Research Methodology - Dr. Krishan K. Pandey

Testing of Hypotheses-II 297
Illustration 6
Two samples with values 90, 94, 36 and 44 in one case and the other with values 53, 39, 6, 24, and 33
are given. Test applying Wilcoxon test whether the two samples come from populations with the
same mean at 10% level against the alternative hypothesis that these samples come from populations
with different means.
Solution: Let us first assign ranks as stated earlier and we get:
Table 12.6
Size of sample item Rank Name of related sample
in ascending order (Sample one as A
Sample two as B)
6 1 B
24 2 B
33 3 B
36 4 A
39 5 B
44 6 A
53 7 B
90 8 A
94 9 A
Sum of ranks assigned to items of sample one = 4 + 6 + 8 + 9 = 27
No. of items in this sample = 4
Sum of ranks assigned to items of sample two = 1 + 2 + 3 + 5 + 7 = 18
No. of items in this sample = 5
As the number of items in the two samples is less than 8, we cannot use the normal curve
approximation technique as stated above and shall use the table giving values of Wilcoxon’s distribution.
To use this table, we denote ‘W s ’ as the smaller of the two sums and ‘W l ’ the larger. Also, let ‘s’ be
the number of items in the sample with smaller sum and let ‘l’ be the number of items in the sample
with the larger sum. Taking these notations we have for our question the following values:
W s = 18; s = 5; W l = 27; l = 4
The value of W s is 18 for sample two which has five items and as such s = 5. We now find the
difference between W s and the minimum value it might have taken, given the value of s. The minimum
value that W s could have taken, given that s = 5, is the sum of ranks 1 through 5 and this comes as
equal to 1 + 2 + 3 + 4 + 5 = 15. Thus, (W s – Minimum W s ) = 18 – 15 = 3. To determine the probability
that a result as extreme as this or more so would occur, we find the cell of the table which is in the
column headed by the number 3 and in the row for s = 5 and l = 4 (the specified values of l are given
in the second column of the table). The entry in this cell is 0.056 which is the required probability of
getting a value as small as or smaller than 3 and now we should compare it with the significance level
of 10%. Since the alternative hypothesis is that the two samples come from populations with different
means, a two-tailed test is appropriate and accordingly 10% significance level will mean 5% in the
left tail and 5% in the right tail. In other words, we should compare the calculated probability with the