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Gandee Vasan/Getty Images
CHAPTER
11
Sampling Distributions
How much on the average do American households earn? The government’s Current Population Survey contacted a sample of 113,146 households in March 2005.
Their mean income in 2004 was x = $60,528.1 That $60,528 describes the sample, but we use it to estimate the mean income of all households. This is an example
of statistical inference: we use information from a sample to infer something about
a wider population.
Because the results of random samples and randomized comparative experiments include an element of chance, we can’t guarantee that our inferences are
correct. What we can guarantee is that our methods usually give correct answers.
We will see that the reasoning of statistical inference rests on asking, “How often
would this method give a correct answer if I used it very many times?” If our data
come from random sampling or randomized comparative experiments, the laws of
probability answer the question “What would happen if we did this many times?”
This chapter presents some facts about probability that help answer this question.
In this chapter we cover...
Parameters and statistics
Statistical estimation and
the law of large numbers
Sampling distributions
The sampling distribution
of x
The central limit theorem
Statistical process control∗
x charts∗
Thinking about process
control∗
Parameters and statistics
As we begin to use sample data to draw conclusions about a wider population, we
must take care to keep straight whether a number describes a sample or a population. Here is the vocabulary we use.
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C H A P T E R 11 • Sampling Distributions
PARAMETER, STATISTIC
A parameter is a number that describes the population. In statistical
practice, the value of a parameter is not known because we cannot examine
the entire population.
A statistic is a number that can be computed from the sample data without
making use of any unknown parameters. In practice, we often use a statistic
to estimate an unknown parameter.
EXAMPLE 11.1
Household income
The mean income of the sample of households contacted by the Current Population
Survey was x = $60,528. The number $60,528 is a statistic because it describes this one
Current Population Survey sample. The population that the poll wants to draw conclusions about is all 113 million U.S. households. The parameter of interest is the mean
income of all of these households. We don’t know the value of this parameter.
population mean μ
sample mean x
Remember: statistics come from samples, and parameters come from populations. As long as we were just doing data analysis, the distinction between population and sample was not important. Now, however, it is essential. The notation
we use must reflect this distinction. We write μ (the Greek letter mu) for the mean
of a population. This is a fixed parameter that is unknown when we use a sample
for inference. The mean of the sample is the familiar x, the average of the observations in the sample. This is a statistic that would almost certainly take a different
value if we chose another sample from the same population. The sample mean x
from a sample or an experiment is an estimate of the mean μ of the underlying
population.
APPLY YOUR KNOWLEDGE
Simon Marcus/CORBIS
11.1 Effects of caffeine. How does caffeine affect our bodies? In a matched pairs
experiment, subjects pushed a button as quickly as they could after taking a
caffeine pill and also after taking a placebo pill. The mean pushes per minute were
283 for the placebo and 311 for caffeine. Is each of the boldface numbers a
parameter or a statistic?
11.2 Indianapolis voters. Voter registration records show that 68% of all voters in
Indianapolis are registered as Republicans. To test a random digit dialing device,
you use the device to call 150 randomly chosen residential telephones in
Indianapolis. Of the registered voters contacted, 73% are registered Republicans.
Is each of the boldface numbers a parameter or a statistic?
11.3 Inspecting bearings. A carload lot of bearings has mean diameter 2.5003
centimeters (cm). This is within the specifications for acceptance of the lot by the
purchaser. By chance, an inspector chooses 100 bearings from the lot that have
mean diameter 2.5009 cm. Because this is outside the specified limits, the lot is
mistakenly rejected. Is each of the boldface numbers a parameter or a statistic?
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Statistical estimation and the law of large numbers
273
Statistical estimation and
the law of large numbers
Statistical inference uses sample data to draw conclusions about the entire population. Because good samples are chosen randomly, statistics such as x are random
variables. We can describe the behavior of a sample statistic by a probability model
that answers the question “What would happen if we did this many times?” Here
is an example that will lead us toward the probability ideas most important for
statistical inference.
EXAMPLE 11.2
Does this wine smell bad?
Sulfur compounds such as dimethyl sulfide (DMS) are sometimes present in wine. DMS
causes “off-odors” in wine, so winemakers want to know the odor threshold, the lowest
concentration of DMS that the human nose can detect. Different people have different
thresholds, so we start by asking about the mean threshold μ in the population of all
adults. The number μ is a parameter that describes this population.
To estimate μ, we present tasters with both natural wine and the same wine spiked
with DMS at different concentrations to find the lowest concentration at which they
identify the spiked wine. Here are the odor thresholds (measured in micrograms of DMS
per liter of wine) for 10 randomly chosen subjects:
28
40
28
33
20
31
29
27
17
21
The mean threshold for these subjects is x = 27.4. It seems reasonable to use the sample result x = 27.4 to estimate the unknown μ. An SRS should fairly represent the
population, so the mean x of the sample should be somewhere near the mean μ of the
population. Of course, we don’t expect x to be exactly equal to μ. We realize that if we
choose another SRS, the luck of the draw will probably produce a different x.
If x is rarely exactly right and varies from sample to sample, why is it nonetheless a reasonable estimate of the population mean μ? Here is one answer: if we
keep on taking larger and larger samples, the statistic x is guaranteed to get closer
and closer to the parameter μ. We have the comfort of knowing that if we can
afford to keep on measuring more subjects, eventually we will estimate the mean
odor threshold of all adults very accurately. This remarkable fact is called the law
of large numbers. It is remarkable because it holds for any population, not just for
some special class such as Normal distributions.
LAW OF LARGE NUMBERS
Draw observations at random from any population with finite mean μ. As
the number of observations drawn increases, the mean x of the observed
values gets closer and closer to the mean μ of the population.
High-tech gambling
There are more than 640,000 slot
machines in the United States.
Once upon a time, you put in a coin
and pulled the lever to spin three
wheels, each with 20 symbols. No
longer. Now the machines are video
games with flashy graphics and
outcomes produced by random
number generators. Machines can
accept many coins at once, can pay
off on a bewildering variety of
outcomes, and can be networked to
allow common jackpots. Gamblers
still search for systems, but in the
long run the law of large numbers
guarantees the house its 5% profit.
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C H A P T E R 11 • Sampling Distributions
The law of large numbers can be proved mathematically starting from the basic
laws of probability. The behavior of x is similar to the idea of probability. In the
long run, the proportion of outcomes taking any value gets close to the probability
of that value, and the average outcome gets close to the population mean. Figure
10.1 (page 248) shows how proportions approach probability in one example. Here
is an example of how sample means approach the population mean.
EXAMPLE 11.3
The law of large numbers in action
In fact, the distribution of odor thresholds among all adults has mean 25. The mean
μ = 25 is the true value of the parameter we seek to estimate. Figure 11.1 shows how the
sample mean x of an SRS drawn from this population changes as we add more subjects
to our sample.
The first subject in Example 11.2 had threshold 28, so the line in Figure 11.1 starts
there. The mean for the first two subjects is
x=
28 + 40
= 34
2
This is the second point on the graph. At first, the graph shows that the mean of the
sample changes as we take more observations. Eventually, however, the mean of the
observations gets close to the population mean μ = 25 and settles down at that value.
If we started over, again choosing people at random from the population, we would
get a different path from left to right in Figure 11.1. The law of large numbers says that
whatever path we get will always settle down at 25 as we draw more and more people.
35
34
33
Mean of first n observations
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31
30
29
28
27
26
25
24
23
22
1
5
10
50
100
500 1000
5000 10,000
Number of observations, n
F I G U R E 1 1 . 1 The law of large numbers in action: as we take more observations, the
sample mean x always approaches the mean μ of the population.
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Sampling distributions
The Law of Large Numbers applet animates Figure 11.1 in a different setting.
