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# Fitting Linear Models to Data

Module by: First Last. E-mail the author

Summary: In this section, you will:

• Draw and interpret scatter plots.
• Find the line of best fit.
• Distinguish between linear and nonlinear relations.
• Use a linear model to make predictions.

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A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot.

## Drawing and Interpreting Scatter Plots

A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure 1 shows a sample scatter plot.

Notice this scatter plot does not indicate a linear relationship. The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam.

### Example 1

#### Problem 1

##### Using a Scatter Plot to Investigate Cricket Chirps

Table 1 shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit1. Plot this data, and determine whether the data appears to be linearly related.

 Chirps 44 35 20.4 33 31 35 18.5 37 26 Temperature 80.5 70.5 57 66 68 72 52 73.5 53
##### Solution

Plotting this data, as depicted in Figure 2 suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so.

## Finding the Line of Best Fit

Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear function?” One way to approximate our linear function is to sketch the line that seems to best fit the data. Then we can extend the line until we can verify the y-intercept. We can approximate the slope of the line by extending it until we can estimate the riserun.riserun.

### Example 2

#### Problem 1

##### Finding a Line of Best Fit

Find a linear function that fits the data in Table 1 by “eyeballing” a line that seems to fit.

##### Solution

On a graph, we could try sketching a line.

Using the starting and ending points of our hand drawn line, points (0, 30) and (50, 90), this graph has a slope of

m=6050=1.2m=6050=1.2

and a y-intercept at 30. This gives an equation of

T(c)=1.2c+30T(c)=1.2c+30

where cc is the number of chirps in 15 seconds, and T(c)T(c) is the temperature in degrees Fahrenheit. The resulting equation is represented in Figure 3.

##### Analysis

This linear equation can then be used to approximate answers to various questions we might ask about the trend.

### Recognizing Interpolation or Extrapolation

While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship will behave outside of the values for which we have data. We use a process known as interpolation when we predict a value inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain and range of the data.

Figure 4 compares the two processes for the cricket-chirp data addressed in Example 2. We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44.

There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer applies after a certain point, it is sometimes called model breakdown. For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when x=50,x=50, that is in 50 years, the model would not apply because we could not account for factors fifty years in the future.

#### A General Note: Interpolation and Extrapolation:

Different methods of making predictions are used to analyze data.

• The method of interpolation involves predicting a value inside the domain and/or range of the data.
• The method of extrapolation involves predicting a value outside the domain and/or range of the data.
• Model breakdown occurs at the point when the model no longer applies.

#### Example 3

##### Problem 1
###### Understanding Interpolation and Extrapolation

Use the cricket data from Table 1 to answer the following questions:

1. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
2. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
###### Solution
1. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model:
T(30)=30+1.2(30)          =66degrees T(30)=30+1.2(30)          =66degrees

Based on the data we have, this value seems reasonable.
2. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model:
40=30+1.2c 10=1.2c  c8.33 40=30+1.2c 10=1.2c  c8.33

We can compare the regions of interpolation and extrapolation using Figure 5.

###### Analysis

Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees.

#### Try It:

##### Exercise 1

According to the data from Table 1, what temperature can we predict it is if we counted 20 chirps in 15 seconds?

54°F54°F

### Finding the Line of Best Fit Using a Graphing Utility

While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values2. One such technique is called least squares regression and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators3. Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression.

#### How To:

Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression.

1. Enter the input in List 1 (L1).
2. Enter the output in List 2 (L2).
3. On a graphing utility, select Linear Regression (LinReg).

#### Example 4

##### Problem 1
###### Finding a Least Squares Regression Line

Find the least squares regression line using the cricket-chirp data in Table 1.

###### Solution
1. Enter the input (chirps) in List 1 (L1).
2. Enter the output (temperature) in List 2 (L2). See Table 2.
 L1 44 35 20.4 33 31 35 18.5 37 26 L2 80.5 70.5 57 66 68 72 52 73.5 53
3. On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation:
T(c)=30.281+1.143cT(c)=30.281+1.143c
###### Analysis

Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to:

T(30)=30.281+1.143(30)          =64.571          64.6 degrees T(30)=30.281+1.143(30)          =64.571          64.6 degrees

The graph of the scatter plot with the least squares regression line is shown in Figure 6.

