Question

The following results were obtained with respect to two variable x and y:

$\sum x=30,\sum y=42,\sum xy=199,\sum x^2=184,\sum y^2=318,n=6.$ The regression coefficients. Find the likely value of y when x = 10.

Answer 1

$b_{yx}=\frac{\sum xy-\frac{1}{n}\sum x.\sum y}{\sum x^2-\frac{1}{n}(\sum x{)}^2}$
$=\frac{199-\frac{1}{6}\left(30\right)\left(42\right)}{184-\frac{1}{6}(30{)}^2}$
$=\frac{199-210}{184-150}=\frac{-11}{34}=-0.323$
$b_{xy}=\frac{\sum xy-\frac{1}{n}\sum x\sum y}{\sum y^2-\frac{1}{n}(\sum y{)}^2}$
$=\frac{199-\frac{1}{6}\left(30\right)\left(42\right)}{318-\frac{1}{6}(42{)}^2}$
$=\frac{199-210}{318-294}=\frac{-11}{24}=-0.458$

$\overline{x}=\frac{\sum x}{6}=\frac{30}{6}=5$
$\overline{y}=\frac{\sum y}{6}=\frac{42}{6}=7$
Regression line of y on $x$, $y-\overline{y}=b_{yx}(x-\overline{x})$
$y-7=\frac{-11}{34}(x-5)$
$34y-238=-11x+55$
$11x+34y-293=0$

$x=10$
$11\times 10+34y-293=0$
$\Rightarrow 34y=183$
$y=\frac{183}{34}$
$y=5.382$