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

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CHAPTER FOURTEEN

Alternative

Approaches

to Inference

Introduction

In this ﬁnal chapter we consider some inferential methods that are different in

important ways from those considered earlier. Recall that many of the conﬁdence

intervals and test procedures developed in Chapters 9–12 were based on some sort

of a normality assumption. As long as such an assumption is at least approximately

satisﬁed, the actual conﬁdence and signiﬁcance levels will be at least approxi-

mately equal to the “nominal” levels, those prescribed by the experimenter through

the choice of particular

critical values. However, if there is a substantial

violation of the normality assumption, the actual levels may differ considerably

from the nominal levels (e.g., the use of t

.025

in a conﬁdence interva l formula may

actually result in a conﬁdence level of only 88% rather than the nominal 95%).

In the ﬁrst three sections of this chapter, we develop distribution-free or non-

parametric procedures that are valid for a wide variety of underlying distributions

rather than being tied to normality. We have actually already introduced several

such methods: the bootstrap intervals and permutation tests are valid without

restrictive assumptions on the underlying distrib ution(s).

Section 14.4 introduces the Bayesian approach to inference. The standard

frequentist view of inference is that the parameter of interest, y, has a ﬁxed but

unknown value. Bayesians, however, regard y as a random variable having a prior

probability distribution that incorporates whatever is known about its value. Then

to learn more about y, a sample from the conditional distribution f (x|y)is

obtained, and Bayes’ theorem is used to produce the posterior distribution of y

given the data x

, ... , x

. All Bayesian methods are based on this posterior

distribution.

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

DOI 10.1007/978-1-4614-0391-3_14,

Springer Science+Business Media, LLC 2012

758

14.1

The Wilcoxon Signed-Rank Test

A research chemist replicated a particular experiment a total of 10 times and obtained

the following values of reaction temperature, ordered from smallest to largest:

.57 .19 .05 .76 1.30 2.02 2.17 2.46 2.68 3.02

The distribution of reaction temperature is of course continuous. Suppose the

investigator is willing to assume that this distribution is symmetric, so that the pdf

satisfies f ð

m þ tÞ¼f ð

m  tÞ for any t >0, where

m is the median of the distribution

(and also the mean m provided that the mean exists). This condition on f (x) simply

says that the height of the density curve above a value any particular distance to the

right of the median is the same as the height that same distance to the left of the

median. The assumption of symmetry may at first thought seem quite bold, but

remember that we have frequently assumed a normal distribution. Since a normal

distribution is symmetric, the assumption of symmetry without any additional

distributional specification is actually a weaker assumption than normality.

Let’s now consider testing the null hypothesis that

m ¼ 0. This amounts to

saying that a temperature of any particular magnitude, say 1.50, is no more likely

to be positive (+1.50) than to be negative (1.50). A glance at the data casts doubt

on this hypot hesis; for example, the sample median is 1.66, which is far larger in

magnitude than any of the three negative observations.

Figure 14.1 shows graphs of two symmetric pdf’s, one for which H

true and the other for which the median of the distribution considerably exceeds 0.

In the first case we expect the magnitudes of the negative observations in the

sample to be comparable to those of the positive sample observations. However,

in the second case observations of large absolute magnitude will tend to be positive

rather than negative.

For the sample of ten reaction temperatures, let’s for the moment disregard

the signs of the observations and rank the absolute magnitudes from 1 to 10, with

the smallest getting rank 1, the second smallest rank 2, and so on. Then apply the

sign of each observation to the corresponding rank (so some signed ranks will be

negative, e.g. 3, whereas others will be positive, e.g. 8). The test statistic will be

¼ the sum of the positively signed ranks.

Absolute Magnitude .05 .19 .57 .76 1.30 2.02 2.17 2.46 2.68 3.02

Rank 12345678910

Signed Rank 1 2 345678910

¼ 4 þ 5 þ 6 þ 7 þ 8 þ 9 þ 10 ¼ 49

Figure 14.1 Distributions for which (a)

m ¼ 0; (b)

m  0

14.1 The Wilcoxon Signed-Rank Test 759

When the median of the distribution is much greater than 0, most of the observa-

tions with large absolute magnitudes should be positive, resulting in positively

signed ranks and a large value of s

. On the other hand, if the median is 0,

magnitudes of positively signed observations should be intermingled with those

of negatively signed observations , in which case s

will not be very large. Thus

we should reject H

m ¼ 0 when s

is “quite large”— the rejection region shoul d

have the form s

 c.

