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 3 Page 369

Make a table of values to determine the residuals.

Sum for y=3x+4: 936
Sum for y=3x+4.1: 947.24
Better Line of Fit: y=3x+4

Practice makes perfect

We have been given the following table.

x 2 4 6 8
y 2 8 4 6
We have also been given two possible lines of fit.
Lines of Fit
y=3x+4 y=3x+4.1

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=3x+4 Residual for y=3x+4 y Predicted by y=3x+4.1 Residual for y=3x+4.1
2 2 y=3( 2)+4= 10 2- 10= -8 y=3( 2)+4.1= 10.1 2- 10.1= -8.1
4 8 y=3( 4)+4= 16 8- 16= -8 y=3( 4)+4.1= 16.1 8- 16= -8.1
6 4 y=3( 6)+4= 22 4- 22= -18 y=3( 6)+4.1= 22.1 4- 22.1= -18.1
8 6 y=3( 8)+4= 28 6- 28= -22 y=3( 8)+4.1= 28.1 6- 28.1= -22.1
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=3x+4.
s=( -8)^2+( -8)^2+( -18)^2+( -22)^2
s=64+64+324+484
s=936
Thus, the sum of the squared residuals for y=3x+4 is 936. Next, we will continue with the line y=3x+4.1.
s=( -8.1)^2+( -8.1)^2+( -18.1)^2+( -22.1)^2
s=65.61+65.61+327.61+488.41
s=947.24
The sum of the squared residuals for y=3x+4.1 is 947.24. As a result, we can say that y=3x+4.1 is the better line of fit because it has a lesser sum.