2. Fitting a Linear Model to Data - 10. Linear Modeling and Regression - Houghton Mifflin Harcourt Algebra 1, 2015

We are given the following table.

x	1	2	3	4
y	4	7	8	6

We are also given two possible lines of fit.

Lines of Fit
y=x+4	y=x+4.2

We are asked to determine which line of fit is better. To do that, let's first analyze the line of fit y=x+4.

y=x+4

We can calculate the predicted values for y=x+4.

x	y (Actual)	y Predicted by y=x+4
1	4	y= 1+4= 5
2	7	y= 2+4= 6
3	8	y= 3+4= 7
4	6	y= 4+4= 8

Now we can calculate the residuals which are the differences between the actual values and the predicted values.

x	y (Actual)	y Predicted by y=x+4	Residual for y=x+4
1	4	y= 1+4= 5	4- 5= -1
2	7	y= 2+4= 6	7- 6= 1
3	8	y= 3+4= 7	8- 7= 1
4	6	y= 4+4= 8	6- 8= -2

Now, we will square the residuals and find their sum s. This sum will represent how good the fit is. The lower the sum, the better fit. Let's do it!

s=( -1)^2+ 1^2+ 1^2+( -2)^2

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s=1+1+1+4

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s=7

We found that the sum for y=x+4 is s=7. Let's repeat this process for y=x+4.2.

y=x+4.2

Let's find the predicted values and the residuals for y=x+4.2.

x	y (Actual)	y Predicted by y=x+4.2	Residual for y=x+4.2
1	4	y= 1+4.2= 5.2	4- 5.2= -1.2
2	7	y= 2+4.2= 6.2	7- 6.2= 0.8
3	8	y= 3+4.2= 7.2	8- 7.2= 0.8
4	6	y= 4+4.2= 8.2	6- 8.2= -2.2

Now, we will square the residuals and find the sum s for the line y=x+4.2.

s=( -1.2)^2+( 0.8)^2+( 0.8)^2+( -2.2)^2

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s=1.44+0.64+0.64+4.84

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s=7.56

We found that the sum s for y=4.2 is equal to 7.56.

Comparison

We found that the sum of the squared residuals for y=x+4.2 is 7.56 and that the sum for y=x+4 is 7. As a result, we can say that y=x+4 is the better line of fit because it has lesser sum.

Exercise 7 Page 366

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y=x+4

y=x+4.2

Comparison