Tutorials in Statistics-chapter 12 New

7/28/2019 Tutorials in Statistics-chapter 12 New

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CHAPTER TWELVE

THE KRUSKAL-WALLIS ONE-WAY

ANALYSIS OF VARIANCE BY RANKS

TEST (H TEST)

Introduction

Method

Example for Small Samples

Example for Large Samples ( 5>jn )

The Problem with Tied Ranks

Multiple Comparisons Following a SignificantHTest

Chapter Summary

Worked Example on the Kruskal-WallisHTest

Exercises on the Kruskal-WallisHTest


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12.1: Introduction

The Kruskal-Wallis One-Way Analysis of Variance by Ranks test is

the nonparametric equivalent of the parametric One-Way analysis of

variance (F) test that we discussed in Chapter 6. This means that the test

is used when we are dealing with three or more independent samples. Itrequires that measurement be at least on an ordinal scale. The Kruskal-

Wallis technique tests the null hypothesis that all the k independent

samples )2( >k are coming from identical populations with respect to

averages.

12.2: Method

In the computation of the statistic in the KruskalWallis test, all the

observations from the k samples are combined and ranked in a singleseries. The lowest score is given a rank of 1, the next lower score a rank

of 2, etc., and the highest score is given a rank of n , the total number of

observations. Assuming that there are equal sample sizes, then the sum

of the ranks for each group is determined. The Kruskal-Wallis test

determines whether these sums of ranks are so different from each other

and therefore are not likely to have come from the same population with

equal averages. If the sample sizes are not equal, then the Kruskal-Wallistest will test the hypothesis whether the average ranks in each of the k

groups are also very different from each other and are therefore not

likely to have come from the same population or populations with equal

averages. If the average ranks in each of the kgroups are about the same,

then the different ranks would be about equally distributed among the k

groups (i.e., ranks in the various groups would have about the same

average) and therefore that 0H is more likely to be true. Rejection of

0H means that the various average ranks are different from one

another.

It can be shown mathematically that if the k samples actually come

from the same population or from identical populations (with respect to

ranks), i.e., if 0H is true, thenH, the statistic used in the Kruskal-Wallis

test is distributed as a Chi Square )(2

with 1k df, where =k number of samples or groups involved in the study, provided that each

sample size is greater than 5.The Kruskal-WallisH is defined as:

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( ))1(3

1

12

1

2

++

= =

NNN

k

j j

j

n

RH , where kis as defined above, jn

is the number of cases in thejth

sample;N is total sample size, i.e.

kk nnnnn 1321 +++++ ; and jR is the sum of the ranks in the jth

sample.

When at least each 5>jn , the value of H may be referred to the2 tables (Appendix 2.6) with 1k df under a two-tailed test. If the

observed value ofH is greater than or equal to the critical value of2

in the tables at a specified level of significance, 0H is rejected in favour

of 1H , and we conclude that a significant difference exists among the

ranked averages in the kgroups, i.e., at least one ranked average is largeror smaller than another.

In the special case where 3=k , and each sample size )( jn in the

three samples is less than or equal to 5, then the2 approximation is not

sufficiently exact to be used. In such a situation, the values of 1n , 2n ,

3n , and H may be referred to the Kruskal-Wallis probability tables

(Appendix 2.9) to determine the probability associated with an observed

value of H, given the values of 1n , 2n , and 3n , the sample sizes ofthe three samples. The first column in the table gives values of 1n , 2n

, 3n . The second column gives various values ofH , and the third

column the exact probability associated with the occurrence of 0H of

values as large as an observed H. For 0H to be rejected, obsH must

be greater than or equal to critH at a given level of significance. Note

that the value ofH cannot be negative.

12.3: Example for Small Samples

Let the data in Table 12.1 represent the scores obtained by three

groups of subjects on a sub-scale of an aptitude test where scores ranged

between 1 and 15.

To compute the value ofH we first of all combine the scores from

the three groups and rank them, but retaining the identity of each group.

This will result in the following table (Table 12.2).

Table 12.1: Scores obtained by three groups of subjects on a sub-scale

of an aptitude test

Group 1 Group 2 Group 3

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Sub. No Score Sub

No.

Score Sub No. Score

1

2

34

5

4

312

1

2

34

6

7

28

1

2

3

9

10

11

Table 12.2 : Scores obtained by three groups of subjects on a sub-scale

of an aptitude test and the ranks assigned to these scores

(Case 1).


Sub

No.

Score Rank Sub

No.

