Research Methodology - Dr. Krishan K. Pandey

Testing of Hypotheses-II 287
Table 12.1
By X 1 0 2 3 1 0 2 2 3 0 1 1 4 1 2 1 3 5 2 1 3 2 4 1 3 2 0 2 4 2
By Y 0 0 1 0 2 0 0 1 1 2 0 1 2 1 1 0 2 2 6 0 2 3 0 2 1 0 1 0 1 0
Sign + 0 + + – 0 + + + – + 0 + 0 + + + + – + + – + – + + – + + +
(X – Y)
Total Number of + signs = 20
Total Number of – signs = 6
Hence, sample size = 26
(Since there are 4 zeros in the sign row and as such four pairs are discarded, we are left with
30 – 4 = 26.)
Thus the observed proportion of pluses (or successes) in the sample is = 20/26 = 0.7692 and the
observed proportion of minuses (or failures) in the sample is = 6/26 = 0.2308.
As we are to test the null hypothesis that the two archaeologists X and Y are equally good and if
that is so, the number of pluses and minuses should be equal and as such p = 1/2 and q = 1/2. Hence,
the standard error of proportion of successes, given the null hypothesis and the size of the sample, we
have:
p ⋅ q ×
σ prop. = = = 0. 0981
n 26
Since the alternative hypothesis is that the archaeologists X is better (or p > 1/2), we find one
tailed test is appropriate. This can be indicated as under, applying normal approximation to binomial
distribution in the given case:
p = 1/2
1
2
Fig. 12.2
1
2
0.49 of area
0.7276
p + 2.32 ( prop)
Limit
(Shaded area represents
rejection region)
0.01 of area