Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications

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14. The level of monoamine oxidase (MAO) activity

in blood platelets (nm/mg protein/h) was deter-

mined for each individual in a sample of 43

chronic schizophrenics, resulting in

x ¼ 2:69 and

¼ 2.30, as well as for 45 normal subjects,

resulting in

y ¼ 6:35 and s

¼ 4.03. Does this

data strongly suggest that true average MAO

activity for normal subjects is more than twice

the activity level for schizophrenics? Derive a

test procedure and carry out the test using

a ¼ .01. [Hint:

and

here have a different

form from the three standard cases. Let m

and

refer to true average MAO activity for schizo-

phrenics and normal subjects, respectively, and

consider the parameter y ¼ 2m

 m

. Write

and

in terms of y, estimate y, and derive

(“Reduced Monoamine Oxidase Activity in Blood

Platelets from Schizophrenic Patients,” Nature,

July 28, 1972: 225–226).]

15. a. Show for the upper-tailed test with s

and

known that as either m or n increases, b

decreases when m

 m

> D

b. For the case of equal sample sizes (m ¼ n)

and fixed a, what happens to the necessary

sample size n as b is decreased, where b is

the desired type II error probability at a fixed

alternative?

16. To decide whether chemistry or physics majors

have higher starting salaries in industry, n B.S.

graduates of each major are surveyed, yielding

the following results (in $1000’s):

Major Sample Average Sample SD

Chemistry 41.5 2.5

Physics 41.0 2.5

Calculate the P-value for the appropriate two-

sample z test, assuming that the data was based

on n ¼ 100. Then repeat the calculation for

n ¼ 400. Is the small P-value for n ¼ 400 indica-

tive of a difference that has practical significance?

Would you have been satisfied with just a report

of the P-value? Comment briefly.

17. Much recent research has focused on comparing

business environment cultures across several coun-

tries. The article “Perception of Internal Factors for

Corporate Entrepreneurship: A Comparison of

Canadian and U.S. Managers” (Entrep. Theory

Pract., 1999: 9–24) presented the following

summary data on hours per week managers spent

thinking about new ideas.

Country

Sample

Size

Sample

Mean

Sample

U.S. 174 5.8 6.0

Canada 353 5.1 4.6

Does it appear that true average time per week that

U.S. managers spend thinking about new ideas

differs from that for Canadian managers? State

and test the relevant hypotheses.

18. Credit card spending and resulting debt pose very

real threats to consumers in general, and the poten-

tial for abuse is especially serious among college

students. It has been estimated that about

of all

college students possess credit cards, and 80% of

these students received cards during their first year

of college. The article “College Students’ Credit

Card Debt and the Role of Parental Involvement:

Implications for Public Policy” (J. Public Policy

Mark., 2001: 105–113) reported that for 209

students whose parents had no involvement what-

soever in credit card acquisition or payments, the

sample mean total account balance was $421 with

a sample standard deviation of $686, whereas for

75 students whose parents assisted with payments

even though they were under no legal obligation to

do so, the sample mean and sample standard devia-

tion were $666 and $1048, respectively. All sam-

pled students were at most 21 years of age.

a. Do you think it is plausible that the distribu-

tions of total debt for these two types of stu-

dents are normal? Why or why not? Is it

necessary to assume normality in order to com-

pare the two groups using an inferential proce-

dure described in this chapter? Explain.

b. Estimate the true average difference between

total balance for noninvolvement students and

postacquisition-involvement students using a

method that incorporates precision into the

estimate. Then interpret the estimate. [Note:

Data was also reported in the article for pre-

acquisition involvement only and for both pre-

and postacquisition involvement.]

19. Returning to the previous exercise, the mean and

standard deviation of the number of credit cards for

the no-involvement group were 2.22 and 1.58,

respectively, whereas the mean and standard devi-

ation for the payment-help group were 2.09 and

1.65, respectively. Does it appear that the true

average number of cards for no-involvement stu-

dents exceeds the average for payment-help stu-

dents? Carry out an appropriate test of significance.

498

CHAPTER 10 Inferences Based on Two Samples

10.2

The Two-Sample t Test

and Confidence Interval

In practice, it is virtually always the case that the values of the population variances

are unknown. In the previous section, we illustrated for large sample sizes the use

of a test procedure and CI in which the sample variances were used in place of

the population variances. In fact, for large samples, the CLT allows us to use these

methods even when the two populations of interest are not normal.

