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

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The likelihood ratio statistic for testing the relevant hypotheses is

ð:5Þ

S x

=½ð:5Þ

S x



. Taking the natural log of the likelihood ratio and multi-

plying by 2 gives the rejection region 2

 2



 w

a;1

for the large-

sample vers ion of the LR test.

Suppose that a sample of n ¼ 30 errors results in

¼ 38:6 and



¼ 37:3. Th en

 2lnðLRÞ¼2





¼ 2:6

Comparing this to w

:05;1

¼ 3:84, we would not reject the null hypothesis at the

5% significance level. It is plausible that the measurement process is indeed

unbiased.

■

Exercises Section 9.5 (60–71)

60. Reconsider the paint-drying problem discussed in

Example 9.2. The hypotheses were H

: m ¼ 75

versus H

: m < 75, with s assumed to have value

9.0. Consider the alternative value m ¼ 74, which

in the context of the problem would presumably

not be a practically significant departure from H

a. For a level .01 test, compute b at this alterna-

tive for sample sizes n ¼ 100, 900, and 2500.

b. If the observed value of

X is x ¼ 74, what can

you say about the resulting P-value when

n ¼ 2500? Is the data statistically significant

at any of the standard values of a?

c. Would you really want to use a sample size

of 2500 along with a level .01 test (disregard-

ing the cost of such an experiment)? Explain.

61. Consider the large-sample level .01 test in Sec-

tion 9.3 for testing H

: p ¼ .2 against H

: p > .2.

a. For the alternative value p ¼ .21, compute

b(.21) for sample sizes n ¼ 100, 2500,

10,000, 40,000, and 90,000.

b. For

p ¼ x=n ¼ :21, compute the P-value when

n ¼ 100, 2500, 10,000, and 40,000.

4.0

3.0

2.5

3.5

5.0

4.5

5.5

−.5

1.51.0

.50

Σ|x

−

Figure 9.11 Determining the mle of the double exponential parameter by minimizing

 yj

478

CHAPTER 9 Tests of Hypotheses Based on a Single Sample

c. In most situations, would it be reasonable to

use a level .01 test in conjunction with a sample

size of 40,000? Why or why not?

62. For a random sample of n individuals taking a

licensing exam, let X

¼ 1 if the ith individual in

the sample passes the exam and X

¼ 0 otherwise

(i ¼ 1, ... , n).

a. With p denoting the proportion of all exam-

takers who pass, show that the most powerful

test of H

: p ¼ .5 versus H

: p ¼ .75 rejects H

when Sx

 c.

b. If n ¼ 20 and you want a  .05 for the test of

(a), would you reject H

if 15 of the 20 indivi-

duals in the sample pass the exam?

c. What is the power of the test you used in (b)

when p ¼ .75 [i.e., what is p(.75)]?

d. Is the test derived in (a) UMP for testing

the hypotheses H

: p ¼ .5 versus H

: p >.5?

Explain your reasoning.

e. Graph the power function p(p) of the test for the

hypotheses of (d) when n ¼ 20 and a  .05.

f. Return to the scenario of (a), and suppose the

test is based on a sample size of 50. If the

probability of a type II error is approximately

.025, what is the approximate significance level

of the test (use a normal approximation)?

63. The error X in a measurement has a normal distri-

bution with mean value 0 and variance s

. Con-

sider testing H

: s

¼ 2 versus H

: s

¼ 3 based

on a random sample X

, ... , X

of errors.

a. Show that a most powerful test rejects H

when

 c:

b. For n ¼ 10, find the value of c for the test in

(a) that results in a ¼ .05.

c. Is the test of (a) UMP for H

: s

¼ 2 versus

: s

> 2? Justify your assertion.

64. Suppose that X, the fraction of a container that is

filled, has pdf f(x;y) ¼ yx

y1

for 0 < x < 1

(where y > 0), and let X

, ... , X

be a random

sample from this distribution.

a. Show that the most powerful test for H

: y ¼ 1

versus H

: y ¼ 2 rejects the null hypothesis if

Sln(x

)  c.

b. Is the test of (a) UMP for testing H

: y ¼ 1

versus H

: y > 1? Explain your reasoning.

c. If n ¼ 50, what is the (approximate) value of c

for which the test has significance level .05?

