Question 146

If the sum of the squares of deviations of ranks of two students in five subjects is 16, then the coefficient of rank correlation is

Solution

Given, Squares of deviation, D^2= 16

No. of subjects, N=5

Now, according to Spearmen's rank correlation

Coeffiecient of rank correlation ,R = 1 - (6 *Summation D^2)/[N(N^2 -1)]

= 1 - (6*16)/[5(25-1)]

= 1 - (6*16/(5*24)

= 1 - (4/5)

=1/5 Ans.

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