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

Подождите немного. Документ загружается.

Example 13.9

(Example 13.8

continued)

Using

l ¼ 2:10, the estimated expected cell counts are computed from np

lÞ,

where n ¼ 48. For example,

lÞ¼48 

2:1

ð2:1Þ

¼ð48Þðe

2:1

Þ¼5:88

Similarly, np

lÞ¼12:34, np

lÞ¼12:96, np

lÞ¼9:07, and np

lÞ¼

48  5:88 9:07 ¼ 7:75. Th en

ð9  5:88Þ

5:88

þþ

ð6  7:75Þ

7:75

¼ 6:31

Since m ¼ 1 and k ¼ 5, at level .05 we need w

:05;3

¼ 7:815 and w

:05;4

¼ 9:488.

Because 6.31  7.815, we do not reject H

; at the 5% level, the Poisson distribu-

tion provides a reasonable fit to the d ata. Notice that w

:10;3

¼ 6:251 and

:10;4

¼ 7:779, so at level .10 we would have to withhold judgment on whether

the Poisson distribution was appropriate.

For comparison we can with a little additional effort maximize Expres-

sion (13.7). Use of a graphing calculator gives

l ¼ 2:047. Because this

differs very little from 2.10, there is little change in the results. Using 2.047, we

get the estimated expected cell counts 6.197, 12.687, 12.985, 8.860, and 7.271,

and the resulting value of w

is 6.230. Comparing this with w

:05;3

¼ 7:815, we do not

reject the Poisson null hypothesis at the .05 level. Because 6.230 does not

quite exceed w

:10;3

¼ 6:251, we also do not reject the null hypothesis at the

10% level. ■

Sometimes even the maximum likelihood estimates based on the full sample

are quite difficult to compute. This is the case, for example, for the two-parameter

(generalized) negative binomial distribution. In such situations, method-of-

moments estimates are often used and the resulting w

compared to w

a;k1m

although it is not known to what extent the use of moments estimators affects the

true critical value.

Goodness of Fit for Continuous Distributions

The chi-squa red test can also be used to test whether the sample comes from a

specified family of continuous distributions, such as the exponential family or the

normal family. The choice of cells (class intervals) is even more arbitrary in the

continuous case than in the discrete case. To ensure that the chi-squared test is

valid, the cells should be chosen independe ntly of the sample observations. Once

the cells are chosen, it is almost always quite difficult to estimate unspecified

parameters (such as m and s in the normal case) from the observed cell counts, so

instead mle’s based on the full sample are computed. The critical value c

again

satisfies (13.8 ), and the test procedure is given by (13.9).

Example 13.10 The Institute of Nutrition of Central America and Panama (INCAP) has carried out

extensive dietary studies and research projects in Central Amer ica. In one study

reported in the November 1964 issue of the American Journal of Clinical Nutrition

(“The Blood Viscosity of Various Socioeconomic Groups in Guatemala”), serum

738 CHAPTER 13 Goodness-of-Fit Tests and Categorical Data Analysis

total cholesterol measurements for a sample of 49 low-incom e rural Indians were

reported as follows (in mg/L):

204 108 140 152 158 129 175 146 157 174 192 194 144

152 135 223 145 231 115 131 129 142 114 173 226 155

166 220 180 172 143 148 171 143 124 158 144 108 189

136 136 197 131 95 139 181 165 142 162

Is it plausible that serum cholesterol level is normally distributed for this popula-

tion? Suppose that prior to sampling, it was believed that plausible values for m

and s were 150 and 30, resp ectively. The seven equiprobable class intervals for

the standard normal distribution are (1, –1.07), (1.07, –.57), (.57, –.18),

(.18, .18), (.18, .57), (.57, 1.07), and (1.07, 1), with each endpoint also giving the

distance in standard deviations from the mean for any other normal distribution.

For m ¼ 150 and s ¼ 30, these intervals become (1, 117.9), (117.9, 132.9),

(132.9, 144.6), (144.6, 155.4), (155.4, 167.1), (167.1, 182.1), and (182.1, 1).

