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.[footnote]Source: Indiana University Center for Studies of Law in Action, 2007[/footnote] 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.

Answer: 1. Using an online graphing tool, create a table by clicking on the + in the upper left and selecting the table icon. Enter the data. Below your table enter [latex]y_1[/latex]~[latex]ab^{x_1}[/latex]. Notice that [latex]{r}^{2}[/latex] is very close to 1 which indicates the model is a good fit to the data. If using an online graphing tool, the model obtained is [latex]y=0.55877{\left(\text{57,700,000,000}\right)}^{x}[/latex] If using a graphing calculator, the model obtained is [latex]y=0.58304829{\left(\text{22,072,021,300}\right)}^{x}[/latex] 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. Using the online graphing tool model:

[latex]\begin{array}{l}y\hfill & =0.55877{\left(\text{57,700,000,000}\right)}^{x}\hfill & \text{Use the regression model found in part (a)}\text{.}\hfill \\ \hfill & =0.55877{\left(\text{57,700,000,000}\right)}^{0.16}\hfill & \text{Substitute 0}\text{.16 for }x\text{.}\hfill \\ \hfill & \approx \text{29}\text{.44}\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 29.44 times more likely to crash than if driving while sober. Using the graphing calculator model:

[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.