Houghton Mifflin Harcourt Algebra 1, 2015
HM
Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Fitting a Linear Model to Data
Continue to next subchapter

Exercise 4 Page 370

Make a table of values to determine the residuals.

Sum for y=x+1: 8
Sum for y=x+0.9: 7.33
Better Line of Fit: y=x+0.9

Practice makes perfect

We have been given the following table.

x 1 2 3 4
y 2 1 4 3
We have also been given two possible lines of fit.
Lines of Fit
y=x+1 y=x+0.9

In order to determine which line of fit is better, let's calculate the residuals for both lines.

x y (Actual) y Predicted by y=x+1 Residual for y=x+1 y Predicted by y=x+0.9 Residual for y=x+0.9
1 2 y= 1+1= 2 2- 2= 0 y= 1+0.9= 1.9 2- 1.9= 0.1
2 1 y= 2+1= 3 1- 3= -2 y= 2+0.9= 2.9 1- 2.9= -1.9
3 4 y= 3+1= 4 4- 4= 0 y= 3+0.9= 3.9 4- 3.9= 0.1
4 3 y= 4+1= 5 3- 5= -2 y= 4+0.9= 4.9 3- 4.9= -1.9
Now, we will square the residuals and find their sum s for each line so that we can compare which line is a better fit. Let's start with the line y=x+1.
s= 0^2+( -2)^2+ 0^2+( -2)^2
s=0+4+0+4
s=8
Thus, the sum of the squared residuals for y=x+1 is 8. Next, we will continue with the line y=x+0.9.
s=( 0.1)^2+( -1.9)^2+( 0.1)^2+( -1.9)^2
s=0.01+3.61+0.1+3.61
s=7.33
The sum of the squared residuals for y=x+0.9 is 7.33. As a result, we can say that y=x+0.9 is the better line of fit because it has a lesser sum.