Chapter 13, Problem 16CT

To determine

To find the correlation coefficient $r$ .

Answer to Problem 16CT

The correlation of the coefficient $r\approx 0.9959$ .

Explanation of Solution

Given information:

Points $\left(0,2\right),\left(3,4\right),\left(4,5\right)$

Table for the steps to find the least squares regression line using $n=3$ ,

$x$	$y$	$xy$	$x^2$	$y^2$
$0$	$2$	$0$	$0$	$4$
$3$	$4$	$12$	$9$	$16$
$4$	$5$	$20$	$16$	$25$
${\displaystyle \sum _{i=1}^n{x}_i=7}$	${\displaystyle \sum _{i=1}^n{y}_i=11}$	${\displaystyle \sum _{i=1}^n{x}_i{y}_i=32}$	${\displaystyle \sum _{i=1}^n{x}_i{}^2=25}$	${\displaystyle \sum _{i=1}^n{y}_i{}^2=45}$

By using the formula,

  $\begin{array}{l}r=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\left({\displaystyle \sum _{i=1}^n{x}_i}\right)\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}{\sqrt{n{\displaystyle \sum _{i=1}^n{x}_i{}^2-}{\left({\displaystyle \sum _{i=1}^n{x}_i{y}_i}\right)}^2}\sqrt{n{\displaystyle \sum _{i=1}^n{y}_i{}^2-{\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}^2}}}\\ r=\frac{3\left(32\right)-7\left(11\right)}{\sqrt{3\left(25\right)-7^2}\sqrt{3\left(45\right)-{11}^2}}\\ r=\frac{19}{\sqrt{26}\sqrt{14}}\\ r\approx 0.9959\end{array}$

  $r$ is approximately near to $1$ , therefore we can conclude that there is a strong positive linear correlation.