To obtain the estimated cell probabilities p

sÞ; ...; p

sÞ, we first need

the mle’s

m and

s.InChapter7,

s was shown to be ½

ðx





xÞ

=n

1=2

(rather than s),

so with s ¼ 31.75,

m ¼



x ¼ 157:02

s ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðx





xÞ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðn  1Þs

¼ 31:42

Each p

sÞ) is then the probability that a normal rv X with mean 157.02 and

standard deviation 31.42 falls in the ith class interval. For example,

sÞ¼Pð117:9  X  132:9Þ¼Pð1:25  Z :77Þ¼:1150

so np

sÞ¼49 :1150ðÞ¼5:64. Observed and estimated expected cell counts are

shown in Table 13.8.

The computed w

is 4.60. With k ¼ 7 cells and m ¼ 2 parameters estimated,

:05;k1

¼ w

:05;6

¼ 12:592 and w

:05;k1m

¼ w

:05;4

¼ 9:488. Since 4.60  9.488,

a normal distribution provides quite a good fit to the data.

■

Example 13.11 The article “Some Studies on Tuft Weight Distribution in the Opening Roo m”

(Textile Res. J., 1976: 567–573) reports the accompanying data on the distribution

of output tuft weight X (mg) of cotton fibers for the input weight x

¼ 70.

Table 13.8 Observed and expected counts for Example 13.10

13.2 Goodness-of-Fit Tests for Composite Hypotheses 739

The authors postulated a truncated exponential distribution:

: f ðxÞ¼

lx

1  e

lx

0  x  x

The mean of this distribution is

m ¼

xf ðxÞdx ¼



lx

1  e

lx

The parameter l was estimated by replacing m by



x ¼ 13:086 and solving the

resulting equation to obtain

l ¼ :0742 (so

l is a method-of-moments estimate

and not an mle). Then with

l replacing l in f(x), the estimated expected cell

frequencies as displayed previously are computed as

lÞ¼40Pða

i1

 X<a

Þ¼40

i1

f ð xÞdx ¼

40ðe



i1

 e



1  e



where [a

i–1

, a

) is the ith class interval. To obtain expected cell counts of at least 5,

the last six cells are combined to yield observed counts 20, 8, 7, 5 and expected

counts of 18.0, 9.9, 5.5, 6.6. The computed value of chi-squared is then w

¼ 1.34.

Because w

:05;2

¼ 5:992, H

is not rejected, so the truncated exponential model

provides a good fit.

■

A Special Test for Normality

Probability plots were introduced in Section 4.7 as an informal method for asses-

sing the plausibility of any specified population distribution as the one from which

the given sample was selected. The straighter the probability plot, the more

plausible is the distribution on which the plot is based. A normal probability plot

is used for checking whether any member of the normal distribution family is

plausible. Let’s denote the sample x

’s when ordered from smallest to largest by

(1)

, x

(2)

, ..., x

(n)

. Then the plot suggested for checking normality was a plot of the

points (x

(i)

, y

), where y

¼ F

–1

[(i – .5)/ n ].

A quantitative measure of the extent to which points cluster about a straight line

is the sample correlation coefficient r introduced in Chapter 12. Consider calculating r

for the n pairs (x

(1)

, y

), ...,(x

(n)

, y

). The y

’s here are not observed values in a random

sample from a y population, so properties of this r are quite different from those

described in Section 12.5. However, it is true that the more r deviates from one, the

less the probability plot resembles a straight line (remember that a probability plot

must slope upward). This idea can be extended to yield a formal test procedure: Reject

the hypothesis of population normality if r  c

, where c

is a critical value chosen to

yield the desired significance level a. That is, the critical value is chosen so that when

the population distribution is actually normal, the probability of obtaining an r value

that is at most c

(and thus incorrectly rejecting H

) is the desired a. The developers of

the MINITAB statistical computer package give critical values for a ¼ .10, .05, and

.01 in combination with different sample sizes. Because no theory exists for the

distribution of r for a normal plot, the critical values are determined by computer

simulation. These critical values are based on a slightly different definition of the y

’s

than that given previously. The new values give slightly better approximations to the

expected values of the ordered normal observations.

MINITAB will also construct a normal probability plot based on these y

’s.

The plot will be almost identical in appearance to that based on the previous y

’s.

