Question

The manager of the local Hamburger Express wishes to estimate the mean time customers spend at the drive-through window. A sample of 20 customers experienced a mean waiting time of 2.65 minutes, with a standard deviation of 0.45 minute. Develop a 90% confidence interval for the mean waiting time.

Answer 1

A sample of 20 customers experienced a mean waiting time of 2.65 minutes with a standard deviation 0.45 minutes. Therefore, the data is

n = 20

X̅ = 2.65

s = 0.45

 

Confidence level = 90%

Degrees of freedom

df = n – 1

    = 20 – 1

     = 19

 

From students t-distribution table, for 90% confidence and 19 df, we get t = 1.729

 

Confidence interval

X ± t s/√n = 2.65 ± 1.729 × 0.45/√20

                 = 2.65 ± 0.174

                 = (2.476, 2.824)

 

Thus, the 90% confidence interval for the mean waiting time is between 2.48 and 2.82 minutes.

Thus, the 90% confidence interval for the mean waiting time is between 2.48 and 2.82 minutes.