Question

A simple random sample of 10 paired values (x,y) yields the following statistical calculations: ∑x=108, ∑y=138, ∑(x2) =1249, ∑(y2) =2280, ∑(xy) =1676. What is the linear correlation coefficient?

Answer 1

The linear correlation coefficient is $r=1.054$

Explanation:

It is given that $\Sigma x=108$,  $\Sigma y=138$ , $\Sigma x^{2} =1249$ , $\Sigma y^{2} =2280$ and $\Sigma(x y)=1676$

Also, the random sample is $n=10$

The formula to determine the correlation coefficient is given by

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

Substituting the values in the formula, we have,

$r=\frac{10(1676)-(108)(138)}{\sqrt{\left[10(1249)-(108)^{2}\right]\left[10(2280)-(138)^{2}\right]}}$

Simplifying the values, we get,

$r=\frac{16760-14904}{\sqrt{\left[12490-11664\right]\left[22800-19044\right]}}$

Subtracting the values in both numerator and denominator, we have,

$r=\frac{1856}{\sqrt{\left[826\right]\left[3756\right]}}$

Multiplying the denominator,

$r=\frac{1856}{\sqrt{3102456}}$

Simplifying, we have,

$r=\frac{1856}{1761.4}$

Dividing, we get,

$r=1.054$

Thus, the linear correlation coefficient is $r=1.054$