Question

The actual data below is shown along with two possible lines of best fit. Use the blank tables to calculate theresiduals for both lines of fit and then find the sum of the squared residuals. Identify the line with better fit.17. (1 point)Xy(actual)PredictedfromEquationResidualsLine of fit A:2Square ofResidualsxy (actual)13213y = 3x + 5312418312523418523Line of fit B: y = 3.5x + 4231312(actual)PredictedfromEquationResidualsPossible Line of fit A: y = 3x + 5Possible Line of fit B: y = 3.5x + 4Square ofResidualsCompare the sum of the squared residuals to determine which line has the better fit.418co523

Answer 1

Let us first complete the first table for the line of fit A. Using the given x-values and the equation y = 3x + 5, we predict the values of x to be:

if x = 2, y = 3(2) + 5 = 11

if x = 3, y = 3(3) + 5 = 14

if x = 4, y = 3(4) + 5 = 17

if x = 5, y = 3(5) + 5 = 20

Then we calculate for the residuals by subtrating the actual y-values and the predicted y-values.

for x = 2, residual = 13 - 11 = 2

for x = 3, residual = 12 - 14 = -2

for x = 4, residual = 18 - 17 = 1

for x = 5, residual = 23 - 20 = 3

Then we square the residuals.

2^2 = 4

(-2)^2 = 4

1^2 = 1

3^2 = 9

We have completed the first table. We repreat the same process to complete the second table for the line of fit B.

Finally, we add the squares of the residuals.

A: 4 + 4 + 1 + 9 =18

B: 4 + 6.25 + 0 + 2.25 = 12.5

Comparing the sums of the squares of the residuals, we see that Line of fit B has the better fit becaue it has a smaller sum of the squares of the residuals, meaning there is less variation.

To answer the follow up question (#18), remember that the smaller the sum of the squares of the residuals, the better the fit.

So, the answers are: true, false, and true.