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If the Rank Correlation Coefficient is 0.6 and the Sum of Squares of Differences of Ranks is 66, - Mathematics and Statistics

Sum

If the rank correlation coefficient is 0.6 and the sum of squares of differences of ranks is 66, then find the number of pairs of observations.

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Solution

 Given : R = 0 .6, `sum "d"_1^2` = 66 

R = 1 - `(6 sum "d"_1^2)/("n"("n"^2 - 1))`

`0.6 = 1 - (6 sum "d"_1^2)/("n"("n"^2 - 1))`

`0.4 = (6 xx 66)/("n"("n"^2 - 1))`

`"n"("n"^2 - 1) = (6 xx 66)/0.4`

`= (6 xx 66 xx 10)/4`

= 15 × 66 = 990

n (n - 1)(n + 1) = 990

= 10 × 9 × 11

∴ n = 10

Concept: Rank Correlation
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