For each exercise, determine the linear correlation coefficient using

a. Definition $4.8$on page $183$

b. Format $4.3$on page $185$,

To find the linear correlation coefficient by the definition.

The linear correlation coefficient using the definition is

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

The standard deviations are

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

Calculation:

Make a table of values

Find the standard deviations

$$\begin{gathered} {\sigma }_x=\sqrt{\frac{1}{n-1}\sum (x-\overline{x}{)}^2} \\ =\sqrt{\frac{1}{3-1}\left(2\right)} \\ =1 \\ {\sigma }_y=\sqrt{\frac{1}{n-1}\sum (y-\overline{y}{)}^2} \\ =\sqrt{\frac{1}{3-1}\left(14\right)} \\ =\sqrt{7} \end{gathered}$$

Find the correlation coefficient

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

$=\frac{\frac{1}{3-1}(-4)}{\left(1\right)\left(\sqrt{7}\right)}$

Hence,

The linear coefficient by the definition$=-0.759$

To find the linear correlation coefficient by the formula.

The linear correlation coefficient using the formula

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

Calculation:

Make a table of values

Find the correlation coefficient

localid="1650908959084" role="math" $$\begin{gathered} r=\frac{\sum xy-\frac{\sum x\sum y}{n}}{\sqrt{\left(\sum x^2-\frac{{\left(\sum x\right)}^2}{n}\right)\left(\sum y^2-\frac{{\left(\sum y\right)}^2}{n}\right)}} \\ =\frac{-22-\frac{6(-9)}{3}}{\sqrt{(14-(36/3\left)\right)(41-(81/3\left)\right)}} \end{gathered}$$

Hence,

The linear coefficient by the formula$=-0.756$