The critical value c should be chosen so that the test has a desired significance

level (type I error probability), such as .05 or .01. This necessitates finding the

distribution of the test statistic S

when the null hypothesis is true. Let’s consider

n ¼ 5, in which case there are 2

¼ 32 ways of applying signs to the five ranks 1, 2, 3,

4, and 5 (each rank could have a  sign or a + sign). The key point is that when H

true, any collection of five signed ranks has the same chance as does any other

collection. That is, the smallest observation in absolute magnitude is equally likely to

be positive or negative, the same is true of the second smallest observation in absolute

magnitude, and so on. Thus the collection 1, 2, 3, 4, 5 of signed ranks is just as likely

as the collection 1, 2, 3, 4, 5, and just as likely as any one of the other 30 possibilities.

Table 14.1 lists the 32 possible signed-rank sequences when n ¼ 5 along

with the value s

for each sequence. This immediately gives the “null distribution”

of S

displayed in Table 14.2. For example, Table 14.1 shows that three of the

32 possible sequences have s

¼ 8, so PS

¼ 8 when H

is trueðÞ¼1=32 þ

1=32 þ 1=32 ¼ 3=32. This null distribution appears in Table 14.2. Notice that it

Table 14.1 Possible signed-rank sequences for n ¼ 5

Sequence s

1 2 3 4 50 1 2 3 þ4 54

þ1 2 3 4 51 þ1 2 3 þ4 55

1 þ2 3 4 52 1 þ2 3 þ4 56

1 2 þ3 4 53 1 2 þ3 þ4 57

þ1 þ2 3 4 53 þ1 þ2 3 þ4 57

1 2 þ3  4 54 þ1 2 þ3 þ4 58

1 þ2 þ3 4 55 1 þ2 þ3 þ4 59

þ1 þ2 þ3 4 56 þ1 þ2 þ3 þ4 510

1 2 3 4 þ55 1 2 3 þ4 þ59

þ1 2 3 4 þ56 þ1 2 3 þ4 þ510

1

þ2 3 4 þ57 1 þ2 3 þ4 þ511

1 2 þ3 4 þ58 1 2 þ3 þ4 þ512

þ1 þ2 3 4 þ58 þ1 þ2 3 þ4 þ512

þ1 2 þ3 4 þ59 þ1 2 þ3 þ4 þ513

1 þ2 þ3 4 þ510 1 þ2 þ3 þ4 þ514

þ1 þ

2 þ3 4 þ511 þ1 þ2 þ3 þ4 þ515

Table 14.2 Null distribution of S

when n ¼ 5

0 1234567

p(s

) 1/32 1/32 1/32 2/32 2/32 3/32 3/32 3/32

8 9 10 11 12 13 14 15

p(s

) 3/32 3/32 3/32 2/32 2/32 1/32 1/32 1/32

760

CHAPTER 14 Alternative Approaches to Inference

is symmetric about 7.5 [more generally, symmetrically distributed over the possible

values 0; 1; 2;:::; nnþ 1ðÞ=2]. This symmetry is important in relating the rejection

region of lower-tailed and two-tailed tests to that of an upper-tailed test.

For n ¼ 10 there are 2

¼ 1024 possible signed rank sequences, so a listing

would involve much effort. Each sequence, though, would have probability 1/1024

when H

is true, from which the distribution of S

when H

is true can be easily obtained.

We are now in a p osition to determine a rejection region for testing H

m ¼ 0

versus H

m > 0 that has a suitably small significance level a. Consider the

rejection region R ¼fs

: s

 13g¼ 13; 14; 15

. Then

a ¼ P reject H

when H

is trueðÞ

¼ PðS

¼ 13; 14; or 15 when H

is trueÞ

¼ 1=32 þ 1=32 þ 1=32 ¼ 3=32

¼ :094

so that R ¼ {13, 14, 15} specifies a test with approximate level .1. For the rejec-

tion region {14, 15}, a ¼ 2/32 ¼ .063. For the sample x

¼ :58; x

¼ 2:50;

¼:21; x

¼ 1:23; x

¼ :97, the signed rank sequence is 1, +2, +3, +4, +5,

so s

¼ 14 and at level .063 H

would be rejected.

