Question

A sample of 30 fast food restaurants were visited the time between pulling up to the order kiosk and receiving the order was recorded. The sample had a mean of 3.8 and a sample standard deviation of 2.257. (Show your work in detail)

A. (10 pts ) construct a 90 %confidence interval estimate of the population mean, and comment on the results

B.(10 pts) construct a 95%  confidence interval estimate of the population mean,

and comment on the results

C (10 pts) Now Assume the population standard deviation is known and is equal to 1.2. construct a 95% confidence interval of the population mean

Answer 1

a) For n - 1 = 29 degrees of freedom, we have from the t distribution tables here:
P( t₂₉ < 1.699) = 0.95

Therefore, due to symmetry, we have here:
P( - 1.699 < t₂₉ < 1.699) = 0.9

Therefore now the confidence inteval here is obtained as:

This is the required 90% confidence interval here.

b) For 95% confidence level now, we have from the t distribution tables:

P( - 2.045 < t₂₉ < 2.045) = 0.95

Therefore the confidence interval here is obtained as:

This is the required 95% confidence interval here.

c) As we are given the population standard deviation here, we use the Z distribution here. From standard normal tables, we have here:

P( -1.96 < Z < 1.96) = 0.95

Therefore the confidence interval here is obtained as:

This is the required 95% confidence interval here.