Question

The manager of a chain of fast-food restaurants believes that the proportion of customers who are served within 2 minutes of placing their order is greater during lunch hours than during dinner hours. To investigate, he selects a random sample of 35 customers who placed an order during lunch hours and a random sample of 75 customers who placed an order during dinner hours. The manager finds that 71.4% of the customers were served within 2 minutes during the lunch hours and 66.7% of the customers were served within 2 minutes during the dinner hours.
(a) Construct and interpret a 99% confidence interval for the difference in the proportion of the population of all dinner customers and all lunch customers who are served within 2 minutes.

(b) Does the confidence interval constructed in part (a) give the manager reason to believe that the proportion of customers who are served within 2 minutes of placing their order is greater during lunch hours than during dinner hours?

Answer 1

Answer:

A. The 99% confidence interval for the difference in the proportion of all dinner customers and all lunch customers who are served within 2 minutes is between -0.195 and 0.289. (0.714 - 0.667) ± 2.576*(√0.714(1- 0.714)/35 + 0.667(1- 0.667)/75).

B. Since the confidence interval contains 0, we cannot conclude that the proportion of customers served within 2 minutes of placing their order is greater during lunch hours than during dinner hours.

Step-by-step explanation:

got this off chegg.