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.

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Solution

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

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Q. If

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then find the rank correlation coefficient.

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i^{t h}

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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.

P=1−6∑di2n(n2−1)0.6=1−6×66n(n2−1)6×66n(n2−1)=0.4n(n2−1)=32×66×10=990n=1010(102−1)=10(99)=990

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.

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