You can use the applet to watch x change as you average more observations until
it eventually settles down at the mean μ.
The law of large numbers is the foundation of such business enterprises as
gambling casinos and insurance companies. The winnings (or losses) of a gambler
on a few plays are uncertain—that’s why gambling is exciting. In Figure 11.1, the
mean of even 100 observations is not yet very close to μ. It is only in the long run
that the mean outcome is predictable. The house plays tens of thousands of times.
So the house, unlike individual gamblers, can count on the long-run regularity
described by the law of large numbers. The average winnings of the house on tens
of thousands of plays will be very close to the mean of the distribution of winnings.
Needless to say, this mean guarantees the house a profit. That’s why gambling can
be a business.
APPLY YOUR KNOWLEDGE
11.4 Means in action. Figure 11.1 shows how the mean of n observations behaves as
we keep adding more observations to those already in hand. The first 10
observations are given in Example 11.2. Demonstrate that you grasp the idea of
Figure 11.1: find the means of the first one, two, three, four, and five of these
observations and plot the successive means against n. Verify that your plot agrees
with the first part of the plot in Figure 11.1.
11.5 Insurance. The idea of insurance is that we all face risks that are unlikely but
carry high cost. Think of a fire destroying your home. Insurance spreads the risk:
we all pay a small amount, and the insurance policy pays a large amount to those
few of us whose homes burn down. An insurance company looks at the records for
millions of homeowners and sees that the mean loss from fire in a year is μ = $250
per person. (Most of us have no loss, but a few lose their homes. The $250 is the
average loss.) The company plans to sell fire insurance for $250 plus enough to
cover its costs and profit. Explain clearly why it would be unwise to sell only
12 policies. Then explain why selling thousands of such policies is a safe business.
Sampling distributions
The law of large numbers assures us that if we measure enough subjects, the statistic
x will eventually get very close to the unknown parameter μ. But our study in
Example 11.2 had just 10 subjects. What can we say about x from 10 subjects as
an estimate of μ? We ask: “What would happen if we took many samples of 10
subjects from this population?” Here’s how to answer this question:
•
•
•
•
Take a large number of samples of size 10 from the population.
Calculate the sample mean x for each sample.
Make a histogram of the values of x.
Examine the distribution displayed in the histogram for shape, center, and
spread, as well as outliers or other deviations.
APPLET
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C H A P T E R 11 • Sampling Distributions
Take many SRSs and collect
their means x.
The distribution of all
the x's is close to Normal.
SRS size 10
SRS size 10
SRS size 10
x = 26.42
x = 24.28
x = 25.22
•
•
•
Population,
mean μ = 25
20
25
30
F I G U R E 1 1 . 2 The idea of a sampling distribution: take many samples from the same
population, collect the x’s from all the samples, and display the distribution of the x’s.
The histogram shows the results of 1000 samples.
simulation
In practice it is too expensive to take many samples from a large population such as
all adult U.S. residents. But we can imitate many samples by using software. Using
software to imitate chance behavior is called simulation.
EXAMPLE 11.4
What would happen in many samples?
Extensive studies have found that the DMS odor threshold of adults follows roughly
a Normal distribution with mean μ = 25 micrograms per liter and standard deviation
σ = 7 micrograms per liter. With this information, we can simulate many repetitions of
Example 11.2 with different subjects drawn at random from the population.
Figure 11.2 illustrates the process of choosing many samples and finding the sample
mean threshold x for each one. Follow the flow of the figure from the population at the
left, to choosing an SRS and finding the x for this sample, to collecting together the
x ’s from many samples. The first sample has x = 26.42. The second sample contains
a different 10 people, with x = 24.28, and so on. The histogram at the right of the
figure shows the distribution of the values of x from 1000 separate SRSs of size 10. This
histogram displays the sampling distribution of the statistic x.
SAMPLING DISTRIBUTION
The sampling distribution of a statistic is the distribution of values taken
by the statistic in all possible samples of the same size from the same
population.
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Sampling distributions
Strictly speaking, the sampling distribution is the ideal pattern that would
emerge if we looked at all possible samples of size 10 from our population. A distribution obtained from a fixed number of trials, like the 1000 trials in Figure 11.2, is
only an approximation to the sampling distribution. One of the uses of probability
theory in statistics is to obtain exact sampling distributions without simulation.
The interpretation of a sampling distribution is the same, however, whether we
obtain it by simulation or by the mathematics of probability.
We can use the tools of data analysis to describe any distribution. Let’s apply
those tools to Figure 11.2. What can we say about the shape, center, and spread of
this distribution?
•
•
•
Shape: It looks Normal! Detailed examination confirms that the distribution
of x from many samples does have a distribution that is very close to Normal.
Center: The mean of the 1000 x ’s is 24.95. That is, the distribution is
centered very close to the population mean μ = 25.
Spread: The standard deviation of the 1000 x ’s is 2.217, notably smaller
than the standard deviation σ = 7 of the population of individual subjects.
Although these results describe just one simulation of a sampling distribution, they
reflect facts that are true whenever we use random sampling.
APPLY YOUR KNOWLEDGE
11.6 Generating a sampling distribution. Let’s illustrate the idea of a sampling
distribution in the case of a very small sample from a very small population. The
population is the scores of 10 students on an exam:
Student
0
1
2
3
4
5
6
7
8
9
Score
82
62
80
58
72
73
65
66
74
62
The parameter of interest is the mean score μ in this population. The sample is
an SRS of size n = 4 drawn from the population. Because the students are
labeled 0 to 9, a single random digit from Table B chooses one student for the
sample.
(a) Find the mean of the 10 scores in the population. This is the population
mean μ.
(b) Use the first digits in row 116 of Table B to draw an SRS of size 4 from this
population. What are the four scores in your sample? What is their mean x? This
statistic is an estimate of μ.
(c) Repeat this process 9 more times, using the first digits in rows 117 to
125 of Table B. Make a histogram of the 10 values of x. You are
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C H A P T E R 11 • Sampling Distributions
constructing the sampling distribution of x. Is the center of your histogram
close to μ?
The sampling distribution of x
Figure 11.2 suggests that when we choose many SRSs from a population, the sampling distribution of the sample means is centered at the mean of the original population and less spread out than the distribution of individual observations. Here
are the facts.
MEAN AND STANDARD DEVIATION OF A SAMPLE MEAN 2
Rigging the lottery
We have all seen televised
lottery drawings in which numbered
balls bubble about and are randomly
popped out by air pressure. How
might we rig such a drawing? In
1980, when the Pennsylvania lottery
used just three balls, a drawing was
rigged by the host and several stagehands. They injected paint into all
balls bearing 8 of the 10 digits. This
weighed them down and guaranteed
that all three balls for the winning
number would have the remaining
2 digits. The perps then bet
on all combinations of these digits.
When 6-6-6 popped out, they won
$1.2 million. Yes, they were caught.
Suppose that x is the mean of an SRS of size n drawn from a large
population with mean μ and standard deviation σ . Then
√the sampling
distribution of x has mean μ and standard deviation σ/ n.
These facts about the mean and the standard deviation of the sampling distribution of x are true for any population, not just for some special class such
as Normal distributions. Both facts have important implications for statistical
inference.
•
unbiased estimator
•
The mean of the statistic x is always equal to the mean μ of the population.
That is, the sampling distribution of x is centered at μ. In repeated sampling,
x will sometimes fall above the true value of the parameter μ and sometimes
below, but there is no systematic tendency to overestimate or underestimate
the parameter. This makes the idea of lack of bias in the sense of “no
favoritism” more precise. Because the mean of x is equal to μ, we say that the
statistic x is an unbiased estimator of the parameter μ.