A General Description of the Wilcoxon Signed-Rank Test

Because the underlying distribution is assumed symmetric, m ¼

m, so we will state

the hypotheses of interest in terms of m rather than

ASSUMPTION

, X

, ... , X

is a random sample from a continuous and symmetric

probability distribution with mean (and median) m.

When the hypothesized value of m is m

, the absolute differences

 m

; :::

 m

j, must be ranked from smallest to largest.

Null hypothesis: H

: m ¼ m

Test statistic value: s

¼ the sum of the ranks associated with positive (x

 m

)’s

Alternative Hypothesis Rejection Region for Level a Test

: m > m

 c

: m < m

 c

½where c

¼ nðn þ 1Þ=2  c



: m 6¼ m

either s

 c or s

 nðn þ 1Þ=2  c

where the critical values c

and c obtained from Appendix Table A.12 satisfy

PðS

 c

Þa and PðS

 cÞa=2 when H

is true.

If the tails of the distribution are “too heavy,” as was the case with the Cauchy distribution of Chapter 7,

then m will not exist. In such cases, the Wilcoxon test will still be valid for tests concerning

14.1 The Wilcoxon Signed-Rank Test

761

Example 14.1 A producer of breakfast cereals wants to verify that a filler machine is operating

correctly. The machine is supposed to fill one-pound boxes with 460 g, on the

average. This is a little above the 453.6 g needed for one pound. When the contents

are weighed, it is found that 15 boxes yield the following measurements:

454.4 470.8 447.5 453.2 462.6 445.0 455.9 458.2

461.6 457.3 452.0 464.3 459.2 453.5 465.8

It is believed that deviations of any magnitude from 460 g are just as likely to be

positive as negative (in accord with the symmetry assumption) but the distribution

may not be normal. Therefore, the Wilcoxon signed-rank test will be used to see if

the filler machine is calibrated correctly.

The hypotheses are H

: m ¼ 460 versus H

: m 6¼ 460, where m is the true

average weight. Subtracting 460 from each measurement gives

5.6 10.8 12.5  6.8 2.6 15.0 4.1 1.8 1.6 2.7

8.0 4.3 .8 6.5 5.8

The ranks are obtained by ordering these from smallest to largest without regard to sign.

Absolute

Magnitude .8 1.6 1.8 2.6 2.7 4.1 4.3 5.6 5.8 6.5 6.8 8.0 10.8 12.5 15.0

Rank 123456789101112131415

Sign  +  + +  +   + 

Thus s

¼ 2 þ 4 þ 7 þ 9 þ 13 ¼ 35. From Appendix Table A.12, PðS

 95Þ¼

PðS

 25Þ¼:024 when H

is true, so the two-tailed test with approximate level

.05 rejects H

when either s

 95 or  25 [the exact a is 2(.024) ¼ .048]. Since

¼ 35 is not in the rejection region, it cannot be concluded at level .05 that m

differs from 460. Even at level .094 (approximately .1), H

cannot be rejected, since

P(S

 30) ¼ P(S

 90) ¼ .047 implies that s

values between 30 and 90 are not

significant at that level. The P-value of the data is thus >.1.

■

Although a theoretical implication of the continuity of the underlying distri-

bution is that ties will not occur, in practice they often do because of the discrete-

ness of measuring instruments. If there are several data values with the same

absolute magnitude, then they would be assigned the average of the ranks they

would receive if they differed very slightly from one another. For example, if in

Example 14.1 x

¼ 458.2 is changed to 458.4, then two different values of

 460) would have absolute magnitude 1.6. The ranks to be averaged would

be 2 and 3, so each would be assigned rank 2.5.

