# One- and Two-Sample Estimation Problems Chapter 5 5.1

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```Chapter 5
One- and Two-Sample
Estimation Problems
5.1
Introduction
In previous chapters, we emphasized sampling properties of the sample mean and
variance. We also emphasized displays of data in various forms. The purpose of
these presentations is to build a foundation that allows us to draw conclusions about
the population parameters from experimental data. For example, the Central Limit
Theorem provides information about the distribution of the sample mean X̄. The
distribution involves the population mean μ. Thus, any conclusions concerning μ
drawn from an observed sample average must depend on knowledge of this sampling
distribution. Similar comments apply to S 2 and σ 2 . Clearly, any conclusions we
draw about the variance of a normal distribution will likely involve the sampling
distribution of S 2 .
In this chapter, we begin by formally outlining the purpose of statistical inference. We follow this by discussing the problem of estimation of population
parameters. We conﬁne our formal developments of speciﬁc estimation procedures to problems involving one and two samples.
5.2
Statistical Inference
In Chapter 1, we discussed the general philosophy of formal statistical inference.
Statistical inference consists of those methods by which one makes inferences or
generalizations about a population. The trend today is to distinguish between the
classical method of estimating a population parameter, whereby inferences are
based strictly on information obtained from a random sample selected from the
population, and the Bayesian method, which utilizes prior subjective knowledge
about the probability distribution of the unknown parameters in conjunction with
the information provided by the sample data. Throughout most of this chapter,
we shall use classical methods to estimate unknown population parameters such as
the mean, the proportion, and the variance by computing statistics from random
195
196
Chapter 5 One- and Two-Sample Estimation Problems
samples and applying the theory of sampling distributions, much of which was
covered in Chapter 4.
Statistical inference may be divided into two major areas: estimation and
tests of hypotheses. We treat these two areas separately, dealing with theory
and applications of estimation in this chapter and hypothesis testing in Chapter
6. To distinguish clearly between the two areas, consider the following examples.
A candidate for public oﬃce may wish to estimate the true proportion of voters
favoring him by obtaining opinions from a random sample of 100 eligible voters.
The fraction of voters in the sample favoring the candidate could be used as an
estimate of the true proportion in the population of voters. A knowledge of the
sampling distribution of a proportion enables one to establish the degree of accuracy
of such an estimate. This problem falls in the area of estimation.
Now consider the case in which one is interested in ﬁnding out whether brand
A ﬂoor wax is more scuﬀ-resistant than brand B ﬂoor wax. He or she might
hypothesize that brand A is better than brand B and, after proper testing, accept or
reject this hypothesis. In this example, we do not attempt to estimate a parameter,
but instead we try to arrive at a correct decision about a prestated hypothesis.
Once again we are dependent on sampling theory and the use of data to provide
us with some measure of accuracy for our decision.
5.3
Classical Methods of Estimation
A point estimate of some population parameter θ is a single value θ̂ of a statistic
Θ̂. For example, the value x̄ of the statistic X̄, computed from a sample of size n,
is a point estimate of the population parameter μ. Similarly, p̂ = x/n is a point
estimate of the true proportion p for a binomial experiment.
An estimator is not expected to estimate the population parameter without
error. We do not expect X̄ to estimate μ exactly, but we certainly hope that it is
not far oﬀ. For a particular sample, it is possible to obtain a closer estimate of μ
by using the sample median X̃ as an estimator. Consider, for instance, a sample
consisting of the values 2, 5, and 11 from a population whose mean is 4 but is
supposedly unknown. We would estimate μ to be x̄ = 6, using the sample mean
as our estimate, or x̃ = 5, using the sample median as our estimate. In this case,
the estimator X̃ produces an estimate closer to the true parameter than does the
estimator X̄. On the other hand, if our random sample contains the values 2, 6,
and 7, then x̄ = 5 and x̃ = 6, so X̄ is the better estimator. Not knowing the true
value of μ, we must decide in advance whether to use X̄ or X̃ as our estimator.
Unbiased Estimator
What are the desirable properties of a “good” decision function that would inﬂuence us to choose one estimator rather than another? Let Θ̂ be an estimator whose
value θ̂ is a point estimate of some unknown population parameter θ. Certainly, we
would like the sampling distribution of Θ̂ to have a mean equal to the parameter
estimated. An estimator possessing this property is said to be unbiased.
5.3 Classical Methods of Estimation
197
Deﬁnition 5.1: A statistic Θ̂ is said to be an unbiased estimator of the parameter θ if
μΘ̂ = E(Θ̂) = θ.
Example 5.1: If a sample X1 , . . . , Xn has an unknown population mean μ, then the sample mean
X̄ is an unbiased estimator for μ.
Solution : This result can be easily shown as
n
n
n
1
1
1
E(X̄) = E
Xi =
E(Xi ) =
μ = μ.
n i=1
n i=1
n i=1
This means that the sample mean is always unbiased to the population mean.
Example 5.2: Show that S 2 is an unbiased estimator of the parameter σ 2 .
Solution : One can show that (see Exercise 5.10)
n
(Xi − X̄)2 =
i=1
Now
n
(Xi − μ)2 − n(X̄ − μ)2 .
i=1
n
1 2
(Xi − X̄)
E(S ) = E
n − 1 i=1
n
n
1
1
2
.
=
E(Xi − μ)2 − nE(X̄ − μ)2 =
σ 2 − nσX̄
n − 1 i=1
n − 1 i=1 Xi
2
However,
2
2
σX
= σ 2 , for i = 1, 2, . . . , n, and σX̄
=
i
σ2
.
n
Therefore,
1
E(S ) =
n−1
2
σ2
2
nσ − n
= σ2 .
n
This result shows that the sample variance is always unbiased to the population
variance σ 2 .
Although S 2 is an unbiased estimator of σ 2 , S, on the other hand, is usually a
biased estimator of σ, with the bias becoming insigniﬁcant for large samples. This
example illustrates why we divide by n − 1 rather than n when the variance is
estimated.
Variance of a Point Estimator
If Θ̂1 and Θ̂2 are two unbiased estimators of the same population parameter θ, we
want to choose the estimator whose sampling distribution has the smaller variance.
Hence, if σθ̂2 < σθ̂2 , we say that Θ̂1 is a more eﬃcient estimator of θ than Θ̂2 .
1
2
198
Chapter 5 One- and Two-Sample Estimation Problems
Deﬁnition 5.2: If we consider all possible unbiased estimators of some parameter θ, the one with
the smallest variance is called the most eﬃcient estimator of θ.
Figure 5.1 illustrates the sampling distributions of three diﬀerent estimators,
Θ̂1 , Θ̂2 , and Θ̂3 , all estimating θ. It is clear that only Θ̂1 and Θ̂2 are unbiased,
since their distributions are centered at θ. The estimator Θ̂1 has a smaller variance
than Θ̂2 and is therefore more eﬃcient. Hence, our choice for an estimator of θ,
among the three considered, would be Θ̂1 .
^
1
^
3
^
2
θ
^
θ
Figure 5.1: Sampling distributions of diﬀerent estimators of θ.
For normal populations, one can show that both X̄ and X̃ are unbiased estimators of the population mean μ, but the variance of X̄ is smaller than the variance
of X̃. Thus, both estimates x̄ and x̃ will, on average, equal the population mean
μ, but x̄ is likely to be closer to μ for a given sample, and thus X̄ is more eﬃcient
than X̃.
Interval Estimation
Even the most eﬃcient unbiased estimator is unlikely to estimate the population
parameter exactly. It is true that estimation accuracy increases with large samples,
but there is still no reason we should expect a point estimate from a given sample
to be exactly equal to the population parameter it is supposed to estimate. There
are many situations in which it is preferable to determine an interval within which
we would expect to ﬁnd the value of the parameter. Such an interval is called an
interval estimate.
An interval estimate of a population parameter θ is an interval of the form
θ̂L < θ < θ̂U , where θ̂L and θ̂U depend on the value of the statistic Θ̂ for a
particular sample and also on the sampling distribution of Θ̂. For example, a
random sample of SAT verbal scores for students in the entering freshman class
might produce an interval from 530 to 550, within which we expect to ﬁnd the
true average of all SAT verbal scores for the freshman class. The values of the
endpoints, 530 and 550, will depend on the computed sample mean x̄ and the
2
sampling distribution of X̄. As the sample size increases, we know that σX̄
= σ 2 /n
decreases, and consequently our estimate is likely to be closer to the parameter μ,
5.4 Single Sample: Estimating the Mean
199
resulting in a shorter interval. Thus, the interval estimate indicates, by its length,
the accuracy of the point estimate. An engineer will gain some insight into the
population proportion defective by taking a sample and computing the sample
Interpretation of Interval Estimates
Since diﬀerent samples will generally yield diﬀerent values of Θ̂ and, therefore,
diﬀerent values for θ̂L and θ̂U , these endpoints of the interval are values of corresponding random variables Θ̂L and Θ̂U . From the sampling distribution of Θ̂ we
shall be able to determine Θ̂L and Θ̂U such that P (Θ̂L < θ < Θ̂U ) is equal to any
positive fractional value we care to specify. If, for instance, we ﬁnd Θ̂L and Θ̂U
such that
P (Θ̂L < θ < Θ̂U ) = 1 − α,
for 0 < α < 1, then we have a probability of 1−α of selecting a random sample that
will produce an interval containing θ. The interval θ̂L < θ < θ̂U , computed from
the selected sample, is called a 100(1 − α)% conﬁdence interval, the fraction
1 − α is called the conﬁdence coeﬃcient or the degree of conﬁdence, and
the endpoints, θ̂L and θ̂U , are called the lower and upper conﬁdence limits.
Thus, when α = 0.05, we have a 95% conﬁdence interval, and when α = 0.01, we
obtain a wider 99% conﬁdence interval. The wider the conﬁdence interval is, the
more conﬁdent we can be that the interval contains the unknown parameter. Of
course, it is better to be 95% conﬁdent that the average life of a certain television
transistor is between 6 and 7 years than to be 99% conﬁdent that it is between 3
and 10 years. Ideally, we prefer a short interval with a high degree of conﬁdence.
Sometimes, restrictions on the size of our sample prevent us from achieving short
intervals without sacriﬁcing some degree of conﬁdence.
In the sections that follow, we pursue the notions of point and interval estimation, with each section presenting a diﬀerent special case. The reader should
notice that while point and interval estimation represent diﬀerent approaches to
gaining information regarding a parameter, they are related in the sense that conﬁdence interval estimators are based on point estimators. In the following section,
for example, we will see that X̄ is a very reasonable point estimator of μ. As a
result, the important conﬁdence interval estimator of μ depends on knowledge of
the sampling distribution of X̄.
5.4
Single Sample: Estimating the Mean
The sampling distribution of X̄ is centered at μ, and in most applications the
variance is smaller than that of any other unbiased estimators of μ. Thus, the
sample mean x̄ will be used as a point estimate for the population mean μ. Recall
2
that σX̄
= σ 2 /n, so a large sample will yield a value of X̄ that comes from a
sampling distribution with a small variance. Hence, x̄ is likely to be a very accurate
estimate of μ when n is large.
200
Chapter 5 One- and Two-Sample Estimation Problems
Let us now consider the interval estimate of μ. If our sample is selected from
a normal population or, failing this, if n is suﬃciently large, we can establish a
conﬁdence interval for μ by considering the sampling distribution of X̄.
According to the Central Limit Theorem, we can expect the sampling distribution of X̄ √to be approximately normal with mean μX̄ = μ and standard deviation
σX̄ = σ/ n. Writing zα/2 for the z-value above which we ﬁnd an area of α/2
under the normal curve, we can see from Figure 5.2 that
P (−zα/2 < Z < zα/2 ) = 1 − α,
where
Z=
Hence,
X̄ − μ
√ .
σ/ n
X̄ − μ
√ < zα/2 = 1 − α.
P −zα/2 <
σ/ n
1−α
α /2
−zα /2
0
α /2
zα /2
z
Figure 5.2: P (−zα/2 < Z < zα/2 ) = 1 − α.
√
Multiplying each term in the inequality by σ/ n and then subtracting X̄ from each
term and multiplying by −1 (reversing the sense of the inequalities), we obtain
σ
σ
P X̄ − zα/2 √ < μ < X̄ + zα/2 √
= 1 − α.
n
n
A random sample of size n is selected from a population whose variance σ 2 is known,
and the mean x̄ is computed to give the 100(1 − α)% conﬁdence interval below. It
is important to emphasize that we have invoked the Central Limit Theorem above.
As a result, it is important to note the conditions for applications that follow.
Conﬁdence
Interval on μ, σ 2
Known
If x̄ is the mean of a random sample of size n from a population with known
variance σ 2 , a 100(1 − α)% conﬁdence interval for μ is given by
σ
σ
x̄ − zα/2 √ < μ < x̄ + zα/2 √ ,
n
n
where zα/2 is the z-value leaving an area of α/2 to the right.