An unbiased estimator is “correct on the average” in many samples. How
close the estimator falls to the parameter in most samples is determined by
the spread of the sampling distribution. If individual observations have
standard deviation σ , √
then sample means x from samples of size n have
standard deviation σ/ n. That is, averages are less variable than individual
observations.
We have described the center and spread of the sampling distribution of a
sample mean x, but not its shape. The shape of the distribution of x depends on
the shape of the population. Here is one important case: if measurements in the
population follow a Normal distribution, then so does the sample mean.
SAMPLING DISTRIBUTION OF A SAMPLE MEAN
If individual observations have the N(μ, σ )√
distribution, then the sample
mean x of an SRS of size n has the N(μ, σ/ n) distribution.
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The sampling distribution of x
The distribution of sample
means is less spread out.
Means x of 10 subjects
σ
= 2.21
10
Observations on 1 subject
σ=7
0
10
20
30
40
50
F I G U R E 1 1 . 3 The distribution of single observations compared with the distribution
of the means x of 10 observations. Averages are less variable than individual observations.
EXAMPLE 11.5
Population distribution, sampling distribution
If we measure the DMS odor thresholds of individual adults, the values follow the Normal distribution with mean μ = 25 micrograms per liter and standard deviation σ = 7
micrograms per liter. We call this the population distribution because it shows how
measurements vary within the population.
Take many SRSs of size 10 from this population and find the sample mean x for
each sample, as in Figure 11.2. The sampling distribution describes how the values of x
vary among samples. That sampling distribution is also Normal, with mean μ = 25 and
standard deviation
population distribution
σ
7
√ = = 2.2136
n
10
Figure 11.3 contrasts these two Normal distributions. Both are centered at the population mean, but sample means are much less variable than individual observations.
Not only is the standard deviation of the distribution of x smaller than the
standard deviation of individual observations, but it gets smaller as we take larger
samples. The results of large samples are less variable than the results of small
samples. If n is large, the standard deviation of x is small, and almost all samples
will give values of x that lie very close to the true parameter μ. That is, the sample
mean from a large sample can be trusted to estimate the population mean accurately. However,
the standard deviation of the sampling distribution gets smaller only at
√
the rate n. To cut the standard deviation of x in half, we must take four times as many
observations, not just twice as many.
CAUTION
UTION
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C H A P T E R 11 • Sampling Distributions
APPLY YOUR KNOWLEDGE
11.7 A sample of teens. A study of the health of teenagers plans to measure the
blood cholesterol level of an SRS of youths aged 13 to 16. The researchers will
report the mean x from their sample as an estimate of the mean cholesterol level
μ in this population.
(a) Explain to someone who knows no statistics what it means to say that x is an
“unbiased” estimator of μ.
(b) The sample result x is an unbiased estimator of the population truth μ no
matter what size SRS the study uses. Explain to someone who knows no statistics
why a large sample gives more trustworthy results than a small sample.
11.8 Measurements in the lab. Juan makes a measurement in a chemistry laboratory
and records the result in his lab report. The standard deviation of students’ lab
measurements is σ = 10 milligrams. Juan repeats the measurement 3 times and
records the mean x of his 3 measurements.
(a) What is the standard deviation of Juan’s mean result? (That is, if Juan kept on
making 3 measurements and averaging them, what would be the standard
deviation of all his x ’s?)
(b) How many times must Juan repeat the measurement to reduce the standard
deviation of x to 5? Explain to someone who knows no statistics the advantage of
reporting the average of several measurements rather than the result of a single
measurement.
11.9 National math scores. The scores of 12th-grade students on the National
Assessment of Educational Progress year 2000 mathematics test have a
distribution that is approximately Normal with mean μ = 300 and standard
deviation σ = 35.
(a) Choose one 12th-grader at random. What is the probability that his or her
score is higher than 300? Higher than 335?
(b) Now choose an SRS of four 12th-graders and calculate their mean score x. If
you did this many times, what would be the mean and standard deviation of all
the x-values?
(c) What is the probability that the mean score for your SRS is higher than 300?
Higher than 335?
The central limit theorem
The facts about the mean and standard deviation of x are true no matter what the
shape of the population distribution may be. But what is the shape of the sampling
distribution when the population distribution is not Normal? It is a remarkable
fact that as the sample size increases, the distribution of x changes shape: it looks
less like that of the population and more like a Normal distribution. When the
sample is large enough, the distribution of x is very close to Normal. This is true
no matter what shape the population distribution has, as long as the population
has a finite standard deviation σ . This famous fact of probability theory is called
the central limit theorem. It is much more useful than the fact that the distribution
of x is exactly Normal if the population is exactly Normal.
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The central limit theorem
CENTRAL LIMIT THEOREM
Draw an SRS of size n from any population with mean μ and finite standard
deviation σ . When n is large, the sampling distribution of the sample mean
x is approximately Normal:
σ
x is approximately N μ, √
n
The central limit theorem allows us to use Normal probability calculations
to answer questions about sample means from many observations even
when the population distribution is not Normal.
More general versions of the central limit theorem say that the distribution of
any sum or average of many small random quantities is close to Normal. This is
true even if the quantities are correlated with each other (as long as they are not
too highly correlated) and even if they have different distributions (as long as no
one random quantity is so large that it dominates the others). The central limit
theorem suggests why the Normal distributions are common models for observed
data. Any variable that is a sum of many small influences will have approximately
a Normal distribution.
How large a sample size n is needed for x to be close to Normal depends on
the population distribution. More observations are required if the shape of the
population distribution is far from Normal. Here are two examples in which the
population is far from Normal.
EXAMPLE 11.6
The central limit theorem in action
In March 2004, the Current Population Survey contacted 98,789 households. Figure
11.4(a) is a histogram of the earnings of the 62,101 households that had earned income
greater than zero in 2003. As we expect, the distribution of earned incomes is strongly
skewed to the right and very spread out. The right tail of the distribution is longer than
the histogram shows because there are too few high incomes for their bars to be visible on this scale. In fact, we cut off the earnings scale at $300,000 to save space—a
few households earned even more than $300,000. The mean earnings for these 62,101
households was $57,085.
Regard these 62,101 households as a population. Take an SRS of 100 households.
The mean earnings in this sample is x = $48,600. That’s less than the mean of the population. Take another SRS of size 100. The mean for this sample is x = $64,766. That’s
higher than the mean of the population. What would happen if we did this many times?
Figure 11.4(b) is a histogram of the mean earnings for 500 samples, each of size 100.
The scales in Figures 11.4(a) and 11.4(b) are the same, for easy comparison. Although
the distribution of individual earnings is skewed and very spread out, the distribution of
sample means is roughly symmetric and much less spread out.
Figure 11.4(c) zooms in on the center part of the axis for another histogram of
the same 500 values of x. Although n = 100 is not a very large sample size and the
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30
20
10
0
Percent of households
40
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0
100
150
200
250
300
Earned income (thousands of dollars)
40
(a)
Because the scales are the
same, you can compare this
distribution directly with
Figure 11.4(a).
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50
100
150
200
250
300
Mean earned income in sample (thousands of dollars)
(b)
40,000
45,000
50,000
55,000
60,000
65,000
70,000
75,000
Mean earned income in sample (dollars)
(c)
F I G U R E 1 1 . 4 The central limit theorem in action. (a) The distribution of earned
income in a population of 62,101 households. (b) The distribution of the mean earnings
for 500 SRSs of 100 households each from this population. (c) The distribution of the
sample means in more detail: the shape is close to Normal.
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The central limit theorem
population distribution is extremely skewed, we can see that the distribution of sample
means is close to Normal.
Comparing Figures 11.4(a) and 11.4(b) illustrates the two most important
ideas of this chapter.