Paired Observations

When the data consisted of pairs ðX

; Y

Þ; ...; ðX

; Y

Þ and the differences

¼ X

 Y

; ...; D

¼ X

 Y

were normally distributed, in Chapter 10 we

used a paired t test for hypotheses about the expected difference m

. If normality

is not assumed, hypotheses about m

can be tested by using the Wilcoxon signed-

rank test on the D

’s provided that the distribution of the differences is continuous

and symmetric. If X

and Y

both have continuous distributions that differ only with

762 CHAPTER 14 Alternative Approaches to Inference

respect to their means (so the Y distribution is the X distribut ion shifted by

 m

¼ m

), then D

will have a continuous symmetric distribution (it is not

necessary for the X and Y distributions to be symmetric individually). The null

hypothesis is H

: m

¼ D

, and the test statistic S

is the sum of the ranks

associated with the positive (D

 D

)’s.

Example 14.2 About 100 years ago an experiment was done to see if drugs could help people with

severe insomnia (“The Action of Optical Isomers, II: Hyoscines,” J. Physiol., 1905:

501–510). There were 10 patients who had trouble sleeping, and each patient

tried several medications. Here we compare just the control (no medication) and

levo-hyoscine. Does the drug offer an improvement in average sleep time? The

relevant hypotheses are H

: m

¼ 0 versus H

: m

< 0. Here are the sleep times,

differences, and signed ranks.

Patient 12345678910

Control 0.6 1.1 2.5 2.8 2.9 3.0 3.2 4.7 5.5 6.2

Drug 2.5 5.7 8.0 4.4 6.3 3.8 7.6 5.8 5.6 6.1

Difference  1.9 4.6 5.5 1.6 3.4 .8 4.4 1.1 .1 .1

Signed rank 6 9 10 5 7 3 8 4 1.5 1.5

Notice that there is a tie for the lowest rank, so the two lowest ranks are split

between observations 9 and 10, and each receives rank 1.5. Appendix Table A.12

shows that for a test with significance level approximately .05, the null hypothesis

should be rejected if s

 10ðÞ11ðÞ=2  44 ¼ 11. The test statistic value is 1.5,

which falls in the rejection region. We therefore reject H

at significance level .05

in favor of the conclusion that the drug gives greater mean sleep time. The

accompanying MINITAB output shows the test statistic value and also the

corresponding P-value, which is P(S

1.5 when H

is true).

Test of median ¼ 0.000000 versus median < 0.000000

for Wilcoxon Estimated

N Test Statistic P Median

diff 10 10 1.5 0.005

2.250

■

Efﬁciency of the Wilcoxon Signed-Rank Test

When the underlying distribution being sampled is normal, either the t test or the

signed-rank test can be used to test a hypothesis about m. The t test is the best test in

such a situation because among all level a tests it is the one having minimum b.Itis

generally agreed that there are many experimental situations in which normality

can be reasonably assumed, as well as some in which it should not be. These two

questions mus t be addressed in an attempt to compare the tests:

1. When the underlying distribution is normal (the “home ground” of the t test),

how much is lost by using the signed-rank test?

2. When the underlying distribution is not normal, can a significant improvement

be achieved by using the signed-rank test?

If the Wilcoxon test does not suffer much with respect to the t test on the “home

ground” of the latter, and performs significantly better than the t test for a large number

of other distributions, then there will be a strong case for using the Wilcoxon test.

14.1 The Wilcoxon Signed-Rank Test 763

Unfortunately, there are no simple answers to the two questions. Upon

reflection, it is not surprising that the t test can perform poorly when the underlying

distribution has “heavy tails” (i. e., when observed values lying far from m are

relatively more likely than they are when the distribution is normal). This is because

the behavior of the t test depends on the sample mean and variance, which are both

unstable in the presence of heavy tails. The difficulty in producing answers to the

two questions is that b for the Wilcoxon test is very difficult to obtain and study for

any underlying distribution, and the same can be said for the t test when the

distribution is not normal. Even if b were easily obtained, any measure of efficiency

would clearly depend on which underlying distribution was assumed. A number of

different efficiency measures have been proposed by statisticians ; one that many

statisticians regard as credible is called asym ptotic relative efficiency (ARE).

The ARE of one test with resp ect to another is essentially the limiting ratio of

sample sizes necessary to obtain identical error probabiliti es for the two tests. Thus

if the ARE of one test with respect to a second equals .5, then when sam ple sizes are

large, twice as large a sample size will be required of the first test to perform as well

as the second test. Although the ARE does not characterize test performance for

small sample sizes, the following results can be shown to hold:

1. When the underlying distribution is normal, the ARE of the Wilcoxon test with

respect to the t test is approximately .95.

