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

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Of course, it is not always necessary to choose one statistic over the other.

Sometimes a case can be made for presenting both the mean and the median. In the

case of salaries, the median salary may be more relevant to an employee, but

the mean may be more useful to the employer because the mean is proportional

to the total payroll.

Exercises Section 8.5 (49–57)

49. In a survey, students gave their study time per

week (h), and here are the 22 values:

15.0 10.0 10.0 15.0 25.0 7.0 3.0 8.0 10.0

10.0 11.0 7.0 5.0 15.0 7.5 7.5 12.0 7.0

10.5 6.0 10.0 7.5

We would like to get a 95% confidence interval

for the population mean.

a. Compute the t-based confidence interval of

Section 8.3.

b. Display a normal plot. Is it apparent that the

data set is not normal, so the t-based interval is

of questionable validity?

c. Generate a bootstrap sample of 999 means.

d. Use the standard deviation for part (c) to get a

95% confidence interval for the population

mean.

e. Investigate the distribution of the bootstrap

means to see if the CI of part (d) is valid.

f. Use part (c) to form the 95% confidence inter-

val using the percentile method.

g. Say which interval should be used and explain

why.

50. We would like to obtain a 95% confidence interval

for the population median of the study hours data

in Exercise 49.

a. Obtain a bootstrap sample of 999 medians.

b. Use the standard deviation for part (a) to get a

95% confidence interval for the population

median.

c. Investigate the distribution of the bootstrap

medians and discuss the validity of part (b).

Does the distribution take on just a few values?

d. Use part (a) to form a 95% confidence interval

for the median using the percentile method.

e. For the study hours data, state your preference

between the median and the mean and explain

your reasoning.

51. Here are 68 weight gains in pounds for pregnant

women from conception to delivery (“Classifying

Data Displays with an Assessment of Displays

Found in Popular Software,” Teach. Statist.,

Autumn 2002: 96–101).

25 14 20 38 21 22 36 38 35 37

35 24 31 28 25 32 23 30 39 26

38 20 21 11 35 42 31 25 59 23

43 38 21 76 22 26 10 19 25 25

15 31 34 36 35 33 24 44 35 43

7 322527311425162547

35 14 65 40 35 45 27 24

We would like to get a 95% confidence interval

for the population mean.

a. Compute the t–based confidence interval of

Section 8.3.

b. Check for normality to see if part (a) is valid. Is

the sample large enough that the interval might

be valid anyway?

c. Generate a bootstrap sample of 999 means.

d. Use the standard deviation for part (c) to get a

95% confidence interval for the population

mean.

e. Investigate the distribution of the bootstrap

means to see if the CI of part (d) is valid.

f. Use part (c) to form the 95% confidence inter-

val using the percentile method.

g. Compare the intervals. If they are all close,

then the bootstrap supports the CI of part (a).

52. We would like to obtain a 95% confidence interval

for the population median weight gain using the

data in Exercise 51.

a. Obtain a bootstrap sample of 999 medians.

Use the standard deviation for part (a) to get a

95% confidence interval for the population

median.

c. Investigate the distribution of the bootstrap

medians and discuss the validity of part (b).

Does the distribution take on just a few values?

d. Use part (a) to form a 95% confidence interval

for the median using the percentile method.

e. For the weight gain data, state your preference

between the median and the mean and explain

your reasoning.

53. Nine Australian soldiers were subjected to

extreme conditions, which involved a 100-min

418

CHAPTER 8 Statistical Intervals Based on a Single Sample

walk with a 25-lb pack when the temperature was



C (104



F). One of them overheated (above



C) and was removed from the study. Here are

the rectal Celsius temperatures of the other eight

at the end of the walk (“Neural Network Training

on Human Body Core Temperature Data,” Com-

batant Protection and Nutrition Branch, Aeronau-

tical and Maritime Research Laboratory of

Australia, DSTO TN-0241, 1999):

38.4 38.7 39.0 38.5 38.5 39.0 38.5 38.6

We would like to get a 95% confidence interval

for the population mean.

a. Compute the t-based confidence interval of

Section 8.3.

b. Check for the validity of part (a).

c. Generate a bootstrap sample of 999 means.

d. Use the standard deviation for part (c) to get a

95% confidence interval for the population

mean.

e. Investigate the distribution of the bootstrap

means to see if part (d) is valid.

f. Use part (c) to form the 95% confidence inter-

val using the percentile method.

g. Compare the intervals and explain your prefer-

ence.

h. Based on your knowledge of normal body tem-

perature, would you say that body temperature

can be influenced by environment?

