3/14/23, 8:51 PM ALEKS Explanation 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 21 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 1.7 standard deviation was hours. Assume that the sample is drawn from a normally distributed population. Find a 95% 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 give its lower limit and upper limit. 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.) Lower limit: Upper limit: EXPLANATION We are asked to find a 95% confidence interval for the population standard σ. We'll find a 95% confidence interval for the population variance σ then we'll take square roots to obtain the confidence interval for σ. deviation The sample variance 2 s is the quantity we use to estimate variance of a random sample of size 2 n−1 s σ 2 σ 2 . When 2 s 2 , and is the n taken from a normal population, then follows approximately a chi-square distribution with n − 1 degrees of 2 freedom. In this problem, n = 21, so 21 − 1 s σ 2 has a chi-square distribution with 21 − 1 = 20 degrees of freedom. We'll use this distribution to find our confidence interval. In particular, following the example of the other confidence interval problems so far, we'll find the values that cut off the "middle 95%" of this 2 distribution, then write an inequality placing then solve for σ 2 21 − 1 s σ 2 between these values, and to obtain the interval. To find the "middle 95%" of the https://www-awu.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNsIkasNW8D8A9PVVRA8T8fqtCZgL85seei3__J9kLCBh2I7eMELFj1f-W5zKntM7qVOpdfBjKoyR… 1/3 3/14/23, 8:51 PM ALEKS chi-square distribution with 20 degrees of freedom, we use the values that cut off the lower 2.5% and the upper 2.5% of this distribution. These values are, 2 2 respectively, χ0.975 and χ0.025 . 0.025 (See Figure 1.) 0.025 95% of random samples of size 21 from this So, about χ population satisfy the following. 2 χ0.975 2 21 − 1 s < σ 2 2 2 2 2 χ0.975 Writing the inequality in terms of 2 The observed value of with 21 − 1 s < σ < 2 χ0.025 2 σ, we get this. 2 21 − 1 s χ0.025 ≈ 34.17 95% of samples of size 21 from this population 21 − 1 s < σ < 2 0.025 < χ0.025 2 χ0.025 0.975 2 χ Figure 1 2 Solving for σ , we see that about satisfy the following. 21 − 1 s 2 2 χ0.975 2 s is given as 1.7, and we can compute χ0.975 ≈ 9.591 and (where these computations are based on the chi-square distribution 20 degrees of freedom). So, here is the 95% confidence interval for σ, based on the sample. 2 21 − 1 s 2 χ0.025 2 , 21 − 1 s 2 χ0.975 ≈ 20 1.7 34.17 2 , 20 1.7 9.591 2 ≈ 1.30 , 2.45 ANSWER Lower limit: 1.30 Upper limit: 2.45 https://www-awu.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNsIkasNW8D8A9PVVRA8T8fqtCZgL85seei3__J9kLCBh2I7eMELFj1f-W5zKntM7qVOpdfBjKoyR… 2/3 3/14/23, 8:51 PM ALEKS n is drawn from a normal population, 1 − α 100% confidence interval for the population standard deviation σ is Note: In general, if a random sample of size then a 2 n−1 s , 2 χα/2 2 2 n−1 s 2 χ1 − α/2 , where s is the sample standard deviation and 2 χ1 − α/2 and χα/2 are the values that cut off areas of α/2 in the left and right tails, respectively, of a chi-square distribution with n − 1 degrees of freedom. https://www-awu.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNsIkasNW8D8A9PVVRA8T8fqtCZgL85seei3__J9kLCBh2I7eMELFj1f-W5zKntM7qVOpdfBjKoyR… 3/3