Vidakovic B. Statistics for Bioengineering Sciences: With Matlab and WinBugs Support

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520 13 Goodness-of-Fit Tests

pval2 = 1 - kscdf(zz) %0.0196

% MATLAB’s built in function

[h,p,k] = kstest2(x,y)

% 1

% 0.0202

% 0.3414

13.5 Moran’s Test*

Before discussing Moran’s test, we note that any goodness-of-ﬁt null hypothe-

sis H

: F = F

, where F

is a ﬁxed continuous distribution, can be restated in

terms of a uniform distribution. This is a simple consequence of the following

result:

Result. Let random variable X have a continuous distribution F. Then,

F(X ) has a uniform distribution on [0,1].

This important fact has a simple proof, since the continuity of F ensures the

existence of F

−1

P(F(X ) ≤ x) =P(X ≤ F

−1

(x)) = F(F

−1

(x)) = x, 0 ≤ x ≤1.

Thus the hypothesis H

: F =F

for observed x

,... , x

is equivalent to H

: the

CDF is uniform on [0

1] for transformed values

(

)

(

)

,... ,

Deﬁne spacings

= y

(i+1)

− y

(i)

, i =0, 1, 2,..., n,

where y

(1)

≤ y

(2)

≤ ··· ≤ y

(n)

is the order statistic of y

,... , y

and y

(0)

= 0 and

(n+1)

=1. Then Moran’s statistic

i=0

−1

has an approximately standard normal distribution. This approximation is

adequate if n

>30.

The hypothesis H

: F = F

is rejected if M > z

. The p-value is

1-normcdf(M).

13.6 Departures from Normality 521

Since Moran’s M is minimized by identical spacing d

= 1/(n +1), small

values of M may indicate nonrandom data.

Moran’s test is complementary to other goodness-of-ﬁt tests since it is sen-

sitive to data clustering and anomalous spacing. The alternatives, consisting

of long-tailed densities that remain undetected by Pearson’s

or KS tests,

may be detected by Moran’s test. The following example demonstrates that

Moran’s test is superior in detecting a long-tailed alternative compared to Kol-

mogorov’s test.

Example 13.10. In the following MATLAB script we repeated 1000 times the

following: (i) a sample of size 200 was generated from a t

distribution and (ii)

the sample was tested for standard normality using Moran’s and Kolmogorov’s

tests. Moran’s test rejected H

203 times while by Kolmogorov’s test H

was

rejected 122 times.The exact number of times H

was rejected varies slightly

depending on the random number seed; in this example we used a combined

multiple recursive generator:

stream=RandStream(’mrg32k3a’);

n=200;

stream = RandStream(’mrg32k3a’);

RandStream.setDefaultStream(stream);

pvalmoran =[]; pvalks=[];

for i = 1:1000

x = trnd(4,[n 1]); %simulate t

y=normcdf(x,0,1); %H0: F=N(0,1)

yy =[0; sort(y); 1];

Sd = sum(diff(yy,1).^2);

M =sqrt(n)

Sd/2 - 1); %Moran’s Stat

p = 1-normcdf(M);

pvalmoran = [pvalmoran p];

[h pv dn] = kstest(x, [x normcdf(x, 0, 1)]);

pvalks = [pvalks pv];

end

%Number of times H

0 was rejected in 1000 runs

sum(pvalmoran < 0.05) %203

sum(pvalks < 0.05) %122

13.6 Departures from Normality

Several tests are available speciﬁcally for the normal distribution. The Jarque–

Bera test (Jarque and Bera, 1980) is a goodness-of-ﬁt measure of departure

from normality, based on the sample kurtosis and skewness. The test statistic

is deﬁned as

522 13 Goodness-of-Fit Tests

(κ

−3)

where n is the sample size,

is the sample skewness,

−X )

3/2

and

is the sample kurtosis,

−X )

as on p. 20.

Note that the ﬁrst four moments are used jointly for the calculation of

. The statistic χ

has an asymptotic chi-square distribution with 2 de-

grees of freedom and is a measure of deviation from the moments of normal

distribution (Sk

= 0 and κ = 3). In MATLAB, the Jarque–Bera test is called

[h,p,jbstat,critval]=jbtest(x,alpha).

