\fYou may need to use the appropriate appendix table or technology to answer this question. Many small restaurants in Portland, Oregon, and other cities across the United States do not take reservations. Owners say that with smaller capacity, no-shows are costly, and they would rather have their staff focused on customer service rather than maintaining a reservation system.t However, it is important to be able to give reasonable estimates of waiting time when customers arrive and put their name on the waiting list. The file RestaurantLine contains 10 observations of number of pecple in line ahead of a customer (independent variable x) and actual waiting time (in minutes) (dependent variable y). The estimated regression equation is: = 4.35 + 8.81x and MSE = 94.42. (a) Develop a point estimate (in min) for a customer who arrives with six people on the wait-list. (Round your answer to two decimal places.) 7 =157.21 ~ min (b) Develop a 95% confidence interval for the mean waiting time (in min) for @ customer who arrives with six customers already in line. (Round your answers to two decimal places.) o xomme[ ]k mn (c) Develop a $5% prediction interval for Roger and Sherry Davy's waiting time (in min) if there are six customers in line when they arrive. (Round your answers to two decimal places.) o x mine[ % mn (d) Discuss the difference between part (b) and part (c). The prediction interval is much than the confidence interval. This is because it is difficult to predict the waiting time for an individual customer arriving with six people in line than it is to estimate the mean waiting time for a customer arriving with six people in line. \f3. Standard Error (SE) for Mean: MSE 94.42 (6 - 4.6)2 SE mean - 1 + 1 + n 10 32 1.96 9.442 . 1 + = V9.442 . 1.06125 - V10.026 ~ 3.167 32 4. Critical t-value: For a 95% confidence level with 8 degrees of freedom: to.025,8 ~ 2.306 5. Confidence Interval: CI - 57.21 + 2.306 - 3.167 - 57.21 + 7.30 - [49.91, 64.51] (c) 95% Prediction Interval for Individual Waiting Time To calculate the 95% prediction interval: PI - y tta/2,n-2 . SEpred1. Standard Error (SE) for Prediction: SEIJIH:l - = 4/94.42 . (1 + 0.1 + 0.06125) = v/94.42 - 1.16125 +/109.66 ~ 10.473 2. Prediction Interval: PI 57.21 +2.306-10473 57.21 =24.15 [33.[)6, 81.36] Revised Final Results s 95% Confidence Interval: [49.91, 64.51 minutes * 95% Prediction Interval: [33.06, 81.36] minutes Summary of Changes 1. The correct 95% confidence interval is [49.91, 64.51 . 2. The correct 95% prediction interval is [33.06, 81.36]