Use Data to Build a Logistic Model

In this section, you will practice building a logistic model from a data set. To perform the regression using Desmos, you will enter the dataset as a table and then define $y_1$ as the standard logistic model.

Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients.

It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities:

$y=\dfrac{c}{1+ae^{−bx}}$

Recall that:

$\frac{c}{1+a}$ is the initial value of the model.
when $b > 0$ , the model increases rapidly at first until it reaches its point of maximum growth rate, $(\dfrac{ln(a)}{b},\dfrac{c}{2})$ . At that point, growth steadily slows and the function becomes asymptotic to the upper bound $y=c$ .
$c$ is the limiting value, sometimes called the carrying capacity, of the model.

Logistic Regression

Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command "Logistic" on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form

$y=\dfrac{c}{1+ae^{−bx}}$

Note that

The initial value of the model is $\frac{c}{1+a}$ .
Output values for the model grow closer and closer to $y=c$ as time increases.

How To

Given a set of data, perform logistic regression using a graphing utility.

Use the STAT then EDIT menu to enter given data.
1. Clear any existing data from the lists.
2. List the input values in the L1 column.
3. List the output values in the L2 column.
Graph and observe a scatter plot of the data using the STATPLOT feature.
1. Use ZOOM [9] to adjust axes to fit the data.
2. Verify the data follow a logistic pattern.
Find the equation that models the data.
1. Select "Logistic" from the STAT then CALC menu.
2. Use the values returned for $a, b,$ and $c$ to record the model, $y=\dfrac{c}{1+ae^{−bx}}$ .
Graph the model in the same window as the scatterplot to verify it is a good fit for the data.

Example 3

Using Logistic Regression to Fit a Model to Data

Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. Table 5 shows the percentage of Americans with cellular service between the years 1995 and 2012.

Year	Americans with Cellular Service (%)	Year	Americans with Cellular Service (%)
1995	12.69	2004	62.852
1996	16.35	2005	68.63
1997	20.29	2006	76.64
1998	25.08	2007	82.47
1999	30.81	2008	85.68
2000	38.75	2009	89.14
2001	45.00	2010	91.86
2002	49.16	2011	95.28
2003	55.15	2012	98.17

Table 5

ⓐ Let

$x$ represent time in years starting with

$x=0$ for the year 1995. Let

$y$ represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.
ⓑ Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.
ⓒ Discuss the value returned for the upper limit,

$c$ . What does this tell you about the model? What would the limiting value be if the model were exact?

Solution

ⓐ Using the STAT then EDIT menu on a graphing utility, list the years using values 0–15 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure 5:

Figure 5

Use the "Logistic" command from the STAT then CALC menu to obtain the logistic model,

$y=\dfrac{105.7379526}{1+6.88328979e^{−0.2595440013x}}$

Next, graph the model in the same window as shown in Figure 6 the scatterplot to verify it is a good fit:

Graph of a scattered plot with an estimation line.

Figure 6

ⓑ To approximate the percentage of Americans with cellular service in the year 2013, substitute

$x=18$ for the in the model and solve for

$y$ :

$\begin{aligned}y &=\frac{105.7379526}{1+6.88328979 e^{-0.2595440013 x}} & & \text { Use the regression model found in part (a). } \\&=\frac{105.7379526}{1+6.88328979 e^{-0.2595400013(18)}} & & \text { Substitute } 18 \text { for } x . \\& \approx 99.3 & & \text { Round to the nearest tenth }\end{aligned}$

According to the model, about 99.3% of Americans had cellular service in 2013.

ⓒ The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a population have cellular service?) If the model were exact, the limiting value would be

$c=100$ and the model's outputs would get very close to, but never actually reach 100%. After all, there will always be someone out there without cellular service!

Try It #3

Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.

Year	Seal Population (Thousands)	Year	Seal Population (Thousands)
1997	3.493	2005	19.590
1998	5.282	2006	21.955
1999	6.357	2007	22.862
2000	9.201	2008	23.869
2001	11.224	2009	24.243
2002	12.964	2010	24.344
2003	16.226	2011	24.919
2004	18.137	2012	25.108

Table 6

ⓐ Let

$x$ represent time in years starting with

$x=0$ for the year 1997. Let

$y$ represent the number of seals in thousands. Use logistic regression to fit a model to these data.
ⓑ Use the model to predict the seal population for the year 2020.
ⓒ To the nearest whole number, what is the limiting value of this model?

Source: Rice University, https://openstax.org/books/college-algebra/pages/6-8-fitting-exponential-models-to-data
This work is licensed under a Creative Commons Attribution 4.0 License.

Last modified: Monday, May 9, 2022, 4:48 PM