54. We would like to obtain a 95% confidence interval

for the population median temperature using the

data in Exercise 53.

a. Obtain a bootstrap sample of 999 medians.

b. Use the standard deviation for part (a) to get a

95% confidence interval for the population

median.

c. Investigate the distribution of the bootstrap

medians and discuss the validity of part (b).

Does the distribution take on just a few values?

d. Use part (a) to form a 95% confidence interval

for the median using the percentile method.

e. Compare all the intervals for the mean and

median. Are they fairly similar? How do you

explain that?

55. If you go to a major league baseball game, how

long do you expect the game to be? From the

2,429 games played in 2001, here is a random

sample of 25 times in minutes:

352 150 164 167 225 159 142 182 229 163

188 197 189 235 161 195 177 166 195 160

154 130 189 188 225

This is one of those rare instances in which we can

do a confidence interval and compare with the true

population mean. The mean of all 2,429 lengths is

178.29 (almost 3 h).

a. Compute the t-based confidence interval of

Section 8.3.

b. Use a normal plot to see if part (a) is valid.

c. Generate a bootstrap sample of 999 means.

d. Use the standard deviation for part (c) to get a

95% confidence interval for the population

mean.

e. Investigate the distribution of the bootstrap

means to see if the CI of part (d) is valid.

f. Use part (c) to form the 95% confidence inter-

val using the percentile method.

g. Say which interval should be used and explain

why. Does your interval include the true value,

178.29?

56. The median might be a more meaningful statistic

for the length-of-game data in Exercise 55. The

median of all 2,429 lengths is 175 min.

a. Obtain a bootstrap sample of 999 medians.

b. Use the standard deviation for part (a) to get a

95% confidence interval for the population

median.

c. Investigate the distribution of the bootstrap

medians and discuss the validity of part (b).

Does the distribution take on just a few

values?

d. Use part (a) to form a 95% confidence interval

for the median using the percentile method.

Compare your answer with the population

median, 175.

e. Comparing the percentile intervals for the

mean and the median, is there much difference

in their widths? If not, and you are forced to

choose between them for the length-of-game

data, which do you choose and why?

57. We would like to obtain a 95% confidence interval

for the study time population standard deviation

using the data in Exercise 49.

a. Obtain a bootstrap sample of 999 standard

deviations and use it to form a 95% confidence

interval for the population standard deviation

using the percentile method.

b. Recalling that it requires normal data, use the

method of Section 8.4 to obtain a 95% confi-

dence interval for the population standard devi-

ation. Discuss normality for the study hours

data. How does this interval compare with the

percentile interval?

8.5 Bootstrap Conﬁdence Intervals 419

Supplementary Exercises (58–79)

58. According to the article “Fatigue Testing of

Condoms” (Polymer Testing, 2009: 567–571),

“tests currently used for condoms are surrogates

for the challenges they face in use”, including a

test for holes, an inflation test, a package seal

test, and tests of dimensions and lubricant quality

(all fertile territory for the use of statistical meth-

odology!). The investigators developed a new

test that adds cyclic strain to a level well below

breakage and determines the number of cycles to

break. A sample of 20 condoms of one particular

type resulted in a sample mean number of 1584

and a sample standard deviation of 607. Calcu-

late and interpret a confidence interval at the

99% confidence level for the true average num-

ber of cycles to break. [Note: The article pre-

sented the results of hypothesis tests based on

the t distribution; the validity of these depends

on assuming normal population distributions.]

59. The reaction time (RT) to a stimulus is the interval

of time commencing with stimulus presentation

and ending with the first discernible movement of

a certain type. The article “Relationship of Reac-

tion Time and Movement Time in a Gross Motor

Skill” (Percept. Motor Skills, 1973: 453–454)

reports that the sample average RT for 16 experi-

enced swimmers to a pistol start was .214 s and the

sample standard deviation was .036 s.

a. Making any necessary assumptions, derive a

90% CI for true average RT for all experi-

enced swimmers.

b. Calculate a 90% upper confidence bound for

the standard deviation of the reaction time

distribution.

c. Predict RT for another such individual in a

way that conveys information about precision

and reliability.

60. For each of 18 preserved cores from oil-wet

carbonate reservoirs, the amount of residual gas

saturation after a solvent injection was measured

at water flood-out. Observations, in percentage

of pore volume, were

23.5 31.5 34.0 46.7 45.6 32.5

41.4 37.2 42.5 46.9 51.5 36.4

44.5 35.7 33.5 39.3 22.0 51.2

(See “Relative Permeability Studies of Gas-Water

Flow Following Solvent Injection in Carbonate

Rocks,” Soc. Petrol. Eng. J., 1976: 23–30.)

a. Construct a boxplot of this data, and comment

on any interesting features.

b. Is it plausible that the sample was selected

from a normal population distribution?

c. Calculate a 98% CI for the true average

amount of residual gas saturation.

61. A manufacturer of college textbooks is interested

in estimating the strength of the bindings pro-

duced by a particular binding machine. Strength

can be measured by recording the force required

to pull the pages from the binding. If this force is

measured in pounds, how many books should be

tested to estimate the average force required to

break the binding to within .1 lb with 95% confi-

dence? Assume that s is known to be .8.

62. The Pew Forum on Religion and Public Life

reported on Dec. 9, 2009 that in a survey of

2003 American adults, 25% said they believed

in astrology.

a. Calculate and interpret a confidence interval at

the 99% confidence level for the proportion of

all adult Americans who believe in astrology.

b. What sample size would be required for the

width of a 99% CI to be at most .05 irrespec-

tive of the value of

c. The upper limit of the CI in (a) gives an upper

confidence bound for the proportion being

estimated. What is the corresponding confi-

dence level?

63. There were 12 first-round heats in the men’s

100-m race at the 1996 Atlanta Summer Olym-

pics. Here are the reaction times in seconds (time

to first movement) of the top four finishers of

each heat. The first 12 are the 12 winners, then

the second-place finishers, and so on.

1st .187 .152 .137 .175 .172 .165

.184 .185 .147 .189 .172 .156

2nd .168 .140 .214 .163 .202 .173

.175 .154 .160 .169 .148 .144

3rd .159 .145 .187 .222 .190 .158

.202 .162 .156 .141 .167 .155

4th .156 .164 .160 .145 .163 .170

.182 .187 .148 .183 .162 .186

Because reaction time has little if any relation-

ship to the order of finish, it is reasonable to view

the times as coming from a single population.

420

CHAPTER 8 Statistical Intervals Based on a Single Sample

a. Estimate the population mean in a way that

conveys information about precision and

reliability. [Note:

¼ 8:08100;

1:37813:] Do the runners seem to react faster

than the swimmers in Exercise 59?

b. Calculate a 95% confidence interval for the

population proportion of reaction times that

are below .15. Reaction times below .10 are

regarded as false starts, meaning that the run-

ner anticipates the starter’s gun, because such

times are considered physically impossible.

Linford Christie, who had a reaction time of

.160 in placing second in his first-round heat,

had two such false starts in the finals and was

disqualified.

64. Aphid infestation of fruit trees can be controlled

either by spraying with pesticide or by inunda-

tion with ladybugs. In a particular area, four

different groves of fruit trees are selected for

experimentation. The first three groves are

sprayed with pesticides 1, 2, and 3, respectively,

and the fourth is treated with ladybugs, with the

following results on yield:

Treatment n

(number of

trees)

(bushels/

tree)

1 100 10.5 1.5

2 90 10.0 1.3

3 100 10.1 1.8

4 120 10.7 1.6

Let m

¼ the true average yield (bushels/tree)

after receiving the ith treatment. Then

y ¼

ðm

þ m

Þm

measures the difference in true average yields

between treatment with pesticides and treatment

with ladybugs. When n

, n

, and n

are all

large, the estimator

y obtained by replacing each

by X

is approximately normal. Use this to

derive a large-sample 100(1  a)% CI for y,

and compute the 95% interval for the given data.

