Chapter 4.1, Problem 20E

In Exercises 17-20, compute the correlation coefficient.

To determine

To calculate: To compute the correlation coefficient.

Answer to Problem 20E

The correlation coefficient is $0.339$

Explanation of Solution

Given information:

The data is,

$x$	5	-8	-2	6	9	-10	13	7
$y$	-1	-3	-6	-7	-1	5	13	22

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:

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{5-8-2+6+9-10+13+7}{8}\\ =\frac{20}{8}\\ =2.5\end{array}$

The mean of y is given by:

  $\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{-1-3-6-7-1+5+13+22}{8}\\ =\frac{22}{8}\\ =2.75\end{array}$

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	-1	2.5	6.25	-3.75	14.0625
-8	-3	-10.5	110.25	-5.75	33.0625
-2	-6	-4.5	20.25	-8.75	76.5625
6	-7	3.5	12.25	-9.75	95.0625
9	-1	6.5	42.25	-3.75	14.0625
-10	5	-12.5	156.25	2.25	5.0625
13	13	10.5	110.25	10.25	105.0625
7	22	4.5	20.25	19.25	370.5625
$\overline{x}=2.5$	$\overline{y}=2.75$		$\begin{array}{l}{\displaystyle \sum {\left(x-\overline{x}\right)}^2}\\ =478\end{array}$		$\begin{array}{l}{\displaystyle \sum {\left(y-\overline{y}\right)}^2}\\ =713.5\end{array}$

Hence, the standard deviation is given by:

  $\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{478}{8}}\\ s_x=7.7298\end{array}$

And,

  $\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{713.5}{8}}\\ s_y=9.4439\end{array}$

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	-1	2.5	-3.75	-9.375
-8	-3	-10.5	-5.75	60.375
-2	-6	-4.5	-8.75	39.375
6	-7	3.5	-9.75	-34.125
9	-1	6.5	-3.75	-24.375
-10	5	-12.5	2.25	-28.125
13	13	10.5	10.25	107.625
7	22	4.5	19.25	86.625
$\overline{x}=2.5$	$\overline{y}=2.75$			${\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}=198$

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,

  $r=\frac{1}{8}\cdot \frac{198}{\left(7.7298\right)\left(9.4439\right)}=\mathrm{0.33904..}\approx 0.339$

Therefore, the correlation coefficient for the given data is 0.339