Problem
Using the weights ( $\mathrm{lb}$ ) 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. A. The slope is 58.9 and the $y$-intercept is 0.007499 . B. The slope is -0.00749 and the $y$-intercept is 58.9 . C. The slope is 58.9 and the $y$-intercept is -0.00749 . 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 weight, which is represented by $y$. C. The predictor variable is highway fuel consumption, which is represented by $y$. D. The predictor variable is highway fuel consumption, which is represented by $x$. 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 $3000 \mathrm{lb}$ ? The best predicted value of highway fuel consumption of a car that weighs $3000 \mathrm{lb}$ is $\square \mathrm{mi} / \mathrm{gal}$. (Round to one decimal place as needed.)
Solution
Step 1 ：The regression equation is given as \(\hat{y}=58.9-0.00749 x\), where \(x\) represents weight.
Step 2 ：From the equation, we can see that the slope is -0.00749 and the \(y\)-intercept is 58.9.
Step 3 ：The predictor variable in this case is weight, which is represented by \(x\).
Step 4 ：To find the best predicted value for a car that weighs 3000 lb, we substitute \(x=3000\) into the regression equation.
Step 5 ：So, \(\hat{y}=58.9-0.00749 \times 3000\).
Step 6 ：After calculating, we find that \(\hat{y}=36.43\).
Step 7 ：Therefore, the best predicted value of highway fuel consumption of a car that weighs 3000 lb is \(\boxed{36.43} \, \mathrm{mi/gal}\).