#### Q&A:

Will there ever be a case where two different lines will serve as the best fit for the data?

No. There is only one best fit line.

## Distinguishing Between Linear and Non-Linear Models

As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the correlation coefficient, which is a measure of how closely the line fits the data. Many graphing calculators require the user to turn a ”diagnostic on” selection to find the correlation coefficient, which mathematicians label as r.r. The correlation coefficient provides an easy way to get an idea of how close to a line the data falls.

We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless. To get a sense for the relationship between the value of rr and the graph of the data, Figure 7 shows some large data sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis shows the output.

### A General Note: Correlation Coefficient:

The correlation coefficient is a value,r,r, between –1 and 1.

• r > 0 suggests a positive (increasing) relationship
• r < 0 suggests a negative (decreasing) relationship
• The closer the value is to 0, the more scattered the data.
• The closer the value is to 1 or –1, the less scattered the data is.

### Example 5

#### Problem 1

##### Finding a Correlation Coefficient

Calculate the correlation coefficient for cricket-chirp data in Table 1.

##### Solution

Because the data appear to follow a linear pattern, we can use technology to calculate r.r. Enter the inputs and corresponding outputs and select the Linear Regression. The calculator will also provide you with the correlation coefficient, r=0.9509.r=0.9509. This value is very close to 1, which suggests a strong increasing linear relationship.

Note: For some calculators, the Diagnostics must be turned "on" in order to get the correlation coefficient when linear regression is performed: [2nd]>[0]>[alpha][x–1], then scroll to DIAGNOSTICSON.

## Predicting with a Regression Line

Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that only one such line is a best fit for the data.

### Example 6

#### Problem 1

##### Using a Regression Line to Make Predictions

Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in Table 34. Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008.

 Year '94 '95 '96 '97 '98 '99 '00 '01 '02 '03 '04 Consumption (billions of gallons) 113 116 118 119 123 125 126 128 131 133 136

The scatter plot of the data, including the least squares regression line, is shown in Figure 8.

##### Solution

We can introduce new input variable, t,t,representing years since 1994.

The least squares regression equation is:

C(t)=113.318+2.209tC(t)=113.318+2.209t

Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend.

Using this to predict consumption in 2008 (t=14),(t=14),

C(14)=113.318+2.209(14)          =144.244 C(14)=113.318+2.209(14)          =144.244

The model predicts 144.244 billion gallons of gasoline consumption in 2008.

### Try It:

#### Exercise 2

Use the model we created using technology in Example 6 to predict the gas consumption in 2011. Is this an interpolation or an extrapolation?

##### Solution

150.871 billion gallons; extrapolation

### Media:

Access these online resources for additional instruction and practice with fitting linear models to data.

Visit this website for additional practice questions from Learningpod.

## Key Concepts

• Scatter plots show the relationship between two sets of data. See Example 1.
• Scatter plots may represent linear or non-linear models.
• The line of best fit may be estimated or calculated, using a calculator or statistical software. See Example 2.
• Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data. See Example 3.
• The correlation coefficient, r,r, indicates the degree of linear relationship between data. See Example 5.
• A regression line best fits the data. See Example 6.
• The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables. See Example 4.

## Section Exercises

### Verbal

#### Exercise 3

Describe what it means if there is a model breakdown when using a linear model.

##### Solution

When our model no longer applies, after some value in the domain, the model itself doesn’t hold.

#### Exercise 4

What is interpolation when using a linear model?

#### Exercise 5

What is extrapolation when using a linear model?

##### Solution

We predict a value outside the domain and range of the data.

#### Exercise 6

Explain the difference between a positive and a negative correlation coefficient.

#### Exercise 7

Explain how to interpret the absolute value of a correlation coefficient.

##### Solution

The closer the number is to 1, the less scattered the data, the closer the number is to 0, the more scattered the data.

