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# 1.3: Models and Data

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Fit data to linear models.
• Fit data to quadratic models.
• Fit data to trigonometric models.
• Fit data to exponential growth and decay models.

## Introduction

In our last lesson we examined functions and learned how to classify and sketch functions. In this lesson we will use some classic functions to model data. The lesson will be a set of examples of each of the models. For each, we will make extensive use of the graphing calculator.

Let’s do a quick review of how to model data on the graphing calculator.

Enter Data in Lists

Press [STAT] and then [EDIT] to access the lists, L1 - L6.

View a Scatter Plot

Press 2nd [STAT PLOT] and choose accordingly.

Then press [WINDOW] to set the limits of the axes.

Compute the Regression Equation

Press [STAT] then choose [CALC] to access the regression equation menu. Choose the appropriate regression equation (Linear, Quad, Cubic, Exponential, Sine).

Graph the Regression Equation Over Your Scatter Plot

Go to Y=> [MENU] and clear equations. Press [VARS], then enter 5\begin{align*}5\end{align*} and EQ and press [ENTER] (This series of entries will copy the regression equation to your Y = screen.) Press [GRAPH] to view the regression equation over your scatter plot

Plotting and Regression in Excel

You can also do regression in an Excel spreadsheet. To start, copy and paste the table of data into Excel. With the two columns highlighted, including the column headings, click on the Chart icon and select XY scatter. Accept the defaults until a graph appears. Select the graph, then click Chart, then Add Trendline. From the choices of trendlines choose Linear.

Now let’s begin our survey of the various modeling situations.

Linear Models

For these kinds of situations, the data will be modeled by the classic linear equation \begin{align*}y = mx + b.\end{align*} Our task will be to find appropriate values of \begin{align*}m\end{align*} and \begin{align*}b\end{align*} for given data.

Example 1:

It is said that the height of a person is equal to his or her wingspan (the measurement from fingertip to fingertip when your arms are stretched horizontally). If this is true, we should be able to take a table of measurements, graph the measurements in an \begin{align*}x-y\end{align*} coordinate system, and verify this relationship. What kind of graph would you expect to see? (Answer: You would expect to see the points on the line \begin{align*}y = x\end{align*}.)

Suppose you measure the height and wingspans of nine of your classmates and gather the following data. Use your graphing calculator to see if the following measurements fit this linear model (the line \begin{align*}y = x\end{align*}).

Height (inches) Wingspan (inches)
\begin{align*}67\end{align*} \begin{align*}65\end{align*}
\begin{align*}64\end{align*} \begin{align*}63\end{align*}
\begin{align*}56\end{align*} \begin{align*}57\end{align*}
\begin{align*}60\end{align*} \begin{align*}61\end{align*}
\begin{align*}62\end{align*} \begin{align*}63\end{align*}
\begin{align*}71\end{align*} \begin{align*}70\end{align*}
\begin{align*}72\end{align*} \begin{align*}69\end{align*}
\begin{align*}68\end{align*} \begin{align*}67\end{align*}
\begin{align*}65\end{align*} \begin{align*}65\end{align*}

We observe that only one of the measurements has the condition that they are equal. Why aren’t more of the measurements equal to each other? (Answer: The data do not always conform to exact specifications of the model. For example, measurements tend to be loosely documented so there may be an error arising in the way that measurements were taken.)

We enter the data in our calculator in L1 and L2. We then view a scatter plot. (Caution: note that the data ranges exceed the viewing window range of \begin{align*}[-10, 10].\end{align*} Change the window ranges accordingly to include all of the data, say \begin{align*}[40, 80].\end{align*})

Here is the scatter plot:

Now let us compute the regression equation. Since we expect the data to be linear, we will choose the linear regression option from the menu. We get the equation \begin{align*}y = .76x + 14.\end{align*}

In general we will always wish to graph the regression equation over our data to see the goodness of fit. Doing so yields the following graph, which was drawn with Excel:

Since our calculator will also allow for a variety of non-linear functions to be used as models, we can therefore examine quite a few real life situations. We will first consider an example of quadratic modeling.

Example 2:

The following table lists the number of Food Stamp recipients (in millions) for each year after 1990.

years after 1990 Participants
\begin{align*}1\end{align*} \begin{align*}22.6\end{align*}
\begin{align*}2\end{align*} \begin{align*}25.4\end{align*}
\begin{align*}3\end{align*} \begin{align*}27.0\end{align*}
\begin{align*}4\end{align*} \begin{align*}27.5\end{align*}
\begin{align*}5\end{align*} \begin{align*}26.6\end{align*}
\begin{align*}6\end{align*} \begin{align*}25.5\end{align*}
\begin{align*}7\end{align*} \begin{align*}22.5\end{align*}
\begin{align*}8\end{align*} \begin{align*}19.8\end{align*}
\begin{align*}9\end{align*} \begin{align*}18.2\end{align*}
\begin{align*}10\end{align*} \begin{align*}17.2\end{align*}

