Question

A random sample of 16 airline passengers at the Ontario Airport showed that the mean time spent in line to check in at the ticket counters was 18 minutes with a standard deviation of 7 minutes. Construct and interpret a 99% confidence interval for the mean time spent waiting in line by all passengers at this airport. Assume that such waiting times

Answer 1

The 99% confidence interval for the mean time spent waiting in line by all passengers at this airport is between 12.84 minutes and 23.16 minutes. It means that we are 99% sure that the truw mean waiting time for all passengers in this airport is in this interval.

Explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 36 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.99)/(2) = 0.995`. So we have T = 2.947

The margin of error is:

`M = T(s)/(√(n)) = 2.947(7)/(√(16)) = 5.16`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 18 - 5.16 = 12.84 minutes

The upper end of the interval is the sample mean added to M. So it is 18 + 5.16 = 23.16 minutes

The 99% confidence interval for the mean time spent waiting in line by all passengers at this airport is between 12.84 minutes and 23.16 minutes. It means that we are 99% sure that the truw mean waiting time for all passengers in this airport is in this interval.