65. It is important that face masks used by firefigh-

ters be able to withstand high temperatures

because firefighters commonly work in tempera-

tures of 200–500



F. In a test of one type of mask,

11 of 55 masks had lenses pop out at 250



Construct a 90% CI for the true proportion of

masks of this type whose lenses would pop out

at 250



66. A journal article reports that a sample of size 5

was used as a basis for calculating a 95% CI for

the true average natural frequency (Hz) of dela-

minated beams of a certain type. The resulting

interval was (229.764, 233.504). You decide that

a confidence level of 99% is more appropriate

than the 95% level used. What are the limits of

the 99% interval? [Hint: Use the center of the

interval and its width to determine

x and s.]

67. Chronic exposure to asbestos fiber is a well-

known health hazard. The article “The Acute

Effects of Chrysotile Asbestos Exposure on

Lung Function” (Envir. Res., 1978: 360–372)

reports results of a study based on a sample of

construction workers who had been exposed to

asbestos over a prolonged period. Among the

data given in the article were the following

(ordered) values of pulmonary compliance

(cm

/cm H

O) for each of 16 subjects 8 months

after the exposure period (pulmonary compliance

is a measure of lung elasticity, or how effectively

the lungs are able to inhale and exhale):

167.9 180.8 184.8 189.8 194.8 200.2

201.9 206.9 207.2 208.4 226.3 227.7

228.5 232.4 239.8 258.6

a. Is it plausible that the population distribution

is normal?

b. Compute a 95% CI for the true average pul-

monary compliance after such exposure.

68. In Example 7.9, we introduced the concept of a

censored experiment in which n components are

put on test and the experiment terminates as soon

as r of the components have failed. Suppose

component lifetimes are independent, each hav-

ing an exponential distribution with parameter l.

Let Y

denote the time at which the first failure

occurs, Y

the time at which the second failure

occurs, and so on, so that T

¼ Y

+ ·· +Y

(n  r)Y

is the total accumulated lifetime at

termination. Then it can be shown that 2lT

has

a chi-squared distribution with 2r df. Use this fact

to develop a 100(1  a)% CI formula for true

average lifetime 1/l. Compute a 95% CI from the

data in Example 7.9.

69. Exercise 63 from Chapter 7 introduced “regres-

sion through the origin” to relate a dependent

variable y to an independent variable x. The

assumption there was that for any fixed x value,

the dependent variable is a random variable Y

with mean value bx and variance s

(so that Y

has mean value zero when x ¼ 0). The data

Supplementary Exercises 421

consists of n independent (x

, Y

) pairs, where

each Y

is normally distributed with mean bx

and variance s

. The likelihood is then a product

of normal pdf’s with different mean values but

the same variance.

a. Show that the mle of b is

b ¼ Sx

=Sx

b. Verify that the mle of (a) is unbiased.

c. Obtain an expression for Vð

bÞ and then for s

d. For purposes of obtaining a precise estimate

of b, is it better to have the x

’s all close to

0 (the origin) or quite far from 0? Explain

your reasoning.

e. The natural prediction of Y

. Let

¼ SðY



=ðn  1Þ which is analo-

gous to our earlier sample variance

¼ SðX

 XÞ

=ðn  1Þ for a univariate

sample X

, ... , X

(in which case X is a

natural prediction for each X

). Then it can

be shown that T ¼ð

b  bÞ= S=

ﬃﬃﬃﬃﬃﬃﬃﬃ



has a

t distribution based on n  1 df. Use this to

obtain a CI formula for estimating b, and

calculate a 95% CI using the data from the

cited exercise.

