Question

A recent report for a regional airline reported that the mean number of hours of flying time for its pilots is 70 hours per month. This mean was based on actual flying times for a sample of 59 pilots and the sample standard deviation was 7.5 hours.

2. Calculate a 99% confidence interval estimate of the population mean flying time for the pilots. Round your result to 4 decimal places.

( ,  )

3.Using the information given, what is the smallest sample size necessary to estimate the mean flying time with a margin of error of 1.25 hour and 99% confidence? Note: For consistency's sake, round your t* value to 3 decimal places before calculating the necessary sample size.

Choose n =

Answer 1

Solution :

Given that,

2) Point estimate = sample mean = = 70 hours

sample standard deviation = s = 7.5 hours

sample size = n = 59

Degrees of freedom = df = n - 1 = 59 - 1 = 58

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_0.005,58 = 2.663

Margin of error = E = t_/2,df * (s /n)

= 2.663 * (7.5 / 59)

Margin of error = E = 2.6002

The 99% confidence interval estimate of the population mean is,

  ± E

= 70  ± 2.6002

= (67.3998, 72.6002)

3) margin of error = E =1.25

sample size = n = (  t/2,df * s / E )²

n = ( 2.663 * 7.5 / 1.25)²

n = 255.29

n = 256