Question

00 Pilots who cannot maintain regular sleep hours due to their work schedule often suffer from insomnia. A recent study on sleeping patterns of pilots focused on quantifying deviations from regular sleep hours. A random sample of 15 commercial airline pilots was interviewed, and the pilots in the sample reported the time at which they went to sleep on their most recent working day. The study gave the sample mean and standard deviation of the times reported by pilots, with these times measured in hours after midnight. (Thus, if the pilot reported going to sleep at 11 p.m., the measurement was -1.) The sample mean was 0.6 hours, and the standard deviation was 1.6 hours. Assume that the sample is drawn from a normally distributed population. Find a 90% confidence interval for the population standard deviation, that is, the standard deviation of the time (hours after midnight) at which pilots go to sleep on their work days. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. 0.13 What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval 3

Answer 1

Let sa denotes the sample variance, s denotes the sample standard deviation and n denotes the sample size. Here, $2 = 2.560000, s = 1.600, n = 15 2 The lower critical value is xłower-critical X0.95,14 6.570631 6.571 The upper critical value is Xupper-critical X0.05,14 = 23.684791 ~ 23.685 a 90 percent confidence interval of population variance o2 is (n − 1) * $2 x; (n-1)*8?) X}{_0.90 1- 1-0.90 ,n-1 n-1 2 (15 – 1) * 2.560000 (15 – 1) * 2.560000 2 2 0.10 X 0.10 ,14 2 (15 – 1) * 2.560000 (15 – 1) * 2.560000 23.684791 6.570631 = (1.513207,5.454575) Now, a 90 percent confidence interval of population standard deviation o is = (V1.513207, V5.454575) = (1.230125, 2.335503) (1.23, 2.34)

The lower limit is = 1.23

The upper limit is = 2.34