chapter 18catalogue.pearsoned.ca/.../samplechapter/0321748298.pdf · 474 chapter 18 sampling...

473

CHAPTER

18 Sampling DistributionModels

Where are we going?A recent poll of 1000 randomly sampled Canadians reported that 91% believe that Canada’shealth care system is better than that of the United States.1 But how much can we trust thisstatistic? After all, only 1000 people were polled, leaving out of the study some 22 million orso other Canadian adults. Maybe the true proportion of all Canadians who feel this way is 95%or 85%, or could it be as little as 75%? Is almost anything possible? Just how reliable areproportions based on random samples? We’ll see in this chapter that we can be surprisinglyprecise about how much we expect proportions from random samples to vary. This will enableus to start generalizing from samples we have at hand to the population at large.

The Canadian military mission in Afghanistan has been controversial. Inlate April 2007, after a rash of troop fatalities earlier that month, anAngus Reid poll of 1009 randomly chosen Canadians reported that 52%were in favour of an early withdrawal of troops (before the February

2009 mission mandate end). At nearly the same time, a Strategic Counsel poll of1000 Canadians reported 64% in favour of an early withdrawal. Why the differ-ence? Should we be surprised to find that we could get proportions this differentfrom properly selected random samples drawn from the same population?You’re probably used to seeing that observations vary, but how much variabilityamong polls should we expect to see?

Why do sample proportions vary at all? How can surveys conducted at essen-tially the same time by organizations asking the same questions get different results?The answer is at the heart of Statistics. The proportions vary from sample to samplebecause the samples are composed of different people.

1 In the same June 2008 study, 1000 Americans were polled, and 45% of them thoughtCanada’s health care system was superior.

The n’s justify the means.

—Apocryphal statistical saying

WHO Canadian adults

WHAT When to bring thetroops home

WHEN April 2007

WHERE Canada

WHY Public attitudes

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474 CHAPTER 18 Sampling Distribution Models

It’s actually pretty easy to predict how much a proportion will vary under cir-cumstances like this. Understanding the variability of our estimates will let usactually use that variability to better understand the world.

The Central Limit Theorem for Sample Proportions

We’ve talked about Think, Show, and Tell. Now we have to add Imagine. In order tounderstand the Strategic Counsel poll, we want to imagine the results from all therandom samples of size 1000 that Strategic Counsel didn’t take. What would thehistogram of all the sample proportions look like?

For Canadians’ support for early withdrawal, where do you expect the centre ofthat histogram to be? Of course, we don’t know the answer to that (and probablynever will). But we know that it will be at the true proportion in the population, andwe can call that p. (See the Notation Alert.) For the sake of discussion here, let’ssuppose that 60% of all Canadian adults supported early withdrawal, so we’ll usep = 0.60.

How about the shape of the histogram? We don’t have to just imagine. We cansimulate. We want to simulate a bunch of random samples that we didn’t reallydraw. To do that, we’ll simulate many independent random samples of size 1000,keeping the same probability of success. We chose 0.60 for p as a reasonable value.Here’s a histogram of the sample proportions (denoted ) supporting earlywithdrawal for 2000 independent samples of 1000 adults when the true proportionis p = 0.60.

pN

Activity: SamplingDistribution of a Proportion. Youdon’t have to imagine—you cansimulate.

200

150

100

50

p ’s from Simulated Samplesˆ

# of

Sam

ples

0.546 0.564 0.582 0.600 0.618 0.636 0.654

FIGURE 18.1A histogram of sampleproportions for 2000simulated samples of 1000adults drawn from apopulation with p = 0.60.The sample proportionsvary, but their distribution is centred at the trueproportion, p.

NOTATION ALERT:We use p as our symbol for thetrue (generally unknown) pro-portion of successes in the pop-ulation.2 This violates our“Greek letters for parameters”rule, but if we stuck to that, ournatural choice would be ,which we prefer to reserve forthe universal constant3.1415926 . . . Recall that inChapter 8 we put a “hat”overthe response variable y, to indi-cate that was a prediction ofthe actual response y, derivedfrom the data. Here, denotesthe sample proportion, and thehat over p plays a similar role,indicating an estimate of p,derived from the data. Likewise,the sample proportion of fail-ures estimates the true failureproportion and is .The “hat”signifies an estimate of what isunderneath.

q̂

pN

yN

p

2 But be careful. We’ve already used capital P for a general probability. And we’ll soon seeanother use of P in Chapter 20. There are a lot of p’s in this course; you’ll need to thinkclearly about the context to keep them straight.3 Well, the fact that we spent most of Chapter 6 on the Normal model might have been a hint.

