# As part of the study of the psychobiological correlates of success in athletes, the following measurements are obtained from members of Nepal national football team. Anger 6 7 5 21 13 5 13 14 Vigor 30 23 29 22 19 19 28 19 Calculate Spearman’s rank correlation coefficient.

Solution:

 Anger(x) Vigor(y) Rank of x(R1) Rank of x(R2) d = R1 + R2 d2 6 30 3 8 -5 25 7 23 4 5 -1 1 5 29 1.5 7 -5.5 30.25 21 22 8 4 4 16 13 19 5.5 2 3.5 12.25 5 19 1.5 2 -0.5 0.25 13 28 5.5 6 -0.5 0.25 14 19 7 2 5 25 Σd = 0 Σd2 = 110

Now,

$$R = 1-\frac{6\left \{ \sum d^2 + \frac{m_1(m_1^2 – 1)}{12} + \frac{m_2(m_2^2 – 1)}{12} + \frac{m_3(m_3^2 – 1)}{12} \right \}}{n(n^2 – 1)}$$

Here,

m1 = 2 (5 is repeated 2 times)

m2 = 2 (13 is repeated 2 times)

m3 = 3 (19 is repeated 3 times)

$$= 1-\frac{6\left \{ 110 + \frac{2(2^2 – 1)}{12} + \frac{2(2^2 – 1)}{12} + \frac{3(3^2 – 1)}{12} \right \}}{8(8^2 – 1)}$$

$$= 1-\frac{6\left \{ 110 + 0.5 + 0.5 + 23 \right \}}{504}$$

= -0.35

Hence, Sperman’s rank correlation cofficient is -0.35