Problem
Using the weights (lb) and highway fuel consumption amounts (mi/gal) of the 48 cars listed in the accompanying data set, one gets this regression equation: $\hat{y}=58.9-0.00749 x$, where $x$ represents weight. Complete parts (a) through (d). Click the icon to view the car data. D. The slope is 0.00749 and the $y$-intercept is 58.9 . c. What is the predictor variable? A. The predictor variable is weight, which is represented by $x$ B. The predictor variable is high hay fuel consumption, which is represented by $y$. C. The predictor variable is highway fuel consumption, which is represented by $x$ D. The predictor variable is weight, which is represented by $y$. d. Assuming that there is a significant linear correlation between weight and highway fuel consumption, what is the best predicted value for a car that weighs $2996 \mathrm{lb}$ ? The best predicted value of highway fuel consumption of a car that weighs $2996 \mathrm{lb}$ is $\square$ mi/gal (Round to one decimal place as needed.)
Solution
Step 1 ：The question is asking for the predicted value of highway fuel consumption for a car that weighs 2996 lb. This can be calculated using the given regression equation \(\hat{y}=58.9-0.00749 x\), where \(x\) represents weight. We just need to substitute \(x\) with 2996 in the equation and calculate the value of \(\hat{y}\).
Step 2 ：Substitute \(x\) with 2996 in the equation \(\hat{y}=58.9-0.00749 x\).
Step 3 ：Calculate the value of \(\hat{y}\) to get \(\hat{y}=36.459959999999995\).
Step 4 ：Round the result to one decimal place to get \(\hat{y}=36.5\).
Step 5 ：Final Answer: The best predicted value of highway fuel consumption of a car that weighs 2996 lb is \(\boxed{36.5}\) mi/gal.