2. For any distribution, the ARE will be at least .86 and for many distributions will

be much greater than 1.

We can summarize these results by saying that, in large-sample problems, the

Wilcoxon test is never very much less efficient than the t test and may be much

more efficient if the underlying distribution is far from normal. Although the issue

is far from resolved in the case of sample sizes obtained in most practical problems,

studies have shown that the Wilcoxon test performs reasonably and is thus a viable

alternative to the t test.

Exercises Section 14.1 (1–8)

1. Reconsider the situation described in Exercise 32 of

Section 9.2, and use the Wilcoxon test with a ¼ .05

to test the relevant hypotheses.

2. Use the Wilcoxon test to analyze the data given in

Example 9.9.

3. The accompanying data is a subset of the data re-

ported in the article “Synovial Fluid pH, Lactate,

Oxygen and Carbon Dioxide Partial Pressure in

Various Joint Diseases” (Arthritis Rheum., 1971:

476–477). The observations are pH values of syno-

vial fluid (which lubricates joints and tendons) taken

from the knees of individuals suffering from arthri-

tis. Assuming that true average pH for non-arthritic

individuals is 7.39, test at level .05 to see whether the

data indicates a difference between average pH

values for arthritic and nonarthritic individuals.

7.02 7.35 7.34 7.17 7.28 7.77 7.09

7.22 7.45 6.95 7.40 7.10 7.32 7.14

4. A random sample of 15 automobile mechanics

certified to work on a certain type of car was

selected, and the time (in minutes) necessary for

each one to diagnose a particular problem was

determined, resulting in the following data:

30.6 30.1 15.6 26.7 27.1 25.4 35.0 30.8

31.9 53.2 12.5 23.2 8.8 24.9 30.2

Use the Wilcoxon test at significance level .10 to

decide whether the data suggests that true average

diagnostic time is less than 30 minutes.

5. Both a gravimetric and a spectrophotometric method

are under consideration for determining phosphate

content of a particular material. Twelve samples of

764

CHAPTER 14 Alternative Approaches to Inference

the material are obtained, each is split in half, and a

determination is made on each half using one of the

two methods, resulting in the following data:

Sample 1234

Gravimetric 54.7 58.5 66.8 46.1

Spectrophotometric 55.0 55.7 62.9 45.5

Sample 5678

Gravimetric 52.3 74.3 92.5 40.2

Spectrophotometric 51.1 75.4 89.6 38.4

Sample 9 101112

Gravimetric 87.3 74.8 63.2 68.5

Spectrophotometric 86.8 72.5 62.3 66.0

Use the Wilcoxon test to decide whether one tech-

nique gives on average a different value than the

other technique for this type of material.

6. The signed-rank statistic can be represented as

¼ W

þ W

þþW

; where W

¼ i if the

sign of the x

 m

with the ith largest absolute

magnitude is positive (in which case i is included

in S

) and W

¼ 0 if this value is negative (i ¼ 1, 2,

3, ... , n). Furthermore, when H

is true, the W

’s

are independent and PðW ¼ iÞ¼PðW ¼ 0Þ¼: 5.

a. Use these facts to obtain the mean and variance

of S

when H

is true. [Hint: The sum of the first

n positive integers is nðn þ 1Þ=2, and the sum of

the squares of the first n positive integers is

nðn þ 1Þð2n þ 1Þ=6.]

b. The W

’s are not identically distributed (e.g.,

possible values of W

are 2 and 0 whereas pos-

sible values of W

are 5 and 0), so our Central

Limit Theorem for identically distributed and

independent variables cannot be used here

when n is large. However, a more general CLT

can be used to assert that when H

is true and

n > 20, S

has approximately a normal distri-

bution with mean and variance obtained in (a).

Use this to propose a large-sample standardized

signed-rank test statistic and then an appropriate

rejection region with level a for each of the three

commonly encountered alternative hypotheses.

