Choosing an Appropriate Model

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.

Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model although other models, such as quadratic models, may also be considered.

In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave down away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up and a logarithmic curve always concave down.

A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.

After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.

Example: Choosing a Mathematical Model

Does a linear, exponential, logarithmic, or logistic model best fit the values listed below? Find the model, and use a graph to check your choice.

x	1	2	3	4	5	6	7	8	9
y	0	1.386	2.197	2.773	3.219	3.584	3.892	4.159	4.394

Show Solution

First, plot the data on a graph as in the graph below. For the purpose of graphing, round the data to two significant digits.

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try [latex]y=a\mathrm{ln}\left(bx\right)[/latex]. Plugging in the first point, [latex]\left(\text{1,0}\right)[/latex], gives [latex]0=a\mathrm{ln}b[/latex]. We reject the case that a = 0 (if it were, all outputs would be 0), so we know

[latex]\mathrm{ln}\left(b\right)=0[/latex]. Thus b = 1 and [latex]y=a\mathrm{ln}\left(\text{x}\right)[/latex]. Next we can use the point [latex]\left(\text{9,4}\text{.394}\right)[/latex] to solve for a:
[latex]\begin{array}{l}y=a\mathrm{ln}\left(x\right)\hfill \\ 4.394=a\mathrm{ln}\left(9\right)\hfill \\ a=\frac{4.394}{\mathrm{ln}\left(9\right)}\hfill \end{array}[/latex]

Because [latex]a=\frac{4.394}{\mathrm{ln}\left(9\right)}\approx 2[/latex], an appropriate model for the data is [latex]y=2\mathrm{ln}\left(x\right)[/latex].

To check the accuracy of the model, we graph the function together with the given points.

The graph of [latex]y=2\mathrm{ln}x[/latex].

We can conclude that the model is a good fit to the data. Compare the figure above to the graph of [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] shown below.

The graph of [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex]

The graphs appear to be identical when x > 0. A quick check confirms this conclusion: [latex]y=\mathrm{ln}\left({x}^{2}\right)=2\mathrm{ln}\left(x\right)[/latex] for x > 0. However, if x < 0, the graph of [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] includes an “extra” branch as shown below. This occurs because while [latex]y=2\mathrm{ln}\left(x\right)[/latex] cannot have negative values in the domain (as such values would force the argument to be negative), the function [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] can have negative domain values.

Expressing an Exponential Model in Base e

While powers and logarithms of any base can be used in modeling, the two most common bases are [latex]10[/latex] and [latex]e[/latex]. In science and mathematics, the base e is often preferred. We can use properties of exponents and properties of logarithms to change any base to base e.

Exponential Regression

As we have learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or decrease that helps us determine whether it is best modeled by an exponential function. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let’s review exponential growth and decay.

Recall that exponential functions have the form [latex]y=a{b}^{x}[/latex] or [latex]y={A}_{0}{e}^{kx}[/latex]. When performing regression analysis, we use the form most commonly used on graphing utilities, [latex]y=a{b}^{x}[/latex]. Take a moment to reflect on the characteristics we’ve already learned about the exponential function [latex]y=a{b}^{x}[/latex] (assume a > 0):

• b must be greater than zero and not equal to one.
• The initial value of the model is a.
  • If b > 1, the function models exponential growth. As x increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.
  • If 0 < b < 1, the function models exponential decay. As x increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the x-axis. In other words, the outputs never become equal to or less than zero.

As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable r or [latex]{r}^{2}[/latex]. (You may have to change the calculator’s settings for these to be shown.) The values are an indication of the “goodness of fit” of the regression equation to the data. We more commonly use the value of [latex]{r}^{2}[/latex] instead of r, but the closer either value is to 1, the better the regression equation approximates the data.

Example: Using Exponential Regression to Fit a Model to Data

In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person’s blood alcohol level (BAC) with the risk of being in an accident. The table below shows results from the study.^[1] The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol.

BAC	0	0.01	0.03	0.05	0.07	0.09
Relative Risk of Crashing	1	1.03	1.06	1.38	2.09	3.54
BAC	0.11	0.13	0.15	0.17	0.19	0.21
Relative Risk of Crashing	6.41	12.6	22.1	39.05	65.32	99.78

1. Let x represent the BAC level and let y represent the corresponding relative risk. Use exponential regression to fit a model to these data.
2. After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth.

Show Solution

1. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown below:

Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model,
[latex]y=0.58304829{\left(\text{2.207202130E10}\right)}^{x}[/latex]

Converting from scientific notation, we have:[latex]y=0.58304829{\left(\text{22,072,021,300}\right)}^{x}[/latex]

Notice that [latex]r2 \approx 0.97 [/latex] which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in below:

2. Use the model to estimate the risk associated with a BAC of 0.16. Substitute 0.16 for x in the model and solve for y.

[latex]\begin{array}{l}y\hfill & =0.58304829{\left(\text{22,072,021,300}\right)}^{x}\hfill & \text{Use the regression model found in part (a)}\text{.}\hfill \\ \hfill & =0.58304829{\left(\text{22,072,021,300}\right)}^{0.16}\hfill & \text{Substitute 0}\text{.16 for }x\text{.}\hfill \\ \hfill & \approx \text{26}\text{.35}\hfill & \text{Round to the nearest hundredth}\text{.}\hfill \end{array}[/latex]

If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober.

 

Key Concepts

• We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data.
• Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero.
• We use the command “ExpReg” on a graphing utility to fit function of the form [latex]y=a{b}^{x}[/latex] to a set of data points.
• Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound or where decay begins rapidly and then slows down to get closer and closer to zero.
• Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time.
• We use the command “LnReg” on a graphing utility to fit a function of the form [latex]y=a+b\mathrm{ln}\left(x\right)[/latex] to a set of data points.
• Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit.
• We use the command “Logistic” on a graphing utility to fit a function of the form [latex]y=\frac{c}{1+a{e}^{-bx}}[/latex] to a set of data points.

Glossary

carrying capacity

in a logistic model, the limiting value of the output

doubling time

the time it takes for a quantity to double

half-life

the length of time it takes for a substance to exponentially decay to half of its original quantity

logistic growth model

a function of the form [latex]f\left(x\right)=\frac{c}{1+a{e}^{-bx}}[/latex] where [latex]\frac{c}{1+a}[/latex] is the initial value, c is the carrying capacity, or limiting value, and b is a constant determined by the rate of growth