Lilliefors’ test (Lilliefors, 1967) for departures from normality is a version

of the KS test, where the theoretical distribution is normal but not fully spec-

iﬁed, as the KS test requires. The test statistic is of the same type as in the

KS test. The MATLAB command for Lilliefors’ test is

[h,p,ksstat,critval] =

lillietest(x, alpha)

, where the data to be tested are in vector x.

Example 13.11. Suppose that we generate a random sample of size 30 from an

F distribution with 5 and 10 degrees of freedom and we want to test that the

sample is normal.

x = frnd(5, 10, [1, 30]);

alpha=0.05;

[h1,p1] = jbtest(x, alpha)

%h1 = 1

%p1 = 1.0000e-003

[h2,p2] = lillietest(x, alpha)

%h2 = 1

%p2 = 0.0084

Both the Jarque–Bera and Lilliefors’ tests reject H

that the sample is nor-

mally distributed. In this example, the Jarque–Bera test resulted in a smaller

p-value.

13.7 Exercises 523

13.7 Exercises

13.1. Q–Q Plot for

2χ

. Simulate N =10,000 χ

random variables with k =40

degrees of freedom. Demonstrate empirically that

2χ

−

2k −1

is approximately standard normal

N (0,1) by plotting a histogram and Q–Q

plot.

Hint:

df=40; zs=sqrt(2

chi2rnd(df,[1 10000]))-sqrt(2

df-1);

13.2. Not at All Like Me. You have a theory that if you ask subjects to sort

one-sentence characteristics of people (e.g., “I like to get up early in the

morning”) into ﬁve groups ranging from not at all like me (group 1) to very

much like me (group 5), the percentage falling in each of the ﬁve groups will

be approximately 10, 20, 40, 20, and 10. You have one of your friends sort

50 statements, and you obtain the following data: 8, 9, 21, 8, and 4.

Do these data support your hypothesis at

α =5%?

13.3. Cell Counts. A student takes the blood cell count of ﬁve random blood

samples from a larger volume of solution to determine if it is well mixed.

She expects the cell counts to be distributed uniformly. The data is given

below:

Sample blood Expected blood

cell count cell count

35 27.2

20 27.2

25 27.2

31 27.2

136 136

Can she depend on the results to be uniformly distributed? Do the chi-

square test with

α =0.05.

13.4. GSS Data. Below is part of the data from the 1984 and 1990 General Social

Survey, conducted annually by the National Opinion Research Center. Ran-

dom samples of 1473 persons in 1984 and 899 persons in 1990 were taken

using multistage cluster sampling. One of the questions in a 67-question-

long questionnaire was: Do you think most people would try to take advan-

tage of you if they got a chance, or would they try to be fair?

524 13 Goodness-of-Fit Tests

1984 survey 1990 survey

1. Would take advantage of you 507 325

2. Would try to be fair 913 515

3. Depends 47 53

4. No answer 6 6

Assuming that 1984 frequencies are theoretical, using Pearson’s χ

test ex-

plore how the 1990 frequencies agree. Use

α = 0.05. State your ﬁndings

clearly.

13.5. Strokes on “Black Monday.” In a long-term study of heart disease, the

day of the week on which 63 seemingly healthy men died was recorded.

These men had no history of disease and died suddenly.

Day of week Mon. Tue. Wed. Thu. Fri. Sat. Sun.

No. of deaths 22 7 6 13 5 4 6

(a) Test the hypothesis that these men were just as likely to die on one day

as on any other. Use

α =0.05.

(b) Explain in words what constitutes the error of the second kind in the

testing in (a).

13.6. Benford’s Law. Benford’s law (Benford, 1938; Hill, 1998) concerns the

relative frequencies of leading digits of various data sets, numerical tables,

accounting data, etc. Benford’s law, also called the ﬁrst digit law, states that

in numbers from many sources, the leading digit is much more often a 1

than it is any other digit (speciﬁcally about 30% of the time). Furthermore,

the higher the digit, the less likely it is to occur as the leading digit of a

number. This applies to ﬁgures related to the natural world or ﬁgures of

social signiﬁcance, be it numbers taken from electricity bills, newspaper

articles, street addresses, stock prices, population numbers, death rates,

areas or lengths of rivers, or physical and mathematical constants.

