Q.Calculate the correlation coefficient from the following data.
$N=9$, $\Sigma X=45$, $\Sigma Y=108$, $\Sigma X^2=285$, $\Sigma Y^2=1356$, $\Sigma XY=597$

Substitute into Karl Pearson's coefficient formula to get $r=\dfrac{513}{540}=0.95$.

Karl Pearson's coefficient of correlation is

$r=\frac{N\Sigma XY-\Sigma X\,\Sigma Y}{\sqrt{N\Sigma X^2-(\Sigma X)^2}\,\sqrt{N\Sigma Y^2-(\Sigma Y)^2}}.$

Given: $N=9,\ \Sigma X=45,\ \Sigma Y=108,\ \Sigma X^2=285,\ \Sigma Y^2=1356,\ \Sigma XY=597$.

Step 1 — Numerator.

$N\Sigma XY-\Sigma X\,\Sigma Y=9(597)-45(108)=5373-4860=513.$

Step 2 — First bracket under the root.

$N\Sigma X^2-(\Sigma X)^2=9(285)-45^2=2565-2025=540.$

Step 3 — Second bracket under the root.

$N\Sigma Y^2-(\Sigma Y)^2=9(1356)-108^2=12204-11664=540.$

Step 4 — Combine. …