Question

(1 point) A 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 49 pilots and the sample standard deviation was 8.5 hours.

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

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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 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

2)

sample mean, xbar = 66
sample standard deviation, s = 8.5
sample size, n = 49
degrees of freedom, df = n - 1 = 48

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.682

ME = tc * s/sqrt(n)
ME = 2.682 * 8.5/sqrt(49)
ME = 3.257

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (66 - 2.682 * 8.5/sqrt(49) , 66 + 2.682 * 8.5/sqrt(49))
CI = (62.7433 , 69.2567)

3)

The following information is provided,
Significance Level, α = 0.01, Margin or Error, E = 1, s =8.5

The critical value for significance level, α = 0.01 is 2.682.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (tc *s/E)^2
n = (2.682* 8.5/1)^2
n = 519.703

Therefore, the sample size needed to satisfy the condition n >= 519.703 and it must be an integer number, we conclude that the minimum required sample size is n = 520
Ans : Sample size, n = 520