740 CHAPTER 13 Goodness-of-Fit Tests and Categorical Data Analysis

When there are several tied x

(i)

’

s, MINITAB computes r by using the average of the

corresponding y

’s as the second number in each pair.

Let y

¼ F

–1

[(i – .375)/(n + .25)] and compute the sample correlation coef-

ficient r for the n pairs (x

(1)

, y

), ...,(x

(n)

, y

). The Ryan–Joiner test of

: the population distributio n is normal

versus

: the population distribution is not normal

consists of rejecting H

when r  c

. Critical values c

are give n in Appendix

Table A.11 for various significance levels a and sample sizes n.

Example 13.12 The following sample of n ¼ 20 observations on dielectric breakdown voltage of a

piece of epoxy resin first appeared in Example 4.36.

1.871 1.404 1.127 .917 .742 .587 .446 .313 .186 .062

(i)

24.46 25.61 26.25 26.42 26.66 27.15 27.31 27.54 27.74 27.94

.062 .186 .313 .446 .587 .742 .917 1.127 1.404 1.871

(i)

27.98 28.04 28.28 28.49 28.50 28.87 29.11 29.13 29.50 30.88

We asked MINITAB to carry out the Ryan–Joiner test, and the result appears in

Figure 13.3. The test statistic value is r ¼ .9881, and Appendix Table A.11 gives

.9600 as the critical value that captures lower-tail area .10 under the r sampling

distribution curve when n ¼ 20 and the underlying distribution is actually normal.

Since .9881 > .9600, the null hypothesis of normality cannot be rejected even for a

significance level as large as .10.

■

Figure 13.3 MINITAB output from the Ryan–Joiner test for the data of Example 13.12

13.2 Goodness-of-Fit Tests for Composite Hypotheses 741

Exercises Section 13.2 (12–22)

12. Consider a large population of families in which

each family has exactly three children. If the

genders of the three children in any family are

independent of one another, the number of male

children in a randomly selected family will have

a binomial distribution based on three trials.

a. Suppose a random sample of 160 families

yields the following results. Test the relevant

hypotheses by proceeding as in Example 13.5.

Number of Male Children 0123

Frequency 14 66 64 16

b. Suppose a random sample of families in a

nonhuman population resulted in observed

frequencies of 15, 20, 12, and 3, respectively.

Would the chi-squared test be based on the

same number of degrees of freedom as the test

in part (a)? Explain.

13. A study of sterility in the fruit fly (“Hybrid

Dysgenesis in Drosophila melanogaster:The

Biology of Female and Male Sterility,” Genetics,

1979: 161–174) reports the following data on the

number of ovaries developed for each female fly

in a sample of size 1,388. One model for unilat-

eral sterility states that each ovary develops with

some probability p independently of the other

ovary. Test the fit of this model using w

x ¼ Number of

Ovaries Developed 012

Observed Count 1212 118 58

14. The article “Feeding Ecology of the Red-Eyed

Vireo and Associated Foliage-Gleaning Birds”

(Ecol. Monogr., 1971: 129–152) presents the

accompanying data on the variable X ¼ the num-

ber of hops before the first flight and preceded by

a flight. The author then proposed and fit a geo-

metric probability distribution [p(x) ¼ P(X ¼ x)

¼ p

x–1

· q for x ¼ 1, 2, ...,whereq ¼ 1–p]to

the data. The total sample size was n ¼ 130.

x 1 2 3 456789101112

Number

of Times x

Observed

483120965421 1 2 1

a. The likelihood is ðp

1

qÞðp

1

qÞ¼

n

q

. Show that the mle of p is given by

p ¼

nðÞ=

, and compute

p for the

given data.

b. Estimate the expected cell counts using

p of

part (a) [expected cell counts ¼ n 

x1



q for

x ¼ 1, 2, ... ], and test the fit of the model

using a w

test by combining the counts for

x ¼ 7, 8, ..., and 12 into one cell (x  7).

15. A certain type of flashlight is sold with the four

batteries included. A random sample of 150

flashlights is obtained, and the number of defec-

tive batteries in each is determined, resulting in

the following data:

Number Defective 01234

Frequency 26 51 47 16 10

Let X be the number of defective batteries in a

randomly selected flashlight. Test the null hypo-

thesis that the distribution of X is Bin(4, y). That

is, with p

¼ P(i defectives), test

: p



ð1  yÞ

4i

i ¼ 0; 1; 2; 3; 4

[Hint: To obtain the mle of y, write the likelihood

(the function to be maximized) as y

(1 – y)

where the exponents u and v are linear functions

of the cell counts. Then take the natural log,

differentiate with respect to y, equate the result

to 0, and solve for

y.]