IMAGINEWe see only the sample thatwe actually drew, but bysimulating or modelling, wecan Imagine what we mighthave seen had we drawnother possible randomsamples.

It should be no surprise that we don’t get the same proportion for each samplewe draw, even though the underlying true value is the same for the population. Each

comes from a different simulated sample. The histogram above is a simulation ofwhat we’d get if we could see all the proportions from all possible samples. That distri-bution has a special name. It is called the sampling distribution of the proportions.

Does it surprise you that the histogram is unimodal? Symmetric? That it iscentred at p? You probably don’t find any of this shocking. Does the shape remindyou of any model we’ve already discussed? It’s an amazing and fortunate fact thata Normal model is just the right one for the histogram of sample proportions.

This fact was proved in 1810 by the great French mathematician Pierre-SimonLaplace as part of a more general result (more about this shortly), though there is noreason you should guess that the Normal model would be the one we need here.3

pN


The Central Limit Theorem for Sample Proportions 475

4 For smaller n, we can just use a Binomial model, which we saw in Chapter 17.

FIGURE 18.2A Normal model centred at p with a

standard deviation of is a good

model for a collection of proportionsfound for many random samples ofsize n from a population with successprobability p.

A pq

n

p–3 pq

n –2 pq n –1 pq

n 1 pq n 2 pq

n 3 pq n

Simulation: SimulatingSampling Distributions. Watchthe Normal model appear fromrandom proportions.

0.5535 0.5690–2s

0.5845–1s

0.6000p

0.61551s

0.63102s

0.6465–3s 3s

FIGURE 18.3Using 0.60 for p gives this Normalmodel for Figure 18.1’s histogram ofthe sample proportions of adultsfavouring early withdrawal (n = 1000).

Although we’ll never know the true proportion of adults who favour earlywithdrawal, we’re supposing it to be 60%. Once we put the centre at p = 0.60, thestandard deviation for the Strategic Counsel poll is

%

Here’s a picture of the Normal model for our simulation histogram:

SD1pN2 = Apq

n= A

10.60210.4021000

= 0.0155, or 1 .55

Modelling how sample proportions vary from sample to sample is one of themost powerful ideas we’ll see in this course. A sampling distribution model forhow a sample proportion varies from sample to sample allows us to quantify thatvariation and to talk about how likely it is that we’d observe a sample proportionin any particular interval.

To use a Normal model, we need to specify two parameters: its mean andstandard deviation. The centre of the histogram is naturally at p, so we’ll put , themean of the Normal, at p.

What about the standard deviation? Usually the mean gives us no informa-tion about the standard deviation. Suppose we told you that a batch of bike hel-mets had a mean diameter of 26 centimetres and asked what the standarddeviation was. If you said, “I have no idea,” you’d be exactly right. There’s noinformation about σ from knowing the value of .

But there’s a special fact about proportions. With proportions we get some-thing for free. Once we know the mean, p, we automatically also know the stan-dard deviation for , the proportion of successes:

When we draw simple random samples of n individuals, the proportions wefind will vary from sample to sample. As long as n is reasonably large,4 we canmodel the distribution of these sample proportions with a probability modelthat is

N¢p, Apq

n≤

5 Äp(1 2 p)

n5 Ä

pqn

s(pN) = SD(pN)

pN

m

m



Simulation: The StandardDeviation of a Proportion. Doyou believe this formula forstandard deviation? Don’t justtake our word for it—convinceyourself with an experiment.