Score Rank Sub.

No.

Score Rank

1

2

3

4

5

4

3

12

4

3

2

11

1

2

3

4

6

7

2

8

5

6

1

7

1

2

3

9

10

11

8

9

10

41 =n 201 =R

00.51 =R

42 =n 192 =R

75.42 =R

33 =n 273 =R

00.93 =R

Substituting the above relevant values into the Kruskal-Wallis H

formula and noting that 11344321 =++=++= nnnN we get,

3

)27(

4

)19(

4

)20()12(11

12)1(3

)1(

12

2223

1

2

++=+

+=

=

NNN j j

j

n

RH

361125.43336)24325.90100(

11

1=++=

386363.336386363.39 == , i.e., 3864.3obs H (corrected to 4

decimal places ).

Referring 3864.3obs =H to the Kruskal-Wallis probability tables

(Appendix 2.9) with 41 =n , 42 =n , and 33 =n , we find that at the

0.049 level of significance ( 049.0=p ), we need an H value of

5.5985 to reject 0H . At the 0.051 level of significance ( 051.0=p )

the critical value of 5758.5=H . This means that at the 0.05 level ofsignificance, obsH must lie between 5.5985 and 5.5758 for 0H to be

rejected. This is because 0.05 lies between 0.049 and 0.051)051.005.0049.0(


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even at the 0.051 level of significance, then 0H should be retained and

we conclude that no significant difference exists between the three

groups of subjects in performance on the subscale of the aptitude test.

Note that for each 5>jn , the H distribution, as we observed earlier, is

distributed as a Chi Square )( 2 and therefore for 0H to be rejectedin the H test, the obsH must be greater than or equal to critH . Also

note from Appendix 2.9 that like the2

test, the smaller the value of

H, the more likely it is that 0H is true and therefore must be retained.

Now let us assume that in the study, instead of the scores and ranks

presented in Table 12.2, Table 12.3 represent the scores and ranks

obtained by subjects in the three groups

Table 12.3: Ranks assigned to scores obtained by three groups of

subjects on a sub-scale of an aptitude test (Case 2)


Sub

No.

Score Rank Sub

No.

Score Rank Sub.

No.

Score Rank

1

23

4

5

42

6

4

31

5

1

23

4

7

103

8

6

92

7

1

23

9

1214

8

1011

41 =n 131 =R

25.31 =R

42 =n 242 =R

00.62 =R

33 =n 293 =R

67.93 =R

Computing the value ofH from the above data, we get,

)12(33

)29(

4

)24(

4

)13()12(11

12

222

++=H

If we simplify the above, we will find that 4167.6obsH (corrected to

4 decimal places). Referring the value of 4167.6=obsH to the Kruskal-

Wallis probability tables with 41 =n , 42 =n , and 33 =n , we find that

our obsH value of 6.4167 is not listed in the tables. However, we can

deduce that the probability associated with this observed value of obsH

which is 6.4167 lies between 0.049 and 0.011. Again this is because at049.0=p , the associated H value 5985.5= and at 011.0=p , the

associated H value 8727.6= . Since our computed 4167.6=obsH

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lies between 5.5985 and 6.8727 i.e., 8727.64167.65985.5 n )

Suppose that in the example given for small sample sizes (Section

12.3), the sample sizes were increased to 6, 7, and 8 respectively for

1n , 2n and 3n , with the data as in Table 12.4.

Table 12.4: Scores obtained by three groups of subjects on a sub-scaleof an aptitude test and the ranks assigned to these scores (Case 3).


Sub

No.

Score Rank Sub

No.

Score Rank Sub.

No.

Score Rank

1

2

3

4

5

6

5

4

3

12

6

2

5

4

3

17

6.5

1.5

1

2

3

4

5

6

7

6

7

2

8

9

13

10

6.5

8

1.5

9.5

11.5

18.5

14

1

2

3

4

5

6

7

8

9

10

11

14

15

13

8

10

11.5

14

16

20

21

18.5

9.5

14

61 =n 00.371 =R

17.61 =R

72 =n 50.692 =R

93.92 =R

83 =n 50.1243 =R

56.153 =R

Noting that 21876321 =++=++= nnnN , and substituting the relevant

values into the H formula,

)1(3)1(

12

3

1

2

++

= =

NNN j j

j

n

RH , we get,

)22(3

8

)50.124(

7

)50.69(

6

)37(

)22(21

12

222

++=H

[ ] 661937.5312690.03571228.1666677

2++

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1748.86674.17489666772855.7335)2( ==

i.e, 17.8obsH (corrected to 2 decimal places).