THINKING ABOUT SAMPLE MEANS
Means of random samples are less variable than individual observations.
Means of random samples are more Normal than individual observations.
EXAMPLE 11.7
The central limit theorem in action
The Central Limit Theorem applet allows you to watch the central limit theorem in action. Figure 11.5 presents snapshots from the applet. Figure 11.5(a) shows the density
curve of a single observation, that is, of the population. The distribution is strongly
0
1
0
1
(a)
0
1
(b)
0
(c)
1
(d)
F I G U R E 1 1 . 5 The central limit theorem in action: the distribution of sample means
x from a strongly non-Normal population becomes more Normal as the sample size
increases. (a) The distribution of 1 observation. (b) The distribution of x for
2 observations. (c) The distribution of x for 10 observations. (d) The distribution of x for
25 observations.
APPLET
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C H A P T E R 11 • Sampling Distributions
right-skewed, and the most probable outcomes are near 0. The mean μ of this distribution is 1, and its standard deviation σ is also 1. This particular distribution is called an
exponential distribution. Exponential distributions are used as models for the lifetime in
service of electronic components and for the time required to serve a customer or repair
a machine.
Figures 11.5(b), (c), and (d) are the density curves of the sample means of 2, 10,
and 25 observations from this population. As n increases, the shape becomes more Normal.
√ The mean remains at μ = 1, and the standard deviation decreases, taking the value
1/ n. The density curve for 10 observations is still somewhat
skewed to the right but
already resembles a Normal curve having μ = 1 and σ = 1/ 10 = 0.32. The density
curve for n = 25 is yet more Normal. The contrast between the shapes of the population distribution and of the distribution of the mean of 10 or 25 observations is
striking.
Let’s use Normal calculations based on the central limit theorem to answer a
question about the very non-Normal distribution in Figure 11.5(a).
4
STEP
EXAMPLE 11.8
Maintaining air conditioners
STATE: The time (in hours) that a technician requires to perform preventive maintenance on an air-conditioning unit is governed by the exponential distribution whose
density curve appears in Figure 11.5(a). The mean time is μ = 1 hour and the standard
deviation is σ = 1 hour. Your company has a contract to maintain 70 of these units in
an apartment building. You must schedule technicians’ time for a visit to this building. Is
it safe to budget an average of 1.1 hours for each unit? Or should you budget an average
of 1.25 hours?
FORMULATE: We can treat these 70 air conditioners as an SRS from all units of this
type. What is the probability that the average maintenance time for 70 units exceeds
1.1 hours? That the average time exceeds 1.25 hours?
SOLVE: The central limit theorem says that the sample mean time x spent working on
70 units has approximately the Normal distribution with mean equal to the population
mean μ = 1 hour and standard deviation
σ
1
= = 0.12 hour
70
70
The distribution of x is therefore approximately N(1, 0.12). This Normal curve is the
solid curve in Figure 11.6.
Using this Normal distribution, the probabilities we want are
P (x > 1.10 hours) = 0.2014
P (x > 1.25 hours) = 0.0182
(Software gives these probabilities immediately, or you can standardize and use Table A.
Don’t forget to use standard deviation 0.12 in your software or when you standardize x.)
CONCLUDE: If you budget 1.1 hours per unit, there is a 20% chance that the technicians will not complete the work in the building within the budgeted time. This chance
drops to 2% if you budget 1.25 hours. You therefore budget 1.25 hours per unit.
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The central limit theorem
Exact density
curve for x.
Normal curve from the
central limit theorem.
1.1
F I G U R E 1 1 . 6 The exact distribution (dashed) and the Normal approximation from
the central limit theorem (solid) for the average time needed to maintain an air
conditioner, for Example 11.8. The probability we want is the area to the right of 1.1.
Using more mathematics, we can start with the exponential distribution and
find the actual density curve of x for 70 observations. This is the dashed curve in
Figure 11.6. You can see that the solid Normal curve is a good approximation. The
exactly correct probability for 1.1 hours is an area to the right of 1.1 under the
dashed density curve. It is 0.1977. The central limit theorem Normal approximation 0.2014 is off by only about 0.004.
APPLY YOUR KNOWLEDGE
11.10 What does the central limit theorem say? Asked what the central limit
theorem says, a student replies, “As you take larger and larger samples from a
population, the histogram of the sample values looks more and more Normal.” Is
the student right? Explain your answer.
11.11 Detecting gypsy moths. The gypsy moth is a serious threat to oak and aspen
trees. A state agriculture department places traps throughout the state to detect
the moths. When traps are checked periodically, the mean number of moths
trapped is only 0.5, but some traps have several moths. The distribution of moth
counts is discrete and strongly skewed, with standard deviation 0.7.
(a) What are the mean and standard deviation of the average number of moths x
in 50 traps?
(b) Use the central limit theorem to find the probability that the average number
of moths in 50 traps is greater than 0.6.
Bruce Coleman/Alamy
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C H A P T E R 11 • Sampling Distributions
4
STEP
11.12 SAT scores. The total SAT scores of high school seniors in recent years have
mean μ = 1026 and standard deviation σ = 209. The distribution of SAT scores
is roughly Normal.
(a) Ramon scored 1100. If scores have a Normal distribution, what percentile of
the distribution is this? (That is, what percent of scores are lower than Ramon’s?)
(b) Now consider the mean x of the scores of 70 randomly chosen students. If
x = 1100, what percentile of the sampling distribution of x is this?
(c) Which of your calculations, (a) or (b), is less accurate because SAT scores do
not have an exactly Normal distribution? Explain your answer.
11.13 More on insurance. An insurance company knows that in the entire
population of millions of homeowners, the mean annual loss from fire is μ = $250
and the standard deviation of the loss is σ = $1000. The distribution of losses is
strongly right-skewed: most policies have $0 loss, but a few have large losses. If
the company sells 10,000 policies, can it safely base its rates on the assumption
that its average loss will be no greater than $275? Follow the four-step process in
your answer.
Statistical process control∗
The sampling distribution of the sample mean x has an immediate application to
statistical process control. The goal of statistical process control is to make a process
stable over time and then keep it stable unless planned changes are made. You
might want, for example, to keep your weight constant over time. A manufacturer
of machine parts wants the critical dimensions to be the same for all parts. “Constant over time” and “the same for all” are not realistic requirements. They ignore
the fact that all processes have variation. Your weight fluctuates from day to day; the
critical dimension of a machined part varies a bit from item to item; the time to
process a college admission application is not the same for all applications. Variation occurs in even the most precisely made product due to small changes in the
raw material, the adjustment of the machine, the behavior of the operator, and
even the temperature in the plant. Because variation is always present, we can’t
expect to hold a variable exactly constant over time. The statistical description
of stability over time requires that the pattern of variation remain stable, not that
there be no variation in the variable measured.
STATISTICAL CONTROL
A variable that continues to be described by the same distribution when
observed over time is said to be in statistical control, or simply in control.
Control charts are statistical tools that monitor a process and alert us when
the process has been disturbed so that it is now out of control. This is a
signal to find and correct the cause of the disturbance.
*The rest of this chapter is optional. A more complete treatment of process control appears in Companion
Chapter 27.
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x charts
Control charts work by distinguishing the natural variation in the process from
the additional variation that suggests that the process has changed. A control chart
sounds an alarm when it sees too much variation. The most common application
of control charts is to monitor the performance of an industrial process. The same
methods, however, can be used to check the stability of quantities as varied as
the ratings of a television show, the level of ozone in the atmosphere, and the gas
mileage of your car. Control charts combine graphical and numerical descriptions
of data with use of sampling distributions. They therefore provide a natural bridge
between exploratory data analysis and formal statistical inference.
x charts∗
The population in the control chart setting is all items that would be produced by
the process if it ran on forever in its present state. The items actually produced form
samples from this population. We generally speak of the process rather than the
population. Choose a quantitative variable, such as a diameter or a voltage, that
is an important measure of the quality of an item. The process mean μ is the longterm average value of this variable; μ describes the center or aim of the process.