16. In a genetics experiment, investigators looked at

300 chromosomes of a particular type and

counted the number of sister-chromatid

exchanges on each (“On the Nature of Sister-

Chromatid Exchanges in 5-Bromodeoxyuridine-

Substituted Chromosomes,” Genetics, 1979:

1251–1264). A Poisson model was hypothesized

for the distribution of the number of exchanges.

Test the fit of a Poisson distribution to the data by

first estimating l and then combining the counts

for x ¼ 8andx ¼ 9 into one cell.

x ¼ Number

of Exchanges

0 123456789

Observed

Counts

6 2442596244411462

17. An article in Annals of Mathematical Statistics

reports the following data on the number of

borers in each of 120 groups of borers. Does

the Poisson pmf provide a plausible model for

the distribution of the number of borers in a

group? [Hint: Add the frequencies for 7, 8, ...,

12 to establish a single category “  7.”]

742

CHAPTER 13 Goodness-of-Fit Tests and Categorical Data Analysis

Number

of Borers

0 1 2 3 456789101112

Frequency 24 16 16 18 15 9 6 5 3 4 3 0 1

18. The article “A Probabilistic Analysis of Dissolved

Oxygen–Biochemical Oxygen Demand Relation-

ship in Streams” (J. Water Resources Control

Fed., 1969: 73–90) reports data on the rate of

oxygenation in streams at 20



C in a certain region.

The sample mean and standard deviation were

computed as



x ¼ :173 and s ¼ .066, respectively.

Based on the accompanying frequency distribu-

tion, can it be concluded that oxygenation rate is a

normally distributed variable? Use the chi-

squared test with a ¼ .05.

Rate (per day) Frequency

Below .100 12

.100–below .150 20

.150–below .200 23

.200–below .250 15

.250 or more 13

19. Each headlight on an automobile undergoing an

annual vehicle inspection can be focused either too

high (H), too low (L), or properly (N). Checking

the two headlights simultaneously (and not distin-

guishing between left and right) results in the six

possible outcomes HH, LL, NN, HL, HN,andLN.

If the probabilities (population proportions) for the

single headlight focus direction are P(H) ¼ y

P(L) ¼ y

,andP(N) ¼ 1–y

– y

and the two

headlights are focused independently of each

other, the probabilities of the six outcomes for a

randomly selected car are the following:

¼ y

¼ð1  y

 y

¼ 2y

ð1  y

 y

¼ 2y

ð1  y

 y

Use the accompanying data to test the null

hypothesis

: p

¼ p

ðy

; y

Þ; :::; p

¼ p

ðy

; y

where the p

, y

)’s are given previously.

Outcome HH LL NN HL HN LN

Frequency 49 26 14 20 53 38

[Hint: Write the likelihood as a function of y

and

, take the natural log, then compute @=@y

and

@=@y

, equate them to 0, and solve for

;

20. The article “Compatibility of Outer and Fusible

Interlining Fabrics in Tailored Garments (Textile

Res. J., 1997: 137–142) gave the following

observations on bending rigidity (mN · m) for

medium-quality fabric specimens, from which

the accompanying MINITAB output was

obtained:

24.6 12.7 14.4 30.6 16.1 9.5 31.5 17.2

46.9 68.3 30.8 116.7 39.5 73.8 80.6 20.3

25.8 30.9 39.2 36.8 46.6 15.6 32.3

.999

.99

.96

.80

.50

.20

.05

.01

.001

Probability

20 70 120

bending

West for Normality

R: 0.9116

pvalue(approx): <0.0100

Average: 37.4217

Std Dev. 25.8101

N of data: 23

Normal Probability Plot

Would you use a one-sample t confidence inter-

val to estimate true average bending rigidity?

Explain your reasoning.

21. The article from which the data in Exercise 20 was

obtained also gave the accompanying data on the

composite mass/outer fabric mass ratio for high-

quality fabric specimens.