Referring the 17.8=obsH to the2

Tables with 1k

213 ==df df, we note that the critical value of2

at the 0.05 levelof significance under a nondirectional test 5.99= . Therefore,

5.99=critH . Since ( )99.5()178 =>= critobs . HH , 0H is rejected and

we conclude that a significant difference exists between the three

groups in performance on the sub-scale of the aptitude test.

12.5: The Problem with Tied Ranks

When there are tied ranks, either within and/or between groups, then

the usual procedure of assigning the average of the ranks to the

observations involved is used. This procedure was used in the

computation in Section 12.4 above. The value of H is somewhat

influenced by ties and when there are too many ties involved, a formula

for correction for ties is used. The effect of correcting for ties is to

increase the value ofH and thus make it more likely for 0H to be

rejected. In most cases, however, the effect of the correction is negligible

and it may be used when the obsH is quite close in value to the critH but does not reach significance. The interested reader is referred to:

Siegel, S. (1956). Nonparametric statistics for the behavioral sciences.

New York: McGraw -Hill Book Company, Inc. Pp 188192.

12.6: Multiple Comparisons following a significantHTest

As we noted in Chapter 6 under the One-Way ANOVA, the

rejection of the ANOVA 0H does not necessarily mean that all the

means being compared differ from each other. Like the parametric F

test, the nonparametric H test will dictate that 0H should be rejected

when an inequality exists between any two or more of the k average

ranks being compared. Unfortunately, unlike the F test, there is no

known method of pairwise comparisons between average ranks based on

data from the H test when H is found to be significant. We may

however, use the Mann-Whitney U test to make such pairwise

comparisons. But, as we noted in Chapter 6, the difficulty with manypairwise comparisons is that we are bound to commit an experimentwise

error, i.e., we increase the chances of any of the comparisons being

significant by chance. However, when we have a directional hypothesis

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and 0H is rejected as in the example under Section 12.4, then we have

no choice but to use the Mann-Whitney U test to make any relevant

pairwise comparisons. Whenever we are compelled to make such

pairwise comparisons with the Mann-Whitney U test, we must bear in

mind the possibility of committing an experimentwise error.Let us now go back to the example on the H test presented under

Section 12.4. In the example, we found that 17.8=obsH was

significant at the 0.05 level. If we wish to make pairwise comparisons

between the three groups using the Mann-Whitney Utest, then we will

have to re-rank the data for any two groups being compared as shown in

Table 12.5. The table also shows the computations of the U values.

Referring the smaller of the 2 Us computed (U in all cases) to theMann-Whitney exact probability tables (Appendix 2.7.1), we observe

that between groups 1 and 2 )7,6( 21 == nn the exact probability

associated with the computed value of 11=U is equal to 0.090. [Thisprobability should be multiplied by 2 to obtain a two-tailed probability.

This is because in multiple comparisons, the direction of the difference

could be opposite to the predicted direction. Thus, the exact probability

associated with 11=U in a two-tailed test 0.180== )2090.0( .Since 180.0=p is greater than the desired 05.0=p , 0H is retained

and we conclude that no significant difference exists between groups 1

and 2 in performance on the sub-scale of the Aptitude Test. In a similar

manner, comparing groups 1 and 3, 0.012== )2006.0(p . Since

05.0012.0 < , 0H will be rejected, meaning that a significant

difference exists between groups 1 and 3 in performance on the sub-

scale of the aptitude test.

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Table 12.5: Pairwise comparisons among three groups using the Mann-