70. Let X

, X

, ... , X

be a random sample from a

uniform distribution on the interval [0, y], so that

f ðxÞ¼

0  x  y

0 otherwise

Then if Y ¼ max(X

), by the first proposition in

Section 5.5, U ¼ Y/y has density function

ðuÞ¼

n1

0  u  1

0 otherwise



a. Use f

(u) to verify that

P ða=2Þ

1=n



ð1  a=2Þ

1=n



¼ 1  a

and use this to derive a 100(1  a)% CI for y.

b. Verify that P(a

1/n

 Y/ y  1) ¼ 1  a, and

derive a 100(1  a)% CI for y based on this

probability statement.

c. Which of the two intervals derived previously

is shorter? If your waiting time for a morning

bus is uniformly distributed and observed wait-

ing times are x

¼ 4.2, x

¼ 3.5, x

¼ 1.7,

¼ 1.2, and x

¼ 2.4, derive a 95% CI for y

by using the shorter of the two intervals.

71. Let 0 < g < a. Then a 100(1  a)% CI for m

when n is large is

x  z



ﬃﬃﬃ

;

x þ z

ag



ﬃﬃﬃ



The choice g ¼ a/2 yields the usual interval

derived in Section 8.2;ifg 6¼ a/2, this confidence

interval is not symmetric about

x. The width of the

interval is w ¼ sðz

þ z

ag

Þ=

ﬃﬃﬃ

. Show that w is

minimized for the choice g ¼ a/2, so that the

symmetric interval is the shortest. [Hints: (a) By

definition of z

, F(z

) ¼ 1  a, so that z

1

(1  a); (b) the relationship between the

derivative of a function y ¼ f(x) and the inverse

function x ¼ f

1

(y)is(d/dy) f

1

(y) ¼ 1/f

(x).]

72. Suppose x

, x

, ..., x

are observed values resulting

from a random sample from a symmetric but possi-

bly heavy-tailed distribution. Let

x and f

denote

the sample median and fourth spread, respectively.

Chapter 11 of Understanding Robust and Explor-

atory Data Analysis (see the bibliography in

Chapter 7) suggests the following robust 95% CI

for the population mean (point of symmetry):

x 

conservative t critical value

1:075





ﬃﬃﬃ

The value of the quantity in parentheses is 2.10 for

n ¼ 10, 1.94 for n ¼ 20, and 1.91 for n ¼ 30.

Compute this CI for the restaurant tip data of

Example 8.15, and compare to the t CI appropriate

for a normal population distribution.

73. a. Use the results of Example 8.5 to obtain a 95%

lower confidence bound for the parameter l of

an exponential distribution, and calculate the

bound based on the data given in the example.

b. If lifetime X has an exponential distribution,

the probability that lifetime exceeds t is given

by P(X > t) ¼ e

lt

. Use the result of part (a) to

obtain a 95% lower confidence bound for

the probability that lifetime exceeds 100 min.

74. Let y

and y

denote the mean weights for animals

of two different species. An investigator wishes to

estimate the ratio y

. Unfortunately the species

are extremely rare, so the estimate will be based

on finding a single animal of each species. Let X

denote the weight of the species i animal (i ¼ 1, 2),

assumed to be normally distributed with mean y

and standard deviation 1.

a. What is the distribution of the variable

hðX

; X

; y

Þ¼ðy

 y

Þ=

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

þ y

Show that this variable depends on y

and

only through y

(divide numerator and

denominator by y

b. Consider Expression (8.7) from the first sec-

tion of this chapter with a ¼1.96 and

b ¼ 1.96. Now replace < by ¼ and solve for

. Then show that a confidence interval

results if x

þ x

 1:96

, whereas if this

422

CHAPTER 8 Statistical Intervals Based on a Single Sample

inequality is not satisfied, the resulting confi-

dence set is the complement of an interval.

75. The one-sample CI for a normal mean and PI for a

single observation from a normal distribution

were both based on the central t distribution.