We enter the data in our calculator in L3 and L4 (that enables us to save the last example’s data). We then will view a scatter plot. Change the window ranges accordingly to include all of the data. Use \begin{align*}[-2, 10]\end{align*} for \begin{align*}x\end{align*} and \begin{align*}[-2, 30]\end{align*} for \begin{align*}y.\end{align*}

Here is the scatter plot:

Now let us compute the regression equation. Since our scatter plot suggests a quadratic model for the data, we will choose Quadratic Regression from the menu. We get the equation:

Let’s graph the equation over our data. We see the following graph:

Trigonometric Models

The following example shows how a trigonometric function can be used to model data.

Example 3:

With the skyrocketing cost of gasoline, more people have looked to mass transit as an option for getting around. The following table uses data from the American Public Transportation Association to show the number of mass transit trips (in billions) between 1992 and 2000.

year Trips (billions)
1992 \begin{align*}8.5\end{align*}
1993 \begin{align*}8.2\end{align*}
1994 \begin{align*}7.93\end{align*}
1995 \begin{align*}7.8\end{align*}
1996 \begin{align*}7.87\end{align*}
1997 \begin{align*}8.23\end{align*}
1998 \begin{align*}8.6\end{align*}
1999 \begin{align*}9.08\end{align*}
2000 \begin{align*}9.4\end{align*}

We enter the data in our calculator in L5 and L6, starting in L5 with the number one for 1992 (the first year). We then will view a scatter plot. Change the window ranges accordingly to include all of the data. Use \begin{align*}[-2, 10]\end{align*} for both \begin{align*}x\end{align*} and \begin{align*}y\end{align*} ranges.

Here is the scatter plot:

Now let us compute the regression equation. Since our scatter plot suggests a sine model for the data, we will choose Sine Regression from the menu. We get the equation:

Let us graph the equation over our data. We see the following graph:

This example suggests that the sine over time \begin{align*}t\end{align*} is a function that is used in a variety of modeling situations.

Caution: Although the fit to the data appears quite good, do we really expect the number of trips to continue to go up and down in the future? Probably not. Here is what the graph looks like when projected an additional ten years:

Exponential Models

Our last class of models involves exponential functions. Exponential models can be used to model growth and decay situations. Consider the following data about the declining number of farms for the years 1980 - 2005.

Example 4:

The number of dairy farms has been declining over the past \begin{align*}20+\end{align*} years. The following table charts the decline:

Year Farms (thousands)
1980 \begin{align*}334\end{align*}
1985 \begin{align*}269\end{align*}
1990 \begin{align*}193\end{align*}
1995 \begin{align*}140\end{align*}
2000 \begin{align*}105\end{align*}
2005 \begin{align*}67\end{align*}

We enter the data in our calculator in L5 (entering the year 1980 as 1, the year 1985 as 2, etc.) and L6. We then will view a scatter plot. Change the window ranges accordingly to include all of the data. For the large \begin{align*}y-\end{align*}values, choose the range \begin{align*}[-50, 350]\end{align*} with a scale of \begin{align*}25.\end{align*}

Here is the scatter plot:

Now let us compute the regression equation. Since our scatter plot suggests an exponential model for the data, we will choose Exponential Regression from the menu. We get the equation: \begin{align*}y = 490.6317 * .7266^x\end{align*}

Let’s graph the equation over our data. We see the following graph:

In the homework we will practice using our calculator extensively to model data.

## Lesson Summary

1. Fit data to linear models.
2. Fit data to quadratic models.
3. Fit data to trigonometric models.
4. Fit data to exponential growth and decay models.

