Question

For the following data, the equation of the line of regression of $y$ on $x$ is

$x$	$2$	$4$	$6$	$8$
$y$	$6$	$8$	$12$	$15$

• A. $y=x+2$
• B. $y=x+2.5$
• C. $y=1.55x+2.5$
• D. $y=1.55x+2$

Answer 1

The correct option is C

$y=1.55x+2.5$

Step 1. Find $\overline{x}$ and $\overline{y}$

We know that the equation of the line of regression of $y$ on $x$ is given by,

$(y-\overline{y)}=r.\left(\frac{y}{x}\right)(x-\overline{x)}$

Here, $r=\frac{\mathrm{n}\sum _{}\mathrm{xy}-\sum _{}\mathrm{x}\sum _{}\mathrm{y}}{\mathrm{n}\sum _{}{\mathrm{x}}^2-{\left(\sum _{}\mathrm{x}\right)}^2}$

We have given,

$x$	$2$	$4$	$6$	$8$
$y$	$6$	$8$	$12$	$15$

There are four value of independent variable $x$.

So, $n=4$

$x$	$y$	$x^2$	$y^2$	$x·y$
$2$	$6$	$4$	$36$	$12$
$4$	$8$	$16$	$64$	$32$
$6$	$12$	$36$	$144$	$72$
$8$	$15$	$64$	$225$	$120$
$\sum _{}x=20$	$\sum _{}y=41$	$\sum _{}{x}^2=120$	$\sum _{}{y}^2=469$	$\sum _{}xy=236$

Now,

$\begin{array}{rcl}\overline{\mathrm{x}}& =& \frac{\sum _{}\mathrm{x}}{\mathrm{n}}\\ & =& \frac{20}{4}\\ & =& 5\end{array}$

And

$\begin{array}{rcl}\overline{\mathrm{y}}& =& \frac{\sum _{}\mathrm{y}}{\mathrm{n}}\\ & =& \frac{41}{4}\\ & =& 10.25\end{array}$

Step 2: Find the slope of the regression line

The slope of the regression line is given by $r=\frac{\mathrm{n}\sum _{}\mathrm{xy}-\sum _{}\mathrm{x}\sum _{}\mathrm{y}}{\mathrm{n}\sum _{}{\mathrm{x}}^2-{\left(\sum _{}\mathrm{x}\right)}^2}$

Therefore, by using the table in step $1$, we get

$\begin{array}{rcl}r& =& \frac{4\times 236-20\times 41}{4\times 120-{\left(20\right)}^2}\\ & =& \frac{124}{80}\\ & =& `1.55\end{array}$

Step 3. Find the equation of the line of regression of $y$ on $x$

The line regression on $x$ is:

$\begin{array}{rcl}(y–10.25)& =& 1.55(x–5)\\ & \Rightarrow & y-10.25=1.55x-7.75\\ & \Rightarrow & y=1.55x-7.75+10.25\\ & \Rightarrow & y=1.55x+2.5\end{array}$

Hence option $\left(C\right)$ is correct answer