MS 95 Question 4

To apply the H test or the Kruskal-Wallis test to this problem, we begin by ranking all the given figures from the highest to the lowest, indicating besides each the number of the method as under.

Marks Obtained Rank Assigned Number of the method

90 1 II
87 2 II
85 3 II
80 4 III
79 5 III
78 6 I
77 7 II
76 8 II
75 9 III
74 10 III
73 11 I
71 12 I
62 13 I
60 14 III
58 15 I
As the 3 samples have 5 items each the sampling distribution approximately closely with X (Chi) square distribution

H0: The teacher equally well with the 3 methods

Now for finding the values of R i , we arrange the above table as under.
Marks obtained with different methods and corresponding rank.

Method I Rank Method II Rank Method III Rank
78 6 90 1 80 4
73 11 87 2 79 5
71 12 85 3 75 9
62 13 77 7 74 10
58 15 76 8 60 14
n1 = 5 R1=57 n2=5 R2=21 n3=5 R3=42

Now we calculate H stastic as under

H = (12/n(n+1)) (Sigma i = 1 to k (Ri square/ni))-3(n+1)

= (12/15(15+1)){(57)*(57)/5 + 21*21/5 + 42*42/5} - 3(15+1)

= 0.05 x 5454/5 - 48
= 54.54 - 48
Therefore H cal =6.54

Now Xsquare tab(3-1,005) = X square tab (2,005) = 5.99

THere X square cal < X square tab

Note : Please note X is Chi in greek, X square is X* X.
COnclusion
There is no difference between teaching methods