Question

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 27 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 0.9 hours, and the standard deviation was 1.9 hours. Assume that the sample is drawn from a normally distributed population. Find a 99% 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. (If necessary, consult a list of formulas.)

What is the lower limit of the 99% confidence interval?

What is the upper limit of the 99% confidence interval?

Answer 1

Sample Size,   n=   27
Sample Standard Deviation,   s=   1.9000
Confidence Level,   CL=   0.99
      
      
Degrees of Freedom,   DF=n-1 =    26
alpha,   α=1-CL=   0.01
alpha/2 ,   α/2=   0.005
Lower Chi-Square Value=   χ²1-α/2 =   11.160
Upper Chi-Square Value=   χ²α/2 =   48.290

confidence interval for std dev is       
lower bound= √[(n-1)s²/χ²α/2] =   √(26*1.9² / 48.2899)=   1.3942
      
      
upper bound= √[(n-1)s²/χ²1-α/2] =   √(26*1.9² / 11.1602)=   2.9000