Question

A recent report for a regional airline reported that the mean number of hours of flying time for its pilots is 65 hours per month. This mean was based on actual flying times for a sample of 51 pilots and the sample standard deviation was 6.5 hours.

2. Calculate a 95% 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 95% confidence? Note: For consistency's sake, round your t* value to 3 decimal places before calculating the necessary sample size.

Choose n =

Answer 1

Here n = 51

sample standard deviation = s = 6.5 hours

standard error =se = s/sqrt(n) = 6.5/sqrt(51) = 0.91 hours

Sample mean = = 65 hrs per month

degree of freedom = dF = 51 - 1 = 50

tcritical = TINV(0.05, 50) = 2.0086

2. 95% confidence interval estimate of the population mean flying time for the pilots = +- tcritical se

= 65 +- 2.0086 * 0.91 = (63.17 hrs, 66.83 hrs)

Question 3

margin of error =  1.25 hour

Confidence level = 95%

standard error = 6.5/sqrt(n)

so here we assume t(critical) as 1.96 as we have to predict n. THen we will use trial and error.

1.25 = tcritical * 6.5/sqrt(n)

sqrt(n) = (1.96 * 6.5)/1.25

sqrt(n) = 10.192

n= 103.89 or 104

Now we will use trial and error and find exact value of n

os at n = 106, we found the exact value of n.

so here n = 106