65. Consider a random sample of n component life-

times, where the distribution of lifetime is expo-

nential with parameter l.

a. Obtain a most powerful test for H

: l ¼ 1

versus H

: l ¼ .5, and express the rejection

region in terms of a “simple” statistic.

b. Is the test of (a) uniformly most powerful for

: l ¼ 1 versus H

: l < 1? Justify your

answer.

66. Consider a random sample of size n from the

“shifted exponential” distribution with pdf

fx; yðÞ¼e

ðxyÞ

for x > y and 0 otherwise (the

graph is that of the ordinary exponential pdf with

l ¼ 1 shifted so that it begins its descent at y rather

than at 0). Let Y

denote the smallest order statistic,

and show that the likelihood ratio test of H

: y  1

versus H

: y > 1 rejects the null hypothesis if y

the observed value of Y

,is c.

67. Suppose that each of n randomly selected indivi-

duals is classified according to his/her genotype

with respect to a particular genetic characteristic

and that the three possible genotypes are AA, Aa,

and aa with long-run proportions (probabilities) y

2y(1y), and (1y)

, respectively (0 < y < 1).

It is then straightforward to show that the

likelihood is

½2yð1  yÞ

ð1  yÞ

where x

, x

, and x

are the number of individuals

in the sample who have the AA, Aa, and aa geno-

types, respectively. Show that the most powerful

test for testing H

: y ¼ .5 versus H

: y ¼ .8

rejects the null hypothesis when 2x

+ x

 c.Is

this test UMP for the alternative H

: y > .5?

Explain. [Note: The fact that the joint distribution

of X

, X

, and X

is multinomial can be used to

obtain the value of c that yields a test with any

desired significance level when n is large.]

68. The error in a measurement is normally dis-

tributed with mean m and standard deviation 1.

Consider a random sample of n errors, and show

that the likelihood ratio test for H

: m ¼ 0 versus

: m 6¼ 0 rejects the null hypothesis when either

x  c or x c. What is c for a test with

a ¼ .05? How does the test change if the standard

deviation of an error is s

(known) and the rele-

vant hypotheses are H

: m ¼ 0 versus H

: m 6¼m

69. Measurement error in a particular situation is

normally distributed with mean value m and

standard deviation 4. Consider testing H

: m ¼ 0

versus H

: m 6¼ 0 based on a sample of n ¼ 16

measurements.

a. Verify that the usual test with significance

level .05 rejects H

if either x  1:96 or

9.5 Some Comments on Selecting a Test Procedure 479

x 1:96. [Note: That this test is unbiased

follows from the fact that the way to capture

the largest area under the z curve above an

interval having width 3.92 is to center that inter-

val at 0 (so it extends from 1.96 to 1.96).]

b. Consider the test that rejects H

if either

x  2:17 or x 1:81. What is a, that is, p(0)?

c. What is the power of the test proposed in (b)

when m ¼ .1 and when m ¼.1? (Note that .1

and .1 are very close to the null value, so one

would not expect large power for such values).

Is the test unbiased?

d. Calculate the power of the usual test when

m ¼ .1 and when m ¼.1. Is the usual test a

most powerful test? [Hint: Refer to your calcu-

lations in (c).] [Note: It can be shown that

the usual test is most powerful among all

unbiased tests.]

70. A test of whether a coin is fair will be based on

n ¼ 50 tosses. Let X be the resulting number of

heads. Consider two rejection regions: R

¼ {x:

either x  17 or x  33} and R

¼ {x: either

x  18 or x  37}.

a. Determine the significance level (type I error

probability) for each rejection region.

b. Determine the power of each test when

p ¼ .49. Is the test with rejection region R

uniformly most powerful level .033 test?

Explain.

c. Is the test with rejection region R

unbiased?

Explain.

d. Sketch the power function for the test with

rejection region R

, and then do so for the test

with the rejection region R

. What does your

intuition suggest about the desirability of using

the rejection region R

71. Consider Example 9.24.

a. With t ¼ð

x  m

Þ=ðs=

ﬃﬃﬃ

Þ, show that the likeli-

hood ratio is equal to l ¼ [1 + t

/(n  1)]

n/2

and therefore the approximate chi-square statis-

tic is 2[ln(l)] ¼ n ln[1 + t

/(n  1)].

b. Apply part (a) to test the hypotheses of

Exercise 55, using the data given there. Com-

pare your results with the answers found in

Exercise 55.

