Q.4.143 - Elementary Statistics

Question

Determine the linear correlation coefficient by using

a. Definition 4.8 on page 183

b. Formula 4.3 on page 185

Step-by-Step Solution

Answer

(a)The linear correlation coefficient by using definition is $-0.685$

(b) The linear correlation coefficient by using Formula is $-0.685$

Part(a) Step 1: Given Information

Given data is

We have to obtain the linear correlation coefficient by using definition 

Part(a) Step 2: Explanation

The linear correlation coefficient using definition is as follows 

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

The standard deviations are as follows 

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

Table of values can be computed on the following table 

The standard deviations are

$$\begin{gathered} {\sigma }_x=\sqrt{\frac{1}{n-1}\sum (x-\overline{x}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(24\right)} \\ =\sqrt{6} \\ {\sigma }_y=\sqrt{\frac{1}{n-1}\sum (y-\overline{y}{)}^2} \\ =\sqrt{\frac{1}{5-1}\left(20\right)} \\ =\sqrt{5} \end{gathered}$$

The linear correlation coeficients is

$$\begin{gathered} r=\frac{\frac{1}{{}_{n-1}}\sum \left(x_i-x\right)\left(y_i-\overline{y}\right)}{{\sigma }_x{\sigma }_y} \\ =\frac{{\displaystyle \frac{1}{5-1}}\left(-15\right)}{\left(\sqrt{30}\right)} \\ =-0.685 \end{gathered}$$

Part(b) Step 1: Given Information

The given table is

We have to obtain the linear correlation coefficient by using  formula 

Part(b) Step 2: Explanation

Table of values can be computed on the following table  

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

The linear correlation coefficient is

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