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LINEAR CORRELATION — Statistics STD 12 Commerce — Question

Gujarat BoardEnglish MediumSTD 12 CommerceStatisticsLINEAR CORRELATION1 Mark
Question
The following information is obtained to study the relationship between the advertisement cost and the sales of electric fans of the companies manufacturing electric fans. Find the correlation coefficient advertisement cost and the sales by Karl Pearson’s method.
Company $A$ $B$ $C$ $D$ $E$ $F$
Advertisement Cost $($lakh $Rs.)$ $140$ $120$ $80$ $100$ $80$ $180$
Sales of electric fans $($Crore $Rs.)$ $35$ $45$ $15$ $40$ $20$ $50$
✓

Answer

Ans: Here, Let $X =$ Advertisement cost and $Y =$ Sales of electric fans $($Crore $Rs.).$
Since means $\bar{x}$ and $\bar{y}$ of variables is in the fraction, we prepare the calculative table as follows :
Company $X$ $Y$ $u =\frac{y-A}{C_x}
A = 550, C_x = 50$		 $v =\frac{y-B}{C_y}
B = 25, C_y = 5$		 $uv$ $u^2$ $v^2$
$A$ $140$ $35$ $3$ $1$ $3$ $9$ $1$
$B$ $120$ $45$ $1$ $3$ $3$ $1$ $9$
$C$ $80$ $15$ $-3$ $-3$ $9$ $9$ $9$
$D$ $100$ $40$ $-1$ $2$ $-2$ $1$ $4$
$E$ $80$ $20$ $-3$ $-2$ $6$ $9$ $4$
$F$ $180$ $50$ $7$ $4$ $28$ $49$ $16$
Total $\Sigma{x} = 700$ $\Sigma{y} = 205$ $\Sigma{u} = 4$ $\Sigma{v} = 5$ $\Sigma{uv} = 47$ $\Sigma{u^2}
= 78$		 $\Sigma{v^2}
= 43$
$\bar{x} {=} \frac{\sum \mathrm{x}}{\mathrm{n}}=\frac{700}{6}= 116.67$ and $\bar{y} {=}\frac{\sum y}{n}=\frac{205}{6}= 34.17$
$r = \frac{n \sum u v-\left(\sum u\right)\left(\sum v\right)}{\sqrt{n \sum u^2-\left(\sum u\right)^2} \times \sqrt{n \sum v^2-\left(\sum v\right)^2}} =$

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