Karl Pearson’s Correlation Coefficient

Learn the step-by-step process of finding the correlation coefficient in statistics.

Problem Statement

Find the Karl Pearson’s coefficient of correlation between $X$ and $Y$ for the given data:

$$\begin{array}{rl}X& :6,2,4,9,1,3,5,8\\ Y& :13,8,12,15,9,10,11,16\end{array}$$

Using the assumed means:

$$u_i=X_i–5,v_i=Y_i–12$$

Tabular Data for Calculation

$$\begin{array}{|ccccccc|}\hline X_i& Y_i& u_i=X_i–5& v_i=Y_i–12& u_i{v}_i& u_i^2& v_i^2\\ 6& 13& 1& 1& 1& 1& 1\\ 2& 8& -3& -4& 12& 9& 16\\ 4& 12& -1& 0& 0& 1& 0\\ 9& 15& -2& 3& 12& 16& 9\\ 1& 9& 4& -3& 12& 16& 9\\ 3& 10& -4& -2& 4& 4& 4\\ 5& 11& 0& -1& 0& 0& 1\\ 8& 16& 3& 4& 12& 9& 16\\ \mathbf{\text{Sum}}& —& \mathbf{\text{-2}}& \mathbf{\text{-2}}& \mathbf{\text{53}}& \mathbf{\text{56}}& \mathbf{\text{56}}\\ \hline\end{array}$$

Formula for Karl Pearson’s Correlation Coefficient

The formula is given by:

$$r(X,Y)=\frac{n\mathrm{\Sigma }{u}_i{v}_i–(\mathrm{\Sigma }{u}_i)(\mathrm{\Sigma }{v}_i)}{\sqrt{[n\mathrm{\Sigma }{u}_i^2–(\mathrm{\Sigma }{u}_i{)}^2][n\mathrm{\Sigma }{v}_i^2–(\mathrm{\Sigma }{v}_i{)}^2]}}$$

Solution

Substituting the values:

$$\begin{array}{rl}n& =8,\mathrm{\Sigma }{u}_i{v}_i=53,\mathrm{\Sigma }{u}_i^2=56,\mathrm{\Sigma }{v}_i^2=56,\\ \mathrm{\Sigma }{u}_i& =-2,\mathrm{\Sigma }{v}_i=-2\end{array}$$

Final Calculation

$$r(X,Y)=\frac{8(53)–(-2)(-2)}{\sqrt{[8(56)–(-2{)}^2][8(56)–(-2{)}^2]}}$$

Simplifying:

$$r(X,Y)=\frac{420}{\sqrt{444\cdot 444}}\approx 0.946$$

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Conclusion

The Karl Pearson’s coefficient of correlation is approximately 0.946, indicating a strong positive correlation between $X$ and $Y$.

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