6. Lines of Fit
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Chapter 4
6.
Lines of Fit
In the lesson, the concept of residuals is introduced as the vertical distance between a data point and a line of fit on a scatter plot. The line of fit is essentially a line that best represents a set of data points. To evaluate how well this line fits the data, the sum of squared residuals is calculated. The lower this sum, the better the line models the data. Scatter plots are used to visualize these concepts, helping you understand the relationship between variables. For example, if you are researching used car prices based on mileage, you can plot this data and draw a line of fit to make better predictions. The sum of squared residuals will tell you how reliable your predictions are likely to be.
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Lesson Settings & Tools
| | 8 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Lines of Fit
Slide of 8
There are several ways to analyze how well a line of fit models a data set. In this lesson, the differences between observed data points and the points predicted by the line of fit will be calculated to determine if the regression line models the data well.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Explore
Scatter Plot of Bivariate Data and Line of Fit
Consider the scatter plot of bivariate data. Move the the line in such a way that it models the data.
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Discussion
Residual
One way to assess how well a line of fit describes the data is to analyze residuals.
Concept
Residual
A residual is the vertical distance between a data point and the line of fit. When a line of fit has been drawn on a scatter plot, not all of the data points lie exactly on the line — some of them are above the line and some below. Therefore, each data point has one residual, which can be positive, negative, or zero.
A residual can also be defined as the observed y-value of a data point minus its predicted y-value, found using the line of fit.
Residual = lObserved y-value - lPredicted y-value
Generally, the smaller the absolute values of the residuals, the more reliable the line of fit is. A scatter plot of the residuals can be used to determine how well a model fits data set. The independent variable and the residuals are graphed as ordered pairs (x,residual).
- If the points in the plot are randomly placed about the x-axis, then a linear model describes the data set well.
- If some kind of pattern appears in the scatter plot, a non-linear model is more appropriate for the data.
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Pop Quiz
Practice Finding Sum of Squared Residuals
The applet generates bivariate data and the equation of a line of fit for the data. Calculate the sum of the squares of the residuals for the given equation.
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Example
Determining Which Line of Fit Is Better
The table below shows the finishing times, in seconds, for the Olympic gold medalist in the men's 100-meter dash for the last six Olympic games. Olympic Game 1 represents the 2000 Olympic games and Olympic Game 6 represents the 2020 Olympic games.
| Olympic Game | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Finishing Time (sec) | 9.87 | 9.85 | 9.69 | 9.63 | 9.81 | 9.80 |
Equation I: & y=- 0.1x+10 Equation II: & y = - 0.05x+10 In these equations, x represents the Olympic Game number as described above, and y represents the finishing time in seconds. For each equation, calculate the sum of squared residuals. Which of the equations is a better fit?
b Make a scatter plot for the residuals.
Answer
a Sum of Squared Residuals for Equation I: 0.2605
Sum of Squared Residuals for Equation II: 0.077
b Graph:
Hint
a Make a table of the residual values for each equation and then find the sum of the squares of the residuals.
b Plot the points (x,residual) on a coordinate plane.
Solution
a For the given data set, there are two possible lines of fit.
Equation I: & y=- 0.1x+10 Equation II: & y = - 0.05x+10 In order to determine which line of fit is better, the residuals for both lines will be calculated first. For a data point, its residual is the difference between the y-value of the data point and the y-value predicted by the line of fit.
| x | y (Actual) | y Predicted by y=- 0.1x+10 | Residual for y=- 0.1x+10 |
|---|---|---|---|
| 1 | 9.87 | y=- 0.1( 1)+10= 9.9 | 9.87- 9.9= - 0.03 |
| 2 | 9.85 | y=- 0.1( 2)+10= 9.8 | 9.85- 9.8= 0.05 |
| 3 | 9.69 | y=- 0.1( 3)+10= 9.7 | 9.69- 9.7= - 0.01 |
| 4 | 9.63 | y=- 0.1( 4)+10= 9.6 | 9.63- 9.6= 0.03 |
| 5 | 9.81 | y=- 0.1( 5)+10= 9.5 | 9.81- 9.5= 0.31 |
| 6 | 9.80 | y=- 0.1( 6)+10= 9.4 | 9.80- 9.4= 0.40 |
The values on the last column of the table will be squared and added.
