Chapter 13.3, Problem 17E

To determine

To find: The correlation coefficient $r$ for the data given in Exercise $13$ and then to discuss the linear model fit.

Answer to Problem 17E

  $r=0.9923$ . The model fits the data well.

Explanation of Solution

Given information: Given points are $(-4,1),(-3,3),(-2,4),(-1,6)$ .

Calculation:

  $n=4$

  $\begin{array}{lllll}x\hfill & y\hfill & xy\hfill & x^2\hfill & y^2\hfill \\ -4\hfill & 1\hfill & -4\hfill & 16\hfill & 1\hfill \\ -3\hfill & 3\hfill & -9\hfill & 9\hfill & 9\hfill \\ -2\hfill & 4\hfill & -8\hfill & 4\hfill & 16\hfill \\ -1\hfill & 6\hfill & -6\hfill & 1\hfill & 36\hfill \\ {\displaystyle \sum _{i=1}^n{x}_i=-10}\hfill & {\displaystyle \sum _{i=1}^n{y}_i=14}\hfill & {\displaystyle \sum _{i=1}^n{x}_i{y}_i=-27}\hfill & {\displaystyle \sum _{i=1}^n{x}_i{}^2=30}\hfill & {\displaystyle \sum _{i=1}^n{y}_i{}^2=62}\hfill \end{array}$

Putting all these values in

  $r=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i-\left({\displaystyle \sum _{i=1}^n{x}_i}\right)\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}}{\sqrt{n{{\displaystyle \sum _{i=1}^n{x}_i{}^2-\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}}^2}\sqrt{n{{\displaystyle \sum _{i=1}^n{y}_i{}^2-\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}}^2}}$ ,

  $\begin{array}{l}r=\frac{4(-27)-(-10)(14)}{\sqrt{4(30)-{(-10)}^2}\sqrt{4(62)-{(14)}^2}}=\frac{32}{\sqrt{20}\sqrt{52}}\\ \\ \Rightarrow r\approx 0.9923\end{array}$

Since $r$ is close to $1$ , so there is a strong positive linear correlation between $xandy$ .

Thus, the model fits the data well.