Houghton Mifflin Harcourt Algebra 1, 2015
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2. Fitting a Linear Model to Data
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Exercise 14 Page 370

Make a table of values to determine the residuals.

Sum for y=2x+3: 49
Sum for y=2.4x+3: 45
Better Line of Fit: y=2.4x+3

Practice makes perfect

We have been given the following table.

x 1 2 3 4
y 4 11 5 15
We have also been given two possible lines of fit.
Lines of Fit
y=2x+3 y=2.4x+3

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 Residual for y=2x+3 y Predicted by y=2.4x+3 Residual for y=2.4x+3
1 4 y=2( 1)+3= 5 4- 5= -1 y=2.4( 1)+3= 5.4 4- 5.4= -1.4
2 11 y=2( 2)+3= 7 11- 7= 4 y=2.4( 2)+3= 7.8 11- 7.8= 3.2
3 5 y=2( 3)+3= 9 5- 9= -4 y=2.4( 3)+3= 10.2 5- 10.2= -5.2
4 15 y=2( 4)+3= 11 15- 11= 4 y=2.4( 4)+3= 12.6 15- 12.6= 2.4
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+3.
s=( -1)^2+ 4^2+( -4)^2+ 4^2
s=1+16+16+16
s=49
Thus, the sum of the squared residuals for y=2x+3 is 49. Next, we will continue with the line y=2.4x+3.
s=( -1.4)^2+( 3.2)^2+( -5.2)^2+( 2.4)^2
s=1.96+10.24+27.04+5.76
s=45
The sum of the squared residuals for y=2.4x+3 is 45. As a result, we can say that y=2.4x+3 is the better line of fit because it has a lesser sum.