Question

The coefficient of correlation (${\mathrm{r}}_{\mathrm{xy}}$) for the following will be approximately

$x$	$2$	$4$	$5$	$6$	$3$	$6$	$8$	$10$
$y$	$5$	$6$	$6$	$8$	$4$	$8$	$12$	$15$

• A. $0.90$
• B. $0.96$
• C. $-0.90$
• D. $-0.96$

Answer 1

The correct option is B

$0.96$

The explanation for the correct option:

Step 1: Construction of the table

Given, the table

$x$	$2$	$4$	$5$	$6$	$3$	$6$	$8$	$10$
$y$	$5$	$6$	$6$	$8$	$4$	$8$	$12$	$15$

Here, $\sum _{}{x}^{}=44,\sum _{}y=64$, $n=8$

Calculating $\overline{x}$ and $\overline{y}$

$$\begin{gathered} \overline{x}=\frac{\sum _{}x}{n} \\ =\frac{44}{8} \\ =5.5 \\ \overline{y}=\frac{\sum _{}y}{n} \\ =\frac{64}{8} \\ =8 \end{gathered}$$

Solving for the required correlation coefficient, we should construct the following

$x$	$y$	$x-\overline{x}$	$y-\overline{y}$	$\left(\mathrm{x}-\overline{\mathrm{x}}\right)\left(\mathrm{y}-\overline{\mathrm{y}}\right)$	${\left(x-\overline{x}\right)}^2$	${\left(y-\overline{y}\right)}^2$
$2$	$5$	$-3.5$	$-3$	$10.5$	$12.25$	$9$
$4$	$6$	$-1.5$	$-2$	$3$	$2.25$	$4$
$5$	$6$	$-0.5$	$-2$	$1$	$0.25$	$4$
$6$	$8$	$0.5$	$0$	$0$	$0.25$	$0$
$3$	$4$	$-2.5$	$-4$	$10$	$6.25$	$16$
$6$	$8$	$0.5$	$0$	$0$	$0.25$	$0$
$8$	$12$	$2.5$	$4$	$10$	$6.25$	$16$
$10$	$15$	$4.5$	$7$	$31.5$	$20.25$	$49$
$\sum _{}{x}^{}=44$	$\sum _{}y=64$			$\sum _{}^{}\left(x-\overline{x}\right)\left(y-\overline{y}\right)=66$	$\sum _{}^{}{(x-\stackrel{-}{x})}^2=48$	$\sum _{}^{}{(\mathrm{y}-\stackrel{-}{\mathrm{y}})}^2=98$

From the obtained table, we get the following values,

$\sum _{}^{}{\left(x-\overline{x}\right)}^2=48$

$\sum _{}^{}{\left(y-\overline{y}\right)}^2=98$

$\sum _{}^{}\left(y-\overline{y}\right)\left(x-\overline{x}\right)=66$

Step 2: Calculating the coefficient of correlation ${\mathrm{r}}_{\mathrm{xy}}$

We also know that, from the formulae, ${\mathrm{r}}_{\mathrm{xy}}=\frac{\sum _{}^{}{\left(\mathrm{y}-\overline{\mathrm{y}}\right)}^2{\left(\mathrm{x}-\overline{\mathrm{x}}\right)}^2}{\sqrt{\sum _{}^{}{\left(\mathrm{x}-\overline{\mathrm{x}}\right)}^2}\sqrt{\sum _{}^{}{\left(\mathrm{y}-\overline{\mathrm{y}}\right)}^2}}$

$\Rightarrow {\mathrm{r}}_{\mathrm{xy}}=\frac{\sum _{}^{}{\left(\mathrm{y}-\overline{\mathrm{y}}\right)}^2{\left(\mathrm{x}-\overline{\mathrm{x}}\right)}^2}{\sqrt{\sum _{}^{}{\left(\mathrm{x}-\overline{\mathrm{x}}\right)}^2}\sqrt{\sum _{}^{}{\left(\mathrm{y}-\overline{\mathrm{y}}\right)}^2}}$

$=\frac{66}{\sqrt{48}\sqrt{98}}$

$=\frac{66}{6.928\times 9.899}$

$=\frac{66}{68.58}$

$=0.96$

Hence, option (B) is the correct answer.