Question

(1 point) A recent report for a regional airline reported that the mean number of hours of flying time for its pilots is 56 hours per month. This mean was based on actual flying times for a sample of 51 pilots and the sample standard deviation was 9.5 hours.2. Calculate a 90% 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 90% confidence

Answer 1

(53.812 ; 58.188) ; 156

Explanation:

Given that :

Sample size (n) = 51

Mean (m) = 56

Standard deviation (σ) = 9.5

α = 90%

Using the relation :

Confidence interval = mean ± Error

Error = Zcritical * (standard deviation / sqrt (n))

Zcritical at 90% = 1.645

Error = 1.645 * (9.5 / sqrt(51))

Error = 1.645 * 1.3302660

Error = 2.1882877

Hence,

Confidence interval :

Lower boundary = 56 - 2.1882877 = 53.8117123

Upper boundary = 56 + 2.1882877 = 58.1882877

Confidence interval = (53.812 ; 58.188)

2.)

Margin of Error (ME) = 1.25

α = 90%

Sample size = ((Zcritical * σ) / ME)^2

Zcritical at 90% = 1.645

Sample size = ((1.645 * 9.5) / 1.25)^2

Sample size = (15.6275 / 1.25)^2

Sample size = 12.502^2 = 156.3000

Sample size = 156