The sample mean x of several items estimates μ and helps us judge whether the
center of the process has moved away from its proper value. The most common
control chart plots the means x of small samples taken from the process at regular
intervals over time.
When you first apply control charts to a process, the process may not be in
control. Even if it is in control, you don’t yet understand its behavior. You must
collect data from the process, establish control by uncovering and removing the
reasons for disturbances, and then set up control charts to maintain control. To
quickly explain the main ideas, we’ll assume that you know the usual behavior of
the process from long experience. Here are the conditions we will work with.
PROCESS-MONITORING CONDITIONS
Measure a quantitative variable x that has a Normal distribution. The
process has been operating in control for a long period, so that we know the
process mean μ and the process standard deviation σ that describe the
distribution of x as long as the process remains in control.
EXAMPLE 11.9
Making computer monitors
A manufacturer of computer monitors must control the tension on the mesh of fine
wires that lies behind the surface of the viewing screen. Too much tension will tear
the mesh, and too little will allow wrinkles. Tension is measured by an electrical device
with output readings in millivolts (mV). The proper tension is 275 mV. Some variation
is always present in the production process. When the process is operating properly, the
standard deviation of the tension readings is σ = 43 mV.
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C H A P T E R 11 • Sampling Distributions
TABLE 11.1
Twenty control chart samples of mesh tension
Sample
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
x
Tension measurements
234.5
311.1
247.1
215.4
327.9
304.3
268.9
282.1
260.8
329.3
266.4
168.8
349.9
235.2
257.3
235.1
286.3
328.1
316.4
296.8
272.3
305.8
205.3
296.8
247.2
236.3
276.2
247.7
259.9
231.8
249.7
330.9
334.2
283.1
218.4
252.7
293.8
272.6
287.4
350.5
234.5
238.5
252.6
274.2
283.3
201.8
275.6
259.8
247.9
307.2
231.5
333.6
292.3
245.9
296.2
300.6
236.2
329.7
373.0
280.6
272.3
286.2
316.1
256.8
232.6
238.5
240.2
272.8
345.3
273.4
265.2
318.3
301.5
263.1
275.2
297.6
275.3
260.1
286.0
259.8
253.4
285.4
255.3
260.8
272.7
245.2
265.2
265.6
278.5
285.4
253.2
287.9
319.5
256.8
261.8
271.5
272.9
297.6
315.7
296.9
The operator measures the tension on a sample of 4 monitors each hour. The mean
x of each sample estimates the mean tension μ for the process at the time of the sample.
Table 11.1 shows the samples and their means for 20 consecutive hours of production.
How can we use these data to keep the process in control?
A time plot helps us see whether or not the process is stable. Figure 11.7 is a
plot of the successive sample means against the order in which the samples were
taken. We have plotted each sample mean from the table against its sample number. For example, the mean of the first sample is 253.4 mV, and this is the value
plotted for sample 1. Because the target value for the process mean is μ = 275 mV,
we draw a center line at that level across the plot. How much variation about this
center line do we expect to see? For example, are samples 13 and 19 so high that
they suggest lack of control?
The tension measurements are roughly Normal, and we know that sample
means are more Normal than individual measurements. So the x-values from successive samples will follow a Normal distribution. If the standard deviation of
the individual screens remains at σ = 43 mV, the standard deviation of x from
4 screens is
43
σ
√ = = 21.5 mV
n
4
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x charts
The control limits mark the
natural variation in the process.
400
Sample mean
350
UCL
300
250
LCL
200
150
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Sample number
F I G U R E 1 1 . 7 x chart for the mesh tension data of Table 11.1. The control limits are
labeled UCL for upper control limit and LCL for lower control limit. No points lie
outside the control limits.
As long as the mean remains at its target value μ = 275 mV, the 99.7 part of the
68–95–99.7 rule says that almost all values of x will lie between
σ
μ − 3 √ = 275 − (3)(21.5) = 210.5
n
σ
μ + 3 √ = 275 + (3)(21.5) = 339.5
n
We therefore draw dashed control limits at these two levels on the plot. The
control limits show the extent of the natural variation of x-values when the process
is in control. We now have an x control chart.
x CONTROL CHART
To evaluate the control of a process with given standards μ and σ , make an
x control chart as follows:
• Plot the means x of regular samples of size n against time.
• Draw a horizontal center line at μ.
√
• Draw horizontal control limits at μ ± 3σ/ n.
Any x that does not fall between the control limits is evidence that the
process is out of control.
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C H A P T E R 11 • Sampling Distributions
EXAMPLE 11.10
Interpreting x charts
Figure 11.7 is a typical x chart for a process in control. The means of the 20 samples do
vary, but all lie within the range of variation marked out by the control limits. We are
seeing the natural variation of a stable process.
Figures 11.8 and 11.9 illustrate two ways in which the process can go out of control. In Figure 11.8, the process was disturbed sometime between sample 12 and sample
13. As a result, the mean tension for sample 13 falls above the upper control limit. It
is common practice to mark all out-of-control points with an “x” to call attention to
them. A search for the cause begins as soon as we see a point out of control. Investigation finds that the mounting of the tension-measuring device has slipped, resulting in
readings that are too high. When the problem is corrected, samples 14 to 20 are again in
control.
Figure 11.9 shows the effect of a steady upward drift in the process center, starting
at sample 11. You see that some time elapses before the x for sample 18 is out of control.
The one-point-out signal works better for detecting sudden large disturbances than for
detecting slow drifts in a process.
This point is out of control
because it is above UCL.
400
x
350
Sample mean
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300
250
LCL
200
150
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Sample number
F I G U R E 1 1 . 8 This x chart is identical to that in Figure 11.7, except that a
disturbance has driven x for sample 13 above the upper control limit. The out-of-control
point is marked with an x.
x chart
An x control chart is often called simply an x chart. Because a control chart
is a warning device, it is not necessary that our probability calculations be exactly
correct. Approximate Normality is good enough. In that same spirit, control charts
use the approximate Normal probabilities given by the 68–95–99.7 rule rather
than more exact calculations using Table A.
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x charts
400
Sample mean
350
x
UCL
x
x
300
250
LCL
200
150
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Sample number
F I G U R E 1 1 . 9 The first 10 points on this x chart are as in Figure 11.7. The process
mean drifts upward after sample 10, and the sample means x reflect this drift. The points
for samples 18, 19, and 20 are out of control.
APPLY YOUR KNOWLEDGE
11.14 Auto thermostats. A maker of auto air conditioners checks a sample of
4 thermostatic controls from each hour’s production. The thermostats are set at
75◦ F and then placed in a chamber where the temperature rises gradually. The
temperature at which the thermostat turns on the air conditioner is recorded. The
process mean should be μ = 75◦ . Past experience indicates that the response
temperature of properly adjusted thermostats varies with σ = 0.5◦ . The mean
response temperature x for each hour’s sample is plotted on an x control chart.
Calculate the center line and control limits for this chart.
11.15 Tablet hardness. A pharmaceutical manufacturer forms tablets by compressing
a granular material that contains the active ingredient and various fillers. The
hardness of a sample from each lot of tablets is measured in order to control the
compression process. The process has been operating in control with mean at the
target value μ = 11.5 and estimated standard deviation σ = 0.2. Table 11.2
gives three sets of data, each representing x for 20 successive samples of n = 4
tablets. One set remains in control at the target value. In a second set, the process
mean μ shifts suddenly to a new value. In a third, the process mean drifts
gradually.
