Examples of Karl Pearson’s Coefficient of Correlation
Example 1:
Determine the Coefficient of Correlation between X and Y.
The summation of the product of deviations of Series X and Y from their respective means is 200.
Solution:
The figures given are:
N = 30, σx = 4, σy = 3, and ∑xy = 200
[Tex]r=\frac{\sum{xy}}{N\times{\sigma_x}\times{\sigma_y}} [/Tex]
[Tex]=\frac{200}{30\times4\times3}=\frac{50}{90}=0.5 [/Tex]
Coefficient of Correlation = 0.5
It means that there is a positive correlation between X and Y.
Example 2:
If the Covariance between two variables X and Y is 9.4 and the variance of Series X and Y are 10.6 and 12.5, respectively, then calculate the coefficient of correlation.
Solution:
Covariance between X and Y = [Tex]\frac{\sum{xy}}{N}=9.4 [/Tex]
Variance of X = σx2 = 10.6
[Tex]\sigma_x=\sqrt{10.6}=3.25 [/Tex]
Variance of Y = σy2 = 12.5
[Tex]\sigma_y=\sqrt{12.5}=3.53 [/Tex]
[Tex]r=\frac{\sum{xy}}{N\times{\sigma_x}\times{\sigma_y}} [/Tex]
[Tex]r=\frac{\sum{xy}}{N}\times{\frac{1}{\sigma_x}}\times{\frac{1}{\sigma_y}} [/Tex]
[Tex]=9.4\times{\frac{1}{3.25}}\times{\frac{1}{3.53}} [/Tex]
r = 9.4 x 0.307 x 0.282 = 0.816
Coefficient of Correlation = 0.816
It means that there is quite a high degree of positive correlation between X and Y.