(- 0.03)^2 +0.05^2 + (- 0.01)^2 + 0.03^2 + 0.31^2 + 0.40^2
CalcPow
Calculate power
0.0009 + 0.0025 + 0.0001 + 0.0009 + 0.0961 + 0.16
AddTerms
Add terms
0.2605
The sum of squared residuals for Equation I, SSR_1, is 0.2605. Similarly, the residuals for Equation II will be calculated.
| x | y (Actual) | y Predicted by y=- 0.05x+10 | Residual for y=- 0.05x+10 |
|---|---|---|---|
| 1 | 9.87 | y=- 0.05( 1)+10= 9.95 | 9.87- 9.95= - 0.08 |
| 2 | 9.85 | y=- 0.05( 2)+10= 9.9 | 9.85- 9.9= - 0.05 |
| 3 | 9.69 | y=- 0.05( 3)+10= 9.85 | 9.69- 9.85= - 0.16 |
| 4 | 9.63 | y=- 0.05( 1)+10= 9.8 | 9.63- 9.8= - 0.17 |
| 5 | 9.81 | y=- 0.05( 5)+10= 9.75 | 9.81- 9.75= 0.06 |
| 6 | 9.80 | y=- 0.05( 6)+10= 9.70 | 9.80- 9.70= 0.10 |
Now that the residuals are found, they can be squared and added.
(- 0.08)^2 + (- 0.05)^2 + (- 0.16)^2 + (- 0.17)^2 + 0.06^2 + 0.10^2
CalcPow
Calculate power
0.0064 + 0.0025 + 0.0256 + 0.0289 + 0.0036 + 0.01
AddTerms
Add terms
0.077
The sum of squared residuals for Equation II, SSR_2, is 0.077. As a result, Equation II is the better line of fit because it has a lesser sum of squared residuals. SSR_2 & & SSR_1 0.077 & < & 0.2605
b Recall the residuals for the equations found in the previous part.
| x | Residual for y=- 0.1x+10 | Residual for y=- 0.05x+10 |
|---|---|---|
| 1 | - 0.03 | - 0.08 |
| 2 | 0.05 | - 0.05 |
| 3 | - 0.01 | - 0.16 |
| 4 | 0.03 | - 0.17 |
| 5 | 0.31 | 0.06 |
| 6 | 0.47 | 0.1 |
The points (x,residual) for each equation will be graphed on a scatter plot.
As can be seen, the residuals for Equation II are close the x-axis and therefore the sum of their squares has a lower value.
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Example
Finding the Sum of Squared Residuals Less Than a Certain Value
Maya is researching used cars similar to the one her older sister drives. She found some data showing the mileages x, in thousands of miles, and the selling prices y, in thousands dollars, of several used cars near her.
| x | 23 | 12 | 18 | 30 | 6 | 26 |
|---|---|---|---|---|---|---|
| y | 15 | 15 | 17 | 12 | 19 | 15 |
Maya wants to write the equation of a line of fit for this data set. She knows that the smaller the values of the residuals, the more reliable the line of fit is. She then decides to write an equation in such a way that the sum of squared residual is less than 10. Help Maya write an equation that satisfies the condition.
Answer
Example Equation: y=- 0.25x + 20
Hint
Start by making a scatter plot for the given data set. Draw a line that passes through two data points and write its equation. Then, calculate the residuals for the equation.
Solution
First, the given data will be shown on a graph. Then, a line that passes close to the points will be drawn.