### Algebraic

#### Exercise 8

A regression was run to determine whether there is a relationship between hours of TV watched per day (x)(x) and number of sit-ups a person can do (y).(y). The results of the regression are given below. Use this to predict the number of sit-ups a person who watches 11 hours of TV can do.

y=ax+b a=−1.341 b=32.234 r=−0.896 y=ax+b a=−1.341 b=32.234 r=−0.896

#### Exercise 9

A regression was run to determine whether there is a relationship between the diameter of a tree (x,x,in inches) and the tree’s age (y,y,in years). The results of the regression are given below. Use this to predict the age of a tree with diameter 10 inches.

y=ax+b a=6.301 b=−1.044 r=−0.970 y=ax+b a=6.301 b=−1.044 r=−0.970
##### Solution

61.966 years

For the following exercises, draw a scatter plot for the data provided. Does the data appear to be linearly related?

#### Exercise 10

 0 2 4 6 8 10 –22 –19 –15 –11 –6 –2

#### Exercise 11

 1 2 3 4 5 6 46 50 59 75 100 136

No.

#### Exercise 12

 100 250 300 450 600 750 12 12.6 13.1 14 14.5 15.2

#### Exercise 13

 1 3 5 7 9 11 1 9 28 65 125 216

No.

#### Exercise 14

For the following data, draw a scatter plot. If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? Eyeball the line, and estimate the answer.

 Year Population 1990 11,500 1995 12,100 2000 12,700 2005 13,000 2010 13,750

#### Exercise 15

For the following data, draw a scatter plot. If we wanted to know when the temperature would reach 28 °F, would the answer involve interpolation or extrapolation? Eyeball the line and estimate the answer.

 Temperature, °F 16 18 20 25 30 Time, seconds 46 50 54 55 62
##### Solution

Interpolation. About 60° F. 60° F.

### Graphical

For the following exercises, match each scatterplot with one of the four specified correlations in Figure 9 and Figure 10.

r=0.95r=0.95

r=0.89r=0.89

C

r=0.26r=0.26

#### Exercise 19

r=0.39r=0.39

##### Solution

B

For the following exercises, draw a best-fit line for the plotted data.

### Numeric

#### Exercise 24

The U.S. Census tracks the percentage of persons 25 years or older who are college graduates. That data for several years is given in Table 105. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the percentage exceed 35%?

 Year Percent Graduates 1990 21.3 1992 21.4 1994 22.2 1996 23.6 1998 24.4 2000 25.6 2002 26.7 2004 27.7 2006 28 2008 29.4

#### Exercise 25

The U.S. import of wine (in hectoliters) for several years is given in Table 11. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will imports exceed 12,000 hectoliters?

 Year Imports 1992 2665 1994 2688 1996 3565 1998 4129 2000 4584 2002 5655 2004 6549 2006 7950 2008 8487 2009 9462
##### Solution

Yes, trend appears linear because r=0.985r=0.985 and will exceed 12,000 near midyear, 2016, 24.6 years since 1992.

#### Exercise 26

Table 12 shows the year and the number of people unemployed in a particular city for several years. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the number of unemployed reach 5?

 Year Number Unemployed 1990 750 1992 670 1994 650 1996 605 1998 550 2000 510 2002 460 2004 420 2006 380 2008 320

### Technology

For the following exercises, use each set of data to calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to 3 decimal places of accuracy.

#### Exercise 27

 xx 8 15 26 31 56 yy 23 41 53 72 103
##### Solution

y=1.640x+13.800y=1.640x+13.800, r=0.987r=0.987

#### Exercise 28

 xx 5 7 10 12 15 yy 4 12 17 22 24

#### Exercise 29

 xx yy xx y y 3 21.9 11 15.76 4 22.22 12 13.68 5 22.74 13 14.1 6 22.26 14 14.02 7 20.78 15 11.94 8 17.6 16 12.76 9 16.52 17 11.28 10 18.54 18 9.1
##### Solution

