Question

Given that the observations are:
$(9,-4),(10,-3),(11,-1),(12,0),(13,1),(14,3),(15,5),(16,8)$.
Find the tow lines of regression and estimate the value of $y$ when $x=\mathrm{13.5.}$

Answer 1

$x$	$x^2$	$y$	$y^2$	$xy$
9	81	-4	16	-36
10	100	-3	9	-30
11	121	-1	1	-11
12	144	0	0	0
13	169	1	1	13
14	196	3	9	42
15	225	5	25	75
16	256	8	64	128
$\sum x=100$	$\sum x^2=1292$	$\sum y=9$	$\sum y^2=125$	$\sum xy=181$

$\overline{x}=\frac{\sum x}{N}=\frac{100}{8}=12.5$ $(\because N=8)$

$\overline{y}=\frac{\sum y}{N}=\frac{9}{8}=1.125$

$b_{yx}=\frac{\sum xy-\frac{1}{N}\sum x\sum y}{\sum x^2-\frac{1}{N}(\sum x{)}^2}$

$=\frac{181-\frac{9\times 100}{8}}{1292-\frac{10000}{8}}=\frac{181-112.5}{1292-1250}=\frac{181-112.5}{42}=\frac{68.5}{42}=1.63$

$b_{xy}=\frac{\sum xy-\frac{1}{N}\sum x\sum y}{\sum y^2-\frac{1}{N}(\sum y{)}^2}$

$=\frac{181-\frac{900}{8}}{125-\frac{81}{8}}=\frac{181-112.5}{125-10.125}=\frac{68.5}{114.875}=0.596$

Regression line of $y$ on $x$ $y-\overline{y}=b_{xy}(x-\overline{x})$
$y-1.125=1.63(x-12.5)$
$\Rightarrow y-1.125=1.63x-20.375$
$\Rightarrow y=1.63x-19.25$

Regression line of $x$ on $y$ $x-\overline{x}=b_{yx}(y-\overline{y})$
$x-12.5=0.596(y-1.125)$
$x-12.5=0.596y-0.67$
$x=0.596y+11.83$

Now, $y=1.63x-19.25$
For $x=13.5$
The extimated value of $y=1.63\times 13.5-19.25$
$=22.005-19.25=2.755=2.8$