The linear correlation coefficient is [tex]r=1.054[/tex]
Explanation:
It is given that [tex]$\Sigma x=108$[/tex], [tex]$\Sigma y=138$[/tex] , [tex]$\Sigma x^{2} =1249$[/tex] , [tex]$\Sigma y^{2} =2280$[/tex] and [tex]$\Sigma(x y)=1676$[/tex]
Also, the random sample is [tex]n=10[/tex]
The formula to determine the correlation coefficient is given by
[tex]$r=\frac{n\left(\sum x y\right)-\left(\sum x\right)(\Sigma y)}{\sqrt{\left[n \sum x^{2}-\left(\sum x\right)^{2}\right]\left[n \Sigma y^{2}-(\Sigma y)^{2}\right]}}$[/tex]
Substituting the values in the formula, we have,
[tex]$r=\frac{10(1676)-(108)(138)}{\sqrt{\left[10(1249)-(108)^{2}\right]\left[10(2280)-(138)^{2}\right]}}$[/tex]
Simplifying the values, we get,
[tex]$r=\frac{16760-14904}{\sqrt{\left[12490-11664\right]\left[22800-19044\right]}}$[/tex]
Subtracting the values in both numerator and denominator, we have,
[tex]$r=\frac{1856}{\sqrt{\left[826\right]\left[3756\right]}}$[/tex]
Multiplying the denominator,
[tex]$r=\frac{1856}{\sqrt{3102456}}$[/tex]
Simplifying, we have,
[tex]$r=\frac{1856}{1761.4}$[/tex]
Dividing, we get,
[tex]r=1.054[/tex]
Thus, the linear correlation coefficient is [tex]r=1.054[/tex]