In many problems, though, at least one sample size is small and the popula-

tion variances have unknown values. In the absence of the CLT, we proceed by

making specific assumptions about the underlying population distributions. The use

of inferential procedures that follow from these assumptions is then restricted to

situations in which the assumptions are at least approximately satisfied.

ASSUMPTIONS

Both populations are normal , so that X

, X

, ..., X

is a random sample from

a normal distribution and so is Y

, ... , Y

(with the X’s and Y’s independent

of each other). The plausibility of these assumptions can be judged by

constructing a normal probability plot of the x

’s and another of the y

’s.

The test statistic and confidence interval formula are based on the same standar-

dized variable developed in Section 10.1, but the relevant distribution is now t

rather than z.

THEOREM

When the population distributions are both normal, the standardized variable

T ¼

X  Y ðm

 m

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð10:2Þ

has approximately a t distribution with df n estimated from the data by

n ¼



ðs

=mÞ

m  1

ðs

=nÞ

n  1

ðse

þðse

ðse

m  1

ðse

n  1

where

ﬃﬃﬃﬃ

ﬃﬃﬃ

(round n down to the nearest integer).

10.2 The Two-Sample t Test and Conﬁdence Interval 499

We can give some justification for the theorem. Dividing numerator and

denominator of (10.2) by the standard deviation of the numerator, we get

½X  Y ðm

 m

Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

r ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

The numerator of this ratio is a standard normal rv because it results from standar-

dizing

X  Y, which is normal ly distribut ed because it is the difference of indepen-

dent normal rv’s. The denom inator is independe nt of the numerator because the

sample variances are independent of the sample means. However, in order for

(10.2)tobeat random variable, the denominator needs to be the square root of a

chi-squared rv over its degrees of freedom, and unfortunately this is not generally

true. However, we can try to write the square of the denominator

½S

=m þ S

=n=½s

=m þ s

=n approximat ely as a chi-squared rv W with n degrees

of freedom, divided by n,so



To determine n we equate the means and variances of both sides, with the help

of E(W) ¼ n, V(W) ¼ 2n, ðm  1ÞS

 w

m1

, ðn  1ÞS

 w

n1

, from

Section 6.4. It follows that EðS

Þ¼s

, VðS

Þ¼2 s

= m  1ðÞ, and similarly for

. The mean of the left-hand side is



which is also the mean of the right-hand side, so the means are equal. The variance

of the left-hand side is



ðm  1Þm

ðn  1Þn

and the variance of the right-hand side is













We then equate the two, substituting sample variances for the unknown population

variances, and solve for n. This gives the n of the theorem. ■

Manipulating T in a probability statement to isolate m

 m

gives a CI,

whereas a test stati stic results from replacing m

 m

by the null value D

TWO-SAMPLE

t PROCEDURES

The two-sample t confidence interval for m

 m

with confidence level

100(1  a)% is then

x  y  t

a=2;n

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

A one-sided confidence bound can be calculated as described earlier.

500 CHAPTER 10 Inferences Based on Two Samples

The two-sampl e t test for testing

: m

 m

¼ D

is as follows:

Test statistic value: t ¼

x  y  D

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

Alternative Hypothesis Rejection Region for Approximate Level a Test

: m

 m

> D

t  t

a,n

(upper-tailed test)

: m

 m

< D

t t

a,n

(lower-tailed test)

: m

 m

6¼ D

either t  t

a/2,n

or t t

a/2,n

(two-tailed test)

A P-value can be computed as described in Section 9.4 for the one-sample t test.

Example 10.6 Which way of dispensing champagne, the traditional vertical method or a tilted

“beer-like” pour, preserves more of the tiny gas bubbles that improve flavor and

aroma? The following data was reported in the article “On the Losses of Dissolved

during Champagne Serving” (J. Agr. Food Chem., 2010: 8768–6775).

Temperature (



C) Type of Pour n Mean (g/L) SD

18 Traditional 4 4.0 .5

18 Slanted 4 3.7 .3

12 Traditional 4 3.3 .2

12 Slanted 4 2.0 .3

Assuming that the sampled distributions are normal, let’s calculate confidence

intervals for the difference between true average dissolved CO

loss for the

traditional pour and that for the slanted pour at each of the two temperatures.

