Question

recent report for a regional airline reported that the mean number of hours of flying time for its pilots is 66 hours per month. This mean was based on actual flying times for a sample of 45 pilots and the sample standard deviation was 9 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 2 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,

= 66 hours

s = 9 hours

n = 45

Degrees of freedom = df = n - 1 = 5 - 1 = 44

a ) At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,44 = 2.692

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

= 2.692 * (9 / 45 )

= 3.6117

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

- E < < + E

66 - 3.6117 < < 66 + 3.6117

62.3883 < < 69.6117

b ) margin of error = E = 2

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= ((2.576 *9) / 2 )²

= 134.37

Sample size = 134