Supplementary Exercises (72–94)

72. A sample of 50 lenses used in eyeglasses yields a

sample mean thickness of 3.05 mm and a sample

standard deviation of .34 mm. The desired true

average thickness of such lenses is 3.20 mm. Does

the data strongly suggest that the true average

thickness of such lenses is something other than

what is desired? Test using a ¼ .05.

73. In Exercise 72, suppose the experimenter had

believed before collecting the data that the value

of s was approximately .30. If the experimenter

wished the probability of a type II error to be .05

when m ¼ 3.00, was a sample size of 50 unneces-

sarily large?

74. It is specified that a certain type of iron should

contain .85 g of silicon per 100 g of iron (.85%).

The silicon content of each of 25 randomly

selected iron specimens was determined, and the

accompanying MINITAB output resulted from a

test of the appropriate hypotheses.

Variable N Mean StDev SE

Mean

sil cont 25 0.8880 0.1807 0.0361 1.05 0.30

a. What hypotheses were tested?

b. What conclusion would be reached for a sig-

nificance level of .05, and why? Answer the

same question for a significance level of .10.

75. One method for straightening wire before coiling

it to make a spring is called “roller straightening.”

The article “The Effect of Roller and Spinner

Wire Straightening on Coiling Performance and

Wire Properties” (Springs, 1987: 27–28) reports

on the tensile properties of wire. Suppose a sample

of 16 wires is selected and each is tested to deter-

mine tensile strength (N/mm

). The resulting sam-

ple mean and standard deviation are 2160 and 30,

respectively.

a. The mean tensile strength for springs made

using spinner straightening is 2150 N/mm

What hypotheses should be tested to determine

whether the mean tensile strength for the roller

method exceeds 2150?

b. Assuming that the tensile strength distribution

is approximately normal, what test statistic

would you use to test the hypotheses in part (a)?

c. What is the value of the test statistic for this

data?

d. What is the P-value for the value of the test

statistic computed in part (c)?

480

CHAPTER 9 Tests of Hypotheses Based on a Single Sample

e. For a level .05 test, what conclusion would you

reach?

76. A new method for measuring phosphorus levels in

soil is described in the article “A Rapid Method to

Determine Total Phosphorus in Soils” (Soil Sci.

Amer. J., 1988: 1301–1304). Suppose a sample of

11 soil specimens, each with a true phosphorus

content of 548 mg/kg, is analyzed using the new

method. The resulting sample mean and standard

deviation for phosphorus level are 587 and 10,

respectively.

a. Is there evidence that the mean phosphorus

level reported by the new method differs sig-

nificantly from the true value of 548 mg/kg?

Use a ¼ .05.

b. What assumptions must you make for the test

in part (a) to be appropriate?

77. The article “Orchard Floor Management Utilizing

Soil-Applied Coal Dust for Frost Protection”

(Agric. Forest Meteorol., 1988: 71–82) reports

the following values for soil heat flux of eight

plots covered with coal dust.

34.7 35.4 34.7 37.7 32.5 28.0 18.4 24.9

The mean soil heat flux for plots covered only

with grass is 29.0. Assuming that the heat-flux

distribution is approximately normal, does the

data suggest that the coal dust is effective in

increasing the mean heat flux over that for

grass? Test the appropriate hypotheses using

a ¼ .05.

78. The article “Caffeine Knowledge, Attitudes, and

Consumption in Adult Women” (J. Nutrit. Ed.,

1992: 179–184) reports the following summary

data on daily caffeine consumption for a sample

of adult women: n ¼ 47,

x ¼ 215 mg, s ¼ 235

mg, and range ¼ 51176.

a. Does it appear plausible that the population

distribution of daily caffeine consumption is

normal? Is it necessary to assume a normal

population distribution to test hypotheses

about the value of the population mean con-

sumption? Explain your reasoning.

b. Suppose it had previously been believed that

mean consumption was at most 200 mg. Does

the given data contradict this prior belief? Test

the appropriate hypotheses at significance level

.10 and include a P-value in your analysis.

79. The accompanying output resulted when MINI-

TAB was used to test the appropriate hypotheses

about true average activation time based on the

data in Exercise 56. Use this information to reach

a conclusion at significance level .05 and also at

level .01.

