The following are the final examination marks of three group of students who were taught computer by three different methods.

Method I 94 88 91 74 87 97
Method II 85 82 79 84 61 72 80
Method III 89 67 72 76 69

Are all three methods equally effective? Use H tests at 5% level of significance.

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Solution:

K = 3,     n1 = 6,     n2 = 7     and     n3 = 5

Therefore, N = n1 + n2 + n3 = 6 + 7 + 5 = 18

Problem: To Test

H0All three method are equally effective

H1All three methods are not equally effective

Calculation: Here,

Method Marks Ri Ri2 \(\frac{R_i^2}{n_i}\)
I 94 88 91 74 87 97
Rank 17 14 16 6 13 18 85 7225 1204.17
II 85 92 79 84 61 72 80
Rank 12 10 8 11 1 4.5 9 55.5 3080.25 440.04
III 89 67 72 76 69
Rank 15 2 4.5 7 3 31.5 992.25 198.45
Total \(\sum{\frac{R_i^2}{n_i}}\) = 1842.66

The two tied ranks of group 4.5, T = t3 – t = 4.53 – 4.5 = 86.626

Therefore, ΣT = 86.626

Test Statistics:

Under H0, test statistics in case of third observation is given by,

\(H = \frac{ \frac{12}{N(N + 1)} \sum_{i=1}^{k} \frac{R_i^2}{n_i} – 3(N + 1) }{1 – \frac{\sum{T}}{N^3 – N}}\)

\(= \frac{ \frac{12}{18(18 + 1)} \times 1842.66 – 3(18 + 1) }{1 – \frac{86.625}{18^3 – 18}}\)

\(= \frac{64.65 – 57}{0.985}\)

= 7.77

Critical Value:

Critical Value at 0.05 level of significance for 2 degree of freedom is 5.99

Decision:

H = 7.77 > X20.05, 2 = 5.99. So, We reject H0. It means that all three methods are not equally effective.

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