y=0.962x+26.86,   r=0.965 y=0.962x+26.86,   r=0.965

#### Exercise 30

 xx yy 4 44.8 5 43.1 6 38.8 7 39 8 38 9 32.7 10 30.1 11 29.3 12 27 13 25.8

#### Exercise 31

 xx yy 21 17 25 11 30 2 31 −1 40 −18 50 −40
##### Solution

y=1.981x+60.197y=1.981x+60.197; r=0.998r=0.998

#### Exercise 32

 xx yy 100 2000 80 1798 60 1589 55 1580 40 1390 20 1202

#### Exercise 33

 xx yy 900 70 988 80 1000 82 1010 84 1200 105 1205 108
##### Solution

y=0.121x38.841,r=0.998y=0.121x38.841,r=0.998

### Extensions

#### Exercise 34

Graph f(x)=0.5x+10f(x)=0.5x+10. Pick a set of 5 ordered pairs using inputs x=−2, 1, 5, 6, 9x=−2, 1, 5, 6, 9 and use linear regression to verify that the function is a good fit for the data.

#### Exercise 35

Graph f(x)=2x10f(x)=2x10. Pick a set of 5 ordered pairs using inputs x=−2, 1, 5, 6, 9x=−2, 1, 5, 6, 9 and use linear regression to verify the function.

##### Solution

(−2,−6),(1,−12),(5,−20),(6,−22),(9,−28)(−2,−6),(1,−12),(5,−20),(6,−22),(9,−28); y=−2x−10y=−2x−10

For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs shows dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years:

(46, 1,600), (48, 1,550), (50, 1,505), (52, 1,540), (54, 1,495)(46, 1,600), (48, 1,550), (50, 1,505), (52, 1,540), (54, 1,495).

#### Exercise 36

Use linear regression to determine a function PP where the profit in thousands of dollars depends on the number of units sold in hundreds.

#### Exercise 37

Find to the nearest tenth and interpret the x-intercept.

##### Solution

(189.8,0) (189.8,0)  If 18,980 units are sold, the company will have a profit of zero dollars.

#### Exercise 38

Find to the nearest tenth and interpret the y-intercept.

### Real-World Applications

For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population and the year over the ten-year span, (population, year) for specific recorded years:

(2500, 2000), (2650, 2001), (3000, 2003), (3500, 2006), (4200, 2010) (2500, 2000), (2650, 2001), (3000, 2003), (3500, 2006), (4200, 2010)

#### Exercise 39

Use linear regression to determine a function y,y, where the year depends on the population. Round to three decimal places of accuracy.

##### Solution

y=0.00587x+1985.41y=0.00587x+1985.41

#### Exercise 40

Predict when the population will hit 8,000.

For the following exercises, consider this scenario: The profit of a company increased steadily over a ten-year span. The following ordered pairs show the number of units sold in hundreds and the profit in thousands of over the ten year span, (number of units sold, profit) for specific recorded years:

(46, 250), (48, 305), (50, 350), (52, 390), (54, 410)(46, 250), (48, 305), (50, 350), (52, 390), (54, 410).

#### Exercise 41

Use linear regression to determine a function y, where the profit in thousands of dollars depends on the number of units sold in hundreds .

##### Solution

y=20.25x671.5y=20.25x671.5

#### Exercise 42

Predict when the profit will exceed one million dollars.

For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs show dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span (number of units sold, profit) for specific recorded years:

(46, 250), (48, 225), (50, 205), (52, 180), (54, 165). (46, 250), (48, 225), (50, 205), (52, 180), (54, 165).

#### Exercise 43

Use linear regression to determine a function y, where the profit in thousands of dollars depends on the number of units sold in hundreds .

##### Solution

y=10.75x+742.50y=10.75x+742.50

### Graphs of Linear Functions

For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular:

#### Exercise 57

2x6y=12 x+3y=1 2x6y=12 x+3y=1

parallel

#### Exercise 58

y= 1 3 x2 3x+y=9 y= 1 3 x2 3x+y=9

For the following exercises, find the x- and y- intercepts of the given equation

#### Exercise 59

7x+9y=−637x+9y=−63

##### Solution

(–9,0);(0,–7)(–9,0);(0,–7)

#### Exercise 60

f(x)=2x1f(x)=2x1

For the following exercises, use the descriptions of the pairs of lines to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither?

