Determine the linear correlation coefficient by formula and definition.

Given in the question is a table

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

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

Calculation:

Make a table of values

Find the standard deviations

$$\begin{gathered} {\sigma }_x=\sqrt{\frac{1}{n-1}\sum (x-\overline{x}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(4\right)} \\ =1 \end{gathered}$$

$$\begin{gathered} {\sigma }_y=\sqrt{\frac{1}{n-1}\sum (y-\overline{y}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(10\right)} \\ =1.58 \end{gathered}$$

Find the correlation coefficient

localid="1650916923675" $$\begin{gathered} r=\frac{\frac{1}{n-1}\sum \left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}{{\sigma }_x{\sigma }_y} \\ =\frac{\frac{1}{5-1}(-4)}{\left(1\right)\left(1.58\right)} \end{gathered}$$

Hence,

The linear coefficient by the definition$=-0.632$

Given in the question that a table,

The linear correlation coefficient using the formula

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

Calculation:

Make a table of values

Find the correlation coefficient

$$\begin{gathered} r=\frac{\sum xy-\frac{\sum ^x \sum _y }{n}}{\sqrt{\left(\sum x^2-\frac{{\left(\sum x\right)}^2}{n}\right)\left(\sum y^2-\frac{{\left({\mathrm{\Sigma }}^n\right)}^2}{n}\right)}} \\ =\frac{26-\frac{150}{5}}{\sqrt{(49-(225/5\left)\right)(30-(100\sqrt{5}\left)\right)}} \end{gathered}$$

Hence,

The linear coefficient by the formula$=-0.632$