TEST OF MU ¼ 25.000 VS MU G.T. 25.000

N MEAN STDEV SE MEAN T P VALUE

time 13 27.923 5.619 1.559 1.88 0.043

80. The true average breaking strength of ceramic

insulators of a certain type is supposed to be at

least 10 psi. They will be used for a particular

application unless sample data indicates conclu-

sively that this specification has not been met.

A test of hypotheses using a ¼ .01 is to be based

on a random sample of ten insulators. Assume

that the breaking-strength distribution is normal

with unknown standard deviation.

a. If the true standard deviation is .80, how

likely is it that insulators will be judged satis-

factory when true average breaking strength is

actually only 9.5? Only 9.0?

b. What sample size would be necessary to have

a 75% chance of detecting that true average

breaking strength is 9.5 when the true stan-

dard deviation is .80?

81. The accompanying observations on residual flame

time (sec) for strips of treated children’s nightwear

were given in the article “An Introduction to Some

Precision and Accuracy of Measurement Pro-

blems” (J. Test. Eval., 1982: 132–140). Suppose a

true average flame time of at most 9.75 had been

mandated. Does the data suggest that this condition

has not been met? Carry out an appropriate test

after first investigating the plausibility of assump-

tions that underlie your method of inference.

9.85 9.93 9.75 9.77 9.67 9.87 9.67

9.94 9.85 9.75 9.83 9.92 9.74 9.99

9.88 9.95 9.95 9.93 9.92 9.89

82. The incidence of a certain type of chromosome

defect in the U.S. adult male population is

believed to be 1 in 75. A random sample of 800

individuals in U.S. penal institutions reveals 16

who have such defects. Can it be concluded that

the incidence rate of this defect among prisoners

differs from the presumed rate for the entire adult

male population?

a. State and test the relevant hypotheses using

a ¼ .05. What type of error might you have

made in reaching a conclusion?

b. What P-value is associated with this test?

Based on this P-value, could H

be rejected at

significance level .20?

83. In an investigation of the toxin produced by a

certain poisonous snake, a researcher prepared 26

Supplementary Exercises 481

different vials, each containing 1 g of the toxin, and

then determined the amount of antitoxin needed to

neutralize the toxin. The sample average amount of

antitoxin necessary was found to be 1.89 mg, and

the sample standard deviation was .42. Previous

research had indicated that the true average neutra-

lizing amount was 1.75 mg/g of toxin. Does the

new data contradict the value suggested by prior

research? Test the relevant hypotheses using the

P-value approach. Does the validity of your analy-

sis depend on any assumptions about the population

distribution of neutralizing amount? Explain.

84. The sample average unrestrained compressive

strength for 45 specimens of a particular type of

brick was computed to be 3107 psi, and the sample

standard deviation was 188. The distribution of

unrestrained compressive strength may be some-

what skewed. Does the data strongly indicate that

the true average unrestrained compressive

strength is less than the design value of 3200?

Test using a ¼ .001.

85. To test the ability of auto mechanics to identify

simple engine problems, an automobile with a

single such problem was taken in turn to 72 differ-

ent car repair facilities. Only 42 of the 72 mechan-

ics who worked on the car correctly identified the

problem. Does this strongly indicate that the true

proportion of mechanics who could identify this

problem is less than .75? Compute the P-value and

reach a conclusion accordingly.

86. When X

, X

, ... , X

are independent Poisson

variables, each with parameter l, and n is large,

the sample mean

X has approximately a normal

distribution with m ¼ Eð

XÞ¼l and s

¼ VðXÞ¼

l=n. This implies that

Z ¼

X  l

ﬃﬃﬃﬃﬃﬃﬃﬃ

l=n

has approximately a standard normal distribution.

For testing H

: l ¼ l

, we can replace l by l

the equation for Z to obtain a test statistic. This

statistic is actually preferred to the large-sample

statistic with denominator S=

ﬃﬃﬃ

(when the X

’s

are Poisson) because it is tailored explicitly to the

Poisson assumption. If the number of requests for

consulting received by a certain statistician during

a 5-day work week has a Poisson distribution and

the total number of consulting requests during a

36-week period is 160, does this suggest that the

true average number of weekly requests exceeds

4.0? Test using a ¼ .02.

