Answer to Question #104254 in Statistics and Probability for Amra Musharraf

Question

The equations of two regression lines obtained in a correlation analysis are as follows:
3X+12Y=19, 3Y+9X=46
Obtain
i) the value of correlation coefficient, ii) mean values of X and Y, and iii) the ratio of the coefficient of variability of X to that of Y.

Answer 1

Regression lines: $y=-\frac{1}{4}y+\frac{19}{12},\;\;x=-\frac{1}{3}y+\frac{46}{9}.$

i) $r=\pm \sqrt{b_{xy}*b_{yx}}=\pm \sqrt{(-\frac{1}{4})(-\frac{1}{3})}=\pm \frac{\sqrt{3}}{6}$.

But regression coefficient are negative, so $r=-\frac{\sqrt{3}}{6}$.

ii) Since the point $(\bar x,\bar y)$ lies on the both regression lines, we get:

$3\bar x+12\bar y=19, \;\;3\bar y+9\bar x=46$

Solving this system of equation, we have: $\bar x=5,\;\bar y=\frac{1}{3}.$

iii) $\frac{V(x)}{V(y)}=\frac{b_{xy}}{b_{yx}}=\frac{\frac{1}{4}}{\frac{1}{3}}=\frac{3}{4}.$