Page 309 - Applied statistics and probability for engineers

Section 8-3/Conidence Interval on the Variance and Standard Deviation of a Normal Distribution 287

8-40. A machine produces metal rods used in an auto- s = 0.08 millimeter. Find a 95% lower conidence bound for
mobile suspension system. A random sample of 15 rods is mean wall thickness. Interpret the interval obtained.
selected, and the diameter is measured. The resulting data (in 8-43. An article in Nuclear Engineering International (Febru-
millimeters) are as follows: ary 1988, p. 33) describes several characteristics of fuel rods
used in a reactor owned by an electric utility in Norway. Meas-
8.24 8.25 8.20 8.23 8.24 urements on the percentage of enrichment of 12 rods were
reported as follows:
8.21 8.26 8.26 8.20 8.25
8.23 8.23 8.19 8.28 8.24 2.94 3.00 2.90 2.75 3.00 2.95
(a) Check the assumption of normality for rod diameter. 2.90 2.75 2.95 2.82 2.81 3.05
(b) Calculate a 95% two-sided conidence interval on mean (a) Use a normal probability plot to check the normality
rod diameter. assumption.
(c) Calculate a 95% upper conidence bound on the mean. (b) Find a 99% two-sided conidence interval on the mean
Compare this bound with the upper bound of the two-sided percentage of enrichment. Are you comfortable with
conidence interval and discuss why they are different. the statement that the mean percentage of enrichment is
8-41. An article in Computers & Electrical Engineering 2.95%? Why?
[“Parallel Simulation of Cellular Neural Networks” (1996, 8-44. Using the data from Exercise 8-22 on adhesion without
Vol. 22, pp. 61–84)] considered the speedup of cellular neu- assuming that the standard deviation is known,
ral networks (CNN) for a parallel general-purpose computing (a) Check the assumption of normality by using a normal
architecture based on six transputers in different areas. The probability plot.
data follow: (b) Find a 95% conidence interval for the mean adhesion.
8-45. A healthcare provider monitors the number of CAT
3.775302 3.350679 4.217981 4.030324 4.639692
scans performed each month in each of its clinics. The most
4.139665 4.395575 4.824257 4.268119 4.584193 recent year of data for a particular clinic follows (the reported
4.930027 4.315973 4.600101 variable is the number of CAT scans each month expressed
as the number of CAT scans per thousand members of the
(a) Is there evidence to support the assumption that speedup of
CNN is normally distributed? Include a graphical display health plan):
in your answer. 2.31, 2.09, 2.36, 1.95, 1.98, 2.25, 2.16, 2.07, 1.88, 1.94, 1.97,
(b) Construct a 95% two-sided conidence interval on the mean 2.02.
speedup. (a) Find a 95% two-sided CI on the mean number of CAT
(c) Construct a 95% lower conidence bound on the mean scans performed each month at this clinic.
speedup. (b) Historically, the mean number of scans performed by all
8-42. The wall thickness of 25 glass 2-liter bottles was clinics in the system has been 1.95. If there any evidence
measured by a quality-control engineer. The sample mean was that this particular clinic performs more CAT scans on
x = .05 millimeters, and the sample standard deviation was average than the overall system average?
4

8-3 Confidence Interval on the Variance and

Standard Deviation of a Normal Distribution
Sometimes conidence intervals on the population variance or standard deviation are needed.
When the population is modeled by a normal distribution, the tests and intervals described
in this section are applicable. The following result provides the basis of constructing these
conidence intervals.
2
b Distribution
Let X , X , …, X be a random sample from a normal distribution with mean μ and
1 2 n
2
variance σ , and let S be the sample variance. Then the random variable
2
( n 1) S 2
−
2
X = (8-17)
σ 2
χ
2
has a chi-square ( ) distribution with n – 1 degrees of freedom.