## Review Questions

1. Consider the following table of measurements of circular objects:
1. Make a scatter plot of the data.
2. Based on your plot, which type of regression will you use?
3. Find the line of best fit.
4. Comment on the values of \begin{align*}m\end{align*} and \begin{align*}b\end{align*} in the equation.
Object Diameter (cm) Circumference (cm)
Glass \begin{align*}8.3\end{align*} \begin{align*}26.5\end{align*}
Flashlight \begin{align*}5.2\end{align*} \begin{align*}16.7\end{align*}
Aztec calendar \begin{align*}20.2\end{align*} \begin{align*}61.6\end{align*}
Tylenol bottle \begin{align*}3.4\end{align*} \begin{align*}11.6\end{align*}
Popcorn can \begin{align*}13\end{align*} \begin{align*}41.4\end{align*}
Salt shaker \begin{align*}6.3\end{align*} \begin{align*}20.1\end{align*}
Coffee canister \begin{align*}11.3\end{align*} \begin{align*}35.8\end{align*}
Cat food bucket \begin{align*}33.5\end{align*} \begin{align*}106.5\end{align*}
Dinner plate \begin{align*}27.3\end{align*} \begin{align*}85.6\end{align*}
Ritz cracker \begin{align*}4.9\end{align*} \begin{align*}15.5\end{align*}
1. Manatees are large, gentle sea creatures that live along the Florida coast. Many manatees are killed or injured by power boats. Here are data on powerboat registrations (in thousands) and the number of manatees killed by boats in Florida from 1987 - 1997.
1. Make a scatter plot of the data.
2. Use linear regression to find the line of best fit.
3. Suppose in the year 2000, powerboat registrations increase to \begin{align*}700,000\end{align*}. Predict how many manatees will be killed.
Year Boats Manatees killed
1987 \begin{align*}447\end{align*} \begin{align*}13\end{align*}
1988 \begin{align*}460\end{align*} \begin{align*}21\end{align*}
1989 \begin{align*}480\end{align*} \begin{align*}24\end{align*}
1990 \begin{align*}497\end{align*} \begin{align*}16\end{align*}
1991 \begin{align*}512\end{align*} \begin{align*}24\end{align*}
1992 \begin{align*}513\end{align*} \begin{align*}21\end{align*}
1993 \begin{align*}526\end{align*} \begin{align*}15\end{align*}
1994 \begin{align*}557\end{align*} \begin{align*}33\end{align*}
1995 \begin{align*}585\end{align*} \begin{align*}34\end{align*}
1996 \begin{align*}614\end{align*} \begin{align*}34\end{align*}
1997 \begin{align*}645\end{align*} \begin{align*}39\end{align*}
1. A passage in Gulliver’s Travels states that the measurement of “Twice around the wrist is once around the neck.” The table below contains the wrist and neck measurements of \begin{align*}10\end{align*} people.
1. Make a scatter plot of the data.
2. Find the line of best fit and comment on the accuracy of the quote from the book.
3. Predict the distance around the neck of Gulliver if the distance around his wrist is found to be \begin{align*}52 \;\mathrm{cm}\end{align*}.
Wrist (cm) Neck (cm)
\begin{align*}17.9\end{align*} \begin{align*}39.5\end{align*}
\begin{align*}16\end{align*} \begin{align*}32.5\end{align*}
\begin{align*}16.5\end{align*} \begin{align*}34.7\end{align*}
\begin{align*}15.9\end{align*} \begin{align*}32\end{align*}
\begin{align*}17\end{align*} \begin{align*}33.3\end{align*}
\begin{align*}17.3\end{align*} \begin{align*}32.6\end{align*}
\begin{align*}16.8\end{align*} \begin{align*}33\end{align*}
\begin{align*}17.3\end{align*} \begin{align*}31.6\end{align*}
\begin{align*}17.7\end{align*} \begin{align*}35\end{align*}
\begin{align*}16.9\end{align*} \begin{align*}34\end{align*}
1. The following table gives women’s average percentage of men’s salaries for the same jobs for each \begin{align*}5\end{align*}-year period from 1960 - 2005.
1. Make a scatter plot of the data.
2. Based on your sketch, should you use a linear or quadratic model for the data?
3. Find a model for the data.
4. Can you explain why the data seems to dip at first and then grow?
Year Percentage
1960 \begin{align*}42\end{align*}
1965 \begin{align*}36\end{align*}
1970 \begin{align*}30\end{align*}
1975 \begin{align*}37\end{align*}
1980 \begin{align*}41\end{align*}
1985 \begin{align*}42\end{align*}
1990 \begin{align*}48\end{align*}
1995 \begin{align*}55\end{align*}
2000 \begin{align*}58\end{align*}
2005 \begin{align*}60\end{align*}
1. Based on the model for the previous problem, when will women make as much as men? Is your answer a realistic prediction?
2. The average price of a gallon of gas for selected years from 1975 - 2008 is given in the following table:
1. Make a scatter plot of the data.
2. Based on your sketch, should you use a linear, quadratic, or cubic model for the data?
3. Find a model for the data.
4. If gas continues to rise at this rate, predict the price of gas in the year 2012.
Year Cost
1975 \begin{align*}1\end{align*}
1976 \begin{align*}1.75\end{align*}
1981 \begin{align*}2\end{align*}
1985 \begin{align*}2.57\end{align*}
1995 \begin{align*}2.45\end{align*}
2005 \begin{align*}2.75\end{align*}
2008 \begin{align*}3.45\end{align*}
1. For the previous problem, use a linear model to analyze the situation. Does the linear method provide a better estimate for the predicted cost for the year 2012? Why or why not?
2. Suppose that you place \begin{align*}\1,000\end{align*} in a bank account where it grows exponentially and is compounded annually over the course of six years. The table below shows the amount of money you have at the end of each year.
1. Find the exponential model.
2. In what year will you triple your original amount?
Year Amount
\begin{align*}0\end{align*} \begin{align*}1000\end{align*}
\begin{align*}1\end{align*} \begin{align*}1127.50\end{align*}
\begin{align*}2\end{align*} \begin{align*}1271.24\end{align*}
\begin{align*}3\end{align*} \begin{align*}1433.33\end{align*}
\begin{align*}4\end{align*} \begin{align*}1616.07\end{align*}
\begin{align*}5\end{align*} \begin{align*}1822.11\end{align*}
\begin{align*}6\end{align*} \begin{align*}2054.43\end{align*}
1. Suppose that in the previous problem, you started with \begin{align*}\3,000\end{align*} but maintained the same interest rate.
1. Give a formula for the exponential model. (Hint: note the coefficient in the previous answer!)
2. How long will it take for the initial amount, \begin{align*}\3,000\end{align*}, to triple? Explain your answer.
2. The following table gives the average daily temperature for Indianapolis, Indiana for each month of the year:
1. Construct a scatter plot of the data.
2. Find the sine model for the data.
Month Avg Temp (F)
Jan \begin{align*}22\end{align*}
Feb \begin{align*}26.3\end{align*}
March \begin{align*}37.8\end{align*}
April \begin{align*}51\end{align*}
May \begin{align*}61.7\end{align*}
June \begin{align*}75.3\end{align*}
July \begin{align*}78.5\end{align*}
Aug \begin{align*}84.3\end{align*}
Sept \begin{align*}68.5\end{align*}
Oct \begin{align*}53.2\end{align*}
Nov \begin{align*}38.7\end{align*}
Dec \begin{align*}26.6\end{align*}