5 The standard deviation always has the same units as the data. Here, our units are percent-age points. But that can be confusing, because the standard deviation is not 1.55% of any-thing; it is 1.55 percentage points. If that’s confusing, write the units as “percentage points”instead of percent.6 We ignore the small difference in sample size, 1000 versus 1009, since the effect on thestandard deviation is negligible.7 While the polls were only a couple of days apart, there were fine differences in the word-ing of the questions, possible slight differences in the sampled populations, and a high “notsure” rate of 14% in the Angus Reid poll. Eliminating the “not sure” makes it 60.5% forearly withdrawal. The lesson here is that in addition to sampling error, there is alwaysnonsampling error affecting results. In this chapter, though, we focus on understandingsampling errors—which can be modelled precisely (what a relief!).

How do we know that sample proportions have a mean of p and a

standard deviation of ?

Remember from Chapter 17 that if we have a sequence of independenttrials where we record a success or failure on each trial (Bernoulli trials), thatthe total number of successes, Y, out of n trials has a mean of np and a standarddeviation of .

The sample proportion is found by taking the total number of successes Yand dividing by the sample size:

So,

the true proportion of the population.When the expected value of a sample statistic equals a population param-

eter, we say that the statistic is an unbiased estimator of the parameter. So wehave just proven that the sample proportion is an unbiased estimator of the truepopulation proportion for simple random samples. Sample proportions willgenerally miss the target (the true proportion), of course, but there will be no

E(p̂) 5 E¢Yn≤ 5

E(Y)

n5

npn

5 p

pN =

Yn

pN1npq

Apq

n

Because we have a Normal model, we can use the 68–95–99.7 Rule or look upother probabilities using a table or technology. For example, we know that 95% ofNormally distributed values are within two standard deviations of the mean, and99.7% are within three standard deviations, so we should expect about 19 out of 20such polls to give results within 2 � 1.55% = 3.1%5 of the true percentage, andnearly all to give results within 3 � 1.55% = 4.65% of the truth. So, if we assumefor the moment that the true percentage is 60%, the Strategic Counsel result of 64%seems possible, if unlikely (falling between two and three standard deviationsfrom the centre), but the other poll result of 52% would be essentially impossible,6

as it reaches out more than five standard deviations from the centre!This very natural and expected variability of sample proportions about the

true population proportion as we imagine going from one random sample toanother is called sampling error, though a better name for it might be samplingvariability. Even considering sampling error, our guess of 60% isn’t looking sogood; neither poll seems likely if the truth is 60%, but is there just one truthhere?7

(continued)

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Assumptions and Conditions 477

systematic tendency to over- or undershoot the target (though bias can still beintroduced via nonsampling errors such as undercoverage, nonresponse, etc.).

And

If we know the mean, p, of the sample proportion, then we know the stan-dard deviation as well. This is a special property of proportions.

SD1pN2 = SD¢Yn≤ =

SD1Y2n

=

1npq

1n2= B

npq

n2= B

pq

n

How Good Is the Normal Model?Stop and think for a minute about what we’ve just said. It’s a remarkable claim.We’ve said that if we draw repeated random samples of the same size, n, fromsome population and measure the proportion, , we see in each sample, then thecollection of these proportions will pile up around the underlying population pro-portion, p, and that a histogram of the sample proportions can be modelled wellby a Normal model.

There must be a catch. Suppose the samples were of size two, for example.Then the only possible proportion values would be 0, 0.5, and 1. There’s no waythe histogram could ever look like a Normal model with only three possible val-ues for the variable.

Well, there is a catch. The claim is only approximately true. (But that’s okay.After all, models are only supposed to be approximately true.) And the modelbecomes a better and better representation of the distribution of the sample pro-portions as the sample size gets bigger.8 Samples of size one or two just aren’tgoing to work very well. But the distributions of proportions of many larger sam-ples do have histograms that are remarkably close to a Normal model.

pN

FIGURE 18.4Proportions from samples of size twocan take on only three possible values.A Normal model does not work well.

0.0 0.5 1.0

1500

1000

500

8 Formally, we say the claim is true in the limit as n grows.

Assumptions and ConditionsTo use a model, we usually must make some assumptions. To use the samplingdistribution model for sample proportions, we need two assumptions:

The Independence Assumption: The sampled values must be independentrandom draws from the population under study.The Sample Size Assumption: The sample size, n, must be large enough.