(a) What are the center line and control limits for an x chart for this process?
(b) Draw a separate x chart for each of the three data sets. Mark any points that
are beyond the control limits.
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C H A P T E R 11 • Sampling Distributions
TABLE 11.2
Three sets of x’s from 20 samples of size 4
Sample
Data set A
Data set B
Data set C
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
11.602
11.547
11.312
11.449
11.401
11.608
11.471
11.453
11.446
11.522
11.664
11.823
11.629
11.602
11.756
11.707
11.612
11.628
11.603
11.816
11.627
11.613
11.493
11.602
11.360
11.374
11.592
11.458
11.552
11.463
11.383
11.715
11.485
11.509
11.429
11.477
11.570
11.623
11.472
11.531
11.495
11.475
11.465
11.497
11.573
11.563
11.321
11.533
11.486
11.502
11.534
11.624
11.629
11.575
11.730
11.680
11.729
11.704
12.052
11.905
(c) Based on your work in (b) and the appearance of the control charts, which
set of data comes from a process that is in control? In which case does the process
mean shift suddenly, and at about which sample do you think that the mean
changed? Finally, in which case does the mean drift gradually?
Thinking about process control∗
CAUTION
UTION
The purpose of a control chart is not to ensure good quality by inspecting most of
the items produced. Control charts focus on the process itself rather than on the
individual products. By checking the process at regular intervals, we can detect
disturbances and correct them quickly. Statistical process control achieves high
quality at a lower cost than inspecting all of the products. Small samples of 4 or
5 items are usually adequate for process control.
A process that is in control is stable over time, but stability alone does not guarantee good quality. The natural variation in the process may be so large that many of
the products are unsatisfactory. Nonetheless, establishing control brings a number
of advantages.
•
In order to assess whether the process quality is satisfactory, we must observe
the process when it is operating in control, free of breakdowns and other
disturbances.
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Chapter 11 Summary
•
•
A process in control is predictable. We can predict both the quantity and the
quality of items produced.
When a process is in control, we can easily see the effects of attempts to
improve the process, which are not hidden by the unpredictable variation
that characterizes lack of statistical control.
A process in control is doing as well as it can in its present state. If the process is
not capable of producing adequate quality even when undisturbed, we must make
some major change in the process, such as installing new machines or retraining
the operators.
If the process is kept in control, we know what to expect in the finished product. The process mean μ and standard deviation σ remain stable over time, so
(assuming Normal variation) the 99.7 part of the 68–95–99.7 rule tells us that almost all measurements on individual products will lie in the range μ ± 3σ . These
are sometimes called the natural tolerances for the product. Be careful to distinguish μ ± 3σ , the
√ range we expect for individual measurements, from the x chart control
limits μ ± 3σ/ n, which mark off the expected range of sample means.
natural tolerances
CAUTION
UTION
EXAMPLE 11.11
Natural tolerances for mesh tension
The process of setting the mesh tension on computer monitors has been operating in
control. The x chart is based on μ = 275 mV and σ = 43 mV.
We are therefore confident that almost all individual monitors will have mesh tension between
μ ± 3σ = 275 ± (3)(43) = 275 ± 129
We expect mesh tension measurements to vary between 146 mV and 404 mV. You see
that the spread of individual measurements is wider than the spread of sample means
used for the control limits of the x chart.
APPLY YOUR KNOWLEDGE
11.16 Auto thermostats. Exercise 11.14 describes a process that produces auto
thermostats. The temperature that turns on the thermostats has remained in
control with mean μ = 75◦ F and standard deviation σ = 0.5◦ . What are the
natural tolerances for this temperature? What range covers the middle 95% of
response temperatures?
C H A P T E R 11 SUMMARY
When we want information about the population mean μ for some variable, we
often take an SRS and use the sample mean x to estimate the unknown
parameter μ.
The law of large numbers states that the actually observed mean outcome x must
approach the mean μ of the population as the number of observations increases.
The sampling distribution of x describes how the statistic x varies in all possible
SRSs of the same size from the same population.
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C H A P T E R 11 • Sampling Distributions
The mean of the sampling distribution is μ, so that x is an unbiased estimator
of μ.
√
The standard deviation of the sampling distribution of x is σ/ n for an SRS of
size n if the population has standard deviation σ . That is, averages are less
variable than individual observations.
If the population has a Normal distribution, so does x.
The central limit theorem states that for large n the sampling distribution of x is
approximately Normal for any population with finite standard deviation σ . That
is, averages
√ are more Normal than individual observations. We can use the
N(μ, σ/ n) distribution to calculate approximate probabilities for events
involving x.
All processes have variation. If the pattern of variation is stable over time, the
process is in statistical control. Control charts are statistical plots intended to
warn when a process is out of control.
An x control chart plots the means x of samples from a process against the time
order in which the samples were taken. If the process has been
√ in control with
mean μ and standard deviation σ , control limits at μ ± 3σ/ n mark off the
range of variation we expect to see in the x-values. Values outside the control
limits suggest that the process has been disturbed.
CHECK YOUR SKILLS
11.17 The Bureau of Labor Statistics announces that last month it interviewed all
members of the labor force in a sample of 60,000 households; 4.9% of the people
interviewed were unemployed. The boldface number is a
(a) sampling distribution.
(b) parameter.
(c) statistic.
11.18 A study of voting chose 663 registered voters at random shortly after an
election. Of these, 72% said they had voted in the election. Election records
show that only 56% of registered voters voted in the election. The boldface
number is a
(a) sampling distribution.
(b) parameter.
(c) statistic.
11.19 Annual returns on the more than 5000 common stocks available to investors vary
a lot. In a recent year, the mean return was 8.3% and the standard deviation of
returns was 28.5%. The law of large numbers says that
(a) you can get an average return higher than the mean 8.3% by investing in a
large number of stocks.
(b) as you invest in more and more stocks chosen at random, your average return
on these stocks gets ever closer to 8.3%.
(c) if you invest in a large number of stocks chosen at random, your average
return will have approximately a Normal distribution.
11.20 Scores on the SAT college entrance test in a recent year were roughly Normal
with mean 1026 and standard deviation 209. You choose an SRS of 100 students
and average their SAT scores. If you do this many times, the mean of the average
scores you get will be close to
(a) 1026.
(b) 1026/100 = 102.6.
(c) 1026/ 100 = 10.26.
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Chapter 11 Exercises
11.21 Scores on the SAT college entrance test in a recent year were roughly Normal
with mean 1026 and standard deviation 209. You choose an SRS of 100 students
and average their SAT scores. If you do this many times, the standard deviation of
the average scores you get will be close to
(a) 209.
(b) 100/ 209 = 6.92.
(c) 209/ 100 = 20.9.
11.22 A newborn baby has extremely low birth weight (ELBW) if it weighs less than
1000 grams. A study of the health of such children in later years examined a
random sample of 219 children. Their mean weight at birth was x = 810 grams.
This sample mean is an unbiased estimator of the mean weight μ in the population
of all ELBW babies. This means that
(a) in many samples from this population, the mean of the many values of x will
be equal to μ.
(b) as we take larger and larger samples from this population, x will get closer and
closer to μ.
(c) in many samples from this population, the many values of x will have a
distribution that is close to Normal.
11.23 The number of hours a light bulb burns before failing varies from bulb to bulb.
The distribution of burnout times is strongly skewed to the right. The central
limit theorem says that
(a) as we look at more and more bulbs, their average burnout time gets ever
closer to the mean μ for all bulbs of this type.
(b) the average burnout time of a large number of bulbs has a distribution
of the same shape (strongly skewed) as the distribution for individual
bulbs.
(c) the average burnout time of a large number of bulbs has a distribution that is
close to Normal.
