Answered

Data were recorded for a car's fuel efficiency, in miles per gallon (mpg), and the corresponding speed, in miles per hour (mph). A regression analysis was completed and the computer output is given.



Regression Analysis: ln(Fuel Efficiency) versus ln(Speed)




Predictor		CoefSE CoefTP

Constant		1.4370.187.98330.000

ln(Speed)		0.5410.018828.77940.000

s = 0.0443		R-sq = 0.989\multicolumn{2}{|l|}{R-sq(adj) = 0.99}






What is the equation of the least-squares regression line?


A$\widehat{\text{Speed}} = 1.437 + 0.541 \ln(\text{Fuel Efficiency})$
B$\widehat{\text{Fuel Efficiency}} = 0.541 + 1.437 \ln(\text{Speed})$
C$\ln(\widehat{\text{Fuel Efficiency}}) = 1.437 + 0.541 \ln(\text{Speed})$
D$\ln(\widehat{\text{Fuel Efficiency}}) = 0.541 + 1.437 \ln(\text{Speed})$

Verified Answer :

  • Identify \(\ln(\text{Fuel Efficiency})\) as the dependent variable and \(\ln(\text{Speed})\) as the independent variable from the table title.
  • Locate the intercept (\(\beta_0 = 1.437\)) and the slope coefficient (\(\beta_1 = 0.541\)) from the table coefficients.
  • Apply the linear regression formula \(\widehat{y} = \beta_0 + \beta_1 x\) to form the equation: \(\ln(\widehat{\text{Fuel Efficiency}}) = 1.437 + 0.541 \ln(\text{Speed})\).
  • Select the matching option: \(\boxed{\ln(\widehat{\text{Fuel Efficiency}}) = 1.437 + 0.541 \ln(\text{Speed})}\).

Explanation

  1. Understanding the Regression Analysis Data

To determine the least-squares regression line, we analyze the provided computer output. A regression analysis typically relates a dependent variable (\(y\)) to an independent variable (\(x\)). The standard format for a simple linear regression equation is \(\widehat{y} = \beta_0 + \beta_1 x\), where \(\beta_0\) is the constant (intercept) and \(\beta_1\) is the coefficient (slope) for the predictor variable \(x\). In the provided table, the title "ln(Speed) versus ln(Fuel Efficiency)" follows the statistical convention of "Independent Variable versus Dependent Variable," identifying \(\ln(\text{Fuel Efficiency})\) as the dependent variable (\(y\)) and \(\ln(\text{Speed})\) as the independent variable (\(x\)).

  1. Constructing the Regression Equation

From the table, we extract the necessary coefficients:

  • The constant term (intercept \(\beta_0\)) is given as 1.437.
  • The coefficient (slope \(\beta_1\)) for the predictor \(\ln(\text{Speed})\) is given as 0.541.

Substituting these values into the linear regression model \(\widehat{y} = \beta_0 + \beta_1 x\), we obtain:
\[\ln(\widehat{\text{Fuel Efficiency}}) = 1.437 + 0.541 \ln(\text{Speed})\]

  1. Conclusion and Selection of the Correct Equation

Comparing our constructed equation \(\ln(\widehat{\text{Fuel Efficiency}}) = 1.437 + 0.541 \ln(\text{Speed})\) with the provided options, we see that it exactly matches the third option. Therefore, the correct equation of the least-squares regression line is \(\ln(\widehat{\text{Fuel Efficiency}}) = 1.437 + 0.541 \ln(\text{Speed})\).

Examples

Regression analysis is widely used in automotive engineering to understand how different factors like speed, engine load, or gear ratios influence fuel efficiency. By modeling these relationships with logarithmic transformations (as seen in this problem), engineers can linearize complex curves to perform simple linear regression. This allows them to predict fuel consumption accurately under varying driving conditions, helping to optimize vehicle performance and design more fuel-efficient cars.

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