Fig. 13.9 Frank Benford (1883–1948).

More precisely, the Benford law states that the leading digit n, (n =1,..., 9)

occurs with probability P(n)

=log

(n+1)−log

(n), or as in the table below

[probabilities P(n) rounded to 4 decimal places]:

13.7 Exercises 525

Digit n 1 2 3 4 5 6 7 8 9

P(n) 0.3010 0.1761 0.1249 0.0969 0.0792 0.0669 0.0580 0.0512 0.0458

An oft-cited data set is the distribution of the leading digit of all 309 num-

bers appearing in a particular issue of Reader’s Digest.

Digit 1 2 3 4 5 6 7 8 9

Count 103 57 38 23 20 23 17 15 13

At level α =0.05, test the hypothesis that the observed distribution of lead-

ing digits in Reader’s Digest numbers follows Benford’s law.

Hint:

To speed up the calculation, some theoretical frequencies rounded to

two decimal places are provided:

93.01 54.41 • 29.94 24.47 • • 15.82 14.15

13.7. Simulational Exercise.

• Generate a sample of size n

=216 from a standard normal distribution.

• Select intervals by cutting the range of simulated values by points

−2.7,

−2.2, −2, −1.7, −1.5, −1.2, −1, −0.8, −0.5, −0.3, 0, 0.2, 0.4, 0.9, 1, 1.4,

1.6, 1.9, 2, 2.5, and 2.8.

• Using a

-test conﬁrm the normality of the sample.

• Repeat the previous test if your sample is contaminated by a Cauchy

Ca(0,1) distribution in the following way:

0.95

normal

sample + 0.05

cauchy

sample

13.8. Deathbed Scenes. Can some people postpone their death until after a spe-

cial event takes place? It is believed that famous people do so with respect

to their birthdays to which they attach some importance. A study by Philips

(1972) seems to be consistent with this notion. Philips obtained data

on the

months of birth and death of 1251 famous Americans; the deaths were clas-

siﬁed by the time period between the birth dates and death dates as shown

in the following table:

b e f o r e birth a f t e r

6 5 4 3 2 1 month 1 2 3 4 5

90 100 87 96 101 86 119 118 121 114 113 106

(a) Clearly formulate the statistical question based on the above observa-

tions.

(b) Provide a solution for the question formulated in (a). Use

α =0.05.

13.9. Grouping in a Vervet Monkey Troop. Struhsaker (1965) recorded, in-

dividual by individual, the composition of sleeping groups of wild verver

348 were people listed in Four Hundred Notable Americans and 903 are listed as the

foremost families in three volumes of Who Was Who for the years 1951–1960, 1943–1950,

and 1897–1942.

526 13 Goodness-of-Fit Tests

monkeys (Cercopithecus aethiops) in East Africa. A particular large troop

was observed for 22 nights.

Size 1 2 3 4 5 6 7 8 9 10 11 ≥12

Freq

19 14 19 11 7 7 3 2 3 1 2 1

It is well known that the sizes of groups among humans in various so-

cial settings (shoppers, playgroups, pedestrians) follows a truncated Pois-

son distribution:

P(X

= k) =

exp{−λ}

k!(1 −exp{−λ})

, k

=1,2,3,....

Using a

goodness-of-ﬁt test, demonstrate that the truncated Poisson is

not a good model for this data at the signiﬁcance level

α = 0.05. To ﬁnd

the theoretical frequencies,

λ can be estimated by the method of moments

using an average size of

X = 3.74 and the equation λ/(1 −exp{−λ}) = 3.74.

In MATLAB

fzero(@(lam) lam - 3.74

(1-exp(-lam)), 3) gives

λ =3.642.

Struhsaker further argues that in this context the truncated negative bino-

mial gives a reasonably good ﬁt.

13.10. Crossing Mushrooms. In a botany experiment the results of crossing

two hybrids of a species of mushrooms (Agaricus bisporus) gave observed

frequencies of 120, 53, 36, and 15. Do these results disagree with theoretical

frequencies that specify a 9:3:3:1 ratio? Use

α =0.05.