[Note: When there are ties in the absolute mag-

nitudes, it is still correct to standardize S

subtracting the mean from (a), but there is a

correction for the variance which can be found

in books on nonparametric statistics.]

c. A particular type of steel beam has been de-

signed to have a compressive strength (lb/in

)

of at least 50,000. An experimenter obtained a

random sample of 25 beams and determined the

strength of each one, resulting in the following

data (expressed as deviations from 50,000):

10 27 36  55 73 77 81

90 95 99 113 127 129 136

150 155 159 165 178 183 192

199 212 217 229

Carry out a test using a significance level of

approximately .01 to see if there is strong evi-

dence that the design condition has been violated.

7. The accompanying 25 observations on fracture

toughness of base plate of 18% nickel maraging

steel were reported in the article “Fracture Testing

of Weldments” (ASTM Special Publ. No. 381,

1965: 328–356). Suppose a company will agree to

purchase this steel for a particular application only

if it can be strongly demonstrated from experimen-

tal evidence that true average toughness exceeds

75. Assuming that the fracture toughness distribu-

tion is symmetric, state and test the appropriate

hypotheses at level .05, and compute a P-value.

[Hint: Use Exercise 6(b).]

69.5 71.9 72.6 73.1 73.3 73.5 74.1 74.2 75.3

75.5 75.7 75.8 76.1 76.2 76.2 76.9 77.0 77.9

78.1 79.6 79.7 80.1 82.2 83.7 93.7

8. Suppose that observations X

, X

, ..., X

are made

on a process at times 1, 2, ..., n. On the basis of this

data, we wish to test

: the X

’s constitute an independent and iden-

tically distributed sequence

versus

: X

i+1

tends to be larger than X

for i ¼ 1, ..., n

(an increasing trend)

Suppose the X

’s are ranked from 1 to n. Then when H

is true, larger ranks tend to occur later in the sequence,

whereas if H

is true, large and small ranks tend

to be mixed together. Let R

be the rank of X

and consider the test statistic D ¼

i¼1

ðR

 iÞ

14.1 The Wilcoxon Signed-Rank Test 765

Then small values of D give support to H

(e.g., the

smallest value is 0 for R

¼ 1; R

¼ 2; :::; R

¼ n),

so H

should be rejected in favor of H

if d  c.When

is true, any sequence of ranks has probability 1/n!.

Use this to find c for which the test has a level as close

to .10 as possible in the case n ¼ 4. [Hint: List the 4!

rank sequences, compute d for each one, and then

obtain the null distribution of D. See the Lehmann

book (in the chapter bibliography), for more infor-

mation.]

14.2

The Wilcoxon Rank-Sum Test

When at least one of the sample sizes in a two-sample problem is small, the t test

requires the assumption of normality (at least approximately). There are situations,

though, in which an investigator would want to use a test that is valid even if the

underlying distributions are quite nonnor mal. We now describe such a test, called

the Wilcoxon rank-sum test. An alternative name for the procedure is the Mann–

Whitney test, although the Mann–Whitney test statistic is sometimes expressed in a

slightly different form from that of the Wilcoxon test. The Wilcoxon test procedure

is distribution-free because it will have the desired level of significance for a very

large class of underlying distributions.

ASSUMPTIONS

, ... , X

and Y

, ... , Y

are two independent random samples from

continuous distributions with means m

and m

, respectively. The X and Y

distributions have the same shape and spread, the only possible difference

between the two being in the values of m

and m

When H

: m

 m

¼ D

is true, the X distribution is shifted by the amount D

to the

right of the Y distribution; whe reas when H

is false, the shift is by an amount other

than D

Development of the Test When m ¼ 3, n ¼ 4

Let’s first test H

: m

 m

¼ 0. If m

is actually much larger than m

, then most of

the observed x’s will fall to the right of the observed y’s. However, if H

is true, then

the observed values from the two samples should be intermingled. The test statistic

will provide a quantification of how much intermingling there is in the two samples.

