Q. In a bivariate data $\Sigma x = 30,\ \Sigma y = 400,\ \Sigma x^{2} = 196,\ \Sigma xy = 850\ and\ n = 10.\$the regression coefficient of $y$ on $x$ is

A) -3.1
B) -3.2
C) -3.3
D) -3.4
Explanation

(c)
Cov ($x,y$)=$\frac{\Sigma_{xy}}{n} - \frac{\Sigma_{x}}{n}.\frac{\Sigma_{y}}{n} = \frac{1}{10}(850) - \left( \frac{30}{10} \right)\left( \frac{400}{10} \right)$
$= 85 - 120 = - 35$
And var $(x) = \sigma_{x}^{2} = \frac{1}{n}\Sigma_{\ }x^{2} - \left( \frac{\Sigma_{x}}{n} \right)^{2}$
$= \frac{196}{10} - \left( \frac{30}{10} \right)^{2} = 10.6$
$\therefore\ \ b_{yx} = \frac{cov(x,y)}{var(x)} = \frac{- 35}{10.6} = - 3.3$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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