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:
\[
\begin{aligned}
X &: 6, 2, 4, 9, 1, 3, 5, 8 \\
Y &: 13, 8, 12, 15, 9, 10, 11, 16
\end{aligned}
\]
Using the assumed means:
\[
u_i = X_i – 5, \quad v_i = Y_i – 12
\]
Tabular Data for Calculation
\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
X_i & Y_i & u_i = X_i – 5 & v_i = Y_i – 12 & u_i v_i & u_i^2 & v_i^2 \\ \hline
6 & 13 & 1 & 1 & 1 & 1 & 1 \\
2 & 8 & -3 & -4 & 12 & 9 & 16 \\
4 & 12 & -1 & 0 & 0 & 1 & 0 \\
9 & 15 & -2 & 3 & 12 & 16 & 9 \\
1 & 9 & 4 & -3 & 12 & 16 & 9 \\
3 & 10 & -4 & -2 & 4 & 4 & 4 \\
5 & 11 & 0 & -1 & 0 & 0 & 1 \\
8 & 16 & 3 & 4 & 12 & 9 & 16 \\ \hline
\textbf{Sum} & — & \textbf{-2} & \textbf{-2} & \textbf{53} & \textbf{56} & \textbf{56} \\ \hline
\end{array}
\]
Formula for Karl Pearson’s Correlation Coefficient
The formula is given by:
\[
r(X, Y) = \frac{n \Sigma u_i v_i – (\Sigma u_i)(\Sigma v_i)}{\sqrt{\left[n \Sigma u_i^2 – (\Sigma u_i)^2\right]\left[n \Sigma v_i^2 – (\Sigma v_i)^2\right]}}
\]
Solution
Substituting the values:
\[
\begin{aligned}
n &= 8, \quad \Sigma u_i v_i = 53, \quad \Sigma u_i^2 = 56, \quad \Sigma v_i^2 = 56, \\
\Sigma u_i &= -2, \quad \Sigma v_i = -2
\end{aligned}
\]
Final Calculation
\[
r(X, Y) = \frac{8(53) – (-2)(-2)}{\sqrt{\left[8(56) – (-2)^2\right]\left[8(56) – (-2)^2\right]}}
\]
Simplifying:
\[
r(X, Y) = \frac{420}{\sqrt{444 \cdot 444}} \approx 0.946
\]
PDF Presentation
corelexample2Video
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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