Question

data were recorded for the temperature of a cup of coffee over a 30 - minute period. a regression analysis was completed and the computer output is given.
regression analysis: in(temp) versus time
predictor coef se coef t p
constant 4.20 0.23 18.261 0.000
time - 0.023 0.001 - 16.139 0.000
o = 0.0378 r - sq = 0.98 r - sq(adj) = 0.995
what is the equation of the least - squares regression line?
○ time = 4.20 - 0.023 in(temp)
○ temp = - 0.023 + 4.20 in(time)
○ in(temp) = 4.20 - 0.023 (time)
○ in(temp) = - 0.023 + 4.20 (time)

Explanation:

Step1: Recall the general form of regression equation

The general form of a simple linear regression equation is \(y = a+bx\), where \(a\) is the intercept and \(b\) is the slope. In the case of a regression of \(\ln(\text{Temp})\) versus \(\text{Time}\), the equation will be of the form \(\ln(\widehat{\text{Temp}})=a + b\times\text{Time}\)

Step2: Identify the intercept and slope from the output

From the regression analysis output, the coefficient (\(\text{Coef}\)) for the constant (\(a\)) is \(4.20\) and the coefficient for \(\text{Time}\) (\(b\)) is \(- 0.023\)

Answer:

\(\ln(\widehat{\text{Temp}})=4.20 - 0.023\times\text{Time}\) (corresponds to the option \(\ln(\widehat{\text{Temp}})=4.20 - 0.023(\text{Time})\))