7. Lines of Best Fit
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Chapter 4
7.
Lines of Best Fit
Understanding the line of best fit is crucial for predicting and modeling data sets. This lesson delves into how to determine the linear function that best represents a scatter plot or data set. By analyzing various examples, learners can grasp the importance of linear correlation and how it impacts the accuracy of predictions. The stronger the correlation, the more precise the predictions are likely to be. Tools like graphing calculators play a pivotal role in this process, aiding in the creation and interpretation of best fit lines. Through this lesson, students can gain a comprehensive view of how to use these lines effectively for predictions, especially when data exhibits a strong linear correlation.
Show less Show more expand_more| Student Learning Objectives: |
|---|
|
| | 8 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Lines of Best Fit
Slide of 8
This lesson aims to show how to find the linear function that better models a scatter plot or data set. Additionally, it will be shown how to use these linear functions to make predictions.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Residual Sum on a Scatter Plot
Given a scatter plot, a line of fit can be used to make good predictions of values that are not known. Since there are many possible lines of fit, finding the one that most accurately represents the given data points is an important goal. The goal is finding the line of fit in which the different sums of the residuals is as close to 0 as possible.
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Discussion
Line of Best Fit
A line of best fit, also known as a regression line, is a line of fit that estimates the relationship between the values of a data set. The equation of the line of best fit has been determined using a strict mathematical method.
One commonly used method to determine a line of best fit is the method of least squares. The methods used to find the line of best fit are usually hard to do by hand. Therefore, a line of best fit can be found by performing a linear regression on a graphing calculator. As an example, consider the data set graphed above.
| x | y |
|---|---|
| 0.6 | 1.5 |
| 1.2 | 3.6 |
| 2.6 | 5.2 |
| 3.6 | 6.3 |
| 4.5 | 8.7 |
| 6 | 10.3 |
| 6.6 | 11.8 |
| 7.1 | 11.7 |
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Example
Finding the Line of Best Fit for Measures of Trees
For a school project, Ramsha wants to investigate if there is a correlation between the width of a tree and its height. To do so, she measured the diameter at chest height and the height of some trees in a local park. Her findings are shown in the following table.
| Diameter at chest (cm) | Height (m) |
|---|---|
| 8 | 7 |
| 10 | 10 |
| 15 | 14 |
| 18 | 15 |
| 20 | 18 |
| 22 | 21 |
| 25 | 15 |
| 30 | 20 |
a
What is the equation of the line of best fit using linear regression? Round the values in the equation to two decimal places.
b
What is the correlation coefficient? Round the value to two decimal places. Are the data correlated?
c
Graph the data points and the line of best fit.
d
Write an interpretation of the y-intercept and the slope.
Answer
a
y = 0.55x + 4.74
b
Correlation Coefficient: r≈ 0.86
Are the Data Correlated? Yes, see solution.
c
Graph:
d
See solution.
Hint
a
Use the linear regression feature on a graphing calculator.
b
The correlation coefficient is r in the linear regression output on a graphing calculator.
c
Use the graphing features on a graphing calculator.
d
What do these measures indicate in a line?
Solution
a
The line of best fit can be found using a graphing calculator. First, the data values need to be introduced into the calculator. This is done by pressing the STAT button and then selecting the option Edit.
Then the data values are written in the columns.
By pressing the STAT button and then selecting the CALC menu, the option LinReg(ax+b)
can be found. This option gives the line of best fit, expressed as a linear function in slope-intercept form.
Rounding the values of a and b to two decimal places, the equation of the line of best fit can be written as follows. y = 0.55x + 4.74
b
The correlation coefficient can be found on the linear regression results screen from Part A.
The correlation coefficient is the value of r on the screen. r ≈ 0.86 The value of r varies from -1 to 1. A value close to -1 indicates a negative correlation, while a value close to 1 indicates a positive correlation. Since the correlation coefficient is close to 1, the data has a strong positive correlation.
c
To graph the line of best fit, first press Y= and write the equation of the line of best fit.
Then, to graph the scatter plot push the buttons 2nd and Y=. Choose one of the plots in the list. Select the option ON,
choose the type to be a scatter plot, and assign L_1 and L_2 as XList
and Ylist,
respectively.
The plot can be made by pressing the button GRAPH. It is possible that after drawing the plot the window-size is not adequate for seeing all the information.
To fix this press ZOOM and select the option ZoomStat.