It appears that the line passes through the points (4,20) and (16,16). Knowing two points on the line, the slope of the line and its equation can be found. To find the slope, the points need to be substituted into the Slope Formula.
m = y_2-y_1/x_2-x_1
SubstitutePoints
Substitute ( 4,20) & ( 16,16)
m =16- 20/16- 4
▼
Evaluate right-hand side
SubTerm
Subtract term
m =- 4/12
ReduceFrac
a/b=.a /4./.b /4.
m =- 1/3
MoveNegNumToFrac
Put minus sign in front of fraction
m =- 1/3
Next, using m and either point, the equation of the line can be written in the point-slope form. Use m=- 13 and (x_1,y_1)=(4,20). y-y_1 = m(x-x_1) ⇓ y-20 = - 1/3(x-4) To write the equation in slope-intercept form, y will be isolated.
y-20 = -1/3(x-4)
▼
Solve for y
Distr
Distribute - 1/3
y-20 = -1/3x+4/3
AddEqn
LHS+20=RHS+20
y= - 1/3x+4/3+20
NumberToFrac
a = 3* a/3
y= - 1/3x+4/3+60/3
AddFrac
Add fractions
y= - 1/3x+64/3
Finally, the residuals for this equation can be calculated. To do so, the difference between the y-value of a data point and the corresponding y-value predicted by the line of fit will be calculated for each data point.
| x | y (Actual) | y Predicted by the equation | Residual |
|---|---|---|---|
| 6 | 19 | y=- 1/3( 6)+64/3= 58/3 | 19- 58/3= - 1/3 |
| 12 | 15 | y=- 1/3( 12)+64/3= 52/3 | 15- 52/3= - 7/3 |
| 18 | 17 | y=- 1/3( 18)+64/3= 46/3 | 17- 46/3= 5/3 |
| 23 | 15 | y=- 1/3( 23)+64/3= 41/3 | 15- 41/3= 4/3 |
| 26 | 15 | y=- 1/3( 26)+64/3= 38/3 | 15- 58/3= 7/3 |
| 30 | 12 | y=- 1/3( 30)+64/3= 34/3 | 12- 34/3= 2/3 |
Next, the sum of squared residuals can be found by adding the squares of the numbers in the last column of the table.
(- 1/3 )^2+(- 7/3)^2+(5/3)^2+(4/3)^2+(7/3)^2+ (2/3)^2
▼
Evaluate
NegBaseToPosBase
(- a)^2=a^2
(1/3 )^2+(7/3)^2+(5/3)^2+(4/3)^2+(7/3)^2+ (2/3)^2
PowQuot
(a/b)^m=a^m/b^m
1^2/3^2+7^2/3^2+5^2/3^2+4^2/3^2+7^2/3^2+2^2/3^2
CalcPow
Calculate power
1/9+49/9+25/9+16/9+49/9+4/9
AddFrac
Add fractions
144/9
CalcQuot
Calculate quotient
16
The sum of squared residuals for the equation is 16. c|c Equation & Sum of Residuals [0.4em] [-0.8em] y = - 1/3x+64/3 & 16 This equation does not satisfy the condition that Maya wants. Another equation can be found by slightly increasing the slope of the above equation and slightly decreasing its y-intercept. Use the applet below to find an equation. For example, y=- 0.25x+20 can be a good candidate.
The residuals for this equation can be found as follows.
| x | y (Actual) | y Predicted by the equation | Residual |
|---|---|---|---|
| 6 | 19 | y=- 0.25( 6)+20= 18.5 | 19- 18.5= 0.5 |
| 12 | 15 | y=- 0.25( 12)+20= 17 | 15- 17= - 2 |
| 18 | 17 | y=- 0.25( 18)+20= 15.5 | 17- 15.5= 1.5 |
| 23 | 15 | y=- 0.25( 23)+20= 14.25 | 15- 14.25= 0.75 |
| 26 | 15 | y=- 0.25( 26)+20= 13.5 | 15- 13.5= 1.5 |
| 30 | 12 | y=- 0.25( 30)+20= 12.5 | 12- 12.5= - 0.5 |
Now, the sum of squared residuals for this equation can be calculated.
