Question

A recent report for a regional airline reported that the mean number of hours of flying time for its pilots is 68 hours per month. This mean was based on actual flying times for a sample of 50 pilots and the sample standard deviation was 8.5 hours.

Calculate a 90% confidence interval estimate of the population mean flying time for the pilots. Round your result to 4 decimal places (__,__)

Using the information given, what is the smallest sample size necessary to estimate the mean flying time with a margin of error of 1.75 hour and 90% confidence? Note: For consistency's sake, round your t* value to 3 decimal places before calculating the necessary sample size. Choose n =

Answer 1

Sample size = n = 50

Sample mean = = 68

Standard deviation = s = 8.5

We have to construct 90% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 50 - 1 = 49

Level of significance = 0.10

tc = 1.677 ( Using t table)

So confidence interval is ( 68 - 2.0159 , 68 + 2.0159) = > ( 65.9841 , 70.0159)

We have to find sample size(n)

Margin of error = E = 1.75

standard deviation = s = 8.5

We have to find sample size (n)