A CI for a particular percentile (e.g., the 1st per-

centile or the 95th percentile) of a normal popula-

tion distribution is based on the noncentral

t distribution. A particular distribution of this

type is specified by both df and the value of the

noncentrality parameter d (d ¼ 0 gives the central

t distribution). The key result is that the variable

T ¼

Xm

ﬃﬃ

ðz percentileÞ

ﬃﬃﬃ

S=s

has a noncentral t distribution with df ¼ n  1

and d ¼z percentileðÞ

ﬃﬃﬃ

Let t

.025,n,d

and t

.975,n,d

denote the critical values

that capture upper-tail area .025 and lower-tail

area .025, respectively, under the noncentral

t curve with n df and noncentrality parameter d

(when d ¼ 0, t

.975

¼t

.025

, since central t distri-

butions are symmetric about 0).

a. Use the given information to obtain a formula

for a 95% confidence interval for the (100p)th

percentile of a normal population distribution.

b. For d ¼ 6.58 and df ¼ 15, t

.975

and t

.025

are

(from MINITAB) 4.1690 and 10.9684, respec-

tively. Use this information to obtain a 95% CI

for the 5th percentile of the beer alcohol distri-

bution considered in Example 8.11.

76. The one-sample t CI for m is also a confidence

interval for the population median

m when the

population distribution is normal. We now

develop a CI for

m that is valid whatever the

shape of the population distribution as long as it

is continuous. Let X

, ..., X

be a random sample

from the distribution and Y

, ... , Y

denote the

corresponding order statistics (smallest observa-

tion, second smallest, and so on).

a. What is PðX

mÞ? What is PðfX

mg\

mgÞ?

b. What is PðY

mÞ? What is PðY

mÞ?[Hint:

What condition involving all of the X

’s is

equivalent to the largest being smaller than

the population median?]

c. What is PðY

m < Y

Þ? What does this imply

about the confidence level associated with the

CI (y

, y

) for

d. An experiment carried out to study the time

(min) necessary for an anesthetic to produce

the desired result yielded the following data:

31.2, 36.0, 31.5, 28.7, 37.2, 35.4, 33.3, 39.3,

42.0, 29.9. Determine the confidence interval

of (c) and the associated confidence level. Also

calculate the one-sample t CI using the same

level and compare the two intervals.

77. Consider the situation described in the previous

exercise.

a. What is PðfX

mg\fX

mg\\

> ~mgÞ, that is, the probability that only the

first observation is smaller than the median?

b. What is the probability that exactly one of the n

observations is smaller than the median?

c. What is Pð

m < Y

Þ?[Hint: The event in par-

entheses occurs if all n of the observations

exceed the median. How else can it occur?

What does this imply about the confidence

level associated with the CI (y

, y

n1

) for

Determine the confidence level and CI for the

data given in the previous exercise.]

78. The previous two exercises considered a CI for a

population median

m based on the n order statistics

from a random sample. Let’s now consider a pre-

diction interval for the next observation X

n+1

a. What is P(X

n+1

< X

)? What is P({X

n+1

< X

}

\ {X

n+1

< X

})?

b. What is P(X

n+1

< Y

)? What is P(X

n+1

> Y

c. What is P(Y

< X

n+1

< Y

)? What does this

say about the prediction level for the PI (y

)? Determine the prediction level and interval

for the data given in the previous exercise.

79. Consider 95% CI’s for two different parameters y

and y

, and let A

(i ¼ 1, 2) denote the event that

the value of y

is included in the random interval

that results in the CI. Thus P(A

) ¼ .95.

a. Suppose that the data on which the CI for y

based is independent of the data used to obtain

the CI for y

(e.g., we might have y

¼ m, the

population mean height for American females,

and y

¼ p, the proportion of all Kodak digital

cameras that don’t need warranty service).

What can be said about the simultaneous (i.e.,

joint) confidence level for the two intervals?

That is, how confident can we be that the first

interval contains the value of y

and that the

second contains the value of y

?[Hint: Con-

sider P(A

\ A

).]

b. Now suppose the data for the first CI is not

independent of that for the second one. What

now can be said about the simultaneous confi-

dence level for both intervals? [Hint: Consider

PðA

[ A

Þ, the probability that at least one

interval fails to include the value of what it

is estimating. Now use the fact that

Supplementary Exercises 423

PðA

[ A

ÞPðA

ÞþPðA

Þ [why?] to show

that the probability that both random intervals

include what they are estimating is at least .90.

