Question

The data shown below ($), for a sample of 20 guest bills for Danvers-Hilton resort hotel for a weekend. Assume the population is approximately normal. Develop a 90% confidence interval for the mean guest bills of Danvers-Hilton resort hotel.

1905, 3250 ,2545, 2529, 2627, 2725, 2854, 2525, 2442, 2857, 2677, 2143 ,2370, 2312 ,2675, 2600 ,2018, 2545, 2981 ,2115

A: Upper Bound: 2692.02, Lower Bound: 2258.25

B:Upper Bound: 2758.95, Lower Bound: 2409.98

C: Upper Bound: 2478.35, Lower Bound: 2147.25

D: Upper Bound: 2689.76, Lower Bound: 2379.74

Answer 1

Answer)

First we need to find mean and standard deviation of the given data

Mean = 2534.75

S.d = 331.2123

As the population standard deviation is unknown here, we need to use t table to construct the interval

Degrees of freedom is = sample size - 1, 20-1, 19

For degrees of freedom 19, and 90% confidence level, critical value t = 1.729

Margin of error is (MOE) = t*(s.d/√n) = 1.729*331.2123/√20

MOE = 128.05202535486

Confidence interval is given by

(Mean - MOE, Mean +MOE)

(2406.6979746451, 2662.8020253548)

Closest option is D