Question

A popular Italian eatery has had a dispute among the wait staff regarding the amount of tips waiters receive on the lunch shift and on the dinner shift. Before making any adjustments in schedules or other policies, the restaurant manager wants to estimate the difference in the average tips earned by waiters on the lunch schedule and on the dinner schedule. To do this, she randomly selected 60 tip reports for lunch servers over the past year and 40 tip reports from dinner servers. The results given below were observed. Assuming equal population variances, develop and interpret a 95% confidence interval estimate for the difference between the population mean tips earned by lunch servers and by dinner servers. Mean Tips St. Dev. Tips Lunch $56.50 $11.20 Dinner $104.10 $13.42 Let sample 1 be the sample from the lunch servers and let sample 2 be the sample from the dinner servers. The 95% confidence interval is Is (41-42) I (Round to two decimal places as needed.)

Answer 1

The statistical software output for this problem is:

Two sample T summary confidence interval: Vi : Mean of Population 1 12 : Mean of Population 2 V1-2Difference between two means (with pooled variances) 95% confidence interval results: Difference Sample Diff. Std. Err. DF L. Limit U. Limit Pi - 2 -47.6 2.4764812 98 -52.514496 -42.685504

Hence,

95% confidence interval will be:

-52.51 < -42.69

< (41-2)<