Of course, assumptions are hard—often impossible—to check. That’s why weassume them. But, as we saw in Chapter 8, we should check to see whether theassumptions are reasonable. To think about the Independence Assumption, weoften wonder whether there is any reason to think that the data values mightaffect each other. Fortunately, we can often check conditions that provide informa-tion about the assumptions. Check these conditions before using the Normalmodel to model the distribution of sample proportions:

Randomization Condition: If your data come from an experiment, subjectsshould have been randomly assigned to treatments. If you have a survey,your sample should be a simple random sample of the population. In practice,other more practical sampling designs are often used (e.g., multistage



samples), and while the sampling distribution will still tend to be Normal forlarge enough n, our formula for the SD of the proportion falls apart, and oftengives an underestimate of the true SD. Statisticians then have to mathematicallyderive the correct formula, or they may decide to “bootstrap” it (a computer-intensive approach, discussed in Chapter 33). Statistics Canada bootstrapsnearly all of their SD’s due to complex sample designs.If not a proper random sample, try to be sure that the sampling method wasnot biased and that the data may reasonably be thought of as representative ofsome relevant population.10% Condition: The sample size, n, must be no larger than 10% of the popula-tion. For national polls, the total population is usually very large, so the sam-ple is a small fraction of the population.Success/Failure Condition: The sample size has to be big enough so that weexpect at least 10 successes and at least 10 failures. When np and nq are at least10, we have enough data for sound conclusions. For the Strategic Counsel sur-vey, a “success” might be favouring early withdrawal. If p � 0.60, we expect1000 � 0.60 � 600 successes and 1000 � 0.40 � 400 failures. Both are at least10, so we certainly expect enough successes and enough failures for the condi-tion to be satisfied.

These last two conditions seem to conflict with each other. The Success/Failure Condition wants sufficient data. How much depends on p. If p is near 0.5,we need a sample of only 20 or so. If p is only 0.01, however, we’d need 1000. Butthe 10% Condition says that a sample can be no larger than 10% of the population.If the sample were more than 10% of the population, we would need to makeadjustments to our method for analyzing the data.9 But this is rare in practice.Often, as in polls that sample from all Canadian adults or industrial samples froma day’s production, the populations are much larger than 10 times the sample size.

9 See optional section: Finite Population Correction.10 The proof is pretty technical. We’re not sure it helps our understanding all that mucheither.

Simulation: Simulate theSampling Distribution Model of aProportion. You probably don’twant to work through the formalmathematical proof; a simulationis far more convincing!

The terms “success”and“failure”for the outcomesthat have probability p and qare common in statistics. Butthey are completely arbitrarylabels. When we say that adisease occurs withprobability p, we certainlydon’t mean that getting sickis a “success”in the ordinarysense of the word.

TERMINOLOGYALERT:

A word of caution. In earlierchapters, we would plot andexamine the distribution of thesample, a display of the actualdata that were collected in onesample. But now we’ve plotted asampling distribution—a displayof summary statistics ( s, forexample) for many differentsamples. “Sample distribution”and “sampling distribution”sound a lot alike, but they referto very different things. (Sorryabout that—we didn’t make upthe terms.) And the distinction iscritical.Whenever you read orwrite something about one ofthese, think very carefully aboutwhat the words signify.

pN

A Sampling Distribution Model for a ProportionWe’ve simulated repeated samples and looked at a histogram of the sample pro-portions. We modelled that histogram with a Normal model. Why do we bother tomodel it? Because this model will give us insight into how much the sample pro-portion will vary from sample to sample. We’ve simulated many of the otherrandom samples we might have gotten. The model is an attempt to show the dis-tribution from all the random samples. But how do we know that a Normal modelwill really work? Is this just an observation based on some simulations that mightbe approximately true some of the time?

It turns out that this model can be justified theoretically and that the larger thesample size, the better the model works. That’s the result Laplace proved. Wewon’t bother you with the math because, in this instance, it really wouldn’t helpyour understanding.10 Nevertheless, the fact that we can think of proportionsfrom random samples as random quantities and then say something this specificabout their distribution is a fundamental insight—one that we will use in each ofthe next four chapters.

