Question

Following are the marks of $10$ students obtained in Physics and Chemistry in an examination. Find the rank-correlation coefficient.

x	43	96	74	38	35	43	22	56	35	80
y	30	94	84	13	30	18	30	41	48	95

• A. $0.3697$
• B. $0.4673$
• C. $0.6303$
• D. $0.7834$

Answer 1

Answer: (C)

$0.6303$

Let the ranks of students obtained in Physics be $x$ and the ranks of students obtained in Chemistry be $y$.

$X$	$Y$	Rank $X$ $\left(x\right)$	Rank $Y$ $\left(y\right)$	$d=x-y$	$d^2$
$43$	$30$	$5.5$	$7$	$-1.5$	$2.25$
$96$	$94$	$1$	$2$	$-1$	$1$
$74$	$84$	$3$	$3$	$0$	$0$
$38$	$13$	$7$	$10$	$-3$	$9$
$35$	$30$	$8.5$	$7$	$1.5$	$2.25$
$43$	$18$	$5.5$	$9$	$-3.5$	$12.25$
$22$	$30$	$10$	$7$	$3$	$9$
$56$	$41$	$4$	$5$	$-1$	$1$
$35$	$48$	$8.5$	$4$	$4.5$	$20.25$
$80$	$95$	$2$	$1$	$1$	$1$
			$\sum$	$0$	$58$

In the $X$ series $43$ has repeated twice and given ranks $5.5$ instead of $5$ and $6$. For this the correction factor is $\frac{2(4-1)}{12}=\frac{1}{2}$.

Also $35$ has repeated twice and given ranks $8.5$ instead of $8$ and $9$. For this the correction factor is $\frac{2(4-1)}{12}=\frac{1}{2}$.

In the $Y$ series $30$ has repeated thrice and given ranks $7$ instead of $6,7,8$. For this the correction factor is $\frac{3(9-1)}{12}=2$.

So the total correction factors $C.F=\frac{1}{2}+\frac{1}{2}+2=3$

The rank correlation coefficient is given by

$r=1-\frac{6(\sum d^2-C.F)}{n(n^2-1)}$

$=1-\frac{6(58+3)}{10(100-1)}$

$=1-\frac{276}{10\times 99}$

$=1-\frac{366}{990}$

$=1-0.3696$

$=0.6303$

Therefore the rank correlation coefficient is $0.6303$