5.4 Single Sample: Estimating the Mean
201
For small samples selected from nonnormal populations, we cannot expect our
degree of conﬁdence to be accurate. However, for samples of size n ≥ 30, with
the shape of the distributions not too skewed, sampling theory guarantees good
results.
Clearly, the values of the random variables Θ̂L and Θ̂U , deﬁned in Section 5.3,
are the conﬁdence limits
σ
σ
and θ̂U = x̄ + zα/2 √ .
θ̂L = x̄ − zα/2 √
n
n
Diﬀerent samples will yield diﬀerent values of x̄ and therefore produce diﬀerent
interval estimates of the parameter μ, as shown in Figure 5.3. The dot at the
center of each interval indicates the position of the point estimate x̄ for that random
sample. Note that all of these intervals are of the same width, since their widths
depend only on the choice of zα/2 once x̄ is determined. The larger the value we
choose for zα/2 , the wider we make all the intervals and the more conﬁdent we
can be that the particular sample selected will produce an interval that contains
the unknown parameter μ. In general, for a selection of zα/2 , 100(1 − α)% of the
intervals will cover μ.
10
9
8
Sample
7
6
5
4
3
2
1
μ
x
Figure 5.3: Interval estimates of μ for diﬀerent samples.
Example 5.3: The average zinc concentration recovered from a sample of measurements taken
in 36 diﬀerent locations in a river is found to be 2.6 grams per milliliter. Find
the 95% and 99% conﬁdence intervals for the mean zinc concentration in the river.
Assume that the population standard deviation is 0.3 gram per milliliter.
Solution : The point estimate of μ is x̄ = 2.6. The z-value leaving an area of 0.025 to the
right, and therefore an area of 0.975 to the left, is z0.025 = 1.96 (Table A.3). Hence,
202
Chapter 5 One- and Two-Sample Estimation Problems
the 95% conﬁdence interval is
0.3
0.3
< μ < 2.6 + (1.96) √
,
2.6 − (1.96) √
36
36
which reduces to 2.50 < μ < 2.70. To ﬁnd a 99% conﬁdence interval, we ﬁnd the
z-value leaving an area of 0.005 to the right and 0.995 to the left. From Table A.3
again, z0.005 = 2.575, and the 99% conﬁdence interval is
0.3
0.3
< μ < 2.6 + (2.575) √
,
2.6 − (2.575) √
36
36
or simply
2.47 < μ < 2.73.
We now see that a longer interval is required to estimate μ with a higher degree of
conﬁdence.
The 100(1−α)% conﬁdence interval provides an estimate of the accuracy of our
point estimate. If μ is actually the center value of the interval, then x̄ estimates
μ without error. Most of the time, however, x̄ will not be exactly equal to μ and
the point estimate will be in error. The size of this error will be the absolute value
of the diﬀerence between μ and x̄, and we can be 100(1 − α)% conﬁdent that this
diﬀerence will not exceed zα/2 √σn . We can readily see this if we draw a diagram of
a hypothetical conﬁdence interval, as in Figure 5.4.
Error
x zα /2σ / n
x
μ
x zα /2 σ / n
Figure 5.4: Error in estimating μ by x̄.
Theorem 5.1: If x̄ is used as an estimate of μ, we can be 100(1 − α)% conﬁdent that the error
will not exceed zα/2 √σn .
In Example 5.3, we are 95% conﬁdent that the sample mean
x̄ = 2.6 diﬀers
√
from the true mean μ by an amount less than (1.96)(0.3)/
36
=
0.1 and 99%
√
conﬁdent that the diﬀerence is less than (2.575)(0.3)/ 36 = 0.13.
Frequently, we wish to know how large a sample is necessary to ensure that
the error in estimating μ will be less than a speciﬁed amount e. By Theorem 5.1,
we must choose n such that zα/2 √σn = e. Solving this equation gives the following
formula for n.
5.4 Single Sample: Estimating the Mean
203
Theorem 5.2: If x̄ is used as an estimate of μ, we can be 100(1 − α)% conﬁdent that the error
will not exceed a speciﬁed amount e when the sample size is
n=
zα/2 σ !2
.
e
When solving for the sample size, n, we round all fractional values up to the
next whole number. By adhering to this principle, we can be sure that our degree
of conﬁdence never falls below 100(1 − α)%.
Strictly speaking, the formula in Theorem 5.2 is applicable only if we know
the variance of the population from which we select our sample. Lacking this
information, we could take a preliminary sample of size n ≥ 30 to provide an
estimate of σ. Then, using s as an approximation for σ in Theorem 5.2, we could
determine approximately how many observations are needed to provide the desired
degree of accuracy.
Example 5.4: How large a sample is required if we want to be 95% conﬁdent that our estimate
of μ in Example 5.3 is oﬀ by less than 0.05?
Solution : The population standard deviation is σ = 0.3. Then, by Theorem 5.2,
n=
(1.96)(0.3)
0.05
2
= 138.3.
Therefore, we can be 95% conﬁdent that a random sample of size 139 will provide
an estimate x̄ diﬀering from μ by an amount less than 0.05.
One-Sided Conﬁdence Bounds
The conﬁdence intervals and resulting conﬁdence bounds discussed thus far are
two-sided (i.e., both upper and lower bounds are given). However, there are many
applications in which only one bound is sought. For example, if the measurement
of interest is tensile strength, the engineer receives better information from a lower
bound only. This bound communicates the worst-case scenario. On the other
hand, if the measurement is something for which a relatively large value of μ is not
proﬁtable or desirable, then an upper conﬁdence bound is of interest. An example
would be a case in which inferences need to be made concerning the mean mercury
concentration in a river. An upper bound is very informative in this case.
One-sided conﬁdence bounds are developed in the same fashion as two-sided
intervals. However, the source is a one-sided probability statement that makes use
of the Central Limit Theorem:
X̄ − μ
√ < zα = 1 − α.
P
σ/ n
One can then manipulate the probability statement much as before and obtain
√
P (μ > X̄ − zα σ/ n) = 1 − α.
204
Chapter 5 One- and Two-Sample Estimation Problems
Similar manipulation of P
X̄−μ
√
σ/ n
!
> −zα = 1 − α gives
√
P (μ < X̄ + zα σ/ n) = 1 − α.
As a result, the upper and lower one-sided bounds follow.
One-Sided
Conﬁdence
Bounds on μ, σ 2
Known
If X̄ is the mean of a random sample of size n from a population with variance
σ 2 , the one-sided 100(1 − α)% conﬁdence bounds for μ are given by
√
upper one-sided bound:
x̄ + zα σ/ n;
√
lower one-sided bound:
x̄ − zα σ/ n.
Example 5.5: In a psychological testing experiment, 25 subjects are selected randomly and their
reaction time, in seconds, to a particular stimulus is measured. Past experience
suggests that the variance in reaction times to these types of stimuli is 4 sec2 and
that the distribution of reaction times is approximately normal. The average time
for the subjects is 6.2 seconds. Give an upper 95% bound for the mean reaction
time.
Solution : The upper 95% bound is given by
√
x̄ + zα σ/ n = 6.2 + (1.645) 4/25 = 6.2 + 0.658
= 6.858 seconds.
Hence, we are 95% conﬁdent that the mean reaction time is less than 6.858
seconds.
The Case of σ Unknown
Frequently, we must attempt to estimate the mean of a population when the variance is unknown. The reader should recall learning in Chapter 4 that if we have a
random sample from a normal distribution, then the random variable
T =
X̄ − μ
√
S/ n
has a Student t-distribution with n − 1 degrees of freedom. Here S is the sample
standard deviation. In this situation, with σ unknown, T can be used to construct
a conﬁdence interval on μ. The procedure is the same as that with σ known except
that σ is replaced by S and the standard normal distribution is replaced by the
t-distribution. Referring to Figure 5.5, we can assert that
P (−tα/2 < T < tα/2 ) = 1 − α,
where tα/2 is the t-value with n−1 degrees of freedom, above which we ﬁnd an area
of α/2. Because of symmetry, an equal area of α/2 will fall to the left of −tα/2 .
Substituting for T , we write
X̄ − μ
√ < tα/2 = 1 − α.
P −tα/2 <
S/ n
5.4 Single Sample: Estimating the Mean
205
√
Multiplying each term in the inequality by S/ n, and then subtracting X̄ from
each term and multiplying by −1, we obtain
S
S
= 1 − α.
P X̄ − tα/2 √ < μ < X̄ + tα/2 √
n
n
For a particular random sample of size n, the mean x̄ and standard deviation s are
computed and the following 100(1 − α)% conﬁdence interval for μ is obtained.
1 −α
α /2
−t α
2
0
α /2
tα 2
t
Figure 5.5: P (−tα/2 < T < tα/2 ) = 1 − α.
Conﬁdence
Interval on μ, σ 2
Unknown
If x̄ and s are the mean and standard deviation of a random sample from a
normal population with unknown variance σ 2 , a 100(1 − α)% conﬁdence interval
for μ is
s
s
x̄ − tα/2 √ < μ < x̄ + tα/2 √ ,
n
n
where tα/2 is the t-value with v = n − 1 degrees of freedom, leaving an area of
α/2 to the right.
We have made a distinction between the cases of σ known and σ unknown in
computing conﬁdence interval estimates. We should emphasize that for σ known
we exploited the Central Limit Theorem, whereas for σ unknown we made use
of the sampling distribution of the random variable T . However, the use of the tdistribution is based on the premise that the sampling is from a normal distribution.
As long as the distribution is approximately bell shaped, conﬁdence intervals can
be computed when σ 2 is unknown by using the t-distribution and we may expect
very good results.
Computed one-sided conﬁdence bounds for μ with σ unknown are as the reader
would expect, namely
s
s
x̄ + tα √
and
x̄ − tα √ .
n
n
They are the upper and lower 100(1 − α)% bounds, respectively. Here tα is the
t-value having an area of α to the right.
Example 5.6: The contents of seven similar containers of sulfuric acid are 9.8, 10.2, 10.4, 9.8,
10.0, 10.2, and 9.6 liters. Find a 95% conﬁdence interval for the mean contents of
all such containers, assuming an approximately normal distribution.
206
Chapter 5 One- and Two-Sample Estimation Problems
Solution : The sample mean and standard deviation for the given data are
x̄ = 10.0
and
s = 0.283.
Using Table A.4, we ﬁnd t0.025 = 2.447 for v = 6 degrees of freedom. Hence, the
95% conﬁdence interval for μ is
0.283
0.283
< μ < 10.0 + (2.447) √
,
10.0 − (2.447) √
7
7
which reduces to 9.74 < μ < 10.26.
Concept of a Large-Sample Conﬁdence Interval
Often statisticians recommend that even when normality cannot be assumed, σ is
unknown, and n ≥ 30, s can replace σ and the conﬁdence interval
s
x̄ ± zα/2 √
n
may be used. This is often referred to as a large-sample conﬁdence interval. The
justiﬁcation lies only in the presumption that with a sample as large as 30 and
the population distribution not too skewed, s will be very close to the true σ and
thus the Central Limit Theorem prevails. It should be emphasized that this is only
an approximation and the quality of the result becomes better as the sample size
grows larger.
Example 5.7: Scholastic Aptitude Test (SAT) mathematics scores of a random sample of 500
high school seniors in the state of Texas are collected, and the sample mean and
standard deviation are found to be 501 and 112, respectively. Find a 99% conﬁdence
interval on the mean SAT mathematics score for seniors in the state of Texas.
Solution : Since the sample size is large, it is reasonable to use the normal approximation.
Using Table A.3, we ﬁnd z0.005 = 2.575. Hence, a 99% conﬁdence interval for μ is
112
√
= 501 ± 12.9,
501 ± (2.575)
500
which yields 488.1 < μ < 513.9.
5.5
Standard Error of a Point Estimate
We have made a rather sharp distinction between the goal of a point estimate
and that of a conﬁdence interval estimate. The former supplies a single number
extracted from a set of experimental data, and the latter provides an interval that
is reasonable for the parameter, given the experimental data; that is, 100(1 − α)%
of such computed intervals “cover” the parameter.
These two approaches to estimation are related to each other. The common
thread is the sampling distribution of the point estimator. Consider, for example,
5.6 Prediction Intervals
207
the estimator X̄ of μ with σ known. We indicated earlier that a measure of the
quality of an unbiased estimator is its variance. The variance of X̄ is
2
σX̄
=
σ2
.
n
√
Thus, the standard deviation of X̄, or standard error of X̄, is σ/ n. Simply put,
the standard error of an estimator is its standard deviation. For X̄, the computed
conﬁdence limit
σ
x̄ ± zα/2 √ is written as x̄ ± zα/2 s.e.(x̄),
n
where “s.e.” is the “standard error.” The important point is that the width of the
conﬁdence interval on μ is dependent on the quality of the point estimator through
its standard error. In the case where σ is unknown and sampling
√ is from a normal
distribution, s replaces σ and the estimated standard error s/ n is involved. Thus,
the conﬁdence limits on μ are as follows.