Whitney U test.Group 1 Vrs. Group 2 Group 1 Vs. Group 3* Group 2 Vs. Group 3*

Group 1 Group 2 Group 1 Group 3 Group 2 Group 3

Score Rank Score Rank Score Rank Score

Rank

Score Rank Score Rank

5 5

4 4

3 3

12 12

6 6.5

2 1.5

6 6.5

7 8

2 1.5

8 9

9 10

13 13

10 11

5 4

4 3

3 2

12 11

6 5

2 1

9 7

10 8.5

11 10

14 13

15 14

13 12

8 6

10 8.5

6 2

7 3

2 1

8 4.5

9 6.5

13 12.5

10 9

9 6.5

10 9

11 11

14 14

15 15

13 12.5

8 4.5

10 9

61 =n

321 =R

33.51 =R

72 =n

592 =R

43.82 =R

61 =n

261 =R

33.41 =R

82 =n

792 =R

88.92 =R

71 =n

50.381 =R

50.51 =R

82 =n

50.812 =R

19.102 =R322/)7(6)76( +=U

==+= 313263322142

===314231)76(U

262/)7(6)86( +=U

==+= 2669262148

==434843)86(U

32/)8(7)87( +=U

3845.382856 =+=+= 50.45

5650.45)87( ==U+=10.50

Comparing groups 2 and 3 )5.10,8,7( 21 === Unn , we

find, that the U value of 10.5 is not listed in the MannWhitney exact

probability tables. However, we can obtain the approximate p value by

interpolation as follows:

For 10=U , 020.0=p ;For 11=U , 027.0=p (see Appendix 2.7.1.).Assuming that there are equal intervals between the above twop-values,

then a U- value of 10.5 must lie about midway between the p-values of

0.020 and 0.027. Thus, the p-value associated with the U- value of0235.02/)027.0020.0(5.10 =+ . Therefore, for a two-tailed test,

the p- value associated with the U- value of

047.0)0235.02(5.10 == . Since 05.0047.0


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rejected and we conclude that a significant difference exists between

groups 2 and 3 in performance on the sub-scale of the aptitude test.

By inspecting the averages of the ranked data for any two groups we

are comparing, we should be able to know which average rank is

greater/less than the other average rank in all situations where 0H

isrejected and decide whether or not our directional prediction is

supported. Thus, the data in Table 12.5 shows that between groups 1 and

3, 33.41 =R and 88.92 =R and we may conclude that the

scores in group 3 were generally higher than the scores in group 1.

Similarly, between groups 2 and 3,

19.10and50.5 21 == RR and we may conclude that the

scores in group 3 were generally higher than the scores in group 2. We

must remember that no significant difference existed between groups 1

and 2 and therefore that the scores in these 2 groups were about the

same. Thus, it is reasonable to say that group 3 performed better on the

aptitude test than groups 1 and 2.

The problem of experimentwise error should always be at the back

of our minds when we are making these pairwise comparisons,

especially when the p-value is quite close to the 0.05 decision rule as in

the comparison between groups 2 and 3 where the p-value was foundto be equal to 0.047, quite close to the 0.05 decision rule.

12.7: Chapter Summary

The Kruskal-Wallis One-Way Analysis of Variance by ranks (H)

Test is the nonparametric equivalent of the parametric OneWay

ANOVA (F) Test. The test is used when 3 or more independent groups

are being compared on a dependent variable measure. The use of the

KruskalWallis H Test requires that measurement on the dependent

variable be at least on an ordinal scale. The KruskalWallis H tests the

null hypothesis that all the k independent samples (k > 2) are coming

from identical populations with respect to averages.

In the computation of H, all the observations from the k

independent samples are combined and ranked in a single series from the

lowest to the highest score. The sum of ranks for each group is then

determined. The Kruskal-Wallis test determines whether the sum of the

ranks in the k groups are very much different from each other and are

therefore not likely to have come from the same population (in the case

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when all the sample sizes are equal), or from populations with unequal

averages (in the case when all the sample sizes are not equal.)

It can be shown mathematically that if the k groups all come from

the same population or from identical populations with respect to

average ranks, then H, the statistic employed in the Kruskal

Wallis testis distributed as a Chi Square )(

2 with 1k df, where =k numberof samples or groups, provided that all the sample sizes are each greater

than 5. The H is defined as:

)1(3)1(

12

1

2

++

= =

Nn

R

NN

k

j j

jH

where =k number of samples, =jn number of cases in the jth

sample, =N total sample size, i.e., kk nnnnn 1321 +++++ , and

=jR sum of the ranks in the jthsample. The computed value of H is

therefore referred to the2 tables with the associated df and a

decision is taken as to whether or not 0H should be rejected under a

specified decision rule. Like the 2 test, the value of H cannot be

negative.In the special case where 3=k , and each 5jn , the2

approximation cannot be used. In such cases, thevalues of 1n , 2n and

3n (the three samples sizes) and the computed value of H may be

referred to the Kruskal-Wallis probability tables to determine the

probability associated with the observed value ofH, given the values of

321 and,, nnn . For 0H to be rejected, obsH must be greater than

or equal to critH

in the tables at a given level of significance.When there are tied ranks, the usual method of assigning the average

rank to the various tied positions is used. The value of H is somewhat

influenced by ties and when there are too many ties involved, a

correction for ties is used. The effect of correcting for ties is to increase

the value ofH and thus make it more likely for 0H to be rejected. In

most cases however, the effect of the correction is negligible.

