.5
1.1
1.6
0.1
4.6
−0.1
3.4
Identify the variable under consideration.
Identify the two populations.
Identify the paired-difference variable.
Are the numbers in the table paired differences? Why or why not?
At the 5% significance level, do the data provide sufficient evidence to conclude that laevohysocyamine hydrobromide is effective in increasing sleep? (Note: d = 2.33 and sd = 2.002.)
f. Repeat part (e) at the 1% significance level.
a.
b.
c.
d.
e.
518
CHAPTER 10 Inferences for Two Population Means
10.157 Anorexia Treatment. Anorexia nervosa is a serious eating
disorder, particularly among young women. The following data provide the weights, in pounds, of 17 anorexic young women before
and after receiving a family therapy treatment for anorexia nervosa.
[SOURCE: D. Hand et al. (ed.) A Handbook of Small Data Sets, London: Chapman & Hall, 1994; raw data from B. Everitt (personal communication)]
Before
After
Before
After
Before
After
83.3
86.0
82.5
86.7
79.6
87.3
94.3
91.5
91.9
100.3
76.7
98.0
76.9
94.2
73.4
80.5
81.6
83.8
76.8
101.6
94.9
75.2
77.8
95.2
82.1
77.6
83.5
89.9
86.0
95.5
90.7
92.5
93.8
91.7
Does family therapy appear to be effective in helping anorexic young
women gain weight? Perform the appropriate hypothesis test at the
5% significance level.
10.158 Measuring Treadwear. R. Stichler et al. compared two
methods of measuring treadwear in their paper “Measurement of
Treadwear of Commercial Tires” (Rubber Age, Vol. 73:2). Eleven
tires were each measured for treadwear by two methods, one based
on weight and the other on groove wear. The data, in thousands of
miles, are as follows.
Weight
method
Groove
method
Weight
method
Groove
method
30.5
30.9
31.9
30.4
27.3
20.4
28.7
25.9
23.3
23.1
23.7
20.9
24.5
20.9
18.9
13.7
11.4
16.1
19.9
15.2
11.5
11.2
At the 5% significance level, do the data provide sufficient evidence
to conclude that, on average, the two measurement methods give different results?
10.159 Glaucoma and Corneal Thickness. Glaucoma is a leading cause of blindness in the United States. N. Ehlers measured the
corneal thickness of eight patients who had glaucoma in one eye but
not in the other. The results of the study were published as the paper
“On Corneal Thickness and Intraocular Pressure, II” (Acta Opthalmologica, Vol. 48, pp. 1107–1112). The data on corneal thickness, in
microns, are shown in the following table.
Patient
Normal
Glaucoma
1
2
3
4
5
6
7
8
484
478
492
444
436
398
464
476
488
478
480
426
440
410
458
460
At the 10% significance level, do the data provide sufficient evidence
to conclude that mean corneal thickness is greater in normal eyes than
in eyes with glaucoma?
10.160 Cooling Down. Cooling down with a cold drink before exercise in the heat is believed to help an athlete perform. Researcher
J. Dugas explored the difference between cooling down with an ice
slurry (slushy) and with cold water in the article “Ice Slurry Ingestion Increases Running Time in the Heat” (Clinical Journal of Sports
Medicine, Vol. 21, No. 6, pp. 541–542). Ten male participants drank
a flavored ice slurry and ran on a treadmill in a controlled hot and humid environment. Days later, the same participants drank cold water
and ran on a treadmill in the same hot and humid environment. The
following table shows the times, in minutes, it took to fatigue on the
treadmill for both the ice slurry and the cold water.
Subject
Cold Water
Ice Slurry
1
2
3
4
5
6
7
8
9
10
52
37
44
51
34
38
41
50
29
38
56
43
52
58
38
45
45
58
34
44
At the 1% significance level, do the data provide sufficient evidence
to conclude that, on average, cold water is less effective than ice slurry
for optimizing athletic performance in the heat? (Note: The mean and
standard deviation of the paired differences are −5.9 minutes and
1.60 minutes, respectively.)
In Exercises 10.161–10.166, apply Procedure 10.7 on page 513 to
obtain the required confidence interval. Interpret your result in each
case.
10.161 Behavioral Genetics. Refer to Exercise 10.155.
a. Determine a 95% confidence interval for the difference between
the mean heights of the twins.
b. Repeat part (a) for a 99% confidence level.
10.162 Sleep. Refer to Exercise 10.156.
a. Determine a 90% confidence interval for the additional sleep that
would be obtained, on average, by using laevohysocyamine hydrobromide.
b. Repeat part (a) for a 98% confidence level.
10.163 Anorexia Treatment. Refer to Exercise 10.157 and find a
90% confidence interval for the weight gain that would be obtained,
on average, by using the family therapy treatment.
10.164 Measuring Treadwear. Refer to Exercise 10.158 and find a
95% confidence interval for the mean difference in measurement by
the weight and groove methods.
10.165 Glaucoma and Corneal Thickness. Refer to Exercise 10.159 and obtain an 80% confidence interval for the difference
between the mean corneal thickness of normal eyes and that of eyes
with glaucoma.
10.166 Cooling Down. Refer to Exercise 10.160 and find a
98% confidence interval for the difference between the mean times
to fatigue on a treadmill in a hot and humid environment after cooling down with cold water and after cooling down with an ice slurry.