After doing so the window will resize to show the important information.
d
In Part A the equation for the line of best fit was found.
y = 0.55x + 4.74 In this equation, the slope is 0.55 and the y-intercept is 4.74.
- The slope indicates that for every centimeter that the tree grows in diameter, about 0.55 meters are gained in height.
- The y-intercept indicates that the minimum height for which this equation is valid is about 4.74 meters. This can be because the diameter at chest height is not a good measure for smaller trees.
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Pop Quiz
Practice Finding the Line of Best Fit
Use the linear regression feature of a graphing calculator to find the equation of the line of best fit for the given data set. Compare the obtained equation with the equations shown in the applet, and choose the closest one.
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Example
Predicting the Pressure at a Certain Altitude
The following table displays some values of atmospheric pressures at different altitudes.
| Altitude (thousand feet) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Pressure (PSI) | 14.71 | 14.18 | 13.75 | 13.21 | 12.69 | 12.20 |
a
Use linear regression to determine the equation of the line of best fit. Round the values in the equation to two decimal places.
b
What is the correlation coefficient? Round the answer to two decimal places. Is the data correlated?
c
Draw a graph of the data points and the line of best fit in the same viewing window.
d
Interpret the slope and the y-intercept of the equation of the line of best fit.
e
Make a prediction for the pressure at 6000 feet. Is this a good prediction?
Answer
a
y = -0.50x + 14.71
b
Correlation Coefficient: r ≈ - 1
Is the Data Correlated? Yes, because the correlation coefficient is really close to -1.
c
Graph
d
See solution.
e
Prediction: About 11.7 PSI
Is This a Good Prediction? Yes, see solution.
Hint
a
Use the linear regression feature on a graphing calculator.
b
The correlation coefficient is r in the linear regression output on a graphing calculator.
c
Use the graphing features on a graphing calculator.
d
What do these measures indicate in a line?
e
Are the data values close to the line of best fit? Does this indicate something?
Solution
a
The line of best fit can be found using a graphing calculator. First, the data values need to be introduced into the calculator. This is done by pressing the STAT button and selecting the option Edit.
Then the data values can be written in the columns.
Finally, by pressing the STAT button and then selecting the menu item CALC, the option LinReg(ax+b)
can be found. This option gives a line of best fit, expressed as a linear function in slope-intercept form.
Rounding to two decimal places, the equation for the line of best fit using linear regression can be written as follows. y = -0.50x + 14.71
b
The correlation coefficient can be found on the linear regression results screen from Part A.
The correlation coefficient is the value of r on the screen. r = - 0.999673... The value of r varies from -1 to 1. A value close to -1 indicates a negative correlation, while a value close to 1 indicates a positive correlation. Since the value is almost -1, the data have a very strong negative correlation.
c
To graph the line of best fit, first press Y= and write the equation of the line of best fit.
To graph the scatter plot, first push the buttons 2nd and Y=. Then, choose one of the plots in the list. Select the option ON,
choose the type to be a scatter plot, and assign L_1 and L_2 as XList
and Ylist,
respectively.
The plot can be made by pressing the button GRAPH. It is possible that after drawing the plot the window-size is not large enough to see all of the information.
To fix this, press ZOOM and select the option ZoomStat.
After doing that, the window will resize to show the important information.
d
In Part A the equation for the line of best fit was found.
y = - 0.50x + 14.71 In this equation, the slope is - 0.50 and the y-intercept is 14.71.
- The slope indicates that every one thousand feet of altitude, the pressure diminishes by about 0.50 PSI.
- The y-intercept indicates that the pressure at sea level is about 14.71 PSI.
e
A graphing calculator can also be used to make predictions. To do so, first the window size should be changed to fit the prediction. Since the x-value is given in thousands of feet, the value that should be included for 6000 feet is x=6. To change the window size, press WINDOW.
To find the value of y when x=6, press CALC (2ND and TRACE). Then press ENTER to insert the value of 6 for x. Finally, press ENTER again.
The value of the pressure at 6000 feet is about 11.7 PSI. Since all the data values are close to the line of best fit and the data is strongly correlated, it can be said that this is a good approximation of the actual value.
