# 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.

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.

There may be error in calculation. Please follow the steps only. If you find any error then please report to us.
If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.