Question

From the following data, determine Karl Pearson's coefficient of correlation between X and Y series for 15 paris:

	X-Series	Y-Series
Mean	80	120
Sum of Squares of deviation from Arithmetic Mean	56	156
Sum of product of deviation of X and Y from their respective Means	92

Answer 1

$$\begin{gathered} \underline{\mathit{G}\mathit{i}\mathit{v}\mathit{e}\mathit{n}\mathit{:}} \\ N\mathit{}\mathit{=}15 \\ \overline{X\mathit{}}\mathit{=}80 \\ \overline{Y}\mathit{}\mathit{=}\mathit{}120 \\ \Sigma x^{\mathit{2}}\mathit{=}56 \\ \Sigma y^2\mathit{=}156 \\ \Sigma xy\mathit{=}92 \\ {\sigma }_x\mathit{=}\sqrt{\frac{\Sigma x^{\mathit{2}}}{N}}\mathit{,}{\sigma }_y\mathit{=}\sqrt{\frac{\Sigma y^{\mathit{2}}}{N}} \\ {\sigma }_x\mathit{=}\sqrt{\frac{56}{15}}=1.93,{\sigma }_y\mathit{=}\sqrt{\frac{156}{15}}\mathit{=}3.22 \\ r\mathit{=}\frac{\Sigma xy}{N{\sigma }_x{\sigma }_y}\mathit{=}\frac{92}{15\times 1.93\times 3.22}\mathit{}\mathit{=}0.98 \end{gathered}$$
Hence, Karl Pearson's coefficient of correlation is +0.98