Question

data was recorded for the temperature, in degrees celsius, of a cup of coffee over a 30 - minute period. given the regression equation, $ln(widehat{temp}) = 4.20 - 0.023(time)$, what is the predicted temperature after 3 minutes?
33.45 $^{circ}$c
62.24 $^{circ}$c
65.17 $^{circ}$c
66.69 $^{circ}$c

Explanation:

Step1: Substitute Time = 3 into the equation

Given the regression equation \(\ln(\text{Temp})=4.20 - 0.023(\text{Time})\), when \(\text{Time} = 3\), we have \(\ln(\text{Temp})=4.20-0.023\times3\).
First, calculate \(0.023\times3 = 0.069\). Then \(\ln(\text{Temp})=4.20 - 0.069=4.131\).

Step2: Solve for Temp

Since \(y=\ln(x)\) implies \(x = e^{y}\) (where \(y = 4.131\) and \(x=\text{Temp}\)), we use the formula \(\text{Temp}=e^{\ln(\text{Temp})}\). So \(\text{Temp}=e^{4.131}\).
Using a calculator, \(e^{4.131}\approx62.24\) (because \(e^{4}=54.598\), \(e^{4.1}=60.34\), \(e^{4.131}\approx62.24\)).

Answer:

\(62.24^{\circ}C\) (the second option)