The generalization of the bound on PðA

[ A

to the probability of a k-fold union is one

version of the Bonferroni inequality.]

c. What can be said about the simultaneous

confidence level if the confidence level for

each interval separately is 100(1  a)%?

What can be said about the simultaneous con-

fidence level if a 100(1 – a)% CI is computed

separately for each of k parameters y

, ..., y

Bibliography

DeGroot, Morris, and Mark Schervish, Probability and

Statistics (3rd ed.), Addison-Wesley, Reading,

MA, 2002. A very good exposition of the general

principles of statistical inference.

Efron, Bradley, and Robert Tibshirani, An Introduc-

tion to the Bootstrap, Chapman and Hall, New

York, 1993. The bible of the bootstrap.

Hahn, Gerald, and William Meeker, Statistical

Intervals, Wiley, New York, 1991. Everything

you ever wanted to know about statistical

intervals (confidence, prediction, tolerance, and

others).

Larsen, Richard, and Morris Marx, Introduction

to Mathematical Statistics (4th ed.), Prentice

Hall, Englewood Cliffs, NJ, 2005. Similar to

DeGroot’s presentation, but slightly less

mathematical.

424

CHAPTER 8 Statistical Intervals Based on a Single Sample

CHAPTER NINE

Tests of

Hypotheses Based

on a Single Sample

Introduction

A parameter can be estimated from sample data either by a single number (a point

estimate) or an entire interval of plausible values (a conﬁdence interval). Fre-

quently, however, the objective of an investigation is not to estimate a parameter

but to decide which of two contradictory claims about the parameter is correct.

Methods for accomplishing this comprise the part of statistical inference called

hypothesis testing . In this chapter, w e ﬁrst discuss some of the basic concepts and

terminology in hypothesis testing and then develop decision procedures for the

most frequently encountered testing problems based on a sample from a single

population.

J.L. Devore and K.N. Berk, Modern Mathematical Statistics with Applications, Springer Texts in Statistics,

DOI 10.1007/978-1-4614-0391-3_9,

Springer Science+Business Media, LLC 2012

425

9.1

Hypotheses and Test Procedures

A statistical hypothesis, or just hypothesis, is a claim or assertion either about the

value of a single parameter (population characteristic or characteristic of a proba-

bility distribution), about the values of several parameters, or about the form of an

entire probability distribution. One example of a hypothesis is the claim m ¼ $311,

where m is the true average one–term textbook expenditure for students at a

university. Another example is the statement p < .50, where p is the proportion of

adults who approve of the job that the President is doing. If m

and m

denote the true

average decreases in systolic blood pressure for two different drugs, one hypothesis

is the assertion that m

 m

¼ 0, and another is the statement m

 m

> 5.

Yet another example of a hypothesis is the assertion that the stop ping distance

for a car under particular conditions has a normal distribution. Hypotheses of this

latter sort will be considered in Chapter 13. In this and the next several chapters, we

concentrate on hypotheses about parame ters.

In any hypothesis-testing problem, there are two contradictory hypotheses

under consideration. One hypothesis might be the claim m ¼ $311 and the other

m 6¼ $311, or the two contradictory statements might be p  .50 and p < .50. The

objective is to decide, based on sample information, which of the two hypotheses is

correct. There is a familiar analogy to this in a criminal trial. One claim is the assertion

that the accused individual is innocent. In the U.S. judicial system, this is the claim that

is initially believed to be true. Only in the face of strong evidence to the contrary

should the jury reject this claim in favor of the alternative assertion that the accused

is guilty. In this sense, the claim of innocence is the favored or protected hypothesis,

and the burden of proof is placed on those who believe in the alternative claim.

Similarly, in testing statistical hypotheses, the problem will be formulated so

that one of the claims is initially favored. This initially favored claim will not be

rejected in favor of the alternative claim unless sample evidence contradicts it and

provides strong support for the alternative assertion.