Conﬁdence
Limits on μ, σ 2
Unknown
s
x̄ ± tα/2 √ = x̄ ± tα/2 s.e.(x̄)
n
Again, the conﬁdence interval is no better (in terms of width) than the quality of
the point estimate, in this case through its estimated standard error. Computer
packages often refer to estimated standard errors simply as “standard errors.”
As we move to more complex conﬁdence intervals, there is a prevailing notion
that widths of conﬁdence intervals become shorter as the quality of the corresponding point estimate becomes better, although it is not always quite as simple as we
have illustrated here. It can be argued that a conﬁdence interval is merely an
augmentation of the point estimate to take into account the precision of the point
estimate.
5.6
Prediction Intervals
The point and interval estimations of the mean in Sections 5.4 and 5.5 provide
good information about the unknown parameter μ of a normal distribution or a
nonnormal distribution from which a large sample is drawn. Sometimes, other
than the population mean, the experimenter may also be interested in predicting
the possible value of a future observation. For instance, in quality control, the
experimenter may need to use the observed data to predict a new observation. A
process that produces a metal part may be evaluated on the basis of whether the
part meets speciﬁcations on tensile strength. On certain occasions, a customer may
be interested in purchasing a single part. In this case, a conﬁdence interval on the
mean tensile strength does not capture the required information. The customer
requires a statement regarding the uncertainty of a single observation. This type
of requirement is nicely fulﬁlled by the construction of a prediction interval.
It is quite simple to obtain a prediction interval for the situations we have
considered so far. Assume that the random sample comes from a normal population
with unknown mean μ and known variance σ 2 . A natural point estimator of a
208
Chapter 5 One- and Two-Sample Estimation Problems
new observation is X̄. It is known, from Section 4.4, that the variance of X̄ is
σ 2 /n. However, to predict a new observation, not only do we need to account
for the variation due to estimating the mean, but also we should account for the
variation of a future observation. From the assumption, we know that the
variance of the random error in a new observation is σ 2 . The development of a
prediction interval is best illustrated by beginning with a normal random variable
x0 − x̄, where x0 is the new observation and x̄ comes from the sample. Since x0
and x̄ are independent, we know that
x0 − x̄
x0 − x̄
= z=
2
2
σ + σ /n
σ 1 + 1/n
is n(z; 0, 1). As a result, if we use the probability statement
P (−zα/2 < Z < zα/2 ) = 1 − α
with the z-statistic above and place x0 in the center of the probability statement,
we have the following event occurring with probability 1 − α:
x̄ − zα/2 σ 1 + 1/n < x0 < x̄ + zα/2 σ 1 + 1/n.
Computation of the prediction interval is formalized as follows.
Prediction
Interval of a
Future
Observation, σ 2
Known
For a normal distribution of measurements with unknown mean μ and known
variance σ 2 , a 100(1 − α)% prediction interval of a future observation x0 is
x̄ − zα/2 σ 1 + 1/n < x0 < x̄ + zα/2 σ 1 + 1/n,
where zα/2 is the z-value leaving an area of α/2 to the right.
Example 5.8: Due to the decrease in interest rates, the First Citizens Bank received a lot of
mortgage applications. A recent sample of 50 mortgage loans resulted in an average
loan amount of \$257,300. Assume a population standard deviation of \$25,000. For
the next customer who ﬁlls out a mortgage application, ﬁnd a 95% prediction
interval for the loan amount.
Solution : The point prediction of the next customer’s loan amount is x̄ = \$257,300. The
z-value here is z0.025 = 1.96. Hence, a 95% prediction interval for the future loan
amount is
257,300 − (1.96)(25,000) 1 + 1/50 < x0 < 257,300 + (1.96)(25,000) 1 + 1/50,
which gives the interval (\$207,812.43, \$306,787.57).
The prediction interval provides a good estimate of the location of a future
observation, which is quite diﬀerent from the estimate of the sample mean value.
It should be noted that the variation of this prediction is the sum of the variation
due to an estimation of the mean and the variation of a single observation. However,
as in the past, we ﬁrst consider the case with known variance. It is also important
to deal with the prediction interval of a future observation in the situation where
the variance is unknown. Indeed a Student t-distribution may be used in this case,
as described in the following result. The normal distribution is merely replaced by
the t-distribution.
5.6 Prediction Intervals
Prediction
Interval of a
Future
Observation, σ 2
Unknown
209
For a normal distribution of measurements with unknown mean μ and unknown
variance σ 2 , a 100(1 − α)% prediction interval of a future observation x0 is
x̄ − tα/2 s 1 + 1/n < x0 < x̄ + tα/2 s 1 + 1/n,
where tα/2 is the t-value with v = n − 1 degrees of freedom, leaving an area of
α/2 to the right.
One-sided prediction intervals can also be constructed. Upper prediction bounds
apply in cases where focus must be placed on future large observations. Concern
over future small observations calls for the use of lower prediction bounds. The
upper bound is given by
x̄ + tα s 1 + 1/n
and the lower bound by
x̄ − tα s
1 + 1/n.
Example 5.9: A meat inspector has randomly selected 30 packs of 95% lean beef. The sample
resulted in a mean of 96.2% with a sample standard deviation of 0.8%. Find a 99%
prediction interval for the leanness of a new pack. Assume normality.
Solution : For v = 29 degrees of freedom, t0.005 = 2.756. Hence, a 99% prediction interval for
a new observation x0 is
"
"
1
1
96.2 − (2.756)(0.8) 1 +
< x0 < 96.2 + (2.756)(0.8) 1 + ,
30
30
which reduces to (93.96, 98.44).
Use of Prediction Limits for Outlier Detection
To this point in the text very little attention has been paid to the concept of
outliers, or aberrant observations. The majority of scientiﬁc investigators are
keenly sensitive to the existence of outlying observations or so-called faulty or
“bad” data. It is certainly of interest here since there is an important relationship
between outlier detection and prediction intervals.
It is convenient for our purposes to view an outlying observation as one that
comes from a population with a mean that is diﬀerent from the mean that governs
the rest of the sample of size n being studied. The prediction interval produces a
bound that “covers” a future single observation with probability 1 − α if it comes
from the population from which the sample was drawn. As a result, a methodology for outlier detection involves the rule that an observation is an outlier if
it falls outside the prediction interval computed without including the
questionable observation in the sample. As a result, for the prediction interval of Example 5.9, if a new pack of beef is measured and its leanness is outside
the interval (93.96, 98.44), that observation can be viewed as an outlier.
210
5.7
Chapter 5 One- and Two-Sample Estimation Problems
Tolerance Limits
As discussed in Section 5.6, the scientist or engineer may be less interested in estimating parameters than in gaining a notion about where an individual observation
or measurement might fall. Such situations call for the use of prediction intervals.
However, there is yet a third type of interval that is of interest in many applications. Once again, suppose that interest centers around the manufacturing of a
component part and speciﬁcations exist on a dimension of that part. In addition,
there is little concern about the mean of the dimension. But unlike in the scenario
in Section 5.6, one may be less interested in a single observation and more interested in where the majority of the population falls. If process speciﬁcations are
important, the manager of the process is concerned about long-range performance,
not the next observation. One must attempt to determine bounds that, in some
probabilistic sense, “cover” values in the population (i.e., the measured values of
the dimension).
One method of establishing the desired bounds is to determine a conﬁdence
interval on a ﬁxed proportion of the measurements. This is best motivated by
visualizing a situation in which we are doing random sampling from a normal
distribution with known mean μ and variance σ 2 . Clearly, a bound that covers the
middle 95% of the population of observations is
μ ± 1.96σ.
This is called a tolerance interval, and indeed its coverage of 95% of measured
observations is exact. However, in practice, μ and σ are seldom known; thus, the
user must apply
x̄ ± ks.
Now, of course, the interval is a random variable, and hence the coverage of a
proportion of the population by the interval is not exact. As a result, a 100(1−γ)%
conﬁdence interval must be used since x̄ ± ks cannot be expected to cover any
speciﬁed proportion all the time. As a result, we have the following deﬁnition.
Tolerance Limits
For a normal distribution of measurements with unknown mean μ and unknown
standard deviation σ, tolerance limits are given by x̄ ± ks, where k is determined such that one can assert with 100(1 − γ)% conﬁdence that the given
limits contain at least the proportion 1 − α of the measurements.
Table A.7 gives values of k for 1 − α = 0.90, 0.95, 0.99; γ = 0.05, 0.01; and
selected values of n from 2 to 300.
Example 5.10: Consider Example 5.9. With the information given, ﬁnd a tolerance interval that
gives two-sided 95% bounds on 90% of the distribution of packages of 95% lean
beef. Assume the data came from an approximately normal distribution.
Solution : Recall from Example 5.9 that n = 30, the sample mean is 96.2%, and the sample
standard deviation is 0.8%. From Table A.7, k = 2.14. Using
x̄ ± ks = 96.2 ± (2.14)(0.8),
5.7 Tolerance Limits
211
we ﬁnd that the lower and upper bounds are 94.5 and 97.9.
We are 95% conﬁdent that the above range covers the central 90% of the distribution of 95% lean beef packages.
Distinction among Conﬁdence Intervals, Prediction Intervals,
and Tolerance Intervals
It is important to reemphasize the diﬀerence among the three types of intervals discussed and illustrated in the preceding sections. The computations are straightforward, but interpretation can be confusing. In real-life applications, these intervals
are not interchangeable because their interpretations are quite distinct.
In the case of conﬁdence intervals, one is attentive only to the population
mean. For example, Exercise 5.11 on page 213 deals with an engineering process
that produces shearing pins. A speciﬁcation will be set on Rockwell hardness,
below which a customer will not accept any pins. Here, a population parameter
must take a backseat. It is important that the engineer know where the majority
of the values of Rockwell hardness are going to be. Thus, tolerance limits should be
used. Surely, when tolerance limits on any process output are tighter than process
speciﬁcations, that is good news for the process manager.
It is true that the tolerance limit interpretation is somewhat related to the
conﬁdence interval. The 100(1−α)% tolerance interval on, say, the proportion 0.95
can be viewed as a conﬁdence interval on the middle 95% of the corresponding
normal distribution. One-sided tolerance limits are also relevant. In the case of
the Rockwell hardness problem, it is desirable to have a lower bound of the form
x̄ − ks such that there is 99% conﬁdence that at least 99% of Rockwell hardness
values will exceed the computed value.
Prediction intervals are applicable when it is important to determine a bound
on a single value. The mean is not the issue here, nor is the location of the
majority of the population. Rather, the location of a single new observation is
required.
Case Study 5.1: Machine Quality: A machine produces metal pieces that are cylindrical in shape.
A sample of these pieces is taken, and the diameters are found to be 1.01, 0.97,
1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. Use these data to calculate
three interval types and draw interpretations that illustrate the distinction between
them in the context of the system. For all computations, assume an approximately
normal distribution. The sample mean and standard deviation for the given data
are x̄ = 1.0056 and s = 0.0246.
(a) Find a 99% conﬁdence interval on the mean diameter.
(b) Compute a 99% prediction interval on a measured diameter of a single metal
piece taken from the machine.
(c) Find the 99% tolerance limits that will contain 95% of the metal pieces produced by this machine.
Solution : (a) The 99% conﬁdence interval for the mean diameter is given by
√
x̄ ± t0.005 s/ n = 1.0056 ± (3.355)(0.0246/3) = 1.0056 ± 0.0275.
212
Chapter 5 One- and Two-Sample Estimation Problems
Thus, the 99% conﬁdence bounds are 0.9781 and 1.0331.
(b) The 99% prediction interval for a future observation is given by
x̄ ± t0.005 s 1 + 1/n = 1.0056 ± (3.355)(0.0246) 1 + 1/9,
with the bounds being 0.9186 and 1.0926.
(c) From Table A.7, for n = 9, 1 − γ = 0.99, and 1 − α = 0.95, we ﬁnd k = 4.550
for two-sided limits. Hence, the 99% tolerance limits are given by
x̄ + ks = 1.0056 ± (4.550)(0.0246),
with the bounds being 0.8937 and 1.1175. We are 99% conﬁdent that the
tolerance interval from 0.8937 to 1.1175 will contain the central 95% of the
distribution of diameters produced.
This case study illustrates that the three types of limits can give appreciably different results even though they are all 99% bounds. In the case of the conﬁdence
interval on the mean, 99% of such intervals cover the population mean diameter.
Thus, we say that we are 99% conﬁdent that the mean diameter produced by the
process is between 0.9781 and 1.0331 centimeters. Emphasis is placed on the mean,
with less concern about a single reading or the general nature of the distribution
of diameters in the population. In the case of the prediction limits, the bounds
0.9186 and 1.0926 are based on the distribution of a single “new” metal piece
taken from the process, and again 99% of such limits will cover the diameter of
a new measured piece. On the other hand, the tolerance limits, as suggested in
the previous section, give the engineer a sense of where the “majority,” say the
central 95%, of the diameters of measured pieces in the population reside. The
99% tolerance limits, 0.8937 and 1.1175, are numerically quite diﬀerent from the
other two bounds. If these bounds appear alarmingly wide to the engineer, it reﬂects negatively on process quality. On the other hand, if the bounds represent a
desirable result, the engineer may conclude that a majority (95% in this case) of
the diameters are in a desirable range. Again, a conﬁdence interval interpretation
may be used: namely, 99% of such calculated bounds will cover the middle 95% of
the population of diameters.