For the 18



C temperature, the number of degrees of freedom for the interval is

df ¼



ð:5

=4Þ

ð:3

=4Þ

:007225

:00147083

¼ 4:91

Rounding down, the CI will be based on 4 df. For a confidence level of 99%,

we need t

.005,4

¼ 4.604. The desired interval is

4:0  3:7 ð4:604Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

¼ :3 ð4:604Þð:2915Þ¼:3  1:3 ¼ð1:0; 1:6Þ

Thus we can be highly confident that  1:0 < m

 m

< 1:6, where m

and

are true average losses for the traditional and slant methods, respectively. Notice

that this CI contains 0, so at the 99% confidence level, it is plausible that

 m

¼ 0, that is, that m

¼ m

The df formula for the 12



Ccomparisonyieldsdf¼ .00105625/

.00020208 ¼ 5.23, necessitating the use of t

.005,5

¼ 4.032 for a 99% CI. The result-

ing interval is (.6, 2.0). Thus 0 is not a plausible value for this difference. It appears

from the CI that the true average loss when the slant method is used is smaller than

that when the traditional method is used, so that the slant method is better at this

temperature. This in fact was the conclusion reported in the popular media.

■

10.2 The Two-Sample t Test and Conﬁdence Interval 501

Example 10.7 The deterioration of many municipal pipeline networks across the country is a

growing concern. One technology proposed for pipeline rehabilitation uses a

flexible liner threaded through existing pipe. Th e article “Effect of Welding on a

High-Density Polyethylene Liner” (J. Mater. Civil Eng., 1996: 94–100) reported

the following data on tensile strength (psi) of liner specimens both when a certain

fusion process was used and when this process was not used.

No fusion 2748 2700 2655 2822 2511

3149 3257 3213 3220 2753

m ¼ 10

x ¼ 2902:8 s

¼ 277.3

Fused 3027 3356 3359 3297 3125 2910 2889 2902

n ¼ 8

y ¼ 3108:1 s

¼ 205.9

Figure 10.3 shows normal probability plots from MINITAB. The linear pattern in

each plot supports the assumption that the tensile strength distributions under the

two conditions are both normal.

The authors of the article stated that the fusion process increased the average

tensile strength. The message from the comparative boxplot of Figure 10.4 is not all

that clear. Let’s carry ou t a test of hypotheses to see whether the data supports this

conclusion.

1. Let m

be the tru e average tensile strength of specimens when the no-fusion

treatment is used and m

denote the true average tensile strength when the fusion

treatment is used.

: m

 m

¼ 0 (no difference in the true average tensile strengths for the two

treatments)

: m

 m

< 0 (true average tensile strength for the no-fusion treatment is less

than that for the fusion treatment, so that the investigators’

conclusion is correct)

Figure 10.3 Normal probability plots from MINITAB for the tensile strength data

502

CHAPTER 10 Inferences Based on Two Samples

4. Th e null value is D

¼ 0, so the test statistic is

t ¼

x  y

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

5. We now compute both the test statistic value and the df for the test:

t ¼

2902:8  3108:1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

277:3

205:9

205:3

113:97

¼1:8

Using s

=m ¼ 7689:529 and s

=n ¼ 5299:351,

n ¼

7689:529 þ 5299:351ðÞ

ð7689:529Þ

ð5299:351Þ

168;711;004

10;581;747

¼ 15:94

so the test will be based on 15 df.

6. Appendix Table A.7 shows that the area under the 15 df t curve to the right of 1.8

is .046, so the P-value for a lower-tailed test is also .046. The following

MINITAB output summarizes all the computations:

Twosample T for nofusion vs. fused

N Mean StDev SE Mean

No fusion 10 2903 277 88

Fused 8 3108 206 73

95% C.I. for mu nofusion-mu fused: (488, 38)

T-Test mu nofusion

¼ mu fused (vs <): T ¼ 1.80 P ¼ 0.046 DF ¼ 15

7. Using a significance level of .05, we can barely reject the null hypothesis in

favor of the alternative hypothesis, confirming the conclusion stated in the

article. However, someone demanding more compelling evidence might select

a ¼ .01, a level for which

cannot be rejected.

270026002500

Type 1

Type 2

32003100300029002800 3300 3400

Strength

Figure 10.4 A comparative boxplot of the tensile strength data

10.2 The Two-Sample t Test and Conﬁdence Interval 503

If the question posed had been whether fusing increased true average strength by more

than 100 psi, then the relevant hypotheses would have been

: m

 m

¼100

versus

: m

 m

< 100; that is, the null value would have been D

¼100. ■

Pooled t Procedures

Alternatives to the two-sample t procedures just described result from assuming not

only that the two population distributions are normal but also that they have equal

variances (s

¼ s

). That is, the two population distribution curves are assumed

normal with equal spreads, the only possible difference between them being where

they are centered.

