7/28/2019 Tutorials in Statistics-chapter 12 New
1/18
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
7/28/2019 Tutorials in Statistics-chapter 12 New
2/18
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:
206
7/28/2019 Tutorials in Statistics-chapter 12 New
3/18
( ))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
207
7/28/2019 Tutorials in Statistics-chapter 12 New
4/18
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).
Group 1 Group 2 Group 3
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(
7/28/2019 Tutorials in Statistics-chapter 12 New
5/18
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)
Group 1 Group 2 Group 3
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
209
7/28/2019 Tutorials in Statistics-chapter 12 New
6/18
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).
Group 1 Group 2 Group 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++
210
7/28/2019 Tutorials in Statistics-chapter 12 New
7/18
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
211
7/28/2019 Tutorials in Statistics-chapter 12 New
8/18
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.
212
7/28/2019 Tutorials in Statistics-chapter 12 New
9/18
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
7/28/2019 Tutorials in Statistics-chapter 12 New
10/18
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
214
7/28/2019 Tutorials in Statistics-chapter 12 New
11/18
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
215
7/28/2019 Tutorials in Statistics-chapter 12 New
12/18
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)
216
7/28/2019 Tutorials in Statistics-chapter 12 New
13/18
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
7/28/2019 Tutorials in Statistics-chapter 12 New
14/18
)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=
218
7/28/2019 Tutorials in Statistics-chapter 12 New
15/18
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.
219
7/28/2019 Tutorials in Statistics-chapter 12 New
16/18
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
7/28/2019 Tutorials in Statistics-chapter 12 New
17/18
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
221
7/28/2019 Tutorials in Statistics-chapter 12 New
18/18
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
Top Related