1 answer

(3 points) A recent report for a regional airline reported that the mean number of hours...

Question:

(3 points) 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 55 pilots and the sample standard deviation was 8 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 2 hour and 95% confidence? Note: For consistencys sake, round your t* value to 3 decimal places before calculating the necessary sample size. Choosen-

(3 points) 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 55 pilots and the sample standard deviation was 8 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 2 hour and 95% confidence? Note: For consistency's sake, round your t* value to 3 decimal places before calculating the necessary sample size. Choosen-

Answers

Given that, sample size ( n ) = 55

sample mean (\bar x) = 56 hours per month

sample standard deviation ( s ) = 8 hours

Since, population standard deviation we used t-critical value for finding confidence interval.

t-critical value at \alpha = 0.05 with degrees of freedom = n - 1= 55 -1 = 54 is, t_{\alpha/2} = 2.005

2) The 95% confidence interval for population mean flying time for the pilots is,

\bar x - t_{\alpha/2} * \frac { s }{\sqrt { n }} < \mu < \bar x + t_{\alpha/2} * \frac { s }{\sqrt { n }}

56 - 2.005* \frac { 8 }{\sqrt { 55 }} < \mu < 56 + 2.005 * \frac { 8 }{\sqrt { 55 }}

56 - 2.1628 < \mu < 56 + 2.1628

53.8372 < \mu < 58.1628

Answer: ( 53.8372, 58.1628)

3) given that, margin of error ( E ) = 2 hours

We want to find, the sample size ( n )

n = (t_{\alpha/2} * \frac {s }{E})^2 = (2.005 *\frac {8}{2})^2 = 64.3204

n \approx 64

Required sample size = 64

Note: If you want sample size is nearest to next whole number then it should be n = 65

.

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