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 12 Page 370

Make a table of values to determine the residuals.

Sum for y=1.6x+4: 41.64
Sum for y=1.8x+4: 63.96
Better Line of Fit: y=1.6x+4

Practice makes perfect

We have been given the following table.

x 1 3 5 7
y 2 6 8 13
We have also been given two possible lines of fit.
Lines of Fit
y=1.6x+4 y=1.8x+4

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=1.6x+4 Residual for y=1.6x+4 y Predicted by y=1.8x+4 Residual for y=1.8x+4
1 2 y=1.6( 1)+4= 5.6 2- 5.6= -3.6 y=1.8( 1)+4= 5.8 2- 5.8= -3.8
3 6 y=1.6( 3)+4= 8.8 6- 8.8= -2.8 y=1.8( 3)+4= 9.4 6- 9.4= -3.4
5 8 y=1.6( 5)+4= 12 8- 12= -4 y=1.8( 5)+4= 13 8- 13= -5
7 13 y=1.6( 7)+4= 15.2 13- 15.2= -2.2 y=1.8( 7)+4= 16.6 13- 16.6= -3.6
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=1.6x+4.
s=( -3.6)^2+( -2.8)^2+( -4)^2+( -2.2)^2
s=12.96+7.84+16+4.84
s=41.64
Thus, the sum of the squared residuals for y=1.6x+4 is 41.64. Next, we will continue with the line y=1.8x+4.
s=( -3.8)^2+( -3.4)^2+( -5)^2+( -3.6)^2
s=14.44+11.56+25+12.96
s=63.96
The sum of the squared residuals for y=1.8x+4 is 63.96. As a result, we can say that y=1.6x+4 is the better line of fit because it has a lesser sum.