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 8 Page 370

Make a table of values to determine the residuals.

Sum for y=x+2: 10
Sum for y=x+2.2: 10.16
Better Line of Fit: y=x+2

Practice makes perfect

We have been given the following table.

x 1 2 3 4
y 5 3 6 4
We have also been given two possible lines of fit.
Lines of Fit
y=x+2 y=x+2.2

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=x+2 Residual for y=x+2 y Predicted by y=x+2.2 Residual for y=x+2.2
1 5 y= 1+2= 3 5- 3= 2 y= 1+2.2= 3.2 5- 3.2= 1.8
2 3 y= 2+2= 4 3- 4= -1 y= 2+2.2= 4.2 3- 4.2= -1.2
3 6 y= 3+2= 5 6- 5= 1 y= 3+2.2= 5.2 6- 5.2= 0.8
4 4 y= 4+2= 6 4- 6= -2 y= 4+2.2= 6.2 4- 6.2= -2.2
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=x+2.
s= 2^2+( -1)^2+ 1^2+( -2)^2
s=4+1+1+4
s=10
Thus, the sum of the squared residuals for y=x+2 is 10. Next, we will continue with the line y=x+2.2.
s=( 1.8)^2+( -1.2)^2+( 0.8)^2+( -2.2)^2
s=3.24+1.44+0.64+4.84
s=10.16
The sum of the squared residuals for y=x+2.2 is 10.16. As a result, we can say that y=x+2 is the better line of fit because it has a lesser sum.