Exercises
5.1 A UCLA researcher claims that the life span of
mice can be extended by as much as 25% when the
calories in their diet are reduced by approximately 40%
from the time they are weaned. The restricted diet
is enriched to normal levels by vitamins and protein.
Assuming that it is known from previous studies that
σ = 5.8 months, how many mice should be included
in our sample if we wish to be 99% conﬁdent that the
mean life span of the sample will be within 2 months
of the population mean for all mice subjected to this
reduced diet?
5.2 An electrical ﬁrm manufactures light bulbs that
have a length of life that is approximately normally
distributed with a standard deviation of 40 hours. If
a sample of 30 bulbs has an average life of 780 hours,
ﬁnd a 96% conﬁdence interval for the population mean
of all bulbs produced by this ﬁrm.
5.3 Many cardiac patients wear an implanted pacemaker to control their heartbeat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of 0.0015 inch and an approximately normal distribution, ﬁnd a 95% conﬁdence
Exercises
interval for the mean of the depths of all connector
modules made by a certain manufacturing company.
A random sample of 75 modules has an average depth
of 0.310 inch.
213
5.10 For a random sample X1 , . . . , Xn , show that
n
(Xi − μ)2 =
i=1
5.4 The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters.
(a) Construct a 98% conﬁdence interval for the mean
height of all college students.
(b) What can we assert with 98% conﬁdence about the
possible size of our error if we estimate the mean
height of all college students to be 174.5 centimeters?
5.5 A random sample of 100 automobile owners in the
state of Virginia shows that an automobile is driven on
average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of
measurements to be approximately normal.
(a) Construct a 99% conﬁdence interval for the average number of kilometers an automobile is driven
annually in Virginia.
(b) What can we assert with 99% conﬁdence about the
possible size of our error if we estimate the average number of kilometers driven by car owners in
Virginia to be 23,500 kilometers per year?
5.6 How large a sample is needed in Exercise 5.2 if we
wish to be 96% conﬁdent that our sample mean will be
within 10 hours of the true mean?
5.7 How large a sample is needed in Exercise 5.3 if we
wish to be 95% conﬁdent that our sample mean will be
within 0.0005 inch of the true mean?
5.8 An eﬃciency expert wishes to determine the average time that it takes to drill three holes in a certain
metal clamp. How large a sample will she need to be
95% conﬁdent that her sample mean will be within 15
seconds of the true mean? Assume that it is known
from previous studies that σ = 40 seconds.
5.9 Regular consumption of presweetened cereals contributes to tooth decay, heart disease, and other degenerative diseases, according to studies conducted by Dr.
W. H. Bowen of the National Institute of Health and
Dr. J. Yudben, Professor of Nutrition and Dietetics at
the University of London. In a random sample consisting of 20 similar single servings of Alpha-Bits, the
average sugar content was 11.3 grams with a standard
deviation of 2.45 grams. Assuming that the sugar contents are normally distributed, construct a 95% conﬁdence interval for the mean sugar content for single
servings of Alpha-Bits.
n
(Xi − X̄)2 + n(X̄ − μ)2 .
i=1
5.11 A random sample of 12 shearing pins is taken
in a study of the Rockwell hardness of the pin head.
Measurements on the Rockwell hardness are made for
each of the 12, yielding an average value of 48.50 with
a sample standard deviation of 1.5. Assuming the measurements to be normally distributed, construct a 90%
conﬁdence interval for the mean Rockwell hardness.
5.12 The following measurements were recorded for
the drying time, in hours, of a certain brand of latex
paint:
3.4
2.8
4.4
2.5
3.3
4.0
4.8
5.6
5.2
2.9
3.7
3.0
3.6
2.8
4.8
Assuming that the measurements represent a random
sample from a normal population, ﬁnd a 95% prediction interval for the drying time for the next trial of
the paint.
5.13 Referring to Exercise 5.5, construct a 99% prediction interval for the kilometers traveled annually by
an automobile owner in Virginia.
5.14 Consider Exercise 5.9. Compute a 95% prediction interval for the sugar content of the next single
serving of Alpha-Bits.
5.15 A random sample of 25 tablets of buﬀered aspirin contains, on average, 325.05 mg of aspirin per
tablet, with a standard deviation of 0.5 mg. Find the
95% tolerance limits that will contain 90% of the tablet
contents for this brand of buﬀered aspirin. Assume
that the aspirin content is normally distributed.
5.16 Referring to Exercise 5.11, construct a 95% tolerance interval containing 90% of the measurements.
5.17 In a study conducted by the Zoology department
at Virginia Tech, ﬁfteen samples of water were collected
from a certain station in the James River in order to
gain some insight regarding the amount of orthophosphorus in the river. The concentration of the chemical
is measured in milligrams per liter. Let us suppose
that the mean at the station is not as important as the
upper extreme of the distribution of the concentration
of the chemical at the station. Concern centers around
whether the concentration at the extreme is too large.
Readings for the ﬁfteen water samples gave a sample
mean of 3.84 milligrams per liter and a sample standard deviation of 3.07 milligrams per liter. Assume
214
Chapter 5 One- and Two-Sample Estimation Problems
that the readings are a random sample from a normal distribution. Calculate a prediction interval (upper 95% prediction limit) and a tolerance limit (95%
upper tolerance limit that exceeds 95% of the population of values). Interpret both; that is, tell what each
communicates about the upper extreme of the distribution of orthophosphorus at the sampling station.
5.18 Consider the situation of Case Study 5.1 on page
211. Estimation of the mean diameter, while important, is not nearly as important as trying to pin down
the location of the majority of the distribution of diameters. Find the 95% tolerance limits that contain
95% of the diameters.
5.19 Consider the situation of Case Study 5.1 with
a larger sample of metal pieces. The diameters are as
follows: 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 1.01, 1.03,
0.99, 1.00, 1.00, 0.99, 0.98, 1.01, 1.02, 0.99 centimeters. Once again the normality assumption may be
those of the case study. Discuss how they are diﬀerent
and why.
(a) Compute a 99% conﬁdence interval on the mean
diameter.
(b) Compute a 99% prediction interval on the next diameter to be measured.
(c) Compute a 99% tolerance interval for coverage of
the central 95% of the distribution of diameters.
5.20 A type of thread is being studied for its tensile strength properties. Fifty pieces were tested under
similar conditions, and the results showed an average
5.8
tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilograms. Assuming a normal distribution
of tensile strengths, give a lower 95% prediction limit
on a single observed tensile strength value. In addition, give a lower 95% tolerance limit that is exceeded
by 99% of the tensile strength values.
5.21 Refer to Exercise 5.20. Why are the quantities
requested in the exercise likely to be more important to
the manufacturer of the thread than, say, a conﬁdence
interval on the mean tensile strength?
5.22 Refer to Exercise 5.20 again. Suppose that speciﬁcations by a buyer of the thread are that the tensile
strength of the material must be at least 62 kilograms.
The manufacturer is satisﬁed if at most 5% of the manufactured pieces have tensile strength less than 62 kilograms. Is there cause for concern? Use a one-sided 99%
tolerance limit that is exceeded by 95% of the tensile
strength values.
5.23 Consider the drying time measurements in Exercise 5.12. Suppose the 15 observations in the data
set are supplemented by a 16th value of 6.9 hours. In
the context of the original 15 observations, is the 16th
value an outlier? Show work.
5.24 Consider the data in Exercise 5.11. Suppose the
manufacturer of the shearing pins insists that the Rockwell hardness of the product be less than or equal to
44.0 only 5% of the time. What is your reaction? Use
a tolerance limit calculation as the basis for your judgment.
Two Samples: Estimating the Diﬀerence
between Two Means
If we have two populations with means μ1 and μ2 and variances σ12 and σ22 , respectively, a point estimator of the diﬀerence between μ1 and μ2 is given by the
statistic X̄1 − X̄2 . Therefore, to obtain a point estimate of μ1 − μ2 , we shall select
two independent random samples, one from each population, of sizes n1 and n2 ,
and compute x̄1 − x̄2 , the diﬀerence of the sample means. Clearly, we must consider
the sampling distribution of X̄1 − X̄2 .
According to Theorem 4.3, we can expect the sampling distribution of X̄1 − X̄2
to be approximately normal with mean μX̄1 −X̄2 = μ1 − μ2 and standard deviation
σX̄1 −X̄2 = σ12 /n1 + σ22 /n2 . Therefore, we can assert with a probability of 1 − α
that the standard normal variable
Z=
(X̄1 − X̄2 ) − (μ1 − μ2 )
σ12 /n1 + σ22 /n2
5.8 Two Samples: Estimating the Diﬀerence between Two Means
215
will fall between −zα/2 and zα/2 . Referring once again to Figure 5.2 on page 200,
we write
P (−zα/2 < Z < zα/2 ) = 1 − α.
Substituting for Z, we state equivalently that
(X̄1 − X̄2 ) − (μ1 − μ2 )
< zα/2 = 1 − α,
P −zα/2 <
σ12 /n1 + σ22 /n2
which leads to the following 100(1 − α)% conﬁdence interval for μ1 − μ2 .
Conﬁdence
Interval for
μ1 − μ2 , σ12 and
σ22 Known
If x̄1 and x̄2 are means of independent random samples of sizes n1 and n2
from populations with known variances σ12 and σ22 , respectively, a 100(1 − α)%
conﬁdence interval for μ1 − μ2 is given by
#
#
σ12
σ12
σ22
σ2
(x̄1 − x̄2 ) − zα/2
+
< μ1 − μ2 < (x̄1 − x̄2 ) + zα/2
+ 2,
n1
n2
n1
n2
where zα/2 is the z-value leaving an area of α/2 to the right.
The degree of conﬁdence is exact when samples are selected from normal populations. For nonnormal populations, the Central Limit Theorem allows for a good
approximation for reasonable size samples.
Variances Unknown but Equal
Consider the case where σ12 and σ22 are unknown. If σ12 = σ22 = σ 2 , we obtain a
standard normal variable of the form
(X̄1 − X̄2 ) − (μ1 − μ2 )
.
Z= σ 2 [(1/n1 ) + (1/n2 )]
According to Theorem 4.4, the two random variables
(n1 − 1)S12
σ2
and
(n2 − 1)S22
σ2
have chi-squared distributions with n1 − 1 and n2 − 1 degrees of freedom, respectively. Furthermore, they are independent chi-squared variables, since the random
samples were selected independently. Consequently, their sum
V =
(n1 − 1)S12
(n2 − 1)S22
(n1 − 1)S12 + (n2 − 1)S22
+
=
2
2
σ
σ
σ2
has a chi-squared distribution with v = n1 + n2 − 2 degrees of freedom.
Since the preceding expressions for Z and V can be shown to be independent,
it follows from Theorem 4.5 that the statistic
\$#
(X̄1 − X̄2 ) − (μ1 − μ2 )
(n1 − 1)S12 + (n2 − 1)S22
T = σ 2 (n1 + n2 − 2)
σ 2 [(1/n1 ) + (1/n2 )]
216
Chapter 5 One- and Two-Sample Estimation Problems
has the t-distribution with v = n1 + n2 − 2 degrees of freedom.
A point estimate of the unknown common variance σ 2 can be obtained by
pooling the sample variances. Denoting the pooled estimator by Sp2 , we have the
following.
Pooled Estimate
of Variance
Sp2 =
(n1 − 1)S12 + (n2 − 1)S22
n1 + n2 − 2
Substituting Sp2 in the T statistic, we obtain the less cumbersome form
T =
(X̄1 − X̄2 ) − (μ1 − μ2 )
.
Sp (1/n1 ) + (1/n2 )
Using the T statistic, we have
P (−tα/2 < T < tα/2 ) = 1 − α,
where tα/2 is the t-value with n1 + n2 − 2 degrees of freedom, above which we ﬁnd
an area of α/2. Substituting for T in the inequality, we write
(X̄1 − X̄2 ) − (μ1 − μ2 )
P −tα/2 <
< tα/2 = 1 − α.
Sp (1/n1 ) + (1/n2 )
After the usual mathematical manipulations, the diﬀerence of the sample means
x̄1 − x̄2 and the pooled variance are computed and then the following 100(1 − α)%
conﬁdence interval for μ1 − μ2 is obtained. The value of s2p is easily seen to be a
weighted average of the two sample variances s21 and s22 , where the weights are the
degrees of freedom.