13.11. Renner’s Honey Data Revisited. In Example 6.10 we argued that a

lognormal distribution with parameters

µ = and σ

= 1.0040

provides a

good ﬁt for Renner’s honey data. Find

-statistics and the p-value of the

ﬁt. State decision at level

α = 0.05. The midintervals in the ﬁrst column of

data set

renner.mat|dat, 0.1250,0.3750,0.6250, ..., 7.1250 correspond to

the intervals (0,0.25],(0.25,0.5], (0.5,0.75],. . ., (7,7.5].

13.12. PCB in Yolks of Pelican Eggs. A well-known data set for testing agree-

ment with a normal distribution comes from Risebrough (1972) who ex-

plored concentrations of polychlorinated biphenyl (PCB) in yolk lipids of

pelican eggs. For n

= 65 pelicans from Anacapa Island (northwest of Los

Angeles), the concentrations of PCB were as follows:

452 184 115 315 139 177 214 356 166 246 177 289 175

296 205 324 260 188 208 109 204 89 320 256 138 198

191 193 316 122 305 203 396 250 230 214 46 256 204

150 218 261 143 229 173 132 175 236 220 212 119 144

147 171 216 232 216 164 185 216 199 236 237 206 87

Test the hypothesis that PCB concentrations are consistent with the normal

distribution at the level

α =0.05.

13.7 Exercises 527

13.13. Number of Leaves per Whorl in Ceratophyllum demersum. After spend-

ing a year with Karl Pearson in London, biometrician Raymond Pearl pub-

lished his work on mathematical and statistical modeling of growth of an

aquatic plant, Ceratophyllum demersum (Pearl, 1907). Among several in-

teresting data sets, Pearl gives the table of frequencies of numbers of leaves

per whorl.

Leaves 5 6 7 8 9 10 11 12 Total

Whorls 6 169 258 495 733 617 48 2 2328

Is the binomial distribution B in(12, p) a good ﬁt for the data?

Hint:

The

probability p can be estimated from the data; the average of Leaves should

be close to 12p.

13.14. From 1998–2002 U.S. National Health Interview Survey (NHIS).

Stansﬁeld and Carlton (2007) analyzed data in sibships of size 2 and 3 from

the 1998–2002 U.S. National Health Interview Survey (NHIS).

For the 25,468 sibships of size 2, among all 50,936 children, 10 years of age

and younger, in the NHIS data set (the youngest age cohort available), they

found that 51.38% were boys (B) and 48.62% were girls (G); the B/G ratio

was 1.0566, or about 106 boys for every 100 girls. The number of boys is

given in the following table:

Number of boys 0 1 2

Observed sibships

5,844 13,079 6,545

For the 7,541 sibships of size 3 from the same NHIS data set the number of

boys among the ﬁrst two children is given in the following table:

Number of boys 0 1 2

Observed sibships

1,941 3,393 2,207

(a) If the number of boys among the ﬁrst two children in sibships of size 3

is binomial

B in(2,0.515), ﬁnd what theoretical frequencies are expected in

the table for sibships of size 3. Are the observed and theoretical frequencies

close? Comment.

Hint:

Find the binomial probabilities and multiply them

by n

=7,541.

(b) For the 25,468 sibships of size 2, test the hypothesis that the probability

of a boy is 1/2 versus the one-sided alternative, that is,

: p =0.5 vs H

: p >0.5,

at the level

α =0.05.

Hint:

Be careful about the n here. The rejection region

of this test is RR

=[1.645,∞).

13.15. Neuron Fires Revisited. The 989 ﬁring times of a cell culture of neurons

have been recorded. The records are time instances when a neuron sends a

528 13 Goodness-of-Fit Tests

signal to another linked neuron (a spike) and the largest time instance is

1001.

