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
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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
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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.
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