Question

Calculate Karl Pearson's coefficient of correlation by actual mean method:

$\begin{array}{cccccc}\text{Price (Rs)}& 10& 12& 14& 16& 18\\ \text{Quatity (Units)}& 20& 29& 21& 22& 28\end{array}$

Answer 1

Let X denotes price and Y denotes quantity

$\begin{array}{ccccccc}\text{X}& \overline{X}=14& (X-\overline{X}{)}^2& Y& \overline{Y}=24& (Y-\overline{Y}{)}^2& (X-\overline{X})(Y-\overline{Y})\\ & X-\overline{X}& & & Y-\overline{Y}& & \\ 10& -4& 16& 20& -4& 16& +16\\ 12& -2& 4& 29& +5& 25& -10\\ 14& 0& 0& 21& -3& 9& 0\\ 16& +2& 4& 22& -2& 4& -4\\ 18& +4& 16& 28& +4& 16& +16\\ \sum X=70& \sum (X-\overline{X})& \sum (X-\overline{X}{)}^2& \sum Y=120& \sum (Y-\overline{Y})& \sum (Y-\overline{Y}{)}^2& \sum (X-\overline{X})(Y-\overline{Y})\\ N=5& =0& =40& & =0& =70& =18\end{array}\overline{X}=\frac{\sum X}{N}=\frac{70}{5}=14;\overline{Y}=\frac{\sum Y}{N}=\frac{120}{5}=24r=\frac{\sum (X-\overline{X})(Y-\overline{Y})}{\sqrt{\sum (X-\overline{X}{)}^2}\sqrt{\sum (Y-\overline{Y}{)}^2}}=\frac{18}{\sqrt{40}\sqrt{70}}=\frac{18}{\sqrt{2800}}=\frac{18}{52.91}=0.34$