Solving Using the Equation
This prediction can also be found by substituting 6 for x into the equation for the line of best fit.
y =- 0.50x + 14.71
▼
Solve for y
y = 11.71
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Example
Predicting the Time It Takes to Finish an Exercise
Davontay has a math assignment that consists of eight different exercises. He registered the time (in minutes) in which he completed the first seven exercises.
| Exercise | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Time (minutes) | 4 | 15 | 7 | 16 | 8 | 15 | 5 |
a
What is the equation for the line of best fit using linear regression? Round the values to two decimal places.
b
What is the correlation coefficient? Round the answer to two decimal places. Are the data correlated?
c
Draw a graph of the data points and the line of best fit in the same viewing window.
d
Interpret the slope and the y-intercept of the equation of the line of best fit.
e
Find a prediction for the for the time it will take Davontay to complete the eighth exercise. Is it a good prediction?
Answer
a
y = 0.14x + 9.43
b
Correlation Coefficient: r ≈ 0.06
Are the Data Correlated? No, see solution.
c
Graph:
d
See solution.
e
Prediction: About 10.57 minutes
Is It a Good Prediction? No, see solution.
Hint
a
Use the linear regression feature on a graphing calculator or computer.
b
The correlation coefficient is r in the linear regression output on a graphing calculator.
c
Use the graphing features on a graphing calculator.
d
What do these measures indicate in a line?
e
Are the data values close to the line of best fit? Does this indicate something?
Solution
a
The line of best fit can be found using a graphing calculator. First, the data values need to be introduced into the calculator. This is done by pressing the STAT button and then selecting the option Edit.
The data values can be written in the columns.
Finally, by pressing the STAT button and then selecting the menu item CALC, the option LinReg(ax+b)
can be found. This option gives a line of best fit, expressed as a linear function in slope-intercept form.
Rounding to two decimal places, the equation for the line of best fit using linear regression can be written as follows. y = 0.14x + 9.43 It should be noted that in this equation x is the number of the exercise and y is the time in minutes in which Davontay completed that exercise.
b
The correlation coefficient can be found on the linear regression results screen from Part A.
The correlation coefficient is the value of r on the screen. r = 0.059761... The value of r varies from -1 to 1. A value close to -1 indicates a negative correlation, while a value close to 1 indicates a positive correlation. But the value of r is close to 0, this means that the data have no correlation.
c
To graph the line of best fit, first press Y= and write the equation of the line of best fit.
To graph the scatter plot, first push the buttons 2nd and Y=. Then, choose one of the plots in the list. Select the option ON,
choose the type to be a scatter plot, and assign L_1 and L_2 as XList
and Ylist,
respectively.
The plot can be made by pressing the button GRAPH. It is possible that after drawing the plot the window-size is not large enough to see all of the information.
To fix this, press ZOOM and select the option ZoomStat.
After doing that, the window will resize to show the important information.
Looking at the graph, it can be seen that the line of best fit is not close to any of the provided data points.
d
In Part A the equation for the line of best fit was found.
y = 0.14x + 9.43 In this equation, the slope is 0.14 and the y-intercept is 9.43.
- The slope indicates that after every exercise, the next one takes about an additional 0.14 minutes to complete.
- The y-intercept indicates that the minimum time to complete an exercise is about 9.43 minutes.
From Part C it can be noted that the line is not representative of the given data points. This means that these measures do not reflect the reality of the exercises.
e
A graphing calculator can also be used to make predictions. To do so, first the window size should be changed to fit the prediction value of x=8. To change the window size, press WINDOW.
Then, to find the value of y when x=8, press CALC (2ND and TRACE). Then press ENTER to insert the value of 8 for x. Finally, press ENTER again.
The value of y when x=8 is about 10.57. This means that Davontay will finish the eighth exercise in less than 11 minutes. Since none of the given data values are really close to the line of best fit and the data is not correlated, it can be said that this is a not good approximation for the actual value.
Solving Using the Equation of the Line of Best Fit
This prediction can also be found by substituting 8 for x into the equation for the line of best fit.
y =0.14x + 9.43
▼
Solve for y
y = 10.55
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Closure
Are Lines of Best Fit Always Useful?
In this lesson it was shown how to find the line of best fit for data sets and how to make predictions using these lines. Considering the examples discussed throughout the lesson, it is possible to make two conclusions.
- The lines of best fit are good to make predictions only if the data have a linear correlation.
- The stronger the correlation is, the more accurate the predictions are expected to be.
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Lines of Best Fit
Exercise 2.1
Consider the following graph of a data set and line of best fit.
Of the following options, which is likely the correlation coefficient for the data?
{"type":"choice","form":{"alts":["r = 0","r=0.95","r=0.64","r=-0.92"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}
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To solve this exercise, we will begin by analyzing the given graph. After we make out the characteristics of the graph, we can see which of the given answer options is the best. Remember that the correlation coefficient indicates how close the data points are to the line of best fit.
