Chapter 13.3, Problem 22E

To determine

To find: The least squares regression line for the given data and the sum of the squared differences.

Answer to Problem 22E

The equation is $y=0.0682x+4.9843$ and the sum of the squared differences is 0.0002378.

Explanation of Solution

Given:

The data points are $\left(17,6.137\right)$ , $\left(18,6.217\right)$ , $\left(19,6.290\right)$ and $\left(20,6.340\right)$ .

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	$xy$	$x^2$
17	6.137	104.329	289
18	6.217	111.906	324
19	6.290	119.510	361
20	6.340	126.800	400
${\displaystyle \sum _{i=1}^4{x}_i}=74$	${\displaystyle \sum _{i=1}^4{y}_i}=24.984$	${\displaystyle \sum _{i=1}^4{x}_i{y}_i}=462.545$	${\displaystyle \sum _{i=1}^4{x}_i{}^2}=1374$

Table 1

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

  $\begin{array}{l}a=\frac{4\left(462.545\right)-\left(74\right)\left(24.984\right)}{4\left(1374\right)-{\left(74\right)}^2}\\ b=\frac{1}{4}\left(24.984-a\left(74\right)\right)\end{array}$

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

  $\begin{array}{c}a=0.0682\\ b=4.9843\end{array}$

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

  $\begin{array}{c}f\left(x\right)=\left(0.0682\right)x+\left(4.9843\right)\\ y=0.0682x+4.9843\end{array}$

In order to calculate the sum of the differences, determine the value of $y^{*}=0.0682x+4.9843$ for each value of y, find the difference $y-y^{*}$ , square it and then add all the squares, as shown in the table below:

x	y	$y^{*}$	$y-y^{*}$	${\left(y-y^{*}\right)}^2$
17	6.137	6.1437	-0.0067	0.00004489
18	6.217	6.2119	0.0051	0.00002601
19	6.290	6.2801	0.0099	0.00009801
20	6.340	6.3483	-0.0083	0.00006889
				${\displaystyle \sum _{i=1}^4{\left(y-y^{*}\right)}^2}=0.0002378$

Table 2

From table 2, the sum of the differences is 0.0002378.