Conﬁdence
Interval for
μ1 − μ2 , σ12 = σ22
but Both
Unknown
If x̄1 and x̄2 are the means of independent random samples of sizes n1 and n2 ,
respectively, from approximately normal populations with unknown but equal
variances, a 100(1 − α)% conﬁdence interval for μ1 − μ2 is given by
"
"
1
1
1
1
(x̄1 − x̄2 ) − tα/2 sp
+
< μ1 − μ2 < (x̄1 − x̄2 ) + tα/2 sp
+
,
n1
n2
n1
n2
where sp is the pooled estimate of the population standard deviation and tα/2
is the t-value with v = n1 + n2 − 2 degrees of freedom, leaving an area of α/2
to the right.
Example 5.11: The article “Macroinvertebrate Community Structure as an Indicator of Acid
Mine Pollution,” published in the Journal of Environmental Pollution, reports on
an investigation undertaken in Cane Creek, Alabama, to determine the relationship
between selected physiochemical parameters and diﬀerent measures of macroinvertebrate community structure. One facet of the investigation was an evaluation of
the eﬀectiveness of a numerical species diversity index to indicate aquatic degradation due to acid mine drainage. Conceptually, a high index of macroinvertebrate
species diversity should indicate an unstressed aquatic system, while a low diversity
index should indicate a stressed aquatic system.
5.8 Two Samples: Estimating the Diﬀerence between Two Means
217
Two independent sampling stations were chosen for this study, one located
downstream from the acid mine discharge point and the other located upstream.
For 12 monthly samples collected at the downstream station, the species diversity
index had a mean value x̄1 = 3.11 and a standard deviation s1 = 0.771, while
10 monthly samples collected at the upstream station had a mean index value
x̄2 = 2.04 and a standard deviation s2 = 0.448. Find a 90% conﬁdence interval for
the diﬀerence between the population means for the two locations, assuming that
the populations are approximately normally distributed with equal variances.
Solution : Let μ1 and μ2 represent the population means, respectively, for the species diversity
indices at the downstream and upstream stations. We wish to ﬁnd a 90% conﬁdence
interval for μ1 − μ2 . Our point estimate of μ1 − μ2 is
x̄1 − x̄2 = 3.11 − 2.04 = 1.07.
The pooled estimate, s2p , of the common variance, σ 2 , is
s2p =
(n1 − 1)s21 + (n2 − 1)s22
(11)(0.7712 ) + (9)(0.4482 )
=
= 0.417.
n1 + n2 − 2
12 + 10 − 2
Taking the square root, we obtain sp = 0.646. Using α = 0.1, we ﬁnd in Table A.4
that t0.05 = 1.725 for v = n1 + n2 − 2 = 20 degrees of freedom. Therefore, the 90%
conﬁdence interval for μ1 − μ2 is
"
"
1
1
1
1
1.07 − (1.725)(0.646)
+
< μ1 − μ2 < 1.07 + (1.725)(0.646)
+ ,
12 10
12 10
which simpliﬁes to 0.593 < μ1 − μ2 < 1.547.
Unknown and Unequal Variances
Let us now consider the problem of ﬁnding an interval estimate of μ1 − μ2 when
the unknown population variances are not likely to be equal. The statistic most
often used in this case is
(X̄1 − X̄2 ) − (μ1 − μ2 )
T = 2
,
(S1 /n1 ) + (S22 /n2 )
which has approximately a t-distribution with v degrees of freedom, where
v=
(s21 /n1 + s22 /n2 )2
.
− 1)] + [(s22 /n2 )2 /(n2 − 1)]
[(s21 /n1 )2 /(n1
Since v is seldom an integer, we round it down to the nearest whole number. The
above estimate of the degrees of freedom is called the Satterthwaite approximation
(Satterthwaite, 1946, in the Bibliography).
Using the statistic T , we write
P (−tα/2 < T < tα/2 ) ≈ 1 − α,
where tα/2 is the value of the t-distribution with v degrees of freedom, above which
we ﬁnd an area of α/2. Substituting for T in the inequality and following the
same steps as before, we state the ﬁnal result.
218
Conﬁdence
Interval for
μ1 − μ2 , σ12 = σ22
and Both
Unknown
Chapter 5 One- and Two-Sample Estimation Problems
If x̄1 and s21 and x̄2 and s22 are the means and variances of independent random
samples of sizes n1 and n2 , respectively, from approximately normal populations
with unknown and unequal variances, an approximate 100(1 − α)% conﬁdence
interval for μ1 − μ2 is given by
#
#
s21
s21
s22
s2
(x̄1 − x̄2 ) − tα/2
+
< μ1 − μ2 < (x̄1 − x̄2 ) + tα/2
+ 2,
n1
n2
n1
n2
where tα/2 is the t-value with
v=
(s21 /n1 + s22 /n2 )2
[(s21 /n1 )2 /(n1 − 1)] + [(s22 /n2 )2 /(n2 − 1)]
degrees of freedom, leaving an area of α/2 to the right.
Note that the expression for v above involves random variables, and thus v is
an estimate of the degrees of freedom. In applications, this estimate will not result
in a whole number, and thus the analyst must round down to the nearest integer
to achieve the desired conﬁdence.
Before we illustrate the above conﬁdence interval with an example, we should
point out that all the conﬁdence intervals on μ1 − μ2 are of the same general form
as those on a single mean; namely, they can be written as
point estimate ± tα/2 s%
.e.(point estimate)
or
point estimate ± zα/2 s.e.(point estimate).
For example,in the case where σ1 = σ2 = σ, the estimated standard error of
x̄1 − x̄2 is sp 1/n1 + 1/n2 . For the case where σ12 = σ22 ,
#
s21
s2
+ 2.
s%
.e.(x̄1 − x̄2 ) =
n1
n2
Example 5.12: A study was conducted by the Department of Zoology at Virginia Tech to estimate the diﬀerence in the amounts of the chemical orthophosphorus measured
at two diﬀerent stations on the James River. Orthophosphorus was measured in
milligrams per liter. Fifteen samples were collected from station 1, and 12 samples
were obtained from station 2. The 15 samples from station 1 had an average orthophosphorus content of 3.84 milligrams per liter and a standard deviation of 3.07
milligrams per liter, while the 12 samples from station 2 had an average content of
1.49 milligrams per liter and a standard deviation of 0.80 milligram per liter. Find
a 95% conﬁdence interval for the diﬀerence in the true average orthophosphorus
contents at these two stations, assuming that the observations came from normal
populations with diﬀerent variances.
Solution : For station 1, we have x̄1 = 3.84, s1 = 3.07, and n1 = 15. For station 2, x̄2 = 1.49,
s2 = 0.80, and n2 = 12. We wish to ﬁnd a 95% conﬁdence interval for μ1 − μ2 .
5.9 Paired Observations
219
Since the population variances are assumed to be unequal, we can only ﬁnd an
approximate 95% conﬁdence interval based on the t-distribution with v degrees of
freedom, where
v=
(3.072 /15 + 0.802 /12)2
= 16.3 ≈ 16.
[(3.072 /15)2 /14] + [(0.802 /12)2 /11]
Our point estimate of μ1 − μ2 is
x̄1 − x̄2 = 3.84 − 1.49 = 2.35.
Using α = 0.05, we ﬁnd in Table A.4 that t0.025 = 2.120 for v = 16 degrees of
freedom. Therefore, the 95% conﬁdence interval for μ1 − μ2 is
"
"
3.072
3.072
0.802
0.802
2.35 − 2.120
+
< μ1 − μ2 < 2.35 + 2.120
+
,
15
12
15
12
which simpliﬁes to 0.60 < μ1 − μ2 < 4.10. Hence, we are 95% conﬁdent that the
interval from 0.60 to 4.10 milligrams per liter contains the diﬀerence of the true
average orthophosphorus contents for these two locations.
When two population variances are unknown, the assumption of equal variances
or unequal variances may be precarious.
5.9
Paired Observations
At this point, we shall consider estimation procedures for the diﬀerence of two
means when the samples are not independent and the variances of the two populations are not necessarily equal. The situation considered here deals with a very
special experimental condition, namely that of paired observations. Unlike in the
situations described earlier, the conditions of the two populations are not assigned
randomly to experimental units. Rather, each homogeneous experimental unit receives both population conditions; as a result, each experimental unit has a pair
of observations, one for each population. For example, if we run a test on a new
diet using 15 individuals, the weights before and after going on the diet form the
information for our two samples. The two populations are “before” and “after,”
and the experimental unit is the individual. Obviously, the observations in a pair
have something in common. To determine if the diet is eﬀective, we consider the
diﬀerences d1 , d2 , . . . , dn in the paired observations. These diﬀerences are the values of a random sample D1 , D2 , . . . , Dn from a population of diﬀerences that we
2
shall assume to be normally distributed with mean μD = μ1 − μ2 and variance σD
.
2
2
We estimate σD by sd , the variance of the diﬀerences that constitute our sample.
The point estimator of μD is given by D̄.
Tradeoﬀ between Reducing Variance and Losing Degrees of Freedom
Comparing the conﬁdence intervals obtained with and without pairing makes apparent that there is a tradeoﬀ involved. Although pairing should indeed reduce
variance and hence reduce the standard error of the point estimate, the degrees of
freedom are reduced by reducing the problem to a one-sample problem. As a result,
220
Chapter 5 One- and Two-Sample Estimation Problems
the tα/2 point attached to the standard error is adjusted accordingly. Thus, pairing may be counterproductive. This would certainly be the case if one experienced
2
only a modest reduction in variance (through σD
) by pairing.
Another illustration of pairing involves choosing n pairs of subjects, with each
pair having a similar characteristic such as IQ, age, or breed, and then selecting
one member of each pair at random to yield a value of X1 , leaving the other
member to provide the value of X2 . In this case, X1 and X2 might represent
the grades obtained by two individuals of equal IQ when one of the individuals is
assigned at random to a class using the conventional lecture approach while the
other individual is assigned to a class using programmed materials.
A 100(1 − α)% conﬁdence interval for μD can be established by writing
P (−tα/2 < T < tα/2 ) = 1 − α,
√D and tα/2 , as before, is a value of the t-distribution with n − 1
where T = SD̄−μ
d/ n
degrees of freedom.
It is now a routine procedure to replace T by its deﬁnition in the inequality
above and carry out the mathematical steps that lead to the following 100(1 − α)%
conﬁdence interval for μ1 − μ2 = μD .
μD
Conﬁdence
Interval for
= μ1 − μ2 for
Paired
Observations
If d¯ and sd are the mean and standard deviation, respectively, of the normally
distributed diﬀerences of n random pairs of measurements, a 100(1 − α)% conﬁdence interval for μD = μ1 − μ2 is
sd
sd
d¯ − tα/2 √ < μD < d¯ + tα/2 √ ,
n
n
where tα/2 is the t-value with v = n − 1 degrees of freedom, leaving an area of
α/2 to the right.
Example 5.13: A study was undertaken at Virginia Tech to determine if ﬁre can be used as a
viable management tool to increase the amount of forage available to deer during
the critical months in late winter and early spring. Calcium is a required element
for plants and animals. The amount taken up and stored in plants is closely
correlated to the amount present in the soil. It was hypothesized that a ﬁre may
change the calcium levels present in the soil and thus aﬀect the amount available
to deer. A large tract of land in the Fishburn Forest was selected for a prescribed
burn. Soil samples were taken from 12 plots of equal area just prior to the burn
and analyzed for calcium. Postburn calcium levels were analyzed from the same
plots. These values, in kilograms per plot, are presented in Table 5.1.
Construct a 95% conﬁdence interval for the mean diﬀerence in calcium levels
in the soil prior to and after the prescribed burn. Assume the distribution of
diﬀerences in calcium levels to be approximately normal.
Solution : In this problem, we wish to ﬁnd a 95% conﬁdence interval for the mean diﬀerence
in calcium levels between the preburn and postburn. Since the observations are
paired, we deﬁne μPreburn − μPostburn = μD . The sample size is n = 12, the point estimate of μD is d¯ = 40.58, and the sample standard deviation, sd , of the diﬀerences
Exercises
221
Table 5.1: Data for Example 5.13
Plot
1
2
3
4
5
6
is
Calcium Level (kg/plot)
Preburn
Postburn
9
50
18
50
45
82
18
64
18
82
9
73
&
'
'
sd = (
Plot
7
8
9
10
11
12
Calcium Level (kg/plot)
Preburn
Postburn
32
77
9
54
18
23
9
45
9
36
9
54
1 ¯ 2 = 15.791.
(di − d)
n − 1 i=1
n
Using α = 0.05, we ﬁnd in Table A.4 that t0.025 = 2.201 for v = n − 1 = 11 degrees
of freedom. Therefore, the 95% conﬁdence interval is
15.791
15.791
< μD < 40.58 + (2.201) √
,
40.58 − (2.201) √
12
12
or simply 30.55 < μD < 50.61, from which we can conclude that there is signiﬁcant
diﬀerence in calcium levels between the soil prior and post to the prescribed burns.
Hence there is signiﬁcant reduction in the level of calcium after the burn.
