Chapter 4.1, Problem 19E

In Exercises 17-20, compute the correlation coefficient.

To determine

To calculate: To compute the correlation coefficient.

Answer to Problem 19E

The correlation coefficient is $0.515$

Explanation of Solution

Given information:

The data is,

$x$	5.5	4.2	4.7	5.6	6.0	3.9	6.3	5.7
$y$	4.9	4.8	4.8	4.7	5.5	5.1	5.8	6.5

Formula used:

The correlation coefficient of a data is given by:

  $r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}$

Where, $\overline{x},\overline{y}$ represent the mean of x and y respectively. $s_x,s_y$ represent the standard deviations of x and y. n represents the number of terms.

The standard deviations are given by:

  $s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}},s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}$

Calculation:

The mean of x is given by:

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{5.5+4.2+4.7+5.6+6.0+3.9+6.3+5.7}{8}\\ =\frac{41.9}{8}\\ =5.2375\end{array}$

The mean of y is given by:

  $\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{4.9+4.8+4.8+4.7+5.5+5.1+5.8+6.5}{8}\\ =\frac{42.1}{8}\\ =5.2625\end{array}$

The data can be represented in tabular form as:

x	y	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$	$y-\overline{y}$	${\left(y-\overline{y}\right)}^2$
5.5	4.9	0.2625	0.06891	-0.3625	0.13141
4.2	4.8	-1.0375	1.07641	-0.4625	0.21391
4.7	4.8	-0.5375	0.28891	-0.4625	0.21391
5.6	4.7	0.3625	0.13141	-0.5625	0.31641
6	5.5	0.7625	0.58141	0.2375	0.05641
3.9	5.1	-1.3375	1.78891	-0.1625	0.02641
6.3	5.8	1.0625	1.12891	0.5375	0.28891
5.7	6.5	0.4625	0.21391	1.2375	1.53141
$\overline{x}=5.2375$	$\overline{y}=5.2625$		$\begin{array}{l}{\displaystyle \sum {\left(x-\overline{x}\right)}^2}\\ =5.27875\end{array}$		$\begin{array}{l}{\displaystyle \sum {\left(y-\overline{y}\right)}^2}\\ =2.77875\end{array}$

Hence, the standard deviation is given by:

  $\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{5.27875}{8}}\\ s_x=0.8123\end{array}$

And,

  $\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{2.7788}{8}}\\ s_y=0.58936\end{array}$

Hence, the table for calculating coefficient of correlation is given by:

x	y	$x-\overline{x}$	$y-\overline{y}$	$\left(x-\overline{x}\right)\left(y-\overline{y}\right)$
5.5	4.9	0.2625	-0.3625	-0.09516
4.2	4.8	-1.0375	-0.4625	0.47984
4.7	4.8	-0.5375	-0.4625	0.24859
5.6	4.7	0.3625	-0.5625	-0.20391
6	5.5	0.7625	0.2375	0.18109
3.9	5.1	-1.3375	-0.1625	0.21734
6.3	5.8	1.0625	0.5375	0.57109
5.7	6.5	0.4625	1.2375	0.57234
$\overline{x}=5.2375$	$\overline{y}=5.2625$			${\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}=1.97125$

The correlation coefficient of a data is given by:

  $r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}$

Plugging the values in the formula,

  $\begin{array}{c}r=\frac{1}{8}\cdot \frac{1.97125}{\left(0.8123\right)\left(0.58936\right)}\\ =0.51469\\ \approx 0.515\end{array}$

Therefore, the correlation coefficient for the given data is 0.515