Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Fitting a Linear Model to Data
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Exercise 7 Page 370

Make a table of values to determine the residuals.

Sum for y=2x+1: 238
Sum for y=2x+1.4: 259.44
Better Line of Fit: y=2x+1

Practice makes perfect

We have been given the following table.

x 2 4 6 8
y 3 6 4 5
We have also been given two possible lines of fit.
Lines of Fit
y=2x+1 y=2x+1.4

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=2x+1 Residual for y=2x+1 y Predicted by y=2x+1.4 Residual for y=2x+1.4
2 3 y=2( 2)+1= 5 3- 5= -2 y=2( 2)+1.4= 5.4 3- 5.4= -2.4
4 6 y=2( 4)+1= 9 6- 9= -3 y=2( 4)+1.4= 9.4 6- 9.4= -3.4
6 4 y=2( 6)+1= 13 4- 13= -9 y=2( 6)+1.4= 13.4 4- 13.4= -9.4
8 5 y=2( 8)+1= 17 5- 17= -12 y=2( 8)+1.4= 17.4 5- 17.4= -12.4
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=2x+1.
s=( -2)^2+( -3)^2+( -9)^2+( -12)^2
s=4+9+81+144
s=238
Thus, the sum of the squared residuals for y=2x+1 is 238. Next, we will continue with the line y=2x+1.4.
s=( -2.4)^2+( -3.4)^2+( -9.4)^2+( -12.4)^2
s=5.76+11.56+88.36+153.76
s=259.44
The sum of the squared residuals for y=2x+1.4 is 259.44. As a result, we can say that y=2x+1 is the better line of fit because it has a lesser sum.