11.24 A machine manufactures parts whose diameters vary according to the Normal
distribution with mean μ = 40.150 millimeters (mm) and standard deviation
σ = 0.003 mm. An inspector measures a random sample of 4 parts. The
probability that the average diameter of these 4 parts is less than 40.148 mm is
about
(a) 0.092.
(b) 0.251.
(c) 0.908.
C H A P T E R 11 EXERCISES
11.25 Women’s heights. A random sample of female college students has a mean
height of 65 inches, which is greater than the 64-inch mean height of all young
women. Is each of the bold numbers a parameter or a statistic? Explain your
answer.
11.26 Small classes in school. The Tennessee STAR experiment randomly assigned
children to regular or small classes during their first four years of school. When
these children reached high school, 40.2% of blacks from small classes took the
ACT or SAT college entrance exams. Only 31.7% of blacks from regular classes
took one of these exams. Is each of the boldface numbers a parameter or a
statistic? Explain your answer.
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APPLET
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11.27 Playing the numbers. The numbers racket is a well-entrenched illegal gambling
operation in most large cities. One version works as follows: you choose one of
the 1000 three-digit numbers 000 to 999 and pay your local numbers runner a
dollar to enter your bet. Each day, one three-digit number is chosen at random
and pays off $600. The mean payoff for the population of thousands of bets is
μ = 60 cents. Joe makes one bet every day for many years. Explain what the law
of large numbers says about Joe’s results as he keeps on betting.
11.28 Roulette. A roulette wheel has 38 slots, of which 18 are black, 18 are red, and 2
are green. When the wheel is spun, the ball is equally likely to come to rest in any
of the slots. One of the simplest wagers chooses red or black. A bet of $1 on red
returns $2 if the ball lands in a red slot. Otherwise, the player loses his dollar.
When gamblers bet on red or black, the two green slots belong to the house.
Because the probability of winning $2 is 18/38, the mean payoff from a $1 bet is
twice 18/38, or 94.7 cents. Explain what the law of large numbers tells us about
what will happen if a gambler makes very many bets on red.
11.29 The law of large numbers. Suppose that you roll two balanced dice and look at
the spots on the up-faces. There are 36 possible outcomes, displayed in Figure
10.2 (page 251). Because the dice are balanced, all 36 outcomes are equally likely.
The average number of spots is 7. This is the population mean μ for the idealized
population that contains the results of rolling two dice forever. The law of large
numbers says that the average x from a finite number of rolls gets closer and closer
to 7 as we do more and more rolls.
(a) Click “More dice” once in the Law of Large Numbers applet to get two dice.
Click “Show mean” to see the mean 7 on the graph. Leaving the number of rolls
at 1, click “Roll dice” three times. How many spots did each roll produce? What is
the average for the three rolls? You see that the graph displays at each point the
average number of spots for all rolls up to the last one. Now you understand the
display.
(b) Set the number of rolls to 100 and click “Roll dice.” The applet rolls the two
dice 100 times. The graph shows how the average count of spots changes as we
make more rolls. That is, the graph shows x as we continue to roll the dice. Make
a rough sketch of the final graph.
(c) Repeat your work from (b). Click “Reset” to start over, then roll two dice 100
times. Make a sketch of the final graph of the mean x against the number of rolls.
Your two graphs will often look very different. What they have in common is that
the average eventually gets close to the population mean μ = 7. The law of large
numbers says that this will always happen if you keep on rolling the dice.
11.30 What’s the mean? Suppose that you roll three balanced dice. We wonder what
the mean number of spots on the up-faces of the three dice is. The law of large
numbers says that we can find out by experience: roll three dice many times, and
the average number of spots will eventually approach the true mean. Set up the
Law of Large Numbers applet to roll three dice. Don’t click “Show mean” yet. Roll
the dice until you are confident you know the mean quite closely, then click
“Show mean” to verify your discovery. What is the mean? Make a rough sketch of
the path the averages x followed as you kept adding more rolls.
11.31 Lightning strikes. The number of lightning strikes on a square kilometer of
open ground in a year has mean 6 and standard deviation 2.4. (These values are
typical of much of the United States.) The National Lightning Detection
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Network uses automatic sensors to watch for lightning in a sample of 10 square
kilometers. What are the mean and standard deviation of x, the mean number of
strikes per square kilometer?
11.32 Heights of male students. To estimate the mean height μ of male students on
your campus, you will measure an SRS of students. You know from government
data that the standard deviation of the heights of young men is about 2.8 inches.
How large an SRS must you take to reduce the standard deviation of the sample
mean to one-half inch? Use the four-step process to outline your work.
11.33 Heights of male students, continued. To estimate the mean height μ of male
students on your campus, you will measure an SRS of students. You know from
government data that heights of young men are approximately Normal with
standard deviation about 2.8 inches. You want your sample mean x to estimate μ
with an error of no more than one-half inch in either direction.
(a) What standard deviation must x have so that 99.7% of all samples give an x
within one-half inch of μ? (Use the 68–95–99.7 rule.)
(b) How large an SRS do you need to reduce the standard deviation of x to the
value you found in part (a)?
11.34 More on heights of male students. In Exercise 11.32, you decided to measure
n male students. Suppose that the distribution of heights of all male students on
your campus is Normal with mean 70 inches and standard deviation 2.8 inches.
(a) If you choose one student at random, what is the probability that he is
between 69 and 71 inches tall?
(b) What is the probability that the mean height of your sample is between
69 and 71 inches?
11.35 Durable press fabrics. “Durable press” cotton fabrics are treated to improve
their recovery from wrinkles after washing. Unfortunately, the treatment also
reduces the strength of the fabric. The breaking strength of untreated fabric is
Normally distributed with mean 58 pounds and standard deviation 2.3 pounds.
The same type of fabric after treatment has Normally distributed breaking
strength with mean 30 pounds and standard deviation 1.6 pounds.3 A clothing
manufacturer tests an SRS of 5 specimens of each fabric.
(a) What is the probability that the mean breaking strength of the 5 untreated
specimens exceeds 50 pounds?
(b) What is the probability that the mean breaking strength of the 5 treated
specimens exceeds 50 pounds?
11.36 Glucose testing. Shelia’s doctor is concerned that she may suffer from
gestational diabetes (high blood glucose levels during pregnancy). There is
variation both in the actual glucose level and in the blood test that measures the
level. A patient is classified as having gestational diabetes if the glucose level is
above 140 milligrams per deciliter (mg/dl) one hour after a sugary drink. Shelia’s
measured glucose level one hour after the sugary drink varies according to the
Normal distribution with μ = 125 mg/dl and σ = 10 mg/dl.
(a) If a single glucose measurement is made, what is the probability that Shelia is
diagnosed as having gestational diabetes?
(b) If measurements are made on 4 separate days and the mean result is compared
with the criterion 140 mg/dl, what is the probability that Shelia is diagnosed as
having gestational diabetes?
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11.37 Pollutants in auto exhausts. The level of nitrogen oxides (NOX) in the
exhaust of cars of a particular model varies Normally with mean 0.2 grams per
mile (g/mi) and standard deviation 0.05 g/mi. Government regulations call for
NOX emissions no higher than 0.3 g/mi.
(a) What is the probability that a single car of this model fails to meet the NOX
requirement?
(b) A company has 25 cars of this model in its fleet. What is the probability that
the average NOX level x of these cars is above the 0.3 g/mi limit?
11.38 Glucose testing, continued. Shelia’s measured glucose level one hour after a
sugary drink varies according to the Normal distribution with μ = 125 mg/dl and
σ = 10 mg/dl. What is the level L such that there is probability only 0.05 that
the mean glucose level of 4 test results falls above L? (Hint: This requires a
backward Normal calculation. See page 81 in Chapter 3 if you need to review.)