When H is significant, the Mann-Whitney U test may be used to

make multiple comparisons where necessary, but at the risk of

committing an experimentwise error.

12.8: Worked Example on the Kruskal-Wallis H Test

Question

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A manufacturing company used three different methods of selection

in a particular year to employ twenty-four (24) workers. One group of

workers were selected through interviews only, a second group through a

screening (aptitude) test only, and a third group through both interviews

and a screening test. After one year on the job, the performance of thethree groups of workers were evaluated by different supervisors, one

supervisor each for the Interviews only group, Screening test only

group, and Interviews and screening test group on a scale ranging from

10 to 100. The following data were obtained (higher scores reflect better

performance on the job).

Method of SelectionInterviews only Screening Test only Interviews & Screening

Test

Employe

e No.

Job

Performance Rating

Employe

e No.

Job

PerformanceRating

Employe

e No.

Job

PerformanceRating

12

34

56

78

5535

4560

6040

3050

12

34

56

7

5065

7065

5545

60

12

34

56

78

9

6565

7055

5075

6060

45

(a) All other things being equal, determine whether or not the threemethods of selection led to differences in performance among the

employees in the company.

[Show full working steps.]

(b) Assuming that the first four employees in each group were females,

what conclusions may we draw concerning levels of job performance

in the female population using the three selection methods?

[No working steps needed.]

Solution

(a)

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Step 1: Choice of Statistical Test

We are given three independent groups of employees who have been

randomly and independently sampled from their respective populations

and their job performance evaluated by supervisors. Since different

supervisors rated the employees in the different groups, we cannot besure whether the rating scores assigned by the three supervisors to the

employees in the different groups were objective, and in any case, rating

scales are very subjective by nature. The data may therefore be converted

to ranks (an ordinal scale) and the Kruskal-Wallis H test used to

determine whether or not the three different methods of employee

selection led to differences in job performance among the three groups of

employees.

Step 2: Statement of Hypotheses

We are merely asked to determine whether or not the three selection

methods led to differences in performance among the three groups of

employees. The research hypothesis is thus non-directional. Let

321 and,, RRR represent the average ranks in the population

of employees selected through interviews only, screening test only, and

interviews and screening test respectively. Then the null hypothesis (

0H ) and the alternative hypothesis ( 1H ) may be stated as follows:

0H : No significant difference exists between the three groups of

employees on job performance (i.e., 321 RRR == ).

1H : A significant difference exists between the three groups of

employees on job performance (i.e., Not 0H ; at least two of the

groups differ on job performance.)

Step 3: Decision RulesGiven: 0.05 level of significance, a Kruskal-Wallis H test, 81 =n ,

72 =n , and 93 =n .

Since each 5>n , then H is distributed as a2 with 2131 ==k

df. From the2 Tables, the critical value of

2 with 2 df 99.5= .

Therefore if 99.5


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)1(3)1(

12

1

2

++

= =

Nn

R

NN

k

j j

jH (i)

The data presented is combined, converted to ranks and retabulated as

follows:Methods of Selection

Interviews Only Screening Test Only Interviews & Screening

TestEmployee

No.

Performance

Rating

Rank Employee

No.

Performance

Rating

Rank Employee

No.

Performance

Rating

Rank

1

2

345

67

8

55

35

456060

4030

50

11

2

51515

31

8

1

2

345

67

50

65

706555

4560

8

19.

522.5

19.5

115

15

1

2

345

67

89

65

65

705550

7560

6045

19.5

19.5

22.5118

2415

155

81 =n 601 =R

50.71 =R

72 =n 50.1002 =R

36.142 =R

93 =n 50.1393 =R

50.153 =R

Substituting the above computed values into (i) above and noting that

24978321 =++=++= nnnN , we get,

)25(39

)50.139(7

)50.100(8)60(

)25(2412

222

++=H .

Evaluating, we get,

75)25.162,28928.442,100.450(50

1 ++=H

75)1428.055,4(50

1=

75102856.81 = 102856.6=

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i.e., obs

H 6.102 (corrected to 2 decimal places)

Step 5: Decision

Comparing 6.10=obsH to the decision rules (Step 3), we note that

)5.99()6.10(=>=

critobs HH

0H is rejected at the 0.05 level of significance.

