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 2 Page 369

Make a table of values to determine the residuals.

Sum for y=2x+1: 40
Sum for y=2x+1.1: 41.64
Better Line of Fit: y=2x+1

Practice makes perfect

We have been given the following table.

x 1 2 3 4
y 1 7 3 5
We have also been given two possible lines of fit.
Lines of Fit
y=2x+1 y=2x+1.1

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=2x+1 Residual for y=2x+1 y Predicted by y=2x+1.1 Residual for y=2x+1.1
1 1 y=2( 1)+1= 3 1- 3= -2 y=2( 1)+1.1= 3.1 1- 3.1= -2.1
2 7 y=2( 2)+1= 5 7- 5= 2 y=2( 2)+1.1= 5.1 7- 5.1= 1.9
3 3 y=2( 3)+1= 7 3- 7= -4 y=2( 3)+1.1= 7.1 3- 7.1= -4.1
4 5 y=2( 4)+1= 9 5- 9= -4 y=2( 4)+1.1= 9.1 5- 9.1= -4.1
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=2x+1.
s=( -2)^2+ 2^2+( -4)^2+( -4)^2
s=4+4+16+16
s=40
Thus, the sum of the squared residuals for y=2x+1 is 40. Next, we will continue with the line y=2x+1.1.
s=( -2.1)^2+( 1.9)^2+( -4.1)^2+( -4.1)^2
s=4.41+3.61+16.81+16.81
s=41.64
The sum of the squared residuals for y=2x+1.1 is 41.64. As a result, we can say that y=2x+1 is the better line of fit because it has a lesser sum.