Consider the case m ¼ 3, n ¼ 4. Then if all three observed x’s were to the

right of all four observed y’s, this would provide strong evidence for rejecting H

favor of H

: m

 m

6¼ 0, with a similar conclusion being appropriate if all three

x’s fall below all four of the y’s. Suppose we pool the x’s and y’s into a combined

sample of size m+n¼ 7 and rank these observat ions from smallest to largest,

with the smallest receiving rank 1 and the largest, rank 7. If either most of the

largest ranks or most of the smallest ranks were associated with X observations, we

would begin to doubt H

. This suggests the test statistic

W ¼ the sum of the ranks in the combined sample

associated with X observations

ð14:1Þ

For the values of m and n under consideration, the smallest possible value of W is

w ¼ 1+2+3¼ 6 (if all three x’s are smaller than all four y’s), and the largest

possible value is w ¼ 5+6+7¼ 18 (if all three x ’s are larger than all four y’s).

766 CHAPTER 14 Alternative Approaches to Inference

As an example, suppose x

¼3.10, x

¼ 1.67, x

¼ 2.01, y

¼ 5.27,

¼ 1.89, y

¼ 3.86, and y

¼ .19. Then the pooled ordered sample is 3.10, .19,

1.67, 1.89, 2.01, 3.86, and 5.27. The X ranks for this sample are 1 (for 3.10), 3 (for

1.67), and 5 (for 2.01), so the computed value of W is w ¼ 1+3+5¼ 9.

The test procedure based on the statistic (14.1) is to reject H

if the computed

value w is “too extreme” — that is,  c for an upper-tailed test,  c for a lower-

tailed test, and either  c

or  c

for a two-tailed test. The critical constant(s) c

) should be chosen so that the test has the desired level of significance a.To

see how this should be done, recall that when H

is true, all seven observations

come from the same population. This means that under H

, any possible triple of

ranks associated with the three x’s — such as (1, 4, 5), (3, 5, 6), or (5, 6, 7) — has

the same probability as any other possible rank triple. Since there are



¼ 35

possible rank triples, under H

each rank triple has probability 1/35. From a list of

all 35 rank triples and the w value associated with each, the probability distribution

of W can immediately be determined. For exam ple, there are four rank triples that

have w value 11 — (1, 3, 7), (1, 4, 6), (2, 3, 6), and (2, 4, 5) — so P (W ¼ 11) ¼

4/35. The summary of the listing and computations appears in Table 14.3.

The distribution of Table 14.3 is symmetric about w ¼ (6 + 18)/2 ¼ 12,

which is the middle value in the ordered list of possible W values. This is because

the two rank triples (r, s, t) (with r < s < t) and (8  t,8 s,8 r) have values

of w symmetric about 12, so for each triple with w value below 12, there is a triple

with w value above 12 by the same amount.

If the alternative hypothesis is H

: m

 m

> 0, then H

should be rejected

in favor of H

for large W values. Choosing as the rejection region the set of

W values {17, 18}, a ¼ P type I errorðÞ¼Pð reject H

when H

is trueÞ¼PðW ¼

17 or 18 when H

is trueÞ¼

¼ :057; the region {17, 18} therefore

specifies a test with level of significance approximately .05. Similarly, the region

{6, 7}, which is appropriate for H

: m

 m

< 0, has a ¼ .057  .05. The region

{6, 7, 17, 18}, which is appropriate for the two-sided alternative, has a ¼

¼ :114.

The W value for the data given several paragraphs previously was w ¼ 9, which is

rather close to the middle value 12, so H

would not be rejected at any reasonable

level a for any one of the three H

’s.

General Description of the Rank-Sum Test

The null hypothesis H

: m

 m

¼ D

is handled by subtracting D

from each X

and using the (X

 D

)’s as the X

’s were previously used. Recalling that for any

positive integer K, the sum of the first K integers is K(K + 1)/2, the smallest

possible value of the statistic W is m(m + 1)/2, which occurs when the (X

 D

)’s

are all to the left of the Y sample. The largest possible value of W occurs when the

 D

)’s lie entirely to the right of the Y’s; in this case, W ¼ðn þ 1Þþþ

ðm þ nÞ¼ðsum of first m þ n integersÞðsum of first n integersÞ, which gives

Table 14.3 Probability distribution of W (m ¼ 3, n ¼ 4) when H

is true

w 6 7 8 9 10 11 12 13 14 15 16 17 18

P(W ¼ w)

14.2 The Wilcoxon Rank-Sum Test 767