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

$x=5.574$

$y=14.492$

$x^2=34.361$

$y^2=213.639$

$xy=83.148$

$n=61$

$\Sigma x=5.574\times61=340.014$

$\Sigma y=14.492\times61=884.012$

$\Sigma x^2=34.361\times61=2096 .021$

$\Sigma y^2=213.639\times61=13031.979$

$\Sigma xy=83.148\times61=5072.028$

Following formula is used to calculate sample linear correlation coefficient (r):

$r=\frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2-(\Sigma x)^2][n\Sigma y^2-(\Sigma y)^2]}}$

$r=\frac{61(5072.028)-(340.014)(884.012)}{\sqrt{[(61\times2096.021)-(340.014)^2][(61\times13031.028)-(884.012)^2]}}$

$r=0.69$

To check its significance on level 0.05, there is need to find t-statistic and critical value.

$t=\frac{r\times\sqrt{n-2}}{\sqrt{1-r^2}}$

$t=\frac{0.69\times\sqrt{61-2}}{\sqrt{1-0.69^2}}=7.32$

Degree of freedom = n - 1 = 61- 1 = 59

As per t distribution table, critical value is respect to degree of freedom of 59 and 0.05 level of significance is 1.671.

Since, t-statistic is greater than critical value. This mean simple linear regression is statistically significant.