Question

For a sample of volume 61, average values were found x = 5.574 , y =14.492 , 2 x = 34.361, 2 y = 213.639, xy = 83.148 . Find sample linear correlation coefficient and check its significance on level 0.05.

Answer 1

Solution:

Given,

$n=61,\Sigma x = 5.574 , \Sigma y =14.492 , \Sigma x^2 = 34.361, \Sigma y^2 = 213.639,\Sigma xy = 83.148$

$r=\dfrac{n \Sigma xy-(\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2-(\Sigma x)^2][n\Sigma y^2-(\Sigma y)^2]}} \\ =\dfrac{61 \times 83.148-(5.574)(14.492)}{\sqrt{[61\times 34.361-(5.574)^2][61\times 213.639-(14.492)^2]}} \\=\dfrac{4991.249592}{5145.554175} \\=0.97$

From online calculator of R,

Or we can use R table (Correlation table) to get:

Critical values for r$=\pm0.213$

Our obtained value of r is 0.97 which does not lie in this interval.

So, it falls in rejection region.

Thus, r=0.97 is significant.