#### Exercise 61

• Line 1: Passes through (5,11)(5,11) and (10,1)(10,1)
• Line 2: Passes through (−1,3)(−1,3) and (−5,11)(−5,11)
##### Solution

Line 1: m=2;m=2; Line 2: m=2;m=2; Parallel

#### Exercise 62

• Line 1: Passes through (8,−10)(8,−10) and (0,−26)(0,−26)
• Line 2: Passes through (2,5)(2,5) and (4,4)(4,4)

#### Exercise 63

Write an equation for a line perpendicular to f(x)=5x1f(x)=5x1 and passing through the point (5, 20).

##### Solution

y=0.2x+21y=0.2x+21

#### Exercise 64

Find the equation of a line with a y- intercept of (0, 2)(0, 2) and slope 1212.

#### Exercise 65

Sketch a graph of the linear function f(t)=2t5f(t)=2t5.

#### Exercise 66

Find the point of intersection for the 2 linear functions: x=y+62xy=13x=y+62xy=13

#### Exercise 67

A car rental company offers two plans for renting a car.

• Plan A: 25 dollars per day and 10 cents per mile
• Plan B: 50 dollars per day with free unlimited mileage

How many miles would you need to drive for plan B to save you money?

250.

### Modeling with Linear Functions

#### Exercise 68

Find the area of a triangle bounded by the y axis, the line f(x)=102xf(x)=102x, and the line perpendicular to ff that passes through the origin.

#### Exercise 69

A town’s population increases at a constant rate. In 2010 the population was 55,000. By 2012 the population had increased to 76,000. If this trend continues, predict the population in 2016.

118,000.

#### Exercise 70

The number of people afflicted with the common cold in the winter months dropped steadily by 50 each year since 2004 until 2010. In 2004, 875 people were inflicted.

Find the linear function that models the number of people afflicted with the common cold C as a function of the year, t.t. When will no one be afflicted?

For the following exercises, use the graph in Figure 11 showing the profit, y,y,in thousands of dollars, of a company in a given year, x,x,where xx represents years since 1980.

#### Exercise 71

Find the linear function y, where y depends on x,x, the number of years since 1980.

##### Solution

y=300x+11,500y=300x+11,500

#### Exercise 72

Find and interpret the y-intercept.

For the following exercise, consider this scenario: In 2004, a school population was 1,700. By 2012 the population had grown to 2,500.

#### Exercise 73

Assume the population is changing linearly.

1. How much did the population grow between the year 2004 and 2012?
2. What is the average population growth per year?
3. Find an equation for the population, P, of the school t years after 2004.
##### Solution

a) 800; b) 100 students per year; c) P(t)=100t+1700P(t)=100t+1700

For the following exercises, consider this scenario: In 2000, the moose population in a park was measured to be 6,500. By 2010, the population was measured to be 12,500. Assume the population continues to change linearly.

#### Exercise 74

Find a formula for the moose population, P.P.

#### Exercise 75

What does your model predict the moose population to be in 2020?

##### Solution

18,500

For the following exercises, consider this scenario: The median home values in subdivisions Pima Central and East Valley (adjusted for inflation) are shown in Table 22. Assume that the house values are changing linearly.

Table 22
Year Pima Central East Valley
1970 32,000 120,250
2010 85,000 150,000

#### Exercise 76

In which subdivision have home values increased at a higher rate?

#### Exercise 77

If these trends were to continue, what would be the median home value in Pima Central in 2015?