Let s

denote the common population variance. Then standardizing X  Y gives

Z ¼

X  Y ðm

 m

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

X  Y ðm

 m

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ



which has a standard normal distribution. Before this variable can be used as a basis

for making infere nces about m

 m

, the common variance must be est imated from

sample data. One estimator of s

is S

, the variance of the m observations in the first

sample, and another is S

, the variance of the second sample. Intuitively, a better

estimator than either individual sample variance results from combining the two

sample variances. A first thought might be to use ðS

þ S

Þ=2, the ordinary average

of the two sample variances. However, if m > n, then the first sample contains

more information about s

than does the second sample, and an analogous com-

ment applies if m < n. The following weighted average of the two sample var-

iances, called the pooled (i.e., combined) estimator of s

, adjusts for any

difference between the two sample sizes:

m  1

m þ n  2

n  1

m þ n  2

We can show that S

is proportional to a chi-squared rv with m + n  2 df.

Recall that ðm  1ÞS



 w

m1

; ðn  1ÞS



 w

n1

. Furthermore, S

and S

are independent, so with s

¼ s

ðm þ n  2ÞS

m  1

n  1

is the sum of two independent chi-squared rv’s with m 1 and n 1 df, respectively,

so the sum is a chi-squared rv with (m 1) + (n  1) ¼ m + n  2 df. Furthermore,

it is also independent of

X and Y because the sample means are independent of

the sample variances. Now consider the ratio

X  Y ðm

 m

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1=m þ 1=nðÞ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðm þ n  2ÞS



m þ n  2

X  Y ðm

 m

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ



On the left is the ratio of a standard normal rv to the square root of an

independent chi-squared rv over its degrees of freedom, m + n  2, so the ratio

504 CHAPTER 10 Inferences Based on Two Samples

has the t distribution with m + n  2 degrees of freedom. We see therefore that if S

replaces s

in the expression for Z, the resulting standardized variable has a

t distribution. In the same way that earlier standardized variables were used as

a basis for deriving confidence intervals and test procedures, this t variable imme-

diately leads to the pooled t confidence interval for estimating m

 m

and the

pooled t test for testing hypotheses about a difference between mean s.

In the past, many statisticians recommended these pooled t procedures over

the two-sample t procedures. Th e pooled t test, for example, can be derived from

the likelihood ratio principle, whereas the two-sample t test is not a likelihood

ratio test. Furthermore, the significance level for the pooled t test is exact, whereas

it is only approximate for the two-sample t test. However, recent research has

shown that although the pooled t test does outperform the two-s ample t test by a bit

(smaller b’s for the same a) when s

¼ s

, the former test can easily lead to

erroneous conclusions if applied when the variances are different. Analogous

comments apply to the behavior of the two confidence intervals. That is, the pooled

t procedures are not robust to violations of the equal variance assumption.

It has been suggested that one coul d carry out a preliminary test of

¼ s

and use a pooled t procedure if this null hypothesis is not rejected.

Unfortunately, the usual “F test” of equal variances (Section 10.5) is quite sensitive

to the assumption of normal population distributions, much more so than t proce-

dures. We therefore recommend the conservative approach of using two-sample

t procedures unless there is really compelling evidence for doing otherwise,

particularly when the two sample sizes are different.

Type II Error Probabilities

Determining type II error probabilities (or equivalently, power ¼ 1  b) for the

two-sample t test is complicated. There does not appear to be any simple way to use

the b curves of Appendix Table A.16. The most recent version of MINITAB

(Version 16) will calculate power for the pooled t test but not for the two-sample

t test. However, the UCLA Statistics Department homepage (http://www.stat.ucla.

edu) permits access to a power calculator that will do this. For example, we

specified m ¼ 10, n ¼ 8, s

¼ 300, s

¼ 225 (these are the sample sizes for

Example 10.7, whose sample standard deviations are somewhat smaller than

these values of s

and s

) and asked for the power of a two-tailed level .05 test

: m

 m

¼ 0 when m

 m

¼ 100, 250, and 500. The resulting values of the

power were .1089, .4609, and .9635 (corresponding to b ¼ .89, .54, and .04),

respectively. In general, b will decrease as the sample sizes increase, as a increases,

and as m

 m

moves farther from 0. The software will also calculate sample sizes

necessary to obtain a specified value of power for a particular value of m

 m

Exercises Section 10.2 (20–38)

20. Determine the number of degrees of freedom

for the two-sample t test or CI in each of the

following situations:

a. m ¼ 10, n ¼ 10, s

¼ 5.0, s

¼ 6.0

b. m ¼ 10, n ¼ 15, s

¼ 5.0, s

¼ 6.0

c. m ¼ 10, n ¼ 15, s

¼ 2.0, s

¼ 6.0

d. m ¼ 12, n ¼ 24, s

¼ 5.0, s

¼ 6.0

21. Expert and amateur pianists were compared in a

study “Maintaining Excellence: Deliberate Practice

and Elite Performance in Young and Older

Pianists” (J. Exp. Psychol. Gen., 1996: 331–340).