Exercises
5.25 A study was conducted to determine if a certain treatment has any eﬀect on the amount of metal
removed in a pickling operation. A random sample of
100 pieces was immersed in a bath for 24 hours without
the treatment, yielding an average of 12.2 millimeters
of metal removed and a sample standard deviation of
1.1 millimeters. A second sample of 200 pieces was
exposed to the treatment, followed by the 24-hour immersion in the bath, resulting in an average removal
of 9.1 millimeters of metal with a sample standard deviation of 0.9 millimeter. Compute a 98% conﬁdence
interval estimate for the diﬀerence between the population means. Does the treatment appear to reduce the
mean amount of metal removed?
5.26 Two kinds of thread are being compared for
strength. Fifty pieces of each type of thread are tested
under similar conditions. Brand A has an average tensile strength of 78.3 kilograms with a standard deviation of 5.6 kilograms, while brand B has an average
tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms. Construct a 95% conﬁdence
interval for the diﬀerence of the population means.
5.27 Two catalysts in a batch chemical process are
being compared for their eﬀect on the output of the
process reaction. A sample of 12 batches was prepared
using catalyst 1, and a sample of 10 batches was prepared using catalyst 2. The 12 batches for which catalyst 1 was used in the reaction gave an average yield
of 85 with a sample standard deviation of 4, and the
10 batches for which catalyst 2 was used gave an average yield of 81 and a sample standard deviation of 5.
Find a 90% conﬁdence interval for the diﬀerence between the population means, assuming that the populations are approximately normally distributed with
equal variances.
5.28 In a study conducted at Virginia Tech on the
development of ectomycorrhizal, a symbiotic relationship between the roots of trees and a fungus, in which
minerals are transferred from the fungus to the trees
and sugars from the trees to the fungus, 20 northern
red oak seedlings exposed to the fungus Pisolithus tinctorus were grown in a greenhouse. All seedlings were
planted in the same type of soil and received the same
amount of sunshine and water. Half received no ni-
222
Chapter 5 One- and Two-Sample Estimation Problems
trogen at planting time, to serve as a control, and the
other half received 368 ppm of nitrogen in the form
NaNO3 . The stem weights, in grams, at the end of 140
days were recorded as follows:
No Nitrogen
0.32
0.53
0.28
0.37
0.47
0.43
0.36
0.42
0.38
0.43
Nitrogen
0.26
0.43
0.47
0.49
0.52
0.75
0.79
0.86
0.62
0.46
Construct a 95% conﬁdence interval for the diﬀerence
in the mean stem weight between seedlings that receive no nitrogen and those that receive 368 ppm of
nitrogen. Assume the populations to be normally distributed with equal variances.
5.29 The following data represent the length of time,
in days, to recovery for patients randomly treated with
one of two medications to clear up severe bladder infections:
Medication 1 Medication 2
n1 = 14
n2 = 16
x̄1 = 17
x̄2 = 19
s22 = 1.8
s21 = 1.5
Find a 99% conﬁdence interval for the diﬀerence μ2 −μ1
in the mean recovery times for the two medications, assuming normal populations with equal variances.
5.30 An experiment reported in Popular Science
compared fuel economies for two types of similarly
equipped diesel mini-trucks. Let us suppose that 12
Volkswagen and 10 Toyota trucks were tested in 90kilometer-per-hour steady-paced trials. If the 12 Volkswagen trucks averaged 16 kilometers per liter with a
standard deviation of 1.0 kilometer per liter and the 10
Toyota trucks averaged 11 kilometers per liter with a
standard deviation of 0.8 kilometer per liter, construct
a 90% conﬁdence interval for the diﬀerence between the
average kilometers per liter for these two mini-trucks.
Assume that the distances per liter for the truck models are approximately normally distributed with equal
variances.
5.31 A taxi company is trying to decide whether to
purchase brand A or brand B tires for its ﬂeet of taxis.
To estimate the diﬀerence in the two brands, an experiment is conducted using 12 of each brand. The tires
are run until they wear out. The results are
Brand A:
Brand B:
x̄1 = 36,300 kilometers,
s1 = 5000 kilometers.
x̄2 = 38,100 kilometers,
s2 = 6100 kilometers.
Compute a 95% conﬁdence interval for μA − μB assuming the populations to be approximately normally
distributed. You may not assume that the variances
are equal.
5.32 Referring to Exercise 5.31, ﬁnd a 99% conﬁdence
interval for μ1 − μ2 if tires of the two brands are assigned at random to the left and right rear wheels of
8 taxis and the following distances, in kilometers, are
recorded:
Taxi Brand A Brand B
1
34,400
36,700
2
45,500
46,800
3
36,700
37,700
4
32,000
31,100
5
48,400
47,800
6
32,800
36,400
7
38,100
38,900
8
30,100
31,500
Assume that the diﬀerences of the distances are approximately normally distributed.
5.33 The federal government awarded grants to the
agricultural departments of 9 universities to test the
yield capabilities of two new varieties of wheat. Each
variety was planted on a plot of equal area at each
university, and the yields, in kilograms per plot, were
recorded as follows:
University
Variety 1 2 3 4 5 6 7 8 9
1
38 23 35 41 44 29 37 31 38
2
45 25 31 38 50 33 36 40 43
Find a 95% conﬁdence interval for the mean diﬀerence
between the yields of the two varieties, assuming the
diﬀerences of yields to be approximately normally distributed. Explain why pairing is necessary in this problem.
5.34 The following data represent the running times
of ﬁlms produced by two motion-picture companies.
Company
I
II
Time (minutes)
103 94 110 87 98
97 82 123 92 175 88 118
Compute a 90% conﬁdence interval for the diﬀerence
between the average running times of ﬁlms produced by
the two companies. Assume that the running-time differences are approximately normally distributed with
unequal variances.
5.10 Single Sample: Estimating a Proportion
5.35 Fortune magazine (March 1997) reported the total returns to investors for the 10 years prior to 1996
and also for 1996 for 431 companies. The total returns
for 9 of the companies and the S&P 500 are listed
below. Find a 95% conﬁdence interval for the mean
Company
Coca-Cola
Mirage Resorts
Merck
Microsoft
Johnson & Johnson
Intel
Pﬁzer
Procter & Gamble
Berkshire Hathaway
S&P 500
Total Return
to Investors
1986–96
1996
29.8%
43.3%
27.9%
25.4%
22.1%
24.0%
44.5%
88.3%
22.2%
18.1%
43.8%
131.2%
21.7%
34.0%
21.9%
32.1%
28.3%
6.2%
11.8%
20.3%
5.36 An automotive company is considering two
types of batteries for its automobile. Sample information on battery life is collected for 20 batteries of
type A and 20 batteries of type B. The summary
statistics are x̄A = 32.91, x̄B = 30.47, sA = 1.57,
and sB = 1.74. Assume the data on each battery are
normally distributed and assume σA = σB .
5.10
223
(a) Find a 95% conﬁdence interval on μA − μB .
(b) Draw a conclusion from (a) that provides insight
into whether A or B should be adopted.
5.37 Two diﬀerent brands of latex paint are being
considered for use. Fifteen specimens of each type of
paint were selected, and the drying times, in hours,
were as follows:
Paint B
Paint A
3.5 2.7 3.9 4.2 3.6
4.7 3.9 4.5 5.5 4.0
2.7 3.3 5.2 4.2 2.9
5.3 4.3 6.0 5.2 3.7
4.4 5.2 4.0 4.1 3.4
5.5 6.2 5.1 5.4 4.8
Assume the drying time is normally distributed with
σA = σB . Find a 95% conﬁdence interval on μB − μA ,
where μA and μB are the mean drying times.
5.38 Two levels (low and high) of insulin doses are
given to two groups of diabetic rats to check the insulinbinding capacity, yielding the following data:
Low dose:
High dose:
n1 = 8
n2 = 13
x̄1 = 1.98
x̄2 = 1.30
s1 = 0.51
s2 = 0.35
Assume that the variances are equal. Give a 95% conﬁdence interval for the diﬀerence in the true average
insulin-binding capacity between the two samples.
Single Sample: Estimating a Proportion
A point estimator of the proportion p in a binomial experiment is given by the
statistic P) = X/n, where X represents the number of successes in n trials. Therefore, the sample proportion p̂ = x/n will be used as the point estimate of the
parameter p.
If the unknown proportion p is not expected to be too close to 0 or 1, we can
establish a conﬁdence interval for p by considering the sampling distribution of
P) . Designating a failure in each binomial trial by the value 0 and a success by
the value 1, the number of successes, x, can be interpreted as the sum of n values
consisting only of 0 and 1s, and p̂ is just the sample mean of these n values. Hence,
by the Central Limit Theorem, for n suﬃciently large, P) is approximately normally
distributed with mean
np
X
)
μP = E(P ) = E
=
=p
n
n
and variance
2
σP2 = σX/n
=
2
σX
npq
pq
= 2 =
.
2
n
n
n
Therefore, we can assert that
P) − p
,
P (−zα/2 < Z < zα/2 ) = 1 − α, with Z = pq/n
224
Chapter 5 One- and Two-Sample Estimation Problems
and zα/2 is the value above which we ﬁnd an area of α/2 under the standard normal
curve.
For a random sample of size n, the sample proportion p̂ = x/n is computed, and
the following approximate 100(1 − α)% conﬁdence intervals for p can be obtained.
Large-Sample
Conﬁdence
Intervals for p
If p̂ is the proportion of successes in a random sample of size n and q̂ = 1 − p̂,
an approximate 100(1 − α)% conﬁdence interval for the binomial parameter p
is given by
"
"
p̂q̂
p̂q̂
p̂ − zα/2
< p < p̂ + zα/2
,
n
n
where zα/2 is the z-value leaving an area of α/2 to the right.
When n is small and the unknown proportion p is believed to be close to 0 or to
1, the conﬁdence-interval procedure established here is unreliable and, therefore,
should not be used. To be on the safe side, one should require both np̂ and nq̂
to be greater than or equal to 5. The methods for ﬁnding a conﬁdence interval
for the binomial parameter p are also applicable when the binomial distribution
is being used to approximate the hypergeometric distribution, that is, when n is
small relative to N , as illustrated by Example 5.14.
Example 5.14: In a random sample of n = 500 families owning television sets in the city of Hamilton, Canada, it is found that x = 340 subscribe to HBO. Find a 95% conﬁdence
interval for the actual proportion of families with television sets in this city that
subscribe to HBO.
Solution : The point estimate of p is p̂ = 340/500 = 0.68. Using Table A.3, we ﬁnd that
z0.025 = 1.96. Therefore, the 95% conﬁdence interval for p is
"
"
(0.68)(0.32)
(0.68)(0.32)
0.68 − 1.96
< p < 0.68 + 1.96
,
500
500
which simpliﬁes to 0.6391 < p < 0.7209.
of p, we can be 100(1 − α)% conﬁdent that the error
Theorem 5.3: If p̂ is used as an estimate
will not exceed zα/2 p̂q̂/n.
In Example 5.14, we are 95% conﬁdent that the sample proportion p̂ = 0.68
diﬀers from the true proportion p by an amount not exceeding 0.04.
Choice of Sample Size
Let us now determine how large a sample is necessary to ensure that the error in
estimating p will be less
than a speciﬁed amount e. By Theorem 5.3, we must
choose n such that zα/2 p̂q̂/n = e.
5.10 Single Sample: Estimating a Proportion
225
Theorem 5.4: If p̂ is used as an estimate of p, we can be 100(1 − α)% conﬁdent that the error
will be less than a speciﬁed amount e when the sample size is approximately
n=
2
zα/2
p̂q̂
e2
.
Theorem 5.4 is somewhat misleading in that we must use p̂ to determine the
sample size n, but p̂ is computed from the sample. If a crude estimate of p can
be made without taking a sample, this value can be used to determine n. Lacking
such an estimate, we could take a preliminary sample of size n ≥ 30 to provide
an estimate of p. Using Theorem 5.4, we could determine approximately how
many observations are needed to provide the desired degree of accuracy. Note that
fractional values of n are rounded up to the next whole number.
Example 5.15: How large a sample is required if we want to be 95% conﬁdent that our estimate
of p in Example 5.14 is within 0.02 of the true value?
Solution : Let us treat the 500 families as a preliminary sample, providing an estimate p̂ =
0.68. Then, by Theorem 5.4,
n=
(1.96)2 (0.68)(0.32)
= 2089.8 ≈ 2090.
(0.02)2
Therefore, if we base our estimate of p on a random sample of size 2090, we can be
95% conﬁdent that our sample proportion will not diﬀer from the true proportion
by more than 0.02.
Occasionally, it will be impractical to obtain an estimate of p to be used for
determining the sample size for a speciﬁed degree of conﬁdence. If this happens,
an upper bound for n is established by noting that p̂q̂ = p̂(1 − p̂), which must
be at most 1/4, since p̂ must lie between 0 and 1. This fact may be veriﬁed by
completing the square. Hence
2
1
1
1
1
p̂(1 − p̂) = −(p̂2 − p̂) = − p̂2 − p̂ +
= − p̂ −
,
4
4
4
2
which is always less than 1/4 except when p̂ = 1/2, and then p̂q̂ = 1/4. Therefore,
if we substitute p̂ = 1/2 into the formula for n in Theorem 5.4 when, in fact, p
actually diﬀers from l/2, n will turn out to be larger than necessary for the speciﬁed
degree of conﬁdence; as a result, our degree of conﬁdence will increase.