$91,625 ### Fitting Linear Models to Data #### Exercise 78 Draw a scatter plot for the data in Table 23. Then determine whether the data appears to be linearly related.  0 2 4 6 8 10 –105 –50 1 55 105 160 #### Exercise 79 Draw a scatter plot for the data in Table 24. If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation?  Year Population 1990 5,600 1995 5,950 2000 6,300 2005 6,600 2010 6,900 ##### Solution Extrapolation. #### Exercise 80 Eight students were asked to estimate their score on a 10-point quiz. Their estimated and actual scores are given in Table 25. Plot the points, then sketch a line that fits the data.  Predicted Actual 6 6 7 7 7 8 8 8 7 9 9 10 10 10 10 9 #### Exercise 81 Draw a best-fit line for the plotted data. ##### Solution For the following exercises, consider the data in Table 26, which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year.  Year Percent Graduates 2000 6.5 2002 7.0 2005 7.4 2007 8.2 2010 9.0 #### Exercise 82 Determine whether the trend appears to be linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places. #### Exercise 83 In what year will the percentage exceed 12%? ##### Solution Midway through 2024. #### Exercise 84 Based on the set of data given in Table 27, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places.  xx 17 20 23 26 29 yy 15 25 31 37 40 #### Exercise 85 Based on the set of data given in Table 28, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places.  xx 10 12 15 18 20 yy 36 34 30 28 22 ##### Solution y=1.294x+49.412; r=0.974y=1.294x+49.412; r=0.974 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the population and the year over the ten-year span (population, year) for specific recorded years: (3,600, 2000); (4,000, 2001); (4,700, 2003); (6,000, 2006) (3,600, 2000); (4,000, 2001); (4,700, 2003); (6,000, 2006) #### Exercise 86 Use linear regression to determine a function y,y,where the year depends on the population, to three decimal places of accuracy. #### Exercise 87 Predict when the population will hit 12,000. ##### Solution Early in 2022 #### Exercise 88 What is the correlation coefficient for this model to three decimal places of accuracy? #### Exercise 89 According to the model, what is the population in 2014? ##### Solution 7,660 ## Practice Test ### Exercise 90 Determine whether the following algebraic equation can be written as a linear function. 2x+3y=72x+3y=7 #### Solution Yes. ### Exercise 91 Determine whether the following function is increasing or decreasing. f(x)=2x+5f(x)=2x+5 ### Exercise 92 Determine whether the following function is increasing or decreasing. f(x)=7x+9f(x)=7x+9 #### Solution Increasing ### Exercise 93 Given the following set of information, find a linear equation satisfying the conditions, if possible. Passes through (5, 1) and (3, –9) ### Exercise 94 Given the following set of information, find a linear equation satisfying the conditions, if possible. x intercept at (–4, 0) and y-intercept at (0, –6) #### Solution y=−1.5x6y=−1.5x6 ### Exercise 95 Find the slope of the line in Figure 12. ### Exercise 96 Write an equation for line in Figure 13. #### Solution y=2x1y=2x1 ### Exercise 97 Does Table 29 represent a linear function? If so, find a linear equation that models the data.  xx –6 0 2 4 g(x)g(x) 14 32 38 44 ### Exercise 98 Does Table 30 represent a linear function? If so, find a linear equation that models the data.  x x 1 3 7 11 g(x) g(x) 4 9 19 12 #### Solution No. ### Exercise 99 At 6 am, an online company has sold 120 items that day. If the company sells an average of 30 items per hour for the remainder of the day, write an expression to represent the number of items that were sold nn after 6 am. For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: ### Exercise 100 y= 3 4 x9 4x3y=8 y= 3 4 x9 4x3y=8 #### Solution Perpendicular ### Exercise 101 2x+y=3 3x+ 3 2 y=5 2x+y=3 3x+ 3 2 y=5 ### Exercise 102 Find the x- and y-intercepts of the equation 2x+7y=14.2x+7y=14. #### Solution (7,0)(7,0); (0,2)(0,2) ### Exercise 103 Given below are descriptions of two lines. Find the slopes of Line 1 and Line 2. Is the pair of lines parallel, perpendicular, or neither? Line 1: Passes through (−2,−6)(−2,−6) and (3,14)(3,14) Line 2: Passes through (2,6)(2,6) and (4,14)(4,14) ### Exercise 104 Write an equation for a line perpendicular to f(x)=4x+3f(x)=4x+3 and passing through the point (8,10).(8,10). #### Solution y=0.25x+12y=0.25x+12 ### Exercise 105 Sketch a line with a y-intercept of (0,5)(0,5) and slope 5252. ### Exercise 106 Graph of the linear function f(x)=−x+6f(x)=−x+6. #### Solution ### Exercise 107 For the two linear functions, find the point of intersection: x=y+22x3y=−1x=y+22x3y=−1 ### Exercise 108 A car rental company offers two plans for renting a car. • Plan A:$25 per day and $0.10 per mile • Plan B:$40 per day with free unlimited mileage

How many miles would you need to drive for plan B to save you money?

