For each exercise, determine the linear correlation coefficient by using

a. Definirina $4$on page $183$,

b. Formula $4.3$on page $185$.

Compare year annex an para (a) and (b

To find thelinear 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

$$\begin{gathered} {\sigma }_x=\sqrt{\frac{1}{n-1}\sum (x-\overline{x}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(10\right)} \\ =\sqrt{5/2} \end{gathered}$$

$$\begin{gathered} {\sigma }_y=\sqrt{\frac{1}{n-1}\sum (y-\overline{y}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(14\right)} \\ =\sqrt{7/2} \end{gathered}$$

Find the correlation coefficient

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

Hence,

The linear coefficient by the definition$=0.845$

To find the linear correlation coefficient by the formula.

The linear correlation coefficient using the formula

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

Calculation:

Make a table of values

Find the correlation coefficient

$$\begin{gathered} r=\frac{\sum xy-\frac{{\sum }^x{\sum }_y}{n}}{\sqrt{\left(\sum x^2-\frac{{\left(\sum x\right)}^2}{n}\right)\left(\sum y^2-\frac{{\left(\sum y\right)}^2}{n}\right)}} \\ =\frac{70-\frac{250}{5}}{\sqrt{(30-(100/5\left)\right)(171-(625/5\left)\right)}} \end{gathered}$$

Hence,

The linear coefficient by the formula$=0.845$