The count of ﬁrings in 200 consecutive time intervals of length 5 (= 1000/200)

time units is as follows:

2 7 5 5 1 6 8 4 6 4 5 7 7 5 4 7 4 4 5 9 6 5 6 6 5

5 6 4 10 7 8 8 2 9 5 4 4 4 3 8 3 2 7 6 7 5 6 6 4 6

5 7 3 5 6 5 5 2 7 7 6 4 8 8 9 7 3 3 3 5 6 6 4 6 4

5 4 4 5 2 3 5 1 4 4 3 2 10 4 7 2 1 7 9 4 3 6 8 6 5

2 4 4 3 7 2 2 6 1 3 7 6 6 4 7 5 5 8 7 3 5 5 4 5 6

10 3 6 9 5 7 2 8 4 4 2 3 4 3 3 3 5 4 2 2 7 4 5 4 5

3 5 3 5 5 7 2 5 5 4 8 4 3 5 7 4 4 8 5 4 3 6 4 6 4

1 7 9 4 2 5 4 4 4 4 7 2 4 4 5 2 4 7 5 12 1 5 4 6 7

Are the data consistent with a Poisson distribution?

13.16. Cloudiness in Greenwich. Pearse (1928) cites data describing the degree

of cloudiness in Greenwich, UK for days in the month of July between 1890

and 1904.

Cloudiness 0 1 2 3 4 5 6 7 8 9 10 Total

Days 320 129 74 68 45 45 55 65 90 148 676 1715

Assume that the measure of cloudiness can be modeled as a continuous

random variable C

=10 ·X , where X has a beta B e(α, β) distribution.

Find the best ﬁtting parameters

α and β and assess the goodness of ﬁt.

13.17. Distance Between Spiral Reversals in Cotton Fibers. Tippett (1941)

discusses a frequency histogram for a data set consisting of intervals and

counts for a distance between spiral reversals in cotton ﬁber (Fig. 13.10).

The distances are given in units of mm

−2

, and the sample size was 1117.

Distance [0,2.5) [2.5,4.5) [4.5,6.5) [6.5, 8.5) [8.5,10.5)

Number 7 48 100 106 84

[10.5,12.5) [12.5,16.5) [16.5,20.5) [20.5,24.5) [24.5,28.5)

72 136 94 78 69

[28.5,32.5) [32.5,36.5) [36.5,40.5) [40.5,50.5) [50.5,60.5)

53 45 36 69 40

[60.5,70.5) [70.5,80.5) [80.5,90.5) [90.5,∞) Total

31 21 17 11 1117

Using Pearson’s χ

criterion and signiﬁcance level α =0.05, test the hypoth-

esis that the data are consistent with the

-distribution. Use the midpoints

of intervals to estimate the mean of a theoretical

-distribution. Recall

that the mean in this case is equal to the number of degrees of freedom.

Chapter References 529

Fig. 13.10 Structure of spiral reversals in cotton ﬁber.

MATLAB FILES AND DATA SETS USED IN THIS CHAPTER

http://springer.bme.gatech.edu/Ch13.Goodness/

anacapa.m, consolidator.m, empiricalcdf.m, examples.m,

horsekicks.m, kolsm.m, ks2cdf.m, kscdf.m, moran.m, neuronfiresqq.m,

neurongof.m, notatalllikeme.m, perk.m, plotsimulqq.m, poissonness.m,

qqbivariatenormal.m, qqnorm.m, qqplotsGOF.m, RANDmillion.m,

smirnov.m, smirnov2tests.m, tippett.m, weldon.m

neuronfires.mat, RAND1Millrandomdigits.txt

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Hill, T. (1998). The ﬁrst digit phenomenon. Am. Sci., 86, 358–363.

Hoaglin, D. C. (1980). A poissonness plot. Am. Stat., 34, 146–149.

Jarque, C. and Bera, A. (1980). Efﬁcient tests for normality, homoscedasticity and serial

independence of regression residuals. Econ. Lett., 6, 3, 255–259.

Kemp, W. A. and Kemp, D. C. (1991). Weldon’s dice data revisited. Am. Stat., 45, 3, 216–222.

Kolmogorov, A. N. (1933). Sulla determinazione empirica di una legge di distributione. Gior.

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and Engineering. Wiley, Hoboken.

Lilliefors, H. W. (1967). On the Komogorov-Smirnov test for normality with mean and vari-

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Pearl, R. (1907). Variation and differentiation in Ceratophyllum. Carnegie Inst. Wash. Publ.,

58, 1–136.

Pearse, G. E. (1928). On corrections for the moment-coefﬁcients of frequency distributions.

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