We have changed our point of view in a very important way. No longer is aproportion something we just compute for a set of data. We now see it as a ran-dom quantity that has a (probability) distribution, and thanks to Laplace we havea model for that distribution. We call that the sampling distribution model for theproportion, and we’ll make good use of it.


A Sampling Distribution Model for a Proportion 479

Without the sampling distribution model, the rest of statistics just wouldn’texist. Sampling models are what makes statistics work. They inform us about theamount of variation we should expect when we sample. Suppose we spin a coin100 times in order to decide whether it’s fair. If we get 52 heads, we’re probablynot surprised. Although we’d expect 50 heads, 52 doesn’t seem particularlyunusual for a fair coin. But we would be surprised to see 90 heads; that mightreally make us doubt that the coin is fair. How about 64 heads? Harder to say.That’s a case where we need the sampling distribution model. The samplingmodel quantifies the variability, telling us how surprising any sample proportionis. And it enables us to make informed decisions about how precise our estimateof the true proportion might be. That’s exactly what we’ll be doing for the rest ofthis book.

These sampling distributions are very abstract in nature. Even though wedepend on them, we never actually get to see them. We could take repeated sam-ples to develop the theory and our own intuition about sampling distributionmodels. But, in practice, we only imagine (or simulate) them. They’re importantbecause they act as a bridge from the real world of data to the imaginary model ofthe statistic and enable us to say something about the population when all wehave is data from the real world. This is the huge leap of statistics. Rather thanthinking about the sample proportion as a fixed quantity calculated from our data,we now think of it as a random variable—our value is just one of many we mighthave seen had we chosen a different random sample. By imagining what mighthappen if we were to draw many, many samples from the same population, wecan learn a lot about how statistics computed from our one particular sample arerelated to the corresponding population parameters.

We have now answered thequestion raised at the startof the chapter.To know howvariable a sample proportionis, we need to know theproportion and the size ofthe sample.That’s all.

THE SAMPLING DISTRIBUTION MODEL FOR A PROPORTION

Provided that the sampled values are independent and the sample size is large enough, the sampling distribution of is modelled by a Normal model

with mean and standard deviation SD1pN2 = A pq

n.m1pN2 = p

pN

Using the sampling distribution model for proportions

The U.S. Centers for Disease Control and Prevention report that 22% of 18-year-old women in the UnitedStates have a body mass index (BMI)11 of 25 or more—a value considered by the U.S. National Heart Lungand Blood Institute to be associated with increased health risk.

As part of a routine health check at a large university, the physical education department usually requiresstudents to come in to be measured and weighed. This year, the department decided to try out a self-reportsystem. It asked 200 randomly selected female students to report their heights and weights (from whichtheir BMIs could be calculated). Only 31 of these students had BMIs greater than 25.

Question: Is this proportion of high-BMI students unusually small?

First, check the conditions:

Ç Randomization Condition: The department drew a random sample, so therespondents should be independent and randomly selected from thepopulation.

FOR EXAMPLE

11 BMI = weight in kg/(height in m)2.

(continued)



Sampling distribution models tame the variation in statistics enough toallow us to measure how close our computed statistic values are likely to be tothe underlying parameters they estimate. That’s the path to the margin of erroryou hear about in polls and surveys. We’ll see how to do that in the nextchapter.

We can find a sampling distribution model for any statistic that we can com-pute from sampled data. What isn’t at all clear is whether these sampling distribu-tion models will always be simple probability models like the Normal model. TheNormal is a good model for proportions, and we’ll soon see that it works formeans as well. But it won’t work for every statistic. Fortunately, many of the sta-tistics we work with do have simple sampling distribution models. Usually weneed only a parameter or two to describe them.

Why do we really care about these distributions that we can never see? Theanswer is that once we know the appropriate sampling model, perhaps fromtheory, we can use it to help us understand how the statistic would have beendistributed if we had drawn all possible samples—but without the work andexpense of actually drawing or simulating them. This will help us generalizefrom our data to the world at large.

For Example (continued)

Ç 10% Condition: 200 respondents is less than 10% of all the female students ata “large university.”

