Research Methodology - Dr. Krishan K. Pandey

Testing of Hypotheses-II 303
where n = number of paired observations.
The value of Spearman’s rank correlation coefficient will always vary between ±1 , +1, indicating
a perfect positive correlation and –1 indicating perfect negative correlation between two variables.
All other values of correlation coefficient will show different degrees of correlation.
Suppose we get r = 0.756 which suggests a substantial positive relationship between the concerning
two variables. But how we should test this value of 0.756? The testing device depends upon the
value of n. For small values of n (i.e., n less than 30), the distribution of r is not normal and as such
we use the table showing the values for Spearman’s Rank correlation (Table No. 5 given in Appendix
at the end of the book) to determine the acceptance and rejection regions. Suppose we get r = 0.756
for a problem where n = 15 and want to test at 5% level of significance the null hypothesis that there
is zero correlation in the concerning ranked data. In this case our problem is reduced to test the null
hypothesis that there is no correlation i.e., u r = 0 against the alternative hypothesis that there is a
correlation i.e., μ r ≠ 0 at 5% level. In this case a two-tailed test is appropriate and we look in the
said table in row for n = 15 and the column for a significance level of 0.05 and find that the critical
values for r are ±05179 . i.e., the upper limit of the acceptance region is 0.5179 and the lower limit
of the acceptance region is –0.5179. And since our calculated r = 0.756 is outside the limits of the
acceptance region, we reject the null hypothesis and accept the alternative hypothesis that there is a
correlation in the ranked data.
In case the sample consists of more than 30 items, then the sampling distribution of r is
approximately normal with a mean of zero and a standard deviation of 1/ n − 1 and thus, the
standard error of r is:
σ r
=
1
n − 1
We can use the table of area under normal curve to find the appropriate z values for testing
hypotheses about the population rank correlation and draw inference as usual. We can illustrate it, by
an example.
Illustration 8
Personnel manager of a certain company wants to hire 30 additional programmers for his corporation.
In the past, hiring decisions had been made on the basis of interview and also on the basis of an
aptitude test. The agency doing aptitude test had charged Rs. 100 for each test, but now wants
Rs. 200 for a test. Performance on the test has been a good predictor of a programmer’s ability and
Rs. 100 for a test was a reasonable price. But now the personnel manager is not sure that the test
results are worth Rs. 200. However, he has kept over the past few years records of the scores
assigned to applicants for programming positions on the basis of interviews taken by him. If he
becomes confident (using 0.01 level of significance) that the rank correlation between his interview
scores and the applicants’ scores on aptitude test is positive, then he will feel justified in discontinuing
the aptitude test in view of the increased cost of the test. What decision should he take on the basis
of the following sample data concerning 35 applicants?