87. A hot-tub manufacturer advertises that with its

heating equipment, a temperature of 100



Fcanbe

achieved in at most 15 min. A random sample of 32

tubs is selected, and the time necessary to achieve a

100



F temperature is determined for each tub. The

sample average time and sample standard deviation

are 17.5 min and 2.2 min, respectively. Does this

data cast doubt on the company’s claim? Compute

the P-value and use it to reach a conclusion at level

.05 (assume that the heating-time distribution is

approximately normal).

88. Chapter 8 presented a CI for the variance s

of a

normal population distribution. The key result

there was that the rv w

¼ðn  1ÞS

has a

chi-squared distribution with n  1 df. Consider

the null hypothesis H

: s

¼ s

(equivalently,

s ¼ s

). Then when H

is true, the test statistic

¼ðn  1ÞS

has a chi-squared distribution

with n  1 df. If the relevant alternative is

: s

> s

, rejecting H

if ðn  1ÞS



a;n1

gives a test with significance level a.To

ensure reasonably uniform characteristics for a

particular application, it is desired that the true

standard deviation of the softening point of a

certain type of petroleum pitch be at most .50



The softening points of ten different specimens

were determined, yielding a sample standard devi-

ation of .58



C. Does this strongly contradict the

uniformity specification? Test the appropriate

hypotheses using a ¼ .01.

89. Referring to Exercise 88, suppose an investigator

wishes to test H

: s

¼ .04 versus H

: s

< .04

based on a sample of 21 observations. The com-

puted value of 20s

/.04 is 8.58. Place bounds

on the P-value and then reach a conclusion at

level .01.

90. When the population distribution is normal and n is

large, the sample standard deviation S has approxi-

mately a normal distribution with E(S)  s and

V(S)  s

/(2n). We already know that in this

case, for any n,

X is normal with EðXÞ¼m and

Vð

XÞ¼s

=n.

a. Assuming that the underlying distribution is

normal, what is an approximately unbiased

estimator of the 99th percentile y ¼ m + 2.33s?

b. As discussed in Section 6.4 , when the X

’s are

normal

X and S are independent rv’s (one mea-

sures location whereas the other measures

482

CHAPTER 9 Tests of Hypotheses Based on a Single Sample

spread). Use this to compute Vð

yÞ and s

for

the estimator

y of part (a). What is the esti-

mated standard error

c. Write a test statistic for testing H

: y ¼ y

that

has approximately a standard normal distribu-

tion when H

is true. If soil pH is normally

distributed in a certain region and 64 soil sam-

ples yield

x ¼ 6:33, s ¼ .16, does this provide

strong evidence for concluding that at most

99% of all possible samples would have a pH

of less than 6.75? Test using a ¼ .01.

91. Let X

, X

, ... , X

be a random sample from an

exponential distribution with parameter l. Then it

can be shown that 2lSX

has a chi-squared distri-

bution with n ¼ 2n(by first showing that 2lX

has

a chi-squared distribution with n ¼ 2).

a. Use this fact to obtain a test statistic and rejec-

tion region that together specify a level a test

for H

: m ¼ m

versus each of the three com-

monly encountered alternatives. [Hint: E(X

) ¼

m ¼ 1/l,som ¼ m

is equivalent to l ¼ 1/m

b. Suppose that ten identical components, each

having exponentially distributed time until

failure, are tested. The resulting failure times

are

95 16 11 3 42 71 225 64 87 123

Use the test procedure of part (a) to decide

whether the data strongly suggests that the

true average lifetime is less than the previously

claimed value of 75.

92. Suppose the population distribution is normal with

known s. Let g be such that 0 < g < a. For testing

: m ¼ m

versus H

: m 6¼ m

, consider the test

that rejects H

if either z  z

or z z

ag

, where

the test statistic is Z ¼ð

X  m

Þ=ðs=

ﬃﬃﬃ

Þ:

a. Show that P(type I error) ¼ a.

b. Derive an expression for b(m

). [Hint: Express

the test in the form “reject H

if either

x  c

or  c

.”]

c. Let D > 0. For what values of g (relative to a)

will b(m

+ D) < b(m

 D)?