Step 6: Interpretation

At the 0.05 level of significance, a difference exists in job

performance among the employees selected using the three methods of

selection in that particular year.

[NB: If the question had required us to evaluate the relative effectivenessof the three methods of selection on work performance, then after

rejecting 0H , there would have been the need to compare the average

ranks among the three groups, taking them two at a time, using the

Mann-Whitney U test (see Section 12.6). By inspecting the average

ranks of the three groups in our worked example, it appears that using a

screening test only for selection led to higher job performance than using

interviews only )50.736.14( 12 =>= RR ; and using bothinterviews and screening test also appears to have led to higher job

performance than using interviews only

)50.750.15( 13 =>= RR . However, it also appears that

using screening test only and using both interviews and screening test

did not lead to a difference in job performance between the two groups

)36.14( 2 =R is quite close to )50.15( 3 =R . Try to confirm or

disconfirm these observations by using the Mann-Whitney U test to dothe actual comparisons as in Section 12.6.

(b) If the first 4 subjects in each group were females then we are dealing

with small sample size where 41 =n , 42 =n , and 43 =n . H is

calculated from the given data as follows:

2One would have observed that too many ties occurred in the rankings and a correction

for ties would have been needed, particularly if the obsH value had been found to be

less than but quite close to 5.99. In this example, H (corrected for ties) 6.22= ,which is not much different from the computed H value of 6.10.

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Method of Selection

Interviews only Screening Test Only Interviews &

Screening TestEmployee

No.

Perform

Rating

Rank Employee

No.

Perform

. Rating

Rank Employee

No.

Perform

. Rating

Rank

123

4

553545

60

4.512

6

123

4

506570

65

38.511.5

8.5

123

4

656570

55

8.58.5

11.5

4.5

41 =n , 50.131 =R

375.31 =R

42 =n ; 50.312 =R

875.72 =R

43 =n ; 333 =R

25.83 =R

)1(3)1(

12

1

2

++= = NnR

NN

k

j j

j

H

)13(34

)00.33(

4

)50.31(

4

)50.13()13(12

12222

++=

i.e., 39)25.2720625.2485625.45(13

1 ++=H

39)875.565(13

1=

39528846.43 = 528846.4=

i.e., 4.5288obsH (corrected to 4 decimal places.)

Referring 52884 .obs =H to the KruskalWallis probability tables

(Appendix 2.9) with 41 =n , 42 =n , and 43 =n we find that we need

an H value of 5.6923 or greater to reject 0H at 049.0=p and anH value of 5.6538 or greater to reject 0H at 054.0=p . Therefore

at 05.0=p , we would need an H value greater than 5.6538 but less

than 5.6923 (i.e., .6923)56538.5


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here 4.7272= . This value ofH is however, still not significant at the0.05 level .

12.9: Exercises on the KruskalWallis H Test12.9.1: Three types of drugs, A, B, and C that are believed to lead to

the suppression of opportunistic infection associated with HIV were

clinically tested on volunteer HIV positive patients over a period of two

years. During the two-year period of observation, the number of times

that the patients reported symptoms that were known to be associated

with HIV were recorded. The following data were gathered over the

two-year period of observation.

HIV Treatment with DrugA B C

Patient

No.

No. of HIV-

related

symptoms

reported

Patient

No.

No. of

HIV-

related

symptoms

reported

Patient

No.

No. of

HIV-

related

symptoms

reported

1

2

3

4

20

28

18

40

1

2

3

4

25

28

30

50

1

2

3

10

15

12

Was there a difference in the efficacy levels of the three drugs in the

suppression of opportunistic infections associated with HIV?

12.9.2: Twentyone (21) children with reading disabilities in Englishwere classified by a specialist in Special Education into mild, moderate,

and severe forms of the disability based on an instrument for measuring

reading skill that measured the number of errors (incorrect

pronunciations, improper intonations, etc.) in the reading of a standard

English passage. The following data were recorded for the three groups

of children.

Assuming that the errors recorded represent at best measurement on an

ordinal scale, can the categorization by the specialist be considered valid

in the light of the above data?

Mild Disability Moderate Disability Severe Disability

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Group Group Group

Child

No.

No. of

Errors

Child No. No. of

Errors

Child

No.

No. of

Errors

1

23

4

5

6

10

1211

13

8

7

1

23

4

5

6

7

14

1318

19

22

24

26

1

23

4

5

6

7

8

20

2627

28

32

35

35

38

222

Tutorials in Statistics-chapter 12 New

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