Chapter 13, Problem 17CT

To determine

To find the least squares regression line for the points.

Answer to Problem 17CT

The least square regression line is $y=\frac{3}{2}{x}^2-\frac{11}{10}x+\frac{17}{10}$ .

Explanation of Solution

Given information:

Points $\left(-2,10\right),\left(-1,4\right),\left(0,2\right),\left(1,2\right)$

Table for the steps to find the least squares regression line using $n=4$ ,

$x$	$y$	$x^2$	$x^3$	$x^4$	$xy$	$x^{2}y$
$-2$	$10$	$4$	$-8$	$16$	$-20$	$40$
$-1$	$4$	$1$	$-1$	$1$	$-4$	$4$
$0$	$2$	$0$	$0$	$0$	$0$	$0$
$1$	$2$	$1$	$1$	$1$	$2$	$2$
${\displaystyle \sum _{i=1}^n{x}_i=-2}$	${\displaystyle \sum _{i=1}^n{y}_i=18}$	${\displaystyle \sum _{i=1}^n{x}_i{}^2=6}$	${\displaystyle \sum _{i=1}^n{x}_i{}^3=-8}$	${\displaystyle \sum _{i=1}^n{x}_i{}^4=18}$	${\displaystyle \sum _{i=1}^n{x}_i{y}_i=-22}$	${\displaystyle \sum _{i=1}^n{x}_i{}^2{y}_i=46}$

Finding the least square regression quadratic equation is given as,

  $f\left(x\right)=a{x}^2+bx+c$

  $\begin{array}{c}nc+\left({\displaystyle \sum _{i=1}^n{x}_i}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)a={\displaystyle \sum _{i=1}^n{y}_i}\to 3a-b+2c=9\mathrm{...}\left(1\right)\\ \left({\displaystyle \sum _{i=1}^n{x}_i}\right)c+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^3}\right)a={\displaystyle \sum _{i=1}^n{x}_i{y}_i\to -4a+3b-c=-11}\mathrm{...}\left(2\right)\\ \left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)c+\left({\displaystyle \sum _{i=1}^n{x}_i{}^3}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^4}\right)a={\displaystyle \sum _{i=1}^n{x}_i{}^3{y}_i\to 9a-4b+3c=23\mathrm{...}\left(3\right)}\end{array}$

By solve the equations $1,2,3$

  $a=\frac{3}{2},b=-\frac{11}{10},c=\frac{17}{10}$

The least square regression line is,

  $y=\frac{3}{2}{x}^2-\frac{11}{10}x+\frac{17}{10}$

The above equation is represented by the graph,

  

Thus we can find the least squares regression line for the points.