Question

Find the Karl Pearson's co-efficient of correlation between $x$ and $y$ for the following data:

$x$	$16$	$18$	$21$	$20$	$22$	$26$	$27$	$15$
$y$	$22$	$25$	$24$	$26$	$25$	$30$	$33$	$14$

Answer 1

Assume mean $A=20$ for the $x-$variate and $B=30$ for the y-variate.

$x$	$u=x-20$	$u^2$	$y$	$\nu =y-30$	${\nu }^2$	$u\nu$
$16$	$-4$	$16$	$22$	$-8$	$64$	$32$
$18$	$-2$	$4$	$25$	$-5$	$25$	$10$
$21$	$1$	$1$	$24$	$-6$	$36$	$-6$
$20$	$0$	$0$	$26$	$-4$	$16$	$0$
$22$	$2$	$4$	$25$	$-5$	$25$	$-10$
$26$	$6$	$36$	$30$	$0$	$0$	$0$
$27$	$7$	$49$	$33$	$3$	$9$	$21$
$15$	$-5$	$25$	$14$	$-16$	$256$	$80$
	$\sum u=5$	$\sum u^2=135$		$\sum \nu =-41$	$\sum {\nu }^2=431$	$\sum u\nu =127$

$\therefore r=\frac{\sum uv-\frac{1}{N}\sum u\sum v}{\sqrt{\sum u^2-\frac{(\sum u{)}^2}{N}}.\sqrt{\sum v^2-\frac{(\sum v{)}^2}{N}}}$

$=\frac{127+\frac{205}{8}}{\sqrt{135-\frac{25}{8}}\sqrt{431-\frac{1681}{8}}}$

$=\frac{127+25.625}{\sqrt{135-3.125}\sqrt{431-210.125}}$

$=\frac{152.625}{\sqrt{131.875}\sqrt{220.875}}$

$=\frac{152.625}{11.48\times 14.86}$

$=\frac{152.625}{170.593}=0.894$