Ç Success/Failure Condition: The department expected np � 200(0.22) � 44“successes” and nq � 200(0.78) � 156 “failures,” both at least 10.

It’s okay to use a Normal model to describe the sampling distribution of the pro-portion of respondents with BMIs above 25.

The phys ed department observed .

The department expected , with

, so .

By the 68–95–99.7 Rule, I know that values more than two standard deviationsbelow the mean of a Normal model show up less than 2.5% of the time. Perhapswomen at this university differ from the general population, or self-reporting maynot provide accurate heights and weights.

z =

pN - pSD(pN)

=

0.155 - 0.220.029

= -2.24= 0.029

= A(0.22)(0.78)

200SD(pN) = A

pqn

E(pN) = p = 0.22

pN =

31200

= 0.155

JUST CHECKING1. You want to poll a random sample of 100 students on campus to see ifthey are in favour of the proposed location for the new student centre. Of

course, you’ll get just one number, your sample proportion, . But if youimagined all the possible samples of 100 students you could draw andimagined the histogram of all the sample proportions from these samples, whatshape would it have?

2. Where would the centre of that histogram be?

3. If you think that about half the students are in favour of the plan, what wouldthe standard deviation of the sample proportions be?

pN


A Sampling Distribution Model for a Proportion 481

Working with Sampling Distribution Models for Proportions

I want to find the probability that in a group of90 students, more than 15 will be left-handed.Since 15 out of 90 is 16.7%, I need the probabil-ity of finding more than 16.7% left-handed stu-dents out of a sample of 90 if the proportionof lefties is 13%.

Plan State what we want to know.

Ç Independence Assumption: It is reason-able to assume that the probability thatone student is left-handed is not changedby the fact that another student is right-or left-handed.

Ç Randomization Condition: The 90 stu-dents in the class can be thought of as arandom sample of students.

Ç 10% Condition: Ninety is surely less than10% of the population of all students.(Even if the school itself is small, I’mthinking of the population of all possiblestudents who could have gone to theschool.)

Ç Success/Failure Condition:

nq = 90(0.87) = 78.3 Ú 10

np = 90(0.13) = 11.7 Ú 10

The population proportion is p � 0.13. The con-ditions are satisfied, so I’ll model the samplingdistribution of with a Normal model withmean 0.13 and a standard deviation of

My model for is (0.13, 0.035).pN

SD(pN) = Apqn = A

(0.13)(0.87)90

L 0.035

pN

Model Think about the assumptions andcheck the conditions.

You might be able to think of cases wherethe Independence Assumption is notplausible—for example, if the studentsare all related, or if they were selected forbeing left-or right-handed. But for a ran-dom sample, the assumption of inde-pendence seems reasonable.

State the parameters and the samplingdistribution model.

Suppose that about 13% of the population is left-handed.12 A 200-seat school auditorium has beenbuilt with 15 “lefty seats,”seats that have the built-in desk on the left rather than the right arm ofthe chair. (For the right-handed readers among you, have you ever tried to take notes in a chairwith the desk on the left side?)

Question: In a class of 90 students, what’s the probability that there will not be enough seats forthe left-handed students?

STEP–BY–STEP EXAMPLE

12 Actually, it’s quite difficult to get an accurate estimate of the proportion of lefties in thepopulation. Estimates range from 8% to 15%.



Plot Make a picture. Sketch the modeland shade the area we’re interested in, inthis case the area to the right of 16.7%.

Mechanics Use the standard deviationas a ruler to find the z-score of the cutoffproportion. We see that 16.7% leftieswould be just over one standard devia-tion above the mean.

Find the resulting probability from a tableof Normal probabilities, a computer pro-gram, or a calculator.

P(p̂ . 0.167) 5 P(z . 1.06) 5 0.1446

z =

pN - pSD(pN)

=

0.167 - 0.130.035

= 1.06

0.2353s

0.22s

0.1651s

0.095–1s

0.06–2s

0.025–3s

0.130p

0.167

0.145

There is about a 14.5% chance that there willnot be enough seats for the left-handed stu-dents in the class.

Conclusion Interpret the probability inthe context of the question.

