Question

The amount of time that customers stay in a certain restaurant for lunch is normally distributed with a standard deviation of 14 minutes. A random sample of 15 lunch customers was taken at this restaurant. Construct a 99%confidence interval for the true average amount of time customers spend in the restaurant for lunch. Round your answers to three decimal places and use ascending order.

time 36, 79, 55, 42, 80, 44, 98, 10, 69, 84, 59, 43, 55, 65, 55

Answer 1

Solution:

Given: The amount of time that customers stay in a certain restaurant for lunch is normally distributed with a standard deviation of 14 minutes.

Thus population standard deviation =

Sample size = n = 15

We have to construct a 99%confidence interval for the true average amount of time customers spend in the restaurant for lunch.

Formula:

where

x
36
79
55
42
80
44
98
10
69
84
59
43
55
65
55

Thus

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

Thus

Thus a 99%confidence interval for the true average amount of time customers spend in the restaurant for lunch is between