Question

The values of the same 15 students in two subjects A and B are given below; the two numbers within the brackets denoting the ranks of the same student in A and B respectively.

(1,10) (2,7) (3,2) (4,6) (5,4) (6,8) (7,3) (8,1). (9,11) (10,15) (11,9) (12,5) (13,14) (14,12) (15,13)

Use Spearman’s formula to find the rank Correlation Coefficient.

Answer 1

Explanation:

The values of the same 15 students in two subjects A and B are given below; the two numbers within the brackets denoting the ranks of the same student in A and B respectively.

Answer 2

Answer:

The rank correlation coefficient of the given data is 0.51.

Explanation:

Given the number of students, $n=15$

And the ranks attained by 15 students in two subjects A and B

$\begin{array}{ccccccccccccccccccc}A&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\B&10&7&2&6&4&8&3&1&11&15&9&5&14&12&13\end{array}$

Following is the tabulated data of the ranks of 15 students in two subjects along with the difference in ranks and square of the difference.

$\begin{array}{ccccc}S.No.&R_{A}~&~~R_{B}&d=R_{A} -R_{B} &d^{2} &\\1&1&10&-9&81\\2&2&7&-5&25\\3&3&2&1&1\\4&4&6&-2&4\\5&5&4&1&1\\6&6&8&-2&4\\7&7&3&4&16\\8&8&1&7&49\\9&9&11&-2&4\\10&10&15&-5&25\\11&11&9&2&4\\12&12&5&7&49\\13&13&14&-1&1\\14&14&12&2&4\\15&15&13&2&4\\\end{array}$

The Spearman’s rank correlation coefficient for the observations with the difference between the ranks in each pair of data is given by  

$r=1-\frac{6\Sigma d^{2} }{n(n^{2}-1 )}$

$\Sigma d^{2}$ is the sum of all the $d^{2}$ values =272.

Substituting in the formula,

$\rho=1-\frac{6\times 272 }{15(15^{2}-1 )}$

$=1-\frac{ 1632 }{15(225-1 )}$

$=1-\frac{ 1632 }{15\times224}$

$=1-\frac{1632}{3360}\approx0.51$

Hence, the rank correlation coefficient between the ranks scored by 15 students in two subjects is 0.51.