150

### Exercise 109

Find the area of a triangle bounded by the y axis, the line f(x)=124xf(x)=124x, and the line perpendicular to ff that passes through the origin.

### Exercise 110

A town’s population increases at a constant rate. In 2010 the population was 65,000. By 2012 the population had increased to 90,000. Assuming this trend continues, predict the population in 2018.

165,000

### Exercise 111

The number of people afflicted with the common cold in the winter months dropped steadily by 25 each year since 2002 until 2012. In 2002, 8,040 people were inflicted. Find the linear function that models the number of people afflicted with the common cold CC as a function of the year, t.t. When will less than 6,000 people be afflicted?

For the following exercises, use the graph in Figure 14, showing the profit, yy, in thousands of dollars, of a company in a given year, xx, where xx represents years since 1980.

### Exercise 112

Find the linear function yy, where yy depends on xx, the number of years since 1980.

#### Solution

y=875x+10,675y=875x+10,675

### Exercise 113

Find and interpret the y-intercept.

### Exercise 114

In 2004, a school population was 1250. By 2012 the population had dropped to 875. Assume the population is changing linearly.

1. How much did the population drop between the year 2004 and 2012?
2. What is the average population decline per year?
3. Find an equation for the population, P, of the school t years after 2004.

#### Solution

a) 375; b) dropped an average of 46.875, or about 47 people per year; c) y=46.875t+1250y=46.875t+1250

### Exercise 115

Draw a scatter plot for the data provided in Table 31. Then determine whether the data appears to be linearly related.

 0 2 4 6 8 10 –450 –200 10 265 500 755

### Exercise 116

Draw a best-fit line for the plotted data.

#### Solution

For the following exercises, use Table 32, which shows the percent of unemployed persons 25 years or older who are college graduates in a particular city, by year.

 Year Percent Graduates 2000 8.5 2002 8.0 2005 7.2 2007 6.7 2010 6.4

### Exercise 117

Determine whether the trend appears linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places.

### Exercise 118

In what year will the percentage drop below 4%?

Early in 2018

### Exercise 119

Based on the set of data given in Table 33, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient. Round to three decimal places of accuracy.

 x x 16 18 20 24 26 y y 106 110 115 120 125

For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population (in hundreds) and the year over the ten-year span, (population, year) for specific recorded years:

(4,500, 2000); (4,700, 2001); (5,200, 2003); (5,800, 2006) (4,500, 2000); (4,700, 2001); (5,200, 2003); (5,800, 2006)

### Exercise 120

Use linear regression to determine a function y, where the year depends on the population. Round to three decimal places of accuracy.

#### Solution

y=0.00455x+1979.5y=0.00455x+1979.5

### Exercise 121

Predict when the population will hit 20,000.

### Exercise 122

What is the correlation coefficient for this model?

r=0.999r=0.999

## Footnotes

1. Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010
2. Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the data values.
3. For example, http://www.shodor.org/unchem/math/lls/leastsq.html
4. http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html
5. http://www.census.gov/hhes/socdemo/education/data/cps/historical/index.html. Accessed 5/1/2014.

## Glossary

correlation coefficient:
a value, r,r, between –1 and 1 that indicates the degree of linear correlation of variables, or how closely a regression line fits a data set.
extrapolation:
predicting a value outside the domain and range of the data
interpolation:
predicting a value inside the domain and range of the data
least squares regression:
a statistical technique for fitting a line to data in a way that minimizes the differences between the line and data values
model breakdown:
when a model no longer applies after a certain point

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