Theorem 5.5: If p̂ is used as an estimate of p, we can be at least 100(1 − α)% conﬁdent that
the error will not exceed a speciﬁed amount e when the sample size is
n=
2
zα/2
4e2
.
Example 5.16: How large a sample is required if we want to be at least 95% conﬁdent that our
estimate of p in Example 5.14 is within 0.02 of the true value?
226
Chapter 5 One- and Two-Sample Estimation Problems
Solution : Unlike in Example 5.15, we shall now assume that no preliminary sample has been
taken to provide an estimate of p. Consequently, we can be at least 95% conﬁdent
that our sample proportion will not diﬀer from the true proportion by more than
0.02 if we choose a sample of size
n=
(1.96)2
= 2401.
(4)(0.02)2
Comparing the results of Examples 5.15 and 5.16, we see that information
concerning p, provided by a preliminary sample or from experience, enables us to
choose a smaller sample while maintaining our required degree of accuracy.
5.11
Two Samples: Estimating the Diﬀerence between
Two Proportions
Consider the problem where we wish to estimate the diﬀerence between two binomial parameters p1 and p2 . For example, p1 might be the proportion of smokers
with lung cancer and p2 the proportion of nonsmokers with lung cancer, and the
problem is to estimate the diﬀerence between these two proportions. First, we
select independent random samples of sizes n1 and n2 from the two binomial populations with means n1 p1 and n2 p2 and variances n1 p1 q1 and n2 p2 q2 , respectively;
then we determine the numbers x1 and x2 of people in each sample with lung cancer and form the proportions p̂1 = x1 /n and p̂2 = x2 /n. A point estimator of the
diﬀerence between the two proportions, p1 − p2 , is given by the statistic P)1 − P)2 .
Therefore, the diﬀerence of the sample proportions, p̂1 − p̂2 , will be used as the
point estimate of p1 − p2 .
We use concepts identical to those covered in Section 5.8 in the discussion of
the conﬁdence interval for the diﬀerence μ1 − μ2 . Note that the Z statistic for
μ1 − μ2 has an equivalent in the case of p1 − p2 , namely,
(P̂1 − P̂2 ) − (p1 − p2 )
Z= .
p1 q1 /n1 + p2 q2 /n2
As a result, we have the Large Sample Conﬁdence Interval for p1 − p2 .
Large-Sample
Conﬁdence
Interval for
p1 − p2
If p̂1 and p̂2 are the proportions of successes in random samples of sizes n1 and
n2 , respectively, q̂1 = 1 − p̂1 , and q̂2 = 1 − p̂2 , an approximate 100(1 − α)%
conﬁdence interval for the diﬀerence of two binomial parameters, p1 − p2 , is
given by
"
"
p̂1 q̂1
p̂1 q̂1
p̂2 q̂2
p̂2 q̂2
(p̂1 − p̂2 ) − zα/2
+
< p1 − p2 < (p̂1 − p̂2 ) + zα/2
+
,
n1
n2
n1
n2
where zα/2 is the z-value leaving an area of α/2 to the right.
Example 5.17: A certain change in a process for manufacturing component parts is being considered. Samples are taken from both the existing and the new process so as
to determine if the new process results in an improvement. If 75 of 1500 items
Exercises
227
from the existing process are found to be defective and 80 of 2000 items from the
new process are found to be defective, ﬁnd a 90% conﬁdence interval for the true
diﬀerence in the proportion of defectives between the existing and the new process.
Solution : Let p1 and p2 be the true proportions of defectives for the existing and new processes, respectively. Hence, p̂1 = 75/1500 = 0.05 and p̂2 = 80/2000 = 0.04, and
the point estimate of p1 − p2 is
p̂1 − p̂2 = 0.05 − 0.04 = 0.01.
Using Table A.3, we ﬁnd z0.05 = 1.645. Therefore, substituting into the formula,
with
"
(0.05)(0.95) (0.04)(0.96)
1.645
+
= 0.0117,
1500
2000
we ﬁnd the 90% conﬁdence interval to be −0.0017 < p1 − p2 < 0.0217. Since the
interval contains the value 0, there is no reason to believe that the new process
produces a signiﬁcant decrease in the proportion of defectives over the existing
method.
Up to this point, all conﬁdence intervals presented were of the form
point estimate ± K s.e.(point estimate),
where K is a constant (either t or normal percent point). This form is valid when
the parameter is a mean, a diﬀerence between means, a proportion, or a diﬀerence
between proportions, due to the symmetry of the t- and Z-distributions.
Exercises
5.39 In a random sample of 1000 homes in a certain
city, it is found that 228 are heated by oil. Find 99%
conﬁdence intervals for the proportion of homes in this
city that are heated by oil.
being sold. Of a random sample of 500 MP3 players, 15
failed one or more tests. Find a 90% conﬁdence interval
for the proportion of MP3 players from the population
that pass all tests.
5.40 Compute 95% conﬁdence intervals for the proportion of defective items in a process when it is found
that a sample of size 100 yields 8 defectives.
5.43 A new rocket-launching system is being considered for deployment of small, short-range rockets. The
existing system has p = 0.8 as the probability of a successful launch. A sample of 40 experimental launches
is made with the new system, and 34 are successful.
(a) Construct a 95% conﬁdence interval for p.
(b) Would you conclude that the new system is better?
5.41 (a) A random sample of 200 voters in a town is
selected, and 114 are found to support an annexation suit. Find the 96% conﬁdence interval for the
proportion of the voting population favoring the
suit.
(b) What can we assert with 96% conﬁdence about the
possible size of our error if we estimate the proportion of voters favoring the annexation suit to be
0.57?
5.42 A manufacturer of MP3 players conducts a set
of comprehensive tests on the electrical functions of its
product. All MP3 players must pass all tests prior to
5.44 A geneticist is interested in the proportion of
African males who have a certain minor blood disorder. In a random sample of 100 African males, 24 are
found to be aﬄicted.
(a) Compute a 99% conﬁdence interval for the proportion of African males who have this blood disorder.
(b) What can we assert with 99% conﬁdence about the
possible size of our error if we estimate the propor-
228
Chapter 5 One- and Two-Sample Estimation Problems
tion of African males with this blood disorder to be
0.24?
5.45 How large a sample is needed if we wish to be
96% conﬁdent that our sample proportion in Exercise
5.41 will be within 0.02 of the true fraction of the voting population?
5.46 How large a sample is needed if we wish to be
99% conﬁdent that our sample proportion in Exercise
5.39 will be within 0.05 of the true proportion of homes
in the city that are heated by oil?
5.47 How large a sample is needed in Exercise 5.40 if
we wish to be 98% conﬁdent that our sample proportion will be within 0.05 of the true proportion defective?
5.48 How large a sample is needed to estimate the
percentage of citizens in a certain town who favor having their water ﬂuoridated if one wishes to be at least
99% conﬁdent that the estimate is within 1% of the
true percentage?
5.49 A study is to be made to estimate the proportion of residents of a certain city and its suburbs who
favor the construction of a nuclear power plant near
the city. How large a sample is needed if one wishes to
be at least 95% conﬁdent that the estimate is within
0.04 of the true proportion of residents who favor the
construction of the nuclear power plant?
5.50 Ten engineering schools in the United States
were surveyed. The sample contained 250 electrical en-
5.12
gineers, 80 being women, and 175 chemical engineers,
40 being women. Compute a 90% conﬁdence interval
for the diﬀerence between the proportions of women in
these two ﬁelds of engineering. Is there a signiﬁcant
diﬀerence between the two proportions?
5.51 A certain geneticist is interested in the proportion of males and females in the population who have
a minor blood disorder. In a random sample of 1000
males, 250 are found to be aﬄicted, whereas 275 of
1000 females tested appear to have the disorder. Compute a 95% conﬁdence interval for the diﬀerence between the proportions of males and females who have
the blood disorder.
5.52 In the study Germination and Emergence of
Broccoli, conducted by the Department of Horticulture
at Virginia Tech, a researcher found that at 5◦ C, 10
broccoli seeds out of 20 germinated, while at 15◦ C, 15
out of 20 germinated. Compute a 95% conﬁdence interval for the diﬀerence between the proportions of germination at the two diﬀerent temperatures and decide
if there is a signiﬁcant diﬀerence.
5.53 A clinical trial was conducted to determine if a
certain type of inoculation has an eﬀect on the incidence of a certain disease. A sample of 1000 rats was
kept in a controlled environment for a period of 1 year,
and 500 of the rats were given the inoculation. In the
group not inoculated, there were 120 incidences of the
disease, while 98 of the rats in the inoculated group
contracted it. If p1 is the probability of incidence of
the disease in uninoculated rats and p2 the probability
of incidence in inoculated rats, compute a 90% conﬁdence interval for p1 − p2 .
Single Sample: Estimating the Variance
If a sample of size n is drawn from a normal population with variance σ 2 and
the sample variance s2 is computed, we obtain a value of the statistic S 2 . This
computed sample variance is used as a point estimate of σ 2 . Hence, the statistic
S 2 is called an estimator of σ 2 .
An interval estimate of σ 2 can be established by using the statistic
X2 =
(n − 1)S 2
.
σ2
According to Theorem 4.4, the statistic X 2 has a chi-squared distribution with
n − 1 degrees of freedom when samples are chosen from a normal population. We
may write (see Figure 5.6)
P (χ21−α/2 < X 2 < χ2α/2 ) = 1 − α,
where χ21−α/2 and χ2α/2 are values of the chi-squared distribution with n−1 degrees
of freedom, leaving areas of 1−α/2 and α/2, respectively, to the right. Substituting
5.12 Single Sample: Estimating the Variance
229
for X 2 , we write
(n − 1)S 2
2
P χ21−α/2 <
<
χ
α/2 = 1 − α.
σ2
1α
α /2
α /2
0
2
1
α /2
α2 /2
2
Figure 5.6: P (χ21−α/2 < X 2 < χ2α/2 ) = 1 − α.
Dividing each term in the inequality by (n − 1)S 2 and then inverting each term
(thereby changing the sense of the inequalities), we obtain
(n − 1)S 2
(n − 1)S 2
2
= 1 − α.
P
<σ <
χ2α/2
χ21−α/2
For a random sample of size n from a normal population, the sample variance s2
is computed, and the following 100(1 − α)% conﬁdence interval for σ 2 is obtained.
Conﬁdence
Interval for σ 2
If s2 is the variance of a random sample of size n from a normal population, a
100(1 − α)% conﬁdence interval for σ 2 is
(n − 1)s2
(n − 1)s2
2
<
σ
<
,
χ2α/2
χ21−α/2
where χ2α/2 and χ21−α/2 are χ2 -values with v = n − 1 degrees of freedom, leaving
areas of α/2 and 1 − α/2, respectively, to the right.
An approximate 100(1 − α)% conﬁdence interval for σ is obtained by taking
the square root of each endpoint of the interval for σ 2 .
Example 5.18: The following are the weights, in decagrams, of 10 packages of grass seed distributed by a certain company: 46.4, 46.1, 45.8, 47.0, 46.1, 45.9, 45.8, 46.9, 45.2,
and 46.0. Find a 95% conﬁdence interval for the variance of the weights of all such
packages of grass seed distributed by this company, assuming a normal population.
230
Chapter 5 One- and Two-Sample Estimation Problems
Solution : First we ﬁnd
n
s2 =
n
i=1
x2i −
n
2
xi
i=1
n(n − 1)
(10)(21,273.12) − (461.2)2
=
= 0.286.
(10)(9)
To obtain a 95% conﬁdence interval, we choose α = 0.05. Then, using Table
A.5 with v = 9 degrees of freedom, we ﬁnd χ20.025 = 19.023 and χ20.975 = 2.700.
Therefore, the 95% conﬁdence interval for σ 2 is
(9)(0.286)
(9)(0.286)
< σ2 <
,
19.023
2.700
or simply 0.135 < σ 2 < 0.953.
Exercises
5.54 A random sample of 20 students yielded a mean
of x̄ = 72 and a variance of s2 = 16 for scores on a
college placement test in mathematics. Assuming the
scores to be normally distributed, construct a 98% conﬁdence interval for σ 2 .
that σ 2 = 1 is valid. Assume the population of battery
lives to be approximately normally distributed.
5.55 A manufacturer of car batteries claims that the
batteries will last, on average, 3 years with a variance
of 1 year. If 5 of these batteries have lifetimes of 1.9,
2.4, 3.0, 3.5, and 4.2 years, construct a 95% conﬁdence
interval for σ 2 and decide if the manufacturer’s claim
5.57 Construct a 95% conﬁdence interval for σ 2 in
Exercise 5.9 on page 213.
5.56 Construct a 99% conﬁdence interval for σ 2 in
Case Study 5.1 on page 211.
