Answer to Question #157783 in Statistics and Probability for Ashutosh

Question

Calculate Karl Pearson’s Coefficient of Correlation X 10 10 11 12 12 y 5 6 4 3

Answer 1

The formula for Karl's Pearson's cCoefficient is:

$r=\frac{n\sum_{i=0}^{n}x_iy_i-(\sum_{i=0}^{n}x_i)(\sum_{i=0}^{n}y_i)}{\sqrt{[n\sum_{i=0}^{n}x^2_i-(\sum_{i=0}^{n}x_i)^2][n\sum_{i=0}^{n}y^2_i-(\sum_{i=0}^{n}y_i)^2]}}$

Since there are 4 values fo y and 5 for x, we assume that the last value is for y 0.

We set n=5 and receive:

$n\sum_{i=0}^nx_iy_i=5\cdot(10\cdot5+10\cdot6+11\cdot4+12\cdot3+12\cdot0)=950$

$(\sum_{i=0}^{n}x_i)(\sum_{i=0}^{n}y_i)=(10+10+11+12+12)(5+6+4+3+0)=55\cdot18=990$

$n\sum_{i=0}^{n}x^2_i-(\sum_{i=0}^{n}x_i)^2=5(10^2+10^2+11^2+12^2+12^2)-(10+10+11+12+12)^2=$

$=20$

$n\sum_{i=0}^{n}y^2_i-(\sum_{i=0}^{n}y_i)^2=5(5^2+6^2+4^2+3^2+0^2)-(5+6+4+3+0)^2=106$

$r=\frac{950-990}{\sqrt{20\cdot106}}=-0.8687$

Answer:-0.8687