Question

A sample of 33 airline passengers found that the average check-in time is 2.167 minutes. Based on long-term data, the population standard deviation is known to be 0.48 minutes. Find a 95% confidence interval for the mean check-in time.

Answer 1

Solution :

Given that,

Z_/2 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (0.48 / 33)

= 0.164

At 95% confidence interval estimate of the population mean is,

- E < < + E

2.167 - 0.164 < < 2.167 + 0.164

2.003 < < 2.331

(2.003 , 2.331)