DEFINITION

The null hypothesis, denoted by H

, is the claim that is initially assumed to

be true (the “prior belief” claim). The alternative hypothesis, denoted by H

is the assertion that is contradictory to H

The null hypothesis will be rejected in favor of the alternative hypoth-

esis only if sample evidence suggests that H

is false. If the sample does not

strongly contradict H

, we will continue to believe in the plausibility of the

null hypothesis. The two possible conclusions from a hypothesis-testing

analysis are then reject H

or fail to reject H

A test of hypotheses is a method for using sample data to decide whether the null

hypothesis should be rejected. Thus we might test H

: m ¼ .75 against the alterna-

tive H

: m 6¼ .75. Only if sample data strongly suggests that m is something other

than .75 should the null hypothesis be rejected. In the absence of such evidence, H

should not be rejected, since it is still quite plausible.

Sometimes an investigator does not want to accept a particular assertion unless

and until data can provide strong support for the assertion. As an example, suppose a

company is considering putting a new additive in the dried fruit that it produces.

426 CHAPTER 9 Tests of Hypotheses Based on a Single Sample

The true average shelf life with the current additive is known to be 200 days. With m

denoting the true average life for the new additive, the company would not want to

make a change unless evidence strongly suggested that m exceeds 200. An appropri-

ate problem formulation would involve testing H

: m ¼ 200 against H

: m > 200.

The conclusion that a change is justified is identified with H

, and it would take

conclusive evidence to justify rejecting H

and switching to the new additive.

Scientific research often involves trying to decide whether a current theory

should be replaced by a more plausible and satisfactory explanation of the phenome-

non under investigation. A conservative approach is to identify the current theory

with H

and the researcher’s alternative explanation with H

. Rejection of the current

theory will then occur only when evidence is much more consistent with the new

theory. In many situations, H

is referred to as the “research hypothesis,” since it is

the claim that the researcher would really like to validate. The word null means “of

no value, effect, or consequence,” which suggests that H

should be identified with

the hypothesis of no change (from current opinion), no difference, no improvement,

and so on. Suppose, for example, that 10% of all computer circuit boards produced by

a manufacturer during a recent period were defective. An engineer has suggested a

change in the production process in the belief that it will result in a reduced defective

rate. Let p denote the true proportion of defective boards resulting from the changed

process. Then the research hypothesis, on which the burden of proof is placed, is the

assertion that p < .10. Thus the alternative hypothesis is H

: p < .10.

In our treatment of hypothesis testing, H

will generally be stated as an

equality claim. If y denotes the parameter of interest, the null hypothesis will

have the form H

: y ¼ y

, where y

is a specified number called the null value of

the parameter (value claimed for y by the null hypothesis). As an example, consider

the circuit board situation just discussed. The suggested alternative hypothesis was

: p < .10, the claim that the defective rate is reduced by the process modifica-

tion. A natural choice of H

in this situatio n is the claim that p  .10, according to

which the new process is either no better or worse than the one currently used. We

will instead consi der H

: p ¼ .10 versus H

: p < .10. The rationale for using this

simplified null hypothesis is that any reasonable decision procedure for decidin g

between H

: p ¼ .10 and H

: p < .10 will also be reasonable for deciding between

the claim that p  .10 and H

. The use of a simplified H

is preferred because it has

certain technical benefits, which will be apparent shortly.

The alternative to the null hypothesis H

: y ¼ y

will look like one of the

following three assertions:

1. H

: y > y

(in which case the implicit null hypothesis is y  y

)

2. H

: y < y

(so the implicit null hypothesis states that y  y

)

3. H

: y 6¼ y

For example, let s denote the standard deviation of the distribution of outside diameters

(inches) for an engine piston. If the decision was made to use the piston unless sample

evidence conclusively demonstrated that s > .0001 in., the appropriate hypotheses

would be H

: s ¼ .0001 versus H

: s > .0001. The number y

that appears in both H

and H

(separates the alternative from the null) is called the null value.

Test Procedures

A test procedure is a rule, based on sample data, for deciding whether to reject H

A test of H

: p ¼ .10 versus H

: p < .10 in the circuit board problem might be

9.1 Hypotheses and Test Procedures 427