MINITAB gave r ¼ .9852 as the value of the

Ryan– Joiner test statistic and reported that P-

value > .10. Would you use the one-sample

t test to test hypotheses about the value of the

true average ratio? Why or why not?

22. The article “Nonbloated Burned Clay Aggregate

Concrete” (J. Mater., 1972: 555–563) reports the

following data on 7 day flexural strength of

13.2 Goodness-of-Fit Tests for Composite Hypotheses 743

nonbloated burned clay aggregate concrete samples

(psi):

Test at level .10 to decide whether flexural strength is a

normally distributed variable.

13.3

Two-Way Contingency Tables

In the previous two sections, we discussed inferential problems in which the count

data was displayed in a rectangular table of cells. Each table consisted of one row

and a specified number of columns, where the columns corresponded to categories

into which the population had been divided. We now study problems in which the

data also consists of counts or frequencies, but the data table will now have I rows

(I  2) and J columns, so IJ cells. There are two commonly encountered situations

in which such data arises:

1. Th ere are I populations of interest, each corresponding to a different row of the

table, and each population is divided into the same J categor ies. A sample is

taken from the ith population (i ¼ 1, ..., I), and the counts are entered in the

cells in the ith row of the table. For example, customers of each of I ¼ 3

department store chains might have available the same J ¼ 5 payment

categories: cash, check, store credit card, Visa, and MasterCard.

2. There is a single population of interest, with each individual in the population cate-

gorized with respect to two different factors. There are I categories associated

with the first factor, and J categories associated with the second factor. A single

sample is taken, and the number of individuals belonging in both category i of factor

1 and category j of factor 2 is entered in the cell in row i, column j (i ¼ 1, ..., I;

j ¼ 1, ..., J). As an example, customers making a purchase might be classified

according to both department in which the purchase was made, with I ¼ 6

departments, and according to method of payment, with J ¼ 5asin(1)above.

Let n

denote the number of individuals in the sample (s) falling in the (i, j )th cell

(row i, column j ) of the table—that is, the (i, j )th cell count. The table displaying

the n

’

s is called a two-way contingency table; a prototype is shown in Table 13.9.

Table 13.9 A two-way contingency table

744 CHAPTER 13 Goodness-of-Fit Tests and Categorical Data Analysis

In situations of type 1, we want to investigate whe ther the proportions in the

different categories are the same for all populations. The null hypothesis states that

the populations are homogeneous with respect to these categories . In type 2 situa-

tions, we investigate whethe r the categories of the two factors occur independently

of each other in the population.

Testing for Homogeneity

We assume that each individual in every one of the I populations belongs in exactly

one of J categories. A sample of n

individuals is taken from the ith population; let

n ¼ S n

and

¼ the number of individuals in the ith sample who fall into category j

j

i¼1

the total number of individuals among

the n sampled who fall into category j

The n

’s are recorded in a two-way contingency table with I rows and J columns.

The sum of the n

’s in the ith row is n

, whereas the sum of entries in the jth column

is n

·j

Let

the proportion of the individuals in

population i who fall into category j

Thus, for population 1, the J proportions are p

, p

, ..., p

(which sum to 1) and

similarly for the other populations. The null hypothesis of homogeneity states that

the proportion of individuals in category j is the same for each population and

that this is true for every category; that is, for every j, p

¼ p

¼¼p

When H

is true, we can use p

, p

, ..., p

to denote the population propor-

tions in the J different categories; these proportions are common to all I popula-

tions. The expected number of individuals in the ith sample who fall in the jth

category when H

is true is then E(N

) ¼ n

· p

. To estimate E(N

), we must first

estimate p

, the proportion in category j. Among the total sample of n individuals,

·j

fall into category j, so we use

¼ N

j

=n as the estimator (this can be shown to

be the maximum likelihood estimator of p

). Substitution of the estimate

for p

yields a simple formula for estimated expected counts under H

¼ estimated expected count in cell ði; jÞ¼n



j

ðith row total)ðjth column total)

ð13:10Þ

The test statistic also has the same form as in previous problem situations. The

number of degrees of freedom comes from the general rule of thumb. In each row of

Table 13.9 there are J – 1 freely determined cell counts (each sample size n

fixed), so there are a total of I(J – 1) freely determined cells. Parameters p

, ..., p

are estimated, but because Sp

¼ 1, only J – 1 of these are independent. Thus

df ¼ I(J –1)–(J –1)¼ (J – 1)(I – 1).

