Question

In a contest the competitors are awarded marks out of $20$ by two judges. The scores of the $10$ competitors are given below. Calculate Spearman's rank correlation.

Competitors	$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$	$I$	$J$
Judge $A$	$2$	$11$	$11$	$18$	$6$	$5$	$8$	$16$	$13$	$15$
Judge $B$	$6$	$11$	$16$	$9$	$14$	$2$	$4$	$3$	$13$	$17$

Answer 1

Competitors	Judge $A$ Score	$R_1$	Judge $B$ Score	$R_2$	Difference $d=R_1-R_2$	$d^2$
$A$	$2$	$10$	$6$	$8$	$2$	$4$
$B$	$11$	$5.5$	$11$	$6$	$-0.5$	$0.25$
$C$	$11$	$5.5$	$16$	$3$	$2.5$	$6.25$
$D$	$18$	$1$	$9$	$7$	$-6$	$36$
$E$	$6$	$8$	$14$	$4$	$4$	$16$
$F$	$5$	$9$	$20$	$1$	$8$	$64$
$G$	$8$	$7$	$4$	$9$	$-2$	$4$
$H$	$16$	$2$	$3$	$10$	$-8$	$64$
$I$	$13$	$4$	$13$	$5$	$-1$	$1$
$J$	$15$	$3$	$17$	$2$	$1$	$1$
						$\sum d^2=196.5$

As $B$ and $C$ has same job rating of $5$, their average rank is $\frac{5+6}{2}=5.5$.
As there is one tie in ranks $R_1$ of 2 items, there is one correction in Spearman's rank correlation.
$r=1-\frac{6\left[\begin{array}{c}\sum d^2+\frac{1}{12}(m^3-m)\end{array}\right]}{n(n^2-1)}$

$=1-\frac{6[196.5+\frac{1}{12}(2^3-2\left)\right]}{10({10}^2-1)}$

$=1-\frac{6[196.5+\frac{1}{12}(8-2\left)\right]}{10(100-1)}$

$=1-\frac{6(196.5+0.5)}{990}$

$=1-\frac{6\left(197\right)}{990}=1-\frac{1182}{990}$

$=1-1.193=-0.1939$

$=-0.194$