93. After a period of apprenticeship, an organization

gives an exam that must be passed to be eligible

for membership. Let p ¼ P(randomly chosen

apprentice passes). The organization wishes an

exam that most but not all should be able to pass,

so it decides that p ¼ .90 is desirable. For a par-

ticular exam, the relevant hypotheses are H

p ¼ .90 versus the alternative H

: p 6¼ .90. Sup-

pose ten people take the exam, and let X ¼ the

number who pass.

a. Does the lower-tailed region {0, 1, ... ,5}

specify a level .01 test?

b. Show that even though H

is two-sided, no

two-tailed test is a level .01 test.

c. Sketch a graph of b(p

) as a function of p

for

this test. Is this desirable?

94. A service station has six gas pumps. When no

vehicles are at the station, let p

denote the proba-

bility that the next vehicle will select pump i

(i ¼ 1, 2, ... , 6). Based on a sample of size n,

we wish to test H

: p

¼ ... ¼ p

versus the alter-

native H

: p

¼ p

, p

¼ p

(note

that H

is not a simple hypothesis). Let X be the

number of customers in the sample that select an

even-numbered pump.

a. Show that the likelihood ratio test rejects H

either X  c or X  n  c.[Hint: When H

true, let y denote the common value of p

, p

and p

b. Let n ¼ 10 and c ¼ 9. Determine the power of

the test both when H

is true and also when

¼ p

; p

¼ p

Bibliography

See the bibliographies for Chapters 7 and 8.

Bibliography 483

CHAPTER TEN

Inferences Based

on Two Samples

Introduction

Chapters 8 and 9 presented conﬁdence intervals (CIs) and hypothesis testing

procedures for a single mean m, single proportion

, and a single variance s

Here we extend these methods to situations involving the means, propor tions,

and variances of two different population distributions. For example, let m

and

denote true average decrease in cholesterol for two drugs. Then an investi-

gator might wish to use results from patients assigned at random to two different

groups as a basis for testing the hypothesis

: m

¼ m

versus the alternative

hypothesis

: m

6¼ m

. As another example, let

denote the true proportion of

all Catholics who plan to vote for the Republican candidate in the next presidential

election, and let

represent the true proportion of all Protestants who plan to

vote Republican. Based on a survey of 500 Catholics and 500 Protestants we might

like an interval estimate for the difference



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

DOI 10.1007/978-1-4614-0391-3_10,

Springer Science+Business Media, LLC 2012

484

10.1

z Tests and Confidence Intervals for a

Difference Between Two Population Means

The inferences discussed in this section concern a difference m

 m

between the

means of two different population distributions. An investigator might, for example,

wish to test hypotheses about the difference between the true average weight losses

of two diets. One such hypothesis would state that m

m

¼ 0, that is, that m

¼ m

Alternatively, it may be appropriate to estimate m

 m

by computing a 95% CI.

Such inferences are based on a sample of weight losses for each diet.

BASIC

ASSUMPTIONS

1. X

, X

, ... , X

is a random sample from a population with mean m

and

variance s

2. Y

, Y

, ... , Y

is a random sample from a population with mean m

and

variance s

3. Th e X and Y samples are independent of each other.

The natural estimator of m

m

is X  Y, the difference between the corresponding

sample means. The test statistic results from standardizing this estimator, so we

need expressions for the expected value and standard deviation of

X  Y.

PROPOSITION

The expected value of

X  Y is m

 m

,soX  Y is an unbiased estimator of

 m

. The standard deviation of X  Y is

XY

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

Proof Both these results depend on the rules of expected value and variance

presented in Chapter 6. Since the expected value of a difference is the difference of

expected values,

Eð

X  YÞ¼EðXÞEðYÞ¼m

 m

Because the X and Y samples are independent, X and Y are independent quantities,

so the variance of the difference is the sum of Vð

XÞ and VðYÞ:

Vð

X  YÞ¼VðXÞþVðYÞ¼

The standar d deviation of

X  Y is the squar e root of this expression. ■

If we think of m

 m

as a parameter y, then its estimator is

y ¼ X  Y with

standard deviation s

given by the proposition. When s

and s

both have known

values, the test statistic will have the form ð

y  null valueÞ=s

; this form of a test

10.1 z Tests and Conﬁdence Intervals for a Difference Between Two Population Means 485

statistic was used in several one-sample problems in the previo us chapter. When s

and s

are unknown, the sample variances must be used to estimate s

Test Procedures for Normal Populations

with Known Variances

In Chapters 8 and 9, the first CI and test procedure for a population mean m were

based on the assumption that the population distribution was normal with the

value of the population variance s

known to the investigator. Similarly, we first

assume here that both population distributions are normal and that the values of

both s

and s

are known. Situations in which one or both of these assumptions can

be dispensed with will be present ed shortly.

