Exemplo 1 de Algebra and Trigonometry, 6.8 Fitting Exponential Models to Data

Using Exponential Regression to Fit a Model to Data

Problema e solução em inglês, como no livro original.

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. Table 1 shows results from the study 9. 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.

Source: Indiana University Center for Studies of Law in Action, 2007

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

Table 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.
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.

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

1
Solução
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 in Figure 1:
  
  Figure 1
  
  Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model,
  $y = 0.58304829 {(2.20720213 E 10)}^{x}$
  Converting from scientific notation, we have:
  $y = 0.58304829 {(22,072,021,300)}^{x}$
  Notice that $r^{2} \approx 0.97$ 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 Figure 2:
  
  Figure 2
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 $y .$
  
  $\begin{array}{l} y & = 0.58304829 {(22,072,021,300)}^{x} & Use the regression model found in part (a) . \\ = 0.58304829 {(22,072,021,300)}^{0.16} & Substitute 0 .16 for x . \\ \approx 26 .35 & Round to the nearest hundredth . \end{array}$
  
  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.

Sua vez

O Try It do livro logo depois deste exemplo: mesma ideia, outros números. Só a resposta é dada.

Table 2 shows a recent graduate’s credit card balance each month after graduation.

Month	1	2	3	4	5	6	7	8
Debt ($)	620.00	761.88	899.80	1039.93	1270.63	1589.04	1851.31	2154.92

Table 2

ⓐ Use exponential regression to fit a model to these data.
ⓑ If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating?

Como foi?

Sua vez

O Try It do livro logo depois deste exemplo: mesma ideia, outros números. Só a resposta é dada.

Table 2 shows a recent graduate’s credit card balance each month after graduation.

Month	1	2	3	4	5	6	7	8
Debt ($)	620.00	761.88	899.80	1039.93	1270.63	1589.04	1851.31	2154.92

Table 2

ⓐ Use exponential regression to fit a model to these data.
ⓑ If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating?

Solução

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Solução

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