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Question

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


Solution

$$P = 1 - 6 \dfrac{\sum di^2}{n(n^2 - 1)}$$
$$0.6 = 1 - \dfrac{6 \times 66}{n(n^2 - 1)}$$
$$\dfrac{6 \times 66}{n(n^2 - 1)} = 0.4 $$
$$n(n^2 -1) = \dfrac{3}{2} \times 66 \times 10 = 990$$
$$n = 10 $$
$$10(10^2 - 1) = 10 (99) = 990$$


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