Chapter 13.3, Problem 16E

To determine

To find: The least squares regression line for the given points and verify the equation using a graphing utility.

Answer to Problem 16E

The equation is $y=\frac{21}{26}x-\frac{10}{13}$ and the graph is shown in Figure 1.

Explanation of Solution

Given:

The points are $\left(0,-1\right)$ , $\left(2,1\right)$ , $\left(3,2\right)$ and $\left(5,3\right)$ .

Calculation:

Let the general equation of the least squares regression line be $f\left(x\right)=ax+b$ where the following equations are used to determine the unknowns a and b:

  $\begin{array}{c}a=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-{\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}}{n{\displaystyle \sum _{i=1}^n{x}_i{}^2}-{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}\\ b=\frac{1}{n}\left({\displaystyle \sum _{i=1}^n{y}_i}-a{\displaystyle \sum _{i=1}^n{x}_i}\right)\end{array}$.......(1)

In order to determine value of the coefficients of a and b in the set of equations (1), construct a table as shown below:

x	y	$xy$	$x^2$
0	-1	0	0
2	1	2	4
3	2	6	9
5	3	15	25
${\displaystyle \sum _{i=1}^4{x}_i}=10$	${\displaystyle \sum _{i=1}^4{y}_i}=5$	${\displaystyle \sum _{i=1}^4{x}_i{y}_i}=23$	${\displaystyle \sum _{i=1}^4{x}_i{}^2}=38$

Table 1

Substitute the values from the above table into the set of equations (1).

  $\begin{array}{l}a=\frac{4\left(23\right)-\left(10\right)\left(5\right)}{4\left(38\right)-{\left(10\right)}^2}\\ b=\frac{1}{4}\left(5-a\left(10\right)\right)\end{array}$

Simplify the set of equations to determine the values of a and b.

  $\begin{array}{c}a=\frac{21}{26}\\ b=-\frac{10}{13}\end{array}$

So, the equation of the least squares regression line becomes,

  $\begin{array}{c}f\left(x\right)=\left(\frac{21}{26}\right)x+\left(-\frac{10}{13}\right)\\ y=\frac{21}{26}x-\frac{10}{13}\end{array}$

In order to verify the equation $y=\frac{21}{26}x-\frac{10}{13}$ , plot the points and the equation in a graphing utility. The graph is shown below:

  

Figure 1