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 11 Page 370

Make a table of values to determine the residuals.

Sum for y=2x+3.1: 18.44
Sum for y=2x+3.5: 21
Better Line of Fit: y=2x+3.1

Practice makes perfect

We have been given the following table.

x 1 2 3 4
y 2 9 7 12
We have also been given two possible lines of fit.
Lines of Fit
y=2x+3.1 y=2x+3.5

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+3.1 Residual for y=2x+3.1 y Predicted by y=2x+3.5 Residual for y=2x+3.5
1 2 y=2( 1)+3.1= 5.1 2- 5.1= -3.1 y=2( 1)+3.5= 5.5 2- 5.5= -3.5
2 9 y=2( 2)+3.1= 7.1 9- 7.1= 1.9 y=2( 2)+3.5= 7.5 9- 7.5= 1.5
3 7 y=2( 3)+3.1= 9.1 7- 9.1= -2.1 y=2( 3)+3.5= 9.5 7- 9.5= -2.5
4 12 y=2( 4)+3.1= 11.1 12- 11.1= 0.9 y=2( 4)+3.5= 11.5 12- 11.5= 0.5
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+3.1.
s=( -3.1)^2+( 1.9)^2+( -2.1)^2+( 0.9)^2
s=9.61+3.61+4.41+0.81
s=18.44
Thus, the sum of the squared residuals for y=2x+3.1 is 18.44. Next, we will continue with the line y=2x+3.5.
s=( -3.5)^2+( 1.5)^2+( -2.5)^2+( 0.5)^2
s=12.25+2.25+6.25+0.25
s=21
The sum of the squared residuals for y=2x+3.5 is 21. As a result, we can say that y=2x+3.1 is the better line of fit because it has a lesser sum.