Because the population distributions are normal, both

X and Y have normal

distributions. This implies that

X  Y is normally distributed, with expected value

 m

and standard deviation s

XY

given in the foregoing proposition. Standar-

dizing

X  Y gives the standard normal variable

Z ¼

X  Y ðm

 m

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð10:1Þ

In a hypothesis-testing problem, the null hypothesis will state that m

m

has

a specified value. Denoting this null value by D

, the null hypothesis becomes

 m

¼ D

. Often D

¼ 0, in which case

says that m

¼ m

. A test statistic

results from replacing m

 m

in Expression (10.1) by the null value D

. Because

the test statistic Z is obtained by standardizing

X  Y under the assumption that

is true, it has a standard normal distribution in this case. Consider the alternat ive

hypothesis

: m

 m

> D

. A value x  y that considerably exceeds D

(the

expected value of

X  Y when

is true) provides evidence against

and for

Such a value of

x  y corresponds to a positive and large valu e of z. Thus

should

be rejected in favor of

if z is greater than or equal to an appropriately chosen

critical value. Because the test statistic Z has a standard normal distribution when

is true, the upper-tailed rejection region z  z

gives a test with significance

level (type I error probability) a. Rejection regions for the other alternatives

: m

 m

< D

and

: m

 m

6¼ D

that yield tests with desired significance

level a are lower-tailed and two-tailed, respectively.

Null hypothesis:

: m

 m

¼ D

Test statistic value: z ¼

x  y  D

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

Alternative Hypothesis Rejection Region for Level a Test

: m

 m

> D

z  z

(upper-tailed test)

: m

 m

< D

z z

(lower-tailed test)

: m

 m

6¼ D

either z  z

a/2

or z z

a/2

(two-tailed test)

486 CHAPTER 10 Inferences Based on Two Samples

Because these are z tests, a P-value is computed as it was for the z tests in

Chapter 9 [e.g., P-value ¼ 1  F(z) for an upper-tailed test].

Example 10.1 Each student in a class of 21 responded to a questionnaire that requested their grade

point average (GPA) and the number of hours each week that they studied.

For those who studied less than 10 h/week the GPAs were

2.80 3.40 4.00 3.60 2.00 3.00 3.47 2.80 2.60 2.00

and for those who studied at least 10 h/week the GPAs were

3.00 3.00 2.20 2.40 4.00 2.96 3.41 3.27 3.80 3.10 2.50

Normal plots for both sets are reasonably linear, so the normality assumption is

tenable. Because the standard deviation of GPAs for the whole campus is .6, it is

reasonable to apply that value here. The sample means are 2.97 for the <10 study

hours group and 3.06 for the 10 study hours group. Treating the two samples as

random, is there evidence that true aver age GPA differs for the two study times?

Let’s carry out a test of significance at level .05.

1. Th e parameter of interest is m

 m

, the difference between true mean GPA for

the < 10 (conceptual) population and true mean GPA for the 10 population.

2. Th e null hypothesis is

: m

 m

¼ 0.

3. Th e alternative hypothesis is

: m

 m

6¼ 0; if

is true then m

and m

are

different. Although it would seem unlikely that m

 m

> 0 (those with low

study hours have higher mean GPA) we will allow it as a possibility and do a

two-tailed test.

4. With D

¼ 0, the test statistic value is

z ¼

x  y

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

5. Th e inequal ity in

implies that the test is two-tailed. For a ¼ .05, a/2 ¼ .025

and z

a/2

¼ z

.025

¼ 1.96.

will be rejected if z  1.96 or z 1.96.

6. Substituting m ¼ 10,

x ¼ 2:97, s

¼ :36, n ¼ 11, y ¼ 3:06, and s

¼ :36 into

the formula for z yields

z ¼

2:97  3:06

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

:36

:09

:262

¼:34

That is, the value of

x  y is only one-third of a standard deviation below what

would be expected when

is true.

7. Because the value of z is not even close to the rejection region, there is no reason

to reject the null hypothesis. This test shows no evidence of any relationship

between stud y hours and GPA.

■

10.1 z Tests and Conﬁdence Intervals for a Difference Between Two Population Means 487