1. Not included.
2. Linear.
3. \begin{align*}y = 3.1334x + .3296.\end{align*}
4. \begin{align*}m\end{align*} is an estimate of \begin{align*}\pi\end{align*}, and \begin{align*}b\end{align*} should be zero but due to error in measurement it is not.
1. Not included.
2. Not included.
3. \begin{align*}y=.120546x - 39.0465\end{align*}; about \begin{align*}46\end{align*} manatees will be killed in the year 2000. Note: there were actually \begin{align*}81\end{align*} manatees killed in the year 2000.
1. Not included.
2. Not included.
3. \begin{align*}y=2.0131x - 0.2634\end{align*}; \begin{align*}104.42 \;\mathrm{cm}\end{align*}.
1. Not included.
3. \begin{align*}y = .4848 x^2 - 2.4545x + 39.7333.\end{align*}
4. It might be because the first wave of women into the workforce tended to take whatever jobs they could find without regard for salary.
1. The data suggest that women will reach \begin{align*}100 \%\end{align*} in 2025. To find the answer, you set \begin{align*}y\end{align*} equal to 100 and solve for \begin{align*}x\end{align*}, and \begin{align*}x\end{align*} turns out to be approximately 14. Since the years are in 5-year intervals, \begin{align*}x = 14\end{align*} corresponds to 2025. This is unrealistic based on current reports that women still lag far behind men in equal salaries for equal work.
1. Not included.
2. It appears a cubic model might be appropriate, because of the general upward movement over time, with one dip.
3. \begin{align*}y=0.0003134x^3 - 0.017x^2 + 0.2906x + 0.9227.\end{align*}
4. Since year 2008 was year 34 in our Table 1 input L1, and each unit represents a one year unit, 2012 is represented by \begin{align*}x = 38\end{align*}, and we want to solve for \begin{align*}y\end{align*}. So that gives us \begin{align*}y = 4.61\end{align*}, or \$4.61, as the projected cost per gallon in 2012.
2. Linear \begin{align*}y=0.0516x+1.4923\end{align*}; Predicted cost in 2012 is \begin{align*}\3.45\end{align*}; the linear model seems to be a better predictor of gas prices than the cubic model, at least early in the year 2012. The linear model is also better because the cubic model only allows for one “dip” and gas prices, although generally trending up as the earth’s supply of oil is consumed, could have multiple dips in the future due to temporary fluctuations in demand.
1. Not included.
2. \begin{align*}y=1000*1.1275^x\end{align*}; the amount will triple early in Year \begin{align*}10\end{align*}.
1. \begin{align*}y=3000*1.1275^x\end{align*}
2. The amount will triple early in Year \begin{align*}10\end{align*} as in the last problem because the exponential equations \begin{align*}3000=1000*1.1275^x\end{align*} and \begin{align*}9000=3000*1.1275^x\end{align*} both reduce to the same equation \begin{align*}3=1.1275^x\end{align*} and hence have the same solution.
1. Not included.
2. \begin{align*}y=30.07 * \sin(.5196x-2.1503) + 51.46.\end{align*}

Feb 23, 2012

Oct 30, 2015