Chapter 4, Problem 1CQ

Compute the correlation coefficient for the following data set.

To determine

To compute the correlation coefficient for the given data

Answer to Problem 1CQ

The correlation coefficient is $\gamma =-0.96$

Explanation of Solution

Given:

x	2	5	6	7	11
y	15	9	6	4	1

Concept used:

The correlation coefficient $\gamma =\frac{{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}}{n-1}$ .

First we find the sample means $\overline{x}\text{and}\overline{y}$ .

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{31}{5}=6.2\\ \overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{35}{5}=7\end{array}$

Now to find standard deviations, we construct the table as shown below:

$x$	$y$	${\left(x-\overline{x}\right)}^2$	${\left(y-\overline{y}\right)}^2$
2	15	17.64	64
5	9	1.44	4
6	6	0.04	1
7	4	0.64	9
11	1	23.04	36
		${\displaystyle \sum {\left(x-\overline{x}\right)}^2}=$ 42.8	${\displaystyle \sum {\left(y-\overline{y}\right)}^2}=114$

Therefore, standard deviations are,

  $\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}=\sqrt{\frac{42.8}{5-1}}=\sqrt{\frac{42.8}{4}}=\sqrt{10.7}=3.2710\\ s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n-1}}=\sqrt{\frac{114}{5-1}}=\sqrt{\frac{114}{4}}=\sqrt{28.5}=5.3385\end{array}$

Now to find correlation coefficient, we construct the table as shown below:

$x$	$y$	$\frac{x-\overline{x}}{s_x}$	$\frac{y-\overline{y}}{s_y}$	$\left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)$
2	15	$-$ 1.2840110	1.4985482	$-$ 1.9241523
5	9	$-$ 0.3668602	0.3746370	$-$ 0.1374394
6	6	$-$ 0.0611433	$-$ 0.1873185	0.0114532
7	4	0.2445735	$-$ 0.5619556	$-$ 0.1374394
11	1	1.4674411	$-$ 1.1239112	$-$ 1.6492734
				${\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$ = $-$ 3.8368513

Therefore,

  $\gamma =\frac{{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}}{n-1}=\frac{-3.8368513}{5-1}=\frac{-3.8368513}{4}=-0.9592\approx -0.96$