Use the time/tip data from the table below, which includes data from New York City taxi rides. (The distances are in miles, the times are in minutes, the fares are in dollars, and the tips are in dollars.) Find the regression equation, letting time be the predictor $(\mathrm{x})$ variable. Find the best predicted tip for a ride that takes 20 minutes. How does the result compare to the actual tip amount of $\$ 4.55$ ? \begin{tabular}{l|cccccccc} Distance & 12.71 & 8.51 & 1.02 & 0.49 & 1.40 & 1.80 & 1.65 & 0.68 \\ \hline Time & 27.00 & 31.00 & 8.00 & 2.00 & 18.00 & 25.00 & 11.00 & 6.00 \\ \hline Fare & 36.80 & 31.75 & 7.80 & 4.80 & 12.30 & 16.30 & 9.80 & 6.30 \\ \hline Tip & 0.00 & 2.98 & 2.34 & 0.00 & 2.46 & 1.50 & 1.96 & 1.89 \end{tabular} The regression equation is $\hat{y}=\square+(\square) x$. (Round the $y$-intercept to two decimal places as needed. Round the slope to four decimal places as needed.)
Solution
Step 1 ：First, we need to calculate the slope and y-intercept of the regression line using the given data. The predictor variable is time and the response variable is tip.
Step 2 ：The slope of the regression line is approximately 0.0549 and the y-intercept is approximately 0.6059. This gives us the regression equation \( \hat{y}=0.6059+0.0549x \).
Step 3 ：Using this equation, we can predict the tip for a ride that takes 20 minutes. The predicted tip is approximately $1.72.
Step 4 ：Comparing this predicted tip to the actual tip amount of $4.55, we find that the predicted tip is less than the actual tip.
Step 5 ：Final Answer: The regression equation is \( \hat{y}=0.6059+0.0549x \). The best predicted tip for a ride that takes 20 minutes is approximately \(\boxed{1.72}\). This is less than the actual tip amount of $4.55.