The researchers used a keyboard that allowed mea-

surement of the force applied by a pianist in striking

a key. All 48 pianists played Prelude Number 1

10.2 The Two-Sample t Test and Conﬁdence Interval 505

from Bach’s Well-Tempered Clavier. For 24 ama-

teur pianists the mean force applied was 74.5 with

standard deviation 6.29, and for 24 expert pianists

the mean force was 81.8 with standard deviation

8.64. Do expert pianists hit the keys harder? Assum-

ing normally distributed data, state and test the

relevant hypotheses, and interpret the results.

22. The article “Supervised Exercise Versus Non-

Supervised Exercise for Reducing Weight in

Obese Adults” (J. Sport. Med. Phys. Fit., 2009:

85–90) reported on an investigation in which par-

ticipants were randomly assigned either to a

supervised exercise program or a control group.

Those in the control group were told only that they

should take measures to lose weight. After 4

months, the sample mean decrease in body fat

for the 17 individuals in the experimental group

was 6.2 kg with a sample standard deviation of

4.5 kg, whereas the sample mean and sample

standard deviation for the 17 people in the control

group were 1.7 kg and 3.1 kg, respectively.

Assume normality of the two body fat loss distri-

butions (as did the investigators).

a. Calculate a 99% lower prediction bound for the

body fat loss of a single randomly selected

individual subjected to the supervised exercise

program. Can you be highly confident that such

an individual will actually lose body fat?

b. Does it appear that true average decrease in

body fat is more than 2 kg larger for the exper-

imental condition than for the control condi-

tion? Carry out a test of appropriate hypotheses

using a significance level of .01

23. Fusible interlinings are being used with increasing

frequency to support outer fabrics and improve the

shape and drape of various pieces of clothing. The

article “Compatibility of Outer and Fusible Inter-

lining Fabrics in Tailored Garments” (Textile Res.

J., 1997: 137–142) gave the accompanying data on

extensibility (%) at 100 g/cm for both high-quality

fabric (H) and poor-quality fabric (P) specimens.

H 1.2 .9 .7 1.0 1.7 1.7 1.1 .9 1.7

1.9 1.3 2.1 1.6 1.8 1.4 1.3 1.9 1.6

.8 2.0 1.7 1.6 2.3 2.0

P 1.6 1.5 1.1 2.1 1.5 1.3 1.0 2.6

a. Construct normal probability plots to verify the

plausibility of both samples having been

selected from normal population distributions.

b. Construct a comparative boxplot. Does it sug-

gest that there is a difference between true

average extensibility for high-quality fabric

specimens and that for poor-quality specimens?

c. The sample mean and standard deviation for

the high-quality sample are 1.508 and .444,

respectively, and those for the poor-quality

sample are 1.588 and .530. Use the two-sample

t test to decide whether true average extensibil-

ity differs for the two types of fabric.

24. Low-back pain (LBP) is a serious health problem

in many industrial settings. The article “Isody-

namic Evaluation of Trunk Muscles and Low-

Back Pain Among Workers in a Steel Factory”

(Ergonomics, 1995: 2107–2117) reported the

accompanying summary data on lateral range of

motion (degrees) for a sample of workers without

a history of LBP and another sample with a history

of this malady.

Condition

Sample

Size

Sample

Mean

Sample

No LBP 28 91.5 5.5

LBP 31 88.3 7.8

Calculate a 90% confidence interval for the dif-

ference between population mean extent of lateral

motion for the two conditions. Does the interval

suggest that population mean lateral motion dif-

fers for the two conditions? Is the message differ-

ent if we use a confidence level of 95%?

25. Research has shown that good hip range of motion

and strength in throwing athletes results in

improved performance and decreased body stress.

The article “Functional Hip Characteristics of

Baseball Pitchers and Position Players” ( Am. J.