13.3 Two-Way Contingency Tables 745

Null hypothesis: H

: p

¼ p

¼¼p

j ¼ 1; 2; ...; J

Alternative hypothesis: H

: H

is not true

Test statistic value:

all cells

ðobserved  estimated expectedÞ

estimated expected

i¼1

j¼1

ðn



Rejection region: w

 w

a;I1;J1

P-value information can be obtained as described in Section 13.1. The test

can safely be applied as long as

 5 for all cells.

Example 13.13 A company packages a particula r product in cans of three different sizes, each one

using a different production line. Most cans conform to specifications, but a quality

control engineer has identified the following reasons for nonconformance: (1)

blemish on can; (2) crack in can; (3) improper pull tab location; (4) pull tab

missing; (5) other. A sample of nonconforming units is selected from each of the

three lines, and each unit is categorized according to reason for nonconformity,

resulting in the following contingency table data:

Reason for Nonconformity

Blemish Crack Location Missing Other Sample Size

Production 1 34 65 17 21 13 150

Line 2 23 52 25 19 6 125

3 32 28 16 14 10 100

Total 89 145 58 54 29 375

Does the data suggest that the proportions falling in the various nonconformance

categories are not the same for the three lines? The parameters of interest are the

various proportions, and the relevant hypotheses are

: the production lines are homogeneous with respect to the five non-

conformance categories; that is, p

¼ p

for j ¼ 1, ...,5

: the production lines are not homogeneous with respect to the categories

The estimated expected frequencies (assuming homo geneity) must now be calcu-

lated. Consider the first nonconformance category for the first production line.

When the lines are homogeneous,

estimated expected number among the 150 selected units that are blemished

ðfirst row total)ðfirst column total)

total of sample sizes

ð150Þð189Þ

375

¼ 35:60

The contribution of the cell in the upper-left corner to w

is then

ðobserved  estimated expected)

estimated expected

ð34  35:60Þ

35:60

¼ :072

746 CHAPTER 13 Goodness-of-Fit Tests and Categorical Data Analysis

The other contributions are calculated in a similar manner. Figure 13.4 shows

MINITAB output for the chi-squared test. The observed count is the top number in

each cell, and directly below it is the estimated expected count. The contribution of

each cell to w

appears below the counts, and the test statistic value is w

¼ 14.159.

All estimated expected counts are at least 5, so combining categories is unnecessary.

The test is based on (3 – 1)(5 – 1) ¼ 8 df. Appendix Table A.10 shows that the values

that capture upper-tail areas of .08 and .075 under the 8 df curve are 14.06 and 14.26,

respectively. Thus the P-value is between .075 and .08; MINITAB gives P-value

¼ .079. The null hypothesis of homogeneity should not be rejected at the usual

significance levels of .05 or .01, but it would be rejected for the higher a of .10.

Testing for Independence

We focus now on the relationship between two different factors in a single

population. The number of categories of the first factor will be denoted by I and

the number of categories of the second factor by J. Each individual in the popula-

tion is assumed to belo ng in exactly one of the I categories associated with the first

factor and exactly one of the J categories associated with the second factor. For

example, the population of interest might consist of all individuals who regularly

watch the national news on television, with the first factor being preferred network

(ABC, CBS, NBC, PBS, CNN, or FOX, so I ¼ 6) and the second factor political

philosophy (liberal, moderate, conservative, giving J ¼ 3).

For a sample of n individuals taken from the pop ulation, let n

denote the

number among the n who fall both in category i of the first factor and category j of

the second factor. The n

’s can be displayed in a two-way contingency table with

I rows and J columns. In the case of homogeneity for I populations, the row totals

were fixed in advance, and only the J column totals were random. Now only the

total sample size is fixed, and both the n

i·

’s and n

·j

’s are observed values of random

variables. To state the hypotheses of interest, let

Figure 13.4 MINITAB output for the chi-squared test of Example 13.13 ■

13.3 Two-Way Contingency Tables 747