91Ó°ÊÓ

Given data is

We have to obtain the linear correlation coefficient by using definition

The linear correlation coefficient using definition is as follows

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

The standard deviations are as follows

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

Table of values can be computed on the following table

The standard deviations are

$$\begin{gathered} {\mathrm{σ}}_x=\sqrt{\frac{1}{n-1}∑(x-\stackrel{¯}{x}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(24\right)} \\ =\sqrt{6} \\ {\mathrm{σ}}_y=\sqrt{\frac{1}{n-1}∑(y-\stackrel{¯}{y}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(20\right)} \\ =\sqrt{5} \end{gathered}$$

The linear correlation coeficients is

$$\begin{gathered} r=\frac{\frac{1}{{}_{n-1}}∑\left(x_i-x\right)\left(y_i-\stackrel{¯}{y}\right)}{{\mathrm{σ}}_x{\mathrm{σ}}_y} \\ =\frac{{\displaystyle \frac{1}{5-1}}\left(-15\right)}{\left(\sqrt{30}\right)} \\ =-0.685 \end{gathered}$$

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

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

The linear correlation coefficient is

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