1.
The Kruskal-Wallis H Test
• The Kruskal-Wallis H Test is a
nonparametric procedure that can be used to
compare more than two populations in a
completely randomized design.
• All n = n1+n2+…+nk measurements are jointly
ranked (i.e.treat as one large sample).
• We use the sums of the ranks of the k samples
to compare the distributions.
2.
The Kruskal-Wallis H Test
Here are the steps in doing the Kruskal- Wallis test:
Here are the steps in doing the Kruskal- Wallis test:
a. State the null hypothesis.
a. State the null hypothesis.
Ho: The kk population distributions are
Ho: The population distributions are
identical.
identical.
b. State the alternative.
b. State the alternative.
Ha: at least two of the kk population
Ha: at least two of the population
distributions differ.
distributions differ.
3.
Example
Four groups of students were randomly
assigned to be taught with four different
techniques, and their achievement test scores
were recorded. Are the distributions of test
scores the same, or do they differ in location?
1
2
3
4
65
75
59
94
87
69
78
89
73
83
67
80
79
81
62
88
4.
Teaching Methods
1
65
2
(3) 75
87 (13) 69
3
4
(7) 59(1)
94 (16)
(5) 78 (8) 89 (15)
73
79
Ti
(6) 83 (12) 67 (4) 80 (10)
(9) 81 (11) 62 (2) 88 (14)
31
35
15
55
Rank the 16
H00:the distributions of scores are the same
Rank the 16
H : the distributions of scores are the same
measurements
Ha::the distributions differ in location
measurements
Ha the distributions differ in location
from 1 to 16,
from 1 to 16,
12
Ti 2
and calculate
and calculate
Test statistic: H =
∑
− 3(n + 1)
n(n + 1) ni
the four rank
the four rank
sums.
sums.
12 312 + 352 + 152 + 552
− 3(17) = 8.96
=
16(17)
4
5.
Teaching Methods
H00:the distributions of scores are the same
H : the distributions of scores are the same
Ha::the distributions differ in location
H the distributions differ in location
a
12
Ti 2
Test statistic: H =
∑
− 3(n + 1)
n(n + 1) ni
12 312 + 352 + 152 + 552
− 3(17) = 8.96
=
16(17)
4
Rejection region: For aarightRejection region: For righttailed chi-square test with α = ..
tailed chi-square test with α =
05 and df = 4-1 =3, reject H00if H
05 and df = 4-1 =3, reject H if H
≥ 7.81.
≥ 7.81.
Reject H00.There is sufficient
Reject H . There is sufficient
evidence to indicate that there
evidence to indicate that there
is aadifference in test scores for
is difference in test scores for
the four teaching techniques.
the four teaching techniques.
6.
Activity
Four different teaching techniques in Physics were compared with one
another. Four classes were randomly assigned to undergo one of the four
teaching techniques. After being taught for one month, the students were
given in achievement test. The ff. are their achievement scores.
Teaching Technique
1
2
3
4
65
75
59
94
87
69
78
89
73
83
67
80
79
81
62
88
81
72
83
95
69
79
76
90
Test the null hypothesis that four teaching techniques in Physics do not
differ in effectiveness.
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