Question

Find the line of best fit for the following data, treating x as dependent variable (Regression equation $x$ on $y$):

$X$	14	12	13	14	16	10	13	12
$Y$	14	23	17	24	18	25	23	24

Hence, estimate the value of $x$ when $y=16$.

Answer 1

$X$	$14$	$12$	$13$	$14$	$16$	$10$	$13$	$12$
$Y$	$14$	$23$	$17$	$24$	$18$	$25$	$23$	$24$
$XY$	$196$	$276$	$221$	$336$	$288$	$250$	$299$	$288$
$X^2$	$196$	$144$	$169$	$196$	$256$	$100$	$169$	$144$

$\sum X=104$ $\Rightarrow \stackrel{\_}{X}=\frac{104}{8}=13$

$\sum Y=168$ $\Rightarrow \stackrel{\_}{Y}=\frac{168}{8}=21$

$\sum XY=2154$

$\sum X^2=1374$

Regression line if of form $y=a+bx$

$b=\frac{S_{xy}}{S_{xx}},S_{xy}=\sum xy-\frac{\sum x\sum y}{n},S_{xx}=\sum x^2-\frac{{(\sum x)}^2}{n}$

$\Rightarrow S_{xy}=2154-\frac{\left(104\right)\left(168\right)}{8}=-30\Rightarrow S_{xx}=1374-\frac{{\left(104\right)}^2}{8}=22\Rightarrow b=\frac{-30}{22}=-\frac{15}{11}a=\stackrel{\_}{y}-b\stackrel{\_}{x}\Rightarrow a=21+\frac{15}{11}\left(13\right)\Rightarrow a=\frac{426}{11}y=\frac{426}{11}-\frac{15}{11}x11y+15x=426$

Put $y=16$

$15x=250x=\frac{250}{15}=\frac{50}{3}$