The mean number of hours of flying time for pilots at Continental Airlines is 49 hours per month. Assume that this mean was based on actual flying times for a sample of 100 Continental Pilots and the sample standard deviation was 8.5 hours. a. At 95% confidence, what is the margin of error?
b. What is the 95% confidence interval estimate of the population mean flying time for the Pilots?

a) 17.09 hours

b) The 95% confidence interval estimate of the population mean flying time for the Pilots is between 31.91 hours and 66.09 hours

Step-by-step explanation:

We have the standard deviation of the sample, so we use the t distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 49 - 1 = 48

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 48 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0106

The margin of error is:

M = T*s = 2.0106*8.5 = 17.09

s is the standard deviation of the sample. 17.09 hours is the answer for a.

The lower end of the interval is the sample mean subtracted by M. So it is 49 - 17.09 = 31.91 hours

The upper end of the interval is the sample mean added to M. So it is 49 + 17.09 = 66.09 hours

The 95% confidence interval estimate of the population mean flying time for the Pilots is between 31.91 hours and 66.09 hours

the first one, all those points on the line just make a strait line.

27 + 13x + 15 = (x + 5)( =

step-by-step explanation:

27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( =