Looking at the graph, we can see that the line of best fit has a positive slope. This indicates that the correlation coefficient is positive, which limits our options to 0.64 and 0.95 for the value of r. & r = - 0.92 & r =0 & r = 0.64 & r = 0.95
Since the points are really close to the line of best fit, we can safely say that 0.95 is the better option of the remaining two choices for r.
For clarity, we will look at graphs that fit the other options. First, let's look how a graph with a correlation coefficient of r=0.64 would look.
Looking at the graph, we can notice that even if the slope of the line is really similar, the data points are considerably farther apart from the line of best fit than that in the original graph. Now, let's take a look at a graph with a correlation coefficient of 0.
These graphs have a slope of 0, and the points do not neccesarily follow a pattern. Finally, let's look at a graph with coefficient - 0.92.
Looking carefully, we can notice that the data points are close to the line, but there are some points that are a little farther appart than the points with the coefficient of 0.95. More importantly, the graph has a negative slope, unlike the given data set's line of best fit.
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Solution
For each of the following data sets, its line of best fit is shown. Which data set has a correlation coefficient of 0?
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}
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We are asked which of the data sets has a correlation coefficient of 0. r = 0 We know that a correlation coefficient r lies between -1 and 1.
| Value of r | Correlation |
|---|---|
| r < 0 | Negative correlation |
| r>0 | Positive correlation |
| r=0 | No correlation |
The correlation also gives information about the slope of the regression lines.
| Correlation | Slope of the Regression Line |
|---|---|
| Negative correlation | Negative |
| Positive correlation | Positive |
| No correlation | No slope |
Looking at the table, we can see that we need to look for a data set with no correlation. Let's examine the graphs!
Looking at the graphs, we can see that the graph D is the only one that displays no correlation, since the line has no noticeable slope.
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Solution
| Hours x | GPA y |
|---|---|
| 11 | 3.1 |
| 4 | 3.5 |
| 3 | 3.6 |
| 12 | 2.8 |
| 20 | 2.2 |
| 14 | 2.9 |
| 8 | 3.1 |
| 4 | 3.7 |
| 15 | 2.3 |
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{"type":"choice","form":{"alts":["Students who stream no shows have a GPA of about 3.9","Students with low GPA stream for 3.9 hours.","Each additional hour streaming shows reduces 0.08 from the student's GPA.","Each additional hour streaming shows adds 0.08 to the student's GPA."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
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{"type":"choice","form":{"alts":["No","Yes"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
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We have been given a table with data for hours x spent streaming and student GPA y.
| Hours x | Grade Point Average y |
|---|---|
| 11 | 3.1 |
| 4 | 3.5 |
| 3 | 3.6 |
| 12 | 2.8 |
| 20 | 2.2 |
| 14 | 2.9 |
| 8 | 3.1 |
| 4 | 3.7 |
| 15 | 2.3 |
In order to find a line of fit using our calculator, we need to first enter the values. Let's press the STAT button.
Next, we will choose the first option in the menu Edit,
and we can fill in the values in lists L1 and L2.
We can perform a regression analysis on the data by pressing the STAT button again, followed by using the right-arrow key to select the CALC menu.
This menu lists the various regressions that are available. If we choose the fourth option in the menu LinReg(ax+b)
and press ENTER, the calculator performs a linear regression using the data that was entered.
We can round the values of a and b and substitute them into the equation y= ax+ b. This gives us the equation for the line of best fit. y= -0.08 x+ 3.9 The correlation coefficient is r≈ -0.955. Since r is very close to -1, we know that there is a strong, negative correlation. Therefore, the only true statement is that the equation for the line of best fit is y=-0.08x + 3.9.
In Part A we found that the equation of the line of best fit is y= -0.08x+ 3.9. Let's identify its slope and y-intercept. Slope:& m= -0.08 y-intercept:& b= 3.9 The y-intercept of 3.9 tells us that students who stream for 0 each week should have a GPA around 3.9.
To predict the GPA of a student who streams 15 hours a week, we can substitute x= 15 into our line of best fit.
According to the line of best fit, the GPA of the student is expected to be 2.7.
Let's recall what it means to have causation.
There is causation between two variables when a change in one variable causes a change in the other.
GPA is dependent on how much time is spent on studying. This means that causation will exist if streaming less shows leaves more time for studying. However, the time left by watching less series does not always directly lead to studying more. Therefore, there is no direct causal relationship.
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Solution
Lines of Best Fit
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