Determine the linear correlation coefficient by using

a. Definition 4.8 on page 183

b. Formula 4.3 on page 185

Given data is

We have to obtain the linear correlation coefficient by using definition

The linear correlation coefficient using definition is as follows

$r=\frac{\frac{1}{n-1}\sum \left(x_i-x\right)\left(y_i-\overline{y}\right)}{{\sigma }_x{\sigma }_y}$

The standard deviations are as follows

$$\begin{gathered} {\sigma }_x=\sqrt{\frac{1}{n-1}\sum (x-\overline{x}{)}^2} \\ {\sigma }_y=\sqrt{\frac{1}{n-1}\sum (y-\overline{y}{)}^2} \end{gathered}$$

Table of values can be computed on the following table

The standard deviations are

$$\begin{gathered} {\sigma }_x=\sqrt{\frac{1}{n-1}\sum (x-\overline{x}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(24\right)} \\ =\sqrt{6} \\ {\sigma }_y=\sqrt{\frac{1}{n-1}\sum (y-\overline{y}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(20\right)} \\ =\sqrt{5} \end{gathered}$$

The linear correlation coeficients is

$$\begin{gathered} r=\frac{\frac{1}{{}_{n-1}}\sum \left(x_i-x\right)\left(y_i-\overline{y}\right)}{{\sigma }_x{\sigma }_y} \\ =\frac{{\displaystyle \frac{1}{5-1}}\left(-15\right)}{\left(\sqrt{30}\right)} \\ =-0.685 \end{gathered}$$

The given table is

We have to obtain the linear correlation coefficient by using formula

Table of values can be computed on the following table

$$\begin{gathered} r=\frac{\sum x{y}^{-}\frac{{\sum }_z{\sum }_y}{n}}{\sqrt{\left(\sum x^2-\frac{{\left({\sum }_x\right)}^2}{n}\right)\left(\sum y^2-\frac{{\left({\sum }_y\right)}^2}{n}\right)}} \\ =\frac{0-\frac{75}{5}}{\sqrt{(69-(225/5\left)\right)(25-(25/5\left)\right)}} \\ =-0.685 \end{gathered}$$

The linear correlation coefficient is

$$\begin{gathered} r=\frac{\sum xy-\frac{\sum z\sum y}{n}}{\sqrt{\left(\sum x^2-\frac{{\left({\sum }_x\right)}^2}{n}\right)\left(\sum y^2-\frac{{\left({\sum }_y\right)}^2}{n}\right)}} \\ =\frac{0-\frac{75}{5}}{\sqrt{(69-(225/5\left)\right)(25-(25/5\left)\right)}} \\ =-0.685 \end{gathered}$$