What About Quantitative Data?Proportions summarize categorical variables. And the Normal sampling distribu-tion model looks like it is going to be very useful. But can we do something simi-lar with quantitative data?

Of course we can (or we wouldn’t have asked). Even more remarkable, notonly can we use all of the same concepts, but almost the same model, too.

What are the concepts? We know that when we sample at random or random-ize an experiment, the results we get will vary from sample-to-sample and fromexperiment-to-experiment. The Normal model seems like an incredibly simpleway to summarize all that variation. Could something that simple work formeans? We won’t keep you in suspense. It turns out that means also have a sam-pling distribution that we can model with a Normal model. And it turns out thatLaplace’s theoretical result applies to means, too. As we did with proportions, wecan get some insight from a simulation.

Simulating the Sampling Distribution of a MeanHere’s a simple simulation. Let’s start with one fair die. If we toss this die 10 000times, what should the histogram of the numbers on the face of the die look like?Here are the results of a simulated 10 000 tosses:

1 2 3 4 5 6Die Toss

2000

1500

1000

500

# of

Tos

ses


What About Quantitative Data? 483

Now let’s toss a pair of dice and record the average of the two. If we repeat this(or at least simulate repeating it) 10 000 times, recording the average of each pair,what will the histogram of these 10 000 averages look like? Before you look, thinkabout it fo a minute. Is getting an average of 1 on two dice as likely as getting anaverage of 3 or 3.5?

Let’s see:

1.0 2.0 3.0 4.0 5.0 6.02-Dice Average

2000

1500

1000

500

# of

Tos

ses

We’re much more likely to get an average near 3.5 than we are to get one near1 or 6. Without calculating those probabilities exactly, it’s fairly easy to see that theonly way to get an average of 1 is to get two ones. To get a total of seven (for anaverage of 3.5), though, there are many more possibilities. This distribution evenhas a name: the triangular distribution.

What if we average three dice? We’ll simulate 10 000 tosses of three dice andtake their average:

3 43-Dice Average

1500

1000

500# of

Tos

ses

651 2

What’s happening? First notice that it’s getting harder to have averages nearthe ends. Getting an average of 1 or 6 with three dice requires all three to come up1 or 6, respectively. That’s less likely than for two dice to come up both 1 or both 6.The distribution is being pushed toward the middle. But what’s happening to theshape? (This distribution doesn’t have a name, as far as we know.)

Let’s continue this simulation to see what happens with larger samples.Here’s a histogram of the averages for 10 000 tosses of five dice:

1.05-Dice Average

1500

1000

500# of

Tos

ses

2.0 3.0 4.0 5.0 6.0

The pattern is becoming clearer. Two things continue to happen. The first factwe knew already from the Law of Large Numbers. It says that as the samplesize (number of dice) gets larger, each sample average is more likely to be closer tothe population mean. So we see the shape continuing to tighten around 3.5. But



the shape of the distribution is the surprising part. It’s becoming bell-shaped. Andnot just bell-shaped; it’s approaching the Normal model.

Are you convinced? Let’s skip ahead and try 20 dice. The histogram of aver-ages for 10 000 throws of 20 dice looks like this:

1500

1000

500

1.0 2.0 3.0 4.0 5.0 6.020-Dice Average

# of

Tos

ses

Now we see the Normal shape again (and notice how much smaller the spreadis). But can we count on this happening for situations other than dice throws? Whatkinds of sample means have sampling distributions that we can model with aNormal model? It turns out that Normal models work well amazingly often.

Activity: The SamplingDistribution Model for Means.Don’t just sit there reading aboutthe simulation—do it yourself.

13 Okay, one technical condition. The data must come from a population with a finite vari-ance. You probably can’t imagine a population with an infinite variance, but statisticianscan construct such things, so we have to discuss them in footnotes like this. It really makesno difference in how you think about the important stuff, so you can just forget we men-tioned it.14 The word “central” in the name of the theorem means “fundamental.” It doesn’t refer tothe centre of a distribution, so the title of this section is actually redundant.

“The theory of probabilities is atbottom nothing but commonsense reduced to calculus.”