Sport. Med., 2010: 383–388) reported on a study

involving samples of 40 professional pitchers and

40 professional position players. For the pitchers,

the sample mean trail leg total arc of motion

(degrees) was 75.6 with a sample standard devia-

tion of 5.9, whereas the sample mean and sample

standard deviation for position players were 79.6

and 7.6, respectively. Assuming normality, test

appropriate hypotheses to decide whether true

average range of motion for the pitchers is less

than that for the position players (as hypothesized

by the investigators). In reaching your conclusion,

what type of error might you have committed?

26. Tennis elbow is thought to be aggravated by

the impact experienced when hitting the ball.

The article “Forces on the Hand in the Tennis

One-Handed Backhand” (Int. J. Sport Biomech.

1991: 282–292) reported the force (Newtons) on

the hand just after impact on a one-handed

506

CHAPTER 10 Inferences Based on Two Samples

backhand drive for six advanced players and for

eight intermediate players.

Type of Player

Sample

Size

Sample

Mean

Sample

1. Advanced 6 40.3 11.3

2. Intermediate 8 21.4 8.3

In their analysis of the data, the authors assumed

that both force distributions were normal. Calcu-

late a 95% CI for the difference between true

average force for advanced players (m

) and

true average force for intermediate players (m

Does your interval provide compelling evidence

for concluding that the two m’s are different?

Would you have reached the same conclusion

by calculating a CI for m

 m

(i.e., by reversing

the 1 and 2 labels on the two types of players)?

Explain.

27. As the population ages, there is increasing con-

cern about accident-related injuries to the elderly.

The article “Age and Gender Differences in

Single-Step Recovery from a Forward Fall”

(J Gerontol A Biol Sci Med Sci., 1999 54(1):

M44–50) reported on an experiment in which the

maximum lean angle—the farthest a subject

is able to lean and still recover in one step—

was determined for both a sample of younger

females (21–29 years) and a sample of older

females (67–81 years). The following observations

are consistent with summary data given in the

article:

YF: 29, 34, 33, 27, 28, 32, 31, 34, 32, 27

OF: 18, 15, 23, 13, 12

Does the data suggest that true average maximum

lean angle for older females is more than 10

degrees smaller than it is for younger females?

State and test the relevant hypotheses at signifi-

cance level .10 by obtaining a P-value.

28. The article “Effect of Internal Gas Pressure on the

Compression Strength of Beverage Cans and

Plastic Bottles” (J. Testing Eval., 1993: 129–131)

includes the accompanying data on compression

strength (lb) for a sample of 12-oz aluminum

cans filled with strawberry drink and another sam-

ple filled with cola. Does the data suggest that the

extra carbonation of cola results in a higher average

compression strength? Base your answer on a

P-value. What assumptions are necessary for your

analysis?

Beverage

Sample

Size

Sample

Mean

Sample

Strawberry drink 15 540 21

Cola 15 554 15

29. Which foams more when you pour it, Coke or

Pepsi? Here are measurements by Diane Warfield

on the foam volume (mL) after pouring a 12-oz

can of Coke, based on a sample of 12 cans:

312.2 292.6 331.7 355.1 362.9 331.7

292.6 245.8 280.9 320.0 273.1 288.7

and here are measurements for Pepsi, based on a

sample of 12 cans:

148.3 210.7 152.2 117.1 89.7 140.5

128.8 167.8 156.1 136.6 124.9 136.6

a. Verify graphically that normality is an appro-

priate assumption.

b. Calculate a 99% confidence interval for the

population difference in mean volumes.

c. Does the upper limit of your interval in (b) give

a 99% lower confidence bound for the differ-

ence between the two m’s? If not, calculate

such a bound and interpret it in terms of the

relationship between the foam volumes of

Coke and Pepsi.

d. Summarize in a sentence what you have learned

about the foam volumes of Coke and Pepsi.

30. The accompanying data set gives expenses

(including tuition and fees but not room and

board) for 16 colleges from the 2008 edition of

U.S. News and World Report’s America’s Best

Colleges, which lists 248 national liberal arts col-

leges in four tiers. The first two tiers are combined

in a list of 125 colleges. We drew a random sam-

ple of size 8 from the 62 in the first tier and

another random sample of size 8 from the 63 in

the next tier, excluding non-private colleges.

Tier College Expenses

1 Gettysburg 35760

1 Harvey Mudd 34891

1 Scripps 35850

1 Macalester 33694

1 Hamilton 36860

10.2 The Two-Sample t Test and Conﬁdence Interval 507