5.58 Construct a 90% conﬁdence interval for σ in Exercise 5.11 on page 213.
Review Exercises
5.59 According to the Roanoke Times, in a particular year McDonald’s sold 42.1% of the market share
of hamburgers. A random sample of 75 burgers sold
resulted in 28 of them being from McDonald’s. Use
material in Section 5.10 to determine if this information supports the claim in the Roanoke Times.
5.60 It is claimed that a new diet will reduce a person’s weight by 4.5 kilograms on average in a period
of 2 weeks. The weights of 7 women who followed this
diet were recorded before and after the 2-week period.
Test the claim about the diet by computing a 95% conﬁdence interval for the mean diﬀerence in weights. Assume the diﬀerences of weights to be approximately
normally distributed.
Woman
1
2
3
4
5
6
7
Weight Before
58.5
60.3
61.7
69.0
64.0
62.6
56.7
Weight After
60.0
54.9
58.1
62.1
58.5
59.9
54.4
5.61 A health spa claims that a new exercise program will reduce a person’s waist size by 2 centimeters
on average over a 5-day period. The waist sizes, in
centimeters, of 6 men who participated in this exercise
program are recorded before and after the 5-day period
in the following table:
Review Exercises
231
Man Waist Size Before Waist Size After
91.7
90.4
1
93.9
95.5
2
97.4
98.7
3
112.8
115.9
4
101.3
104.0
5
84.0
85.6
6
By computing a 95% conﬁdence interval for the mean
reduction in waist size, determine whether the health
spa’s claim is valid. Assume the distribution of diﬀerences in waist sizes before and after the program to be
approximately normal.
Endurance Limit (psi)
Polished
Unpolished
0.4% Carbon 0.4% Carbon
85,500
82,600
91,900
82,400
89,400
81,700
84,000
79,500
89,900
79,400
78,700
69,800
87,500
79,900
83,100
83,400
5.62 The Department of Civil Engineering at Virginia
Tech compared a modiﬁed (M-5 hr) assay technique for
recovering fecal coliforms in stormwater runoﬀ from an
urban area to a most probable number (MPN) technique. A total of 12 runoﬀ samples were collected and
analyzed by the two techniques. Fecal coliform counts
per 100 milliliters are recorded in the following table.
5.64 An anthropologist is interested in the proportion
of individuals in two Indian tribes with double occipital hair whorls. Suppose that independent samples are
taken from each of the two tribes, and it is found that
24 of 100 Indians from tribe A and 36 of 120 Indians
from tribe B possess this characteristic. Construct a
95% conﬁdence interval for the diﬀerence pB − pA between the proportions of these two tribes with occipital
hair whorls.
Sample
1
2
3
4
5
6
7
8
9
10
11
12
MPN Count
2300
1200
450
210
270
450
154
179
192
230
340
194
M-5 hr Count
2010
930
400
436
4100
2090
219
169
194
174
274
183
5.65 A manufacturer of electric irons produces these
items in two plants. Both plants have the same suppliers of small parts. A saving can be made by purchasing
thermostats for plant B from a local supplier. A single lot was purchased from the local supplier, and a
test was conducted to see whether or not these new
thermostats were as accurate as the old. The thermostats were tested on tile irons on the 550◦ F setting,
and the actual temperatures were read to the nearest
0.1◦ F with a thermocouple. The data are as follows:
Construct a 90% conﬁdence interval for the diﬀerence
in the mean fecal coliform counts between the M-5 hr
and the MPN techniques. Assume that the count differences are approximately normally distributed.
530.3
549.9
559.1
550.0
5.63 An experiment was conducted to determine
whether surface ﬁnish has an eﬀect on the endurance
limit of steel. There is a theory that polishing increases the average endurance limit (for reverse bending). From a practical point of view, polishing should
not have any eﬀect on the standard deviation of the
endurance limit, which is known from numerous endurance limit experiments to be 4000 psi. An experiment was performed on 0.4% carbon steel using
both unpolished and polished smooth-turned specimens. The data are given below. Find a 95% conﬁdence interval for the diﬀerence between the population means for the two methods, assuming that the
populations are approximately normally distributed.
559.7
550.7
554.5
555.0
New Supplier (◦ F)
559.3 549.4 544.0 551.7
556.9 536.7 558.8 538.8
555.0 538.6 551.1 565.4
554.9 554.7 536.1 569.1
Old Supplier (◦ F)
534.7 554.8 545.0 544.6
563.1 551.1 553.8 538.8
553.0 538.4 548.3 552.9
544.8 558.4 548.7 560.3
566.3
543.3
554.9
538.0
564.6
535.1
Find 95% conﬁdence intervals for σ12 /σ22 and for σ1 /σ2 ,
where σ12 and σ22 are the population variances of the
thermostat readings for the new and old suppliers, respectively.
5.66 It is argued that the resistance of wire A is
greater than the resistance of wire B. An experiment
on the wires shows the results (in ohms) given here. Assuming equal variances, what conclusions do you draw?
232
Chapter 5 One- and Two-Sample Estimation Problems
Wire A
0.140
0.138
0.143
0.142
0.144
0.137
Wire B
0.135
0.140
0.136
0.142
0.138
0.140
5.67 A survey was done with the hope of comparing
salaries of chemical plant managers employed in two
areas of the country, the northern and west central regions. An independent random sample of 300 plant
managers was selected from each of the two regions.
These managers were asked their annual salaries. The
results are as follows
North
West Central
x̄1 = \$102,300
x̄2 = \$98,500
s1 = \$5700
s2 = \$3800
(a) Construct a 99% conﬁdence interval for μ1 − μ2 ,
the diﬀerence in the mean salaries.
(b) What assumption did you make in (a) about the
distribution of annual salaries for the two regions?
Is the assumption of normality necessary? Why or
why not?
(c) What assumption did you make about the two variances? Is the assumption of equality of variances
reasonable? Explain!
5.68 Consider Review Exercise 5.67. Let us assume
that the data have not been collected yet and that previous statistics suggest that σ1 = σ2 = \$4000. Are
the sample sizes in Review Exercise 5.67 suﬃcient to
produce a 95% conﬁdence interval on μ1 − μ2 having a
width of only \$1000? Show all work.
5.69 A labor union is becoming defensive about gross
always claimed that, in a typical month, 95% of its
members were absent less than 10 hours. The union
decided to check this by monitoring a random sample
of 300 of its members. The number of hours absent
was recorded for each of the 300 members. The results
were x̄ = 6.5 hours and s = 2.5 hours. Use the data to
respond to this claim, using a one-sided tolerance limit
and choosing the conﬁdence level to be 99%. Be sure
to interpret what you learn from the tolerance limit
calculation.
5.70 A random sample of 30 ﬁrms dealing in wireless
products was selected to determine the proportion of
such ﬁrms that have implemented new software to improve productivity. It turned out that 8 of the 30 had
implemented such software. Find a 95% conﬁdence interval on p, the true proportion of such ﬁrms that have
implemented new software.
5.71 Refer to Review Exercise 5.70. Suppose there is
concern about whether the point estimate p̂ = 8/30
is accurate enough because the conﬁdence interval
around p is not suﬃciently narrow. Using p̂ as the
estimate of p, how many companies would need to be
sampled in order to have a 95% conﬁdence interval with
a width of only 0.05?
5.72 A manufacturer turns out a product item that is
labeled either “defective” or “not defective.” In order
to estimate the proportion defective, a random sample of 100 items is taken from production, and 10 are
found to be defective. Following implementation of a
quality improvement program, the experiment is conducted again. A new sample of 100 is taken, and this
time only 6 are found to be defective.
(a) Give a 95% conﬁdence interval on p1 − p2 , where
p1 is the population proportion defective before improvement and p2 is the proportion defective after
improvement.
(b) Is there information in the conﬁdence interval
found in (a) that would suggest that p1 > p2 ? Explain.
5.73 A machine is used to ﬁll boxes with product
in an assembly line operation. Much concern centers
around the variability in the number of ounces of product in a box. The standard deviation in weight of product is known to be 0.3 ounce. An improvement is implemented, after which a random sample of 20 boxes is
selected and the sample variance is found to be 0.045
ounce2 . Find a 95% conﬁdence interval on the variance
in the weight of the product. Does it appear from the
range of the conﬁdence interval that the improvement
of the process enhanced quality as far as variability is
concerned? Assume normality on the distribution of
weights of product.
5.74 A consumer group is interested in comparing
operating costs for two diﬀerent types of automobile
engines. The group is able to ﬁnd 15 owners whose
cars have engine type A and 15 whose cars have engine
type B. All 30 owners bought their cars at roughly the
same time, and all have kept good records for a certain 12-month period. In addition, these owners drove
roughly the same number of miles. The cost statistics
are ȳA = \$87.00/1000 miles, ȳB = \$75.00/1000 miles,
sA = \$5.99, and sB = \$4.85. Compute a 95% conﬁdence interval to estimate μA − μB , the diﬀerence in
the mean operating costs. Assume normality and equal
variances.
5.75 A group of human factor researchers are concerned about reaction to a stimulus by airplane pilots
in a certain cockpit arrangement. An experiment was
conducted in a simulation laboratory, and 15 pilots
were used with average reaction time of 3.2 seconds
5.13
Potential Misconceptions and Hazards
with a sample standard deviation of 0.6 second. It is
of interest to characterize the extreme (i.e., worst-case
scenario). To that end, do the following:
(a) Give a particular important one-sided 99% conﬁdence bound on the mean reaction time. What
assumption, if any, must you make on the distribution of reaction times?
(b) Give a 99% one-sided prediction interval and give
an interpretation of what it means. Must you make
an assumption about the distribution of reaction
times to compute this bound?
(c) Compute a one-sided tolerance bound with 99%
conﬁdence that involves 95% of reaction times.
Again, give an interpretation and assumptions
about the distribution, if any. (Note: The onesided tolerance limit values are also included in Table A.7.)
5.13
233
5.76 A certain supplier manufactures a type of rubber mat that is sold to automotive companies. The
material used to produce the mats must have certain
hardness characteristics. Defective mats are occasionally discovered and rejected. The supplier claims that
the proportion defective is 0.05. A challenge was made
by one of the clients who purchased the mats, so an experiment was conducted in which 400 mats were tested
and 17 were found defective.
(a) Compute a 95% two-sided conﬁdence interval on
the proportion defective.
(b) Compute an appropriate 95% one-sided conﬁdence
interval on the proportion defective.
(c) Interpret both intervals from (a) and (b) and comment on the claim made by the supplier.
Potential Misconceptions and Hazards;
Relationship to Material in Other Chapters
The concept of a large-sample conﬁdence interval on a population parameter is
often confusing to the beginning student. It is based on the notion that even when
σ is unknown and one is not convinced that the distribution being sampled is
normal, a conﬁdence interval on μ can be computed from
s
x̄ ± zα/2 √ .
n
In practice, this formula is often used when the sample is too small. The genesis of
this large sample interval is, of course, the Central Limit Theorem (CLT), under
which normality is not necessary in practice. Here the CLT requires a known σ,
of which s is only an estimate. Thus, n must be at least as large as 30 and the
underlying distribution must be close to symmetric, in which case the interval
remains an approximation.
There are instances in which the appropriateness of the practical application
of material in this chapter depends very much on the speciﬁc context. One very
important illustration is the use of the t-distribution for the conﬁdence interval
on μ when σ is unknown. Strictly speaking, the use of the t-distribution requires
that the distribution sampled from be normal. However, it is well known that
any application of the t-distribution is reasonably insensitive (i.e., robust) to the
normality assumption. This represents one of those fortunate situations which
occur often in the ﬁeld of statistics in which a basic assumption does not hold and
yet “everything turns out all right!”
It is our experience that one of the most serious misuses of statistics in practice
evolves from confusion about distinctions in the interpretation of the types of
statistical intervals. Thus, the subsection in this chapter where diﬀerences among
the three types of intervals are discussed is important. It is very likely that in
practice the conﬁdence interval is heavily overused. That is, it is used when
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Chapter 5 One- and Two-Sample Estimation Problems
there is really no interest in the mean; rather, the question is “Where is the next
observation going to fall?” or often, more importantly, “Where is the large bulk of
the distribution?” These are crucial questions that are not answered by computing
an interval on the mean.
The interpretation of a conﬁdence interval is often misunderstood. It is tempting to conclude that the parameter falls inside the interval with probability 0.95.
A conﬁdence interval merely suggests that if the experiment is conducted and data
are observed again and again, about 95% of such intervals will contain the true
parameter. Any beginning student of practical statistics should be very clear on
the diﬀerence among these statistical intervals.
Another potential serious misuse of statistics centers around the use of the χ2 distribution for a conﬁdence interval on a single variance. Again, normality of the
distribution from which the sample is drawn is assumed. Unlike the use of the
t-distribution, the use of the χ2 test for this application is not robust to the
2
normality assumption (i.e., the sampling distribution of (n−1)S
deviates far
σ2
from χ2 if the underlying distribution is not normal).
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