# Calculate Spearman’s rank correlation coefficient for the following ranks given by three judges in a music contest. 1stJudge 2 1 4 6 5 8 9 10 7 3 2nd Judge 4 3 2 5 1 6 8 9 10 7 3rd Judge 5 8 4 7 10 2 1 6 9 3 Indicate which pair of judges has the nearest approaches to music

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Solution:

We know,

$$r_{12} = 1 – \frac{6 \sum d_1^2}{n(n^2 – 1)}$$

$$r_{13} = 1 – \frac{6 \sum d_2^2}{n(n^2 – 1)}$$

$$r_{23} = 1 – \frac{6 \sum d_3^2}{n(n^2 – 1)}$$

Now,

 x y z d1 = x – y d2 = y – z d3 = z – x d12 d22 d32 2 4 5 -2 -1 3 4 1 9 1 3 8 -2 -5 7 4 25 49 4 2 4 2 -2 0 4 4 0 6 5 7 1 -2 1 1 4 1 5 1 10 4 -9 5 16 81 25 8 6 2 2 4 -6 4 16 36 9 8 1 1 7 -8 1 49 64 10 9 6 1 3 -4 1 9 16 7 10 9 -3 1 2 9 1 4 3 7 3 -4 4 0 16 16 0 Σd1 = 0 Σd2 = 0 Σd3 = 0 Σd12 = 60 Σd22 = 206 Σd32 = 204

Now,

r12 = $$1 – \frac{6 \times 60}{10 \times 99}$$ = 0.636

r13 = $$1 – \frac{6 \times 206}{10 \times 99}$$ = -0.248

r23 = $$1 – \frac{6 \times 204}{10 \times 99}$$ = -0.236

Since, r12 > r13 and r23. so, 1st and 2nd judge has nearest approaches to music.

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