(0.5)^2+(- 2)^2+(1.5)^2+(0.75)^2+(1.5)^2+(- 0.5)^2
▼
Evaluate
NegBaseToPosBase
(- a)^2=a^2
(0.5)^2+(2)^2+(1.5)^2+(0.75)^2+(1.5)^2+(0.5)^2
CalcPow
Calculate power
0.25+4+2.25+0.5625+2.25+0.25
AddTerms
Add terms
9.5625
As a result, the sum of squared residuals for y=- 0.25x+20 satisfies the condition. c|c Equation & Sum of Residuals [0.4em] [-0.8em] y = - 1/3x+64/3 & 16 > 10 [0.8em] [-0.8em] y=- 0.25x+20 & 9.5625 < 10 Note that there are numerous equations that satisfy Maya's condition. Here only one of them was shown. The following applet shows the equation of a line fit and the sum of squared residuals. Use it to see how the sum changes as the equation changes.
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Pop Quiz
Practice Determining the Better Model
The applet generates bivariate data and two equations that model the data. To see the coordinates of a data point, move the cursor over it. Use the sum of squared residuals to determine the better fit for the data.
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Closure
Finding the Line That Is the Best Fit
In this lesson, lines of fit and their residuals have been analyzed. When a line models a data set well, the sum of the squared residuals for the line is relatively small. Move the points and the line in the graph to see how the residuals and the sum of their squares change.
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Lines of Fit
Exercise 3.1
| Price, x | 19 | 29 | 35 | 43 | 45 |
|---|---|---|---|---|---|
| Number Sold, y | 125 | 105 | 80 | 62 | 45 |
Maya models the data set with two equations. Equation I: & y=- 2.7x+ 176 Equation II: & y = - 3.5x+ 201
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We are asked to find the sum of the square of the residuals of Equation I. Equation I: & y=- 2.7x+176 Let's first find the residuals. For a data point, its residual is the difference between the y-value of the data point and the y-value predicted by the line of fit.
| x | y (Actual) | y Predicted by y=- 2.7x+176 | Residual |
|---|---|---|---|
| 19 | 125 | - 2.7( 19)+176= 124.7 | 125- 124.7= 0.3 |
| 29 | 105 | - 2.7( 29)+176= 97.7 | 105- 97.7= 7.3 |
| 35 | 80 | - 2.7( 35)+176= 81.5 | 80- 81.5= - 1.5 |
| 43 | 62 | - 2.7( 43)+176= 59.9 | 62- 59.9= 2.1 |
| 45 | 45 | - 2.7( 45)+176= 54.5 | 45- 54.5= - 9.5 |
The values on the last column of the table will be squared and added.
The sum of squared residuals for Equation I, SSR_1, is 150.
In a similar fashion, we will find the residuals for Equation II.
| x | y (Actual) | y Predicted by y=- 3.5x+201 | Residual |
|---|---|---|---|
| 19 | 125 | - 3.5( 19)+201= 134.5 | 125- 134.5= - 9.5 |
| 29 | 105 | - 3.5( 29)+201= 99.5 | 105- 99.5= 5.5 |
| 35 | 80 | - 3.5( 35)+201= 78.5 | 80- 78.5= 1.5 |
| 43 | 62 | - 3.5( 43)+201= 50.5 | 62- 50.5= 11.5 |
| 45 | 45 | - 3.5( 45)+201= 43.5 | 45- 43.5= 1.5 |
Now that the residuals are found, they can be squared and added.
The sum of squared residuals for Equation II, SSR_2, is 257.
The smaller the sum of squared residuals, the better the line of fit. Since the Equation I has a lesser sum of squared residuals, it is the better line of fit.
SSR_1 & & SSR_2
150 & < & 257
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Lines of Fit
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