11.39 Pollutants in auto exhausts, continued. The level of nitrogen oxides (NOX)
in the exhaust of cars of a particular model varies Normally with mean 0.2 g/mi
and standard deviation 0.05 g/mi. A company has 25 cars of this model in its fleet.
What is the level L such that the probability that the average NOX level x for
the fleet is greater than L is only 0.01? (Hint: This requires a backward Normal
calculation. See page 81 in Chapter 3 if you need to review.)
11.40 Returns on stocks. Andrew plans to retire in 40 years. He is thinking of
investing his retirement funds in stocks, so he seeks out information on past
returns. He learns that over the 101 years from 1900 to 2000, the real (that is,
adjusted for inflation) returns on U.S. common stocks had mean 8.7% and
standard deviation 20.2%.4 The distribution of annual returns on common stocks
is roughly symmetric, so the mean return over even a moderate number of years is
close to Normal. What is the probability (assuming that the past pattern of
variation continues) that the mean annual return on common stocks over the
next 40 years will exceed 10%? What is the probability that the mean return will
be less than 5%? Follow the four-step process in your answer.
11.41 Auto accidents. The number of accidents per week at a hazardous intersection
varies with mean 2.2 and standard deviation 1.4. This distribution takes only
whole-number values, so it is certainly not Normal.
(a) Let x be the mean number of accidents per week at the intersection during a
year (52 weeks). What is the approximate distribution of x according to the
central limit theorem?
(b) What is the approximate probability that x is less than 2?
(c) What is the approximate probability that there are fewer than 100 accidents
at the intersection in a year? (Hint: Restate this event in terms of x.)
11.42 Airline passengers get heavier. In response to the increasing weight of airline
passengers, the Federal Aviation Administration in 2003 told airlines to assume
that passengers average 190 pounds in the summer, including clothing and
carry-on baggage. But passengers vary, and the FAA did not specify a standard
deviation. A reasonable standard deviation is 35 pounds. Weights are not
Normally distributed, especially when the population includes both men and
women, but they are not very non-Normal. A commuter plane carries
19 passengers. What is the approximate probability that the total weight of the
passengers exceeds 4000 pounds? Use the four-step process to guide your work.
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Chapter 11 Exercises
(Hint: To apply the central limit theorem, restate the problem in terms of the
mean weight.)
11.43 Generating a sampling distribution. We want to know what percent of
American adults approve of legal gambling. This population proportion p is a
parameter. To estimate p, take an SRS and find the proportion p̂ in the sample
who approve of gambling. If we take many SRSs of the same size, the proportion
p̂ will vary from sample to sample. The distribution of its values in all SRSs is the
sampling distribution of this statistic.
Figure 11.10 is a small population. Each circle represents an adult. The colored
circles are people who disapprove of legal gambling, and the white circles are
people who approve. You can check that 60 of the 100 circles are white, so in this
population the proportion who approve of gambling is p = 60/100 = 0.6.
(a) The circles are labeled 00, 01, . . . , 99. Use line 101 of Table B to draw an SRS
of size 5. What is the proportion p̂ of the people in your sample who approve of
gambling?
(b) Take 9 more SRSs of size 5 (10 in all), using lines 102 to 110 of Table B, a
different line for each sample. You now have 10 values of the sample proportion
p̂. What are they?
(c) Because your samples have only 5 people, the only values p̂ can take are 0/5,
1/5, 2/5, 3/5, 4/5, and 5/5. That is, p̂ is always 0, 0.2, 0.4, 0.6, 0.8, or 1. Mark these
numbers on a line and make a histogram of your 10 results by putting a bar above
each number to show how many samples had that outcome. (You have begun to
construct the sampling distribution of p̂, although just 10 samples is a small start.)
(d) Taking samples of size 5 from a population of size 100 is not a practical
setting, but let’s look at your results anyway. How many of your 10 samples
estimated the population proportion p = 0.6 exactly correctly? Is the true value
0.6 roughly in the center of your sample values?
11.44 A better way to generate a sampling distribution. You can use the Probability
applet to speed up and improve Exercise 11.43. You have a population in which
60% of the individuals approve of legal gambling. You want to take many samples
from this population to observe how the sample proportion who approve of
gambling varies from sample to sample. Set the “Probability of heads” in the
applet to 0.6 and the number of tosses to 40. This simulates an SRS of size 40
from a large population. Each head in the sample is a person who approves of
legal gambling and each tail is a person who disapproves. By alternating between
“Toss” and “Reset” you can take many samples quickly.
(a) Take 50 samples, recording the proportion who approve of gambling in each
sample. (The applet gives this proportion at the top left of its display.) Make a
histogram of the 50 sample proportions.
(b) Another population contains only 20% who approve of legal gambling. Take
50 samples of size 40 from this population, record the number in each sample who
approve, and make a histogram of the 50 sample proportions. How do the centers
of your two histograms reflect the differing truths about the two populations?
The following exercises concern the optional material on statistical process control.
11.45 Dyeing yarn. The unique colors of the cashmere sweaters your firm makes result
from heating undyed yarn in a kettle with a dye liquor. The pH (acidity) of the
Jeff Greenberg/The Image Works
APPLET
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F I G U R E 1 1 . 1 0 A population of 100 people, for Exercise 11.43. The white circles
represent people who approve of legal gambling. The colored circles represent people
who oppose gambling.
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liquor is critical for regulating dye uptake and hence the final color. There are
5 kettles, all of which receive dye liquor from a common source. Twice each day,
the pH of the liquor in each kettle is measured, giving a sample of size 5. The
process has been operating in control with μ = 4.22 and σ = 0.127. Give the
center line and control limits for the x chart.
11.46 Hospital losses. A hospital struggling to contain costs investigates procedures
on which it loses money. Government standards place medical procedures into
Diagnostic Related Groups (DRGs). For example, major joint replacements are
DRG 209. The hospital takes from its records a random sample of 8 DRG 209
patients each month. The losses incurred per patient have been in control, with
mean $6400 and standard deviation $700. Here are the mean losses x for the
samples taken in the next 15 months:
6244
6534
6080
6476
6469
6544
6415
6497
6912
6638
6857
6659
7509
7374
6697
What does an x chart suggest about the hospital’s losses on major joint
replacements? Follow the four-step process in your answer.
11.47 Dyeing yarn, continued. What are the natural tolerances for the pH of an
individual dye kettle in the setting of Exercise 11.45?
11.48 Milling. The width of a slot cut by a milling machine is important to the proper
functioning of a hydraulic system for large tractors. The manufacturer checks the
control of the milling process by measuring a sample of 5 consecutive items
during each hour’s production. The target width for the slot is μ = 0.8750 inch.
The process has been operating in control with center close to the target and
σ = 0.0012 inch. What center line and control limits should be drawn on the
x chart?
11.49 Is the quality OK? Statistical control means that a process is stable. It doesn’t
mean that this stable process produces high-quality items. Return to the
mesh-tensioning process described in Examples 11.9 and 11.11. This process is in
control with mean μ = 275 mV and standard deviation σ = 43 mV.
(a) The current specifications set by customers for mesh tension are 100 to
400 mV. What percent of monitors meet these specifications?
(b) The customers now set tighter specifications, 150 to 350 mV. What percent
meet the new specifications? The process has not changed, but product quality,
measured by percent meeting the specifications, is no longer good.
11.50 Improving the process. The center of the mesh tensions for the process in the
previous exercise is 275 mV. The center of the specifications is 250 mV, so we
should be able to improve the process by adjusting the center to 250 mV. This is
an easy adjustment that does not change the process variation. What percent of
monitors now meet the new specifications?
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