Question
An educationalist has conducted an experiment to know the relation between the usage of Social Media in mobile phone and the result of the examination. A group of $10$ students is selected for this and the following results were obtained regarding, the time spent $x ($in hours$)$ in last week on Social Media and the marks $(2)$ obtained out of $50$ in the examination, taken immediately after it.
$\Sigma x=133, \Sigma y=220, \Sigma x^{2}=2344, \Sigma y^{2}=6500$ and $\Sigma x y=3500$
Later on, it was found that one of the pairs of observations of $X$ and $Y$ was taken as $(13, 20)$ instead of $(15, 25) .$
Find the correct value of the correlation coefficient between $X$ and $Y.$

Answer

Here, $n=10, \Sigma x=133, \Sigma y=220, \Sigma x^{2}=2344, \Sigma y^{2}=6500$ and $\Sigma x y=3500$
Incorrect Pair $: (l 3, 20)$ Correct Pair $: (15,25)$
Now, we find corrected values of these measures as follows :
$\Sigma x=133-13+15=135$
$\Sigma y=220-20+25=225$
$\Sigma x^{2}=2344-(13)^{2}+(15)^{2}=2344-169+225=2400$
$\Sigma y^{2}=6500-(20)^{2}+(25)^{2}=6500-400+625=6725$
$\Sigma x y=3500-(13 \times 20)+(15 \times 25)=3500-260+375=3615$
Substituting these corrected values in the following formula,
$ r =\frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{\sqrt{n \Sigma x^{2}-(\Sigma x)^{2}} \cdot \sqrt{n \Sigma y^{2}-(\Sigma y)^{2}}}$
$ =\frac{10(3615)-(135)(225)}{\sqrt{10(2400)-(135)^{2}} \cdot \sqrt{10(6725)-(225)^{2}}} $
$ =\frac{36150-30375}{\sqrt{24000-18225} \cdot \sqrt{67250-50625}}$
$ =\frac{5775}{\sqrt{5775} \cdot \sqrt{16625}}$
$ =\frac{5775}{\sqrt{96009375}}$
$ =\frac{5775}{9798.4374}$
$ =0.5894$
$\therefore r \simeq 0.59 $

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