91Ó°ÊÓ

Given in the question is a table

The linear correlation coefficient using the definition is

$r=\frac{\frac{1}{n−1}∑\left(x_i−\stackrel{¯}{x}\right)\left(y_i−\stackrel{¯}{y}\right)}{{\mathrm{σ}}_x{\mathrm{σ}}_y}$

The standard deviations are

$\begin{array}{r}{\mathrm{σ}}_x=\sqrt{\frac{1}{n−1}∑(x−\stackrel{¯}{x}{)}^2},\\ {\mathrm{σ}}_y=\sqrt{\frac{1}{n−1}∑(y−\stackrel{¯}{y}{)}^2}\end{array}$

Calculation:

Make a table of values

Find the standard deviations

$$\begin{gathered} {\mathrm{σ}}_x=\sqrt{\frac{1}{n−1}∑(x−\stackrel{¯}{x}{)}^2} \\ =\sqrt{\frac{1}{5−1}\left(4\right)} \\ =1 \end{gathered}$$

$$\begin{gathered} {\mathrm{σ}}_y=\sqrt{\frac{1}{n−1}∑(y−\stackrel{¯}{y}{)}^2} \\ =\sqrt{\frac{1}{5−1}\left(10\right)} \\ =1.58 \end{gathered}$$

Find the correlation coefficient

localid="1650916923675" $$\begin{gathered} r=\frac{\frac{1}{n−1}∑\left(x_i−\stackrel{¯}{x}\right)\left(y_i−\stackrel{¯}{y}\right)}{{\mathrm{σ}}_x{\mathrm{σ}}_y} \\ =\frac{\frac{1}{5−1}(−4)}{\left(1\right)\left(1.58\right)} \end{gathered}$$

Hence,

The linear coefficient by the definition$=-0.632$

Given in the question that a table,

The linear correlation coefficient using the formula

$r=\frac{∑xy−\frac{∑x{\mathrm{Î\pounds }}^y}{n}}{\sqrt{\left(∑x^2−\frac{∑x^2}{n}\right)\left(∑y^2−\frac{∑y^2}{n}\right)}}$

Calculation:

Make a table of values

Find the correlation coefficient

$$\begin{gathered} r=\frac{∑xy−\frac{\stackrel{x}{∑} \underset{y}{∑} }{n}}{\sqrt{\left(∑x^2−\frac{{\left(∑x\right)}^2}{n}\right)\left(∑y^2−\frac{{\left({\mathrm{Î\pounds }}^n\right)}^2}{n}\right)}} \\ =\frac{26−\frac{150}{5}}{\sqrt{(49−(225/5\left)\right)(30−(100\sqrt{5}\left)\right)}} \end{gathered}$$

Hence,

The linear coefficient by the formula$=-0.632$