Research Methodology - Dr. Krishan K. Pandey

Testing of Hypotheses-II 295
Table 12.5
Size of sample item in Rank Name of related sample:
ascending order [A for sample one and
B for sample two]
32 1 B
38 2 A
39 3 A
40 4 B
41 5 B
44 6.5 B
44 6.5 B
46 8 A
48 9 A
52 10 B
53 11.5 B
53 11.5 A
57 13 A
60 14 A
61 15 B
67 16 B
69 17 A
70 18 B
72 19.5 B
72 19.5 B
73 21.5 A
73 21.5 A
74 23 A
78 24 A
From the above we find that the sum of the ranks assigned to sample one items or R 1 = 2 + 3 + 8 +
9 + 11.5 + 13 + 14 + 17 + 21.5 + 21.5 + 23 + 24 = 167.5 and similarly we find that the sum of ranks
assigned to sample two items or R 2 = 1 + 4 + 5 + 6.5 + 6.5 + 10 + 11.5 + 15 + 16 + 18 + 19.5 + 19.5
= 132.5 and we have n 1 = 12 and n 2 = 12
b g
n
Hence, test statistic 1 n1
+ 1
U = n1 ⋅ n2
+ − R1
2
12 12 1
= bgbg b + g
12 12 +
− 167. 5
2
= 144 + 78 – 167.5 = 54.5
Since in the given problem n and n both are greater than 8, so the sampling distribution of U
1 2
approximates closely with normal curve. Keeping this in view, we work out the mean and standard
deviation taking the null hypothesis that the two samples come from identical populations as under: