Question

Calculate Karl Pearson's coefficient of correlation between $x$ and $y$ for the following data and interpret the result:
$(1,6),(2,5),(3,7),(4,9),(5,8),(6,10),(7,11),(8,13),(9,12)$

Answer 1

Let the assumed mean for the first variate $x$ be $5$ and that for the second variate $y$ be $10$.
Here $n=9$

$x$	$y$	$u=x-5$	$v=y-10$	$u^2$	$v^2$	$uv$
$1$	$6$	$-4$	$-4$	$16$	$16$	$16$
$2$	$5$	$-3$	$-5$	$9$	$25$	$15$
$3$	$7$	$-2$	$-3$	$4$	$9$	$6$
$4$	$9$	$-1$	$-1$	$1$	$1$	$1$
$5$	$8$	$0$	$-2$	$0$	$4$	$0$
$6$	$10$	$1$	$0$	$1$	$0$	$0$
$7$	$11$	$2$	$1$	$4$	$1$	$2$
$8$	$13$	$3$	$3$	$9$	$9$	$9$
$9$	$12$	$4$	$2$	$16$	$4$	$8$
		$\sum u=0$	$\sum v=-9$	$\sum u^2=60$	$\sum v^2=69$	$\sum uv=57$

$\rho (x,y)=\frac{\sum uv-\frac{1}{n}(\sum u)(\sum v)}{\sqrt{(\sum u^2-\frac{{(\sum u)}^2}{n}).\sqrt{(\sum v^2-\frac{{(\sum v)}^2}{n})}}}$
$=\frac{57-\frac{1}{9}\left(0\right)(-9)}{\sqrt{(60-\frac{{\left(0\right)}^2}{9})}\sqrt{(60-\frac{{(-9)}^2}{9})}}$
$=\frac{57}{\sqrt{60}\sqrt{60}}=\frac{57}{60}=0.95$