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 10 Page 370

Make a table of values to determine the residuals.

Sum for y=x+1.5: 9
Sum for y=x+1.7: 9.72
Better Line of Fit: y=x+1.5

Practice makes perfect

We have been given the following table.

x 1 2 3 4
y 2 5 4 3
We have also been given two possible lines of fit.
Lines of Fit
y=x+1.5 y=x+1.7

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+1.5 Residual for y=x+1.5 y Predicted by y=x+1.7 Residual for y=x+1.7
1 2 y= 1+1.5= 2.5 2- 2.5= -0.5 y= 1+1.7= 2.7 2- 2.7= -0.7
2 5 y= 2+1.5= 3.5 5- 3.5= 1.5 y= 2+1.7= 3.7 5- 3.7= 1.3
3 4 y= 3+1.5= 4.5 4- 4.5= -0.5 y= 3+1.7= 4.7 4- 4.7= -0.5
4 3 y= 4+1.5= 5.5 3- 5.5= -2.5 y= 4+1.7= 5.7 3- 5.7= -2.7
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+1.5.
s=( -0.5)^2+( 1.5)^2+( -0.5)^2+( -2.5)^2
s=0.25+2.25+0.25+6.25
s=9
Thus, the sum of the squared residuals for y=x+1.5 is 9. Next, we will continue with the line y=x+1.7.
s=( -0.7)^2+( 1.3)^2+( -0.5)^2+( -2.7)^2
s=0.49+1.69+0.25+7.29
s=9.72
The sum of the squared residuals for y=x+1.7 is 9.72. As a result, we can say that y=x+1.5 is the better line of fit because it has a lesser sum.