—Laplace, in Théorie analytique des

probabilités, 1812

The Central Limit Theorem: The FundamentalTheorem of Statistics

The dice simulation may look like a special situation. But it turns out thatwhat we saw with dice is true for means of repeated samples for almost every sit-uation. When we looked at the sampling distribution of a proportion, we had tocheck only a few conditions. For means, the result is even more remarkable. Thereare almost no conditions at all.

Let’s say that again: The sampling distribution of any mean becomes morenearly Normal as the sample size grows. All we need is for the observations to beindependent and collected with randomization. We don’t even care about theshape of the population distribution!13 This surprising fact is the result Laplaceproved in a fairly general form in 1810. At the time, Laplace’s theorem causedquite a stir (at least in mathematics circles) because it is so unintuitive. Laplace’sresult is called the Central Limit Theorem14 (CLT).

Why should the Normal model show up again for the sampling distributionof means as well as proportions? We’re not going to try to persuade you that it isobvious, clear, simple, or straightforward. In fact, the CLT is surprising and a bitweird. Not only does the distribution of means of many random samples getcloser and closer to a Normal model as the sample size grows, this is true regardlessof the shape of the population distribution! Even if we sample from a skewed orbimodal population, the Central Limit Theorem tells us that means of repeatedrandom samples will tend to follow a Normal model as the sample size grows. Ofcourse, you won’t be surprised to learn that it works better and faster the closerthe population distribution is to a Normal model. And it works better for largersamples. If the data come from a population that’s exactly Normal to start with,



then the observations themselves are Normal. If we take samples of size 1, their“means” are just the observations—so, of course, they have Normal sampling dis-tribution. But now suppose the population distribution is very skewed (like theCEO data from Chapter 5, for example). The CLT works, although it may take asample size of dozens or even hundreds of observations for the Normal model towork well.

For example, think about a really bimodal population, one that consists ofonly zeros and ones. The CLT says that even means of samples from this popula-tion will follow a Normal sampling distribution model. But wait. Suppose wehave a categorical variable and we assign a 1 to each individual in the categoryand a 0 to each individual not in the category. And then we find the mean of thesezeros and ones. That’s the same as counting the number of individuals who are inthe category and dividing by n. That mean will be . . . the sample proportion, p, ofindividuals who are in the category (a “success”). So maybe it wasn’t so surprisingafter all that proportions, like means, have Normal sampling distribution models;they are actually just a special case of Laplace’s remarkable theorem. Of course,for such an extremely bimodal population, we’ll need a reasonably large samplesize—and that’s where the special conditions for proportions come in.

THE CENTRAL LIMIT THEOREM (CLT)

The mean15 of a random sample is a random variable whose sampling distri-bution can be approximated by a Normal model if the sample size is bigenough. The larger the sample, the better the approximation will be.

15 And hence the sample total, too, since the mean and the total only differ by the multi-plicative constant n.

Assumptions and ConditionsThe CLT requires essentially the same assumptions as we saw for modelling pro-portions:

Independence Assumption: The sampled values must be independent randomdraws from the population under study.

Sample Size Assumption:The sample size must be sufficiently large.

We can’t check these directly, but we can think about whether the IndependenceAssumption is plausible. We can also check some related conditions:

Randomization Condition: The data values must be sampled randomly, orthe concept of a sampling distribution makes no sense.

10% Condition: When the sample is drawn without replacement (as is usuallythe case), the sample size, n, should be no more than 10% of the population.

Large Enough Sample Condition: Although the CLT tells us that a Normalmodel is useful in thinking about the behaviour of sample means when thesample size is large enough, it doesn’t tell us how large a sample we need. Thetruth is, it depends; there’s no one-size-fits-all rule. If the population is uni-modal and symmetric, even a fairly small sample (like five or ten) is okay. Ifthe population is strongly skewed, like the compensation for CEOs we lookedat in Chapter 5, it can take a pretty large sample (like 50, 100, or even more) toallow use of a Normal model to describe the distribution of sample means.You’ll just need to think about your sample size in the context of what youknow about the population, and then tell whether you believe the LargeEnough Sample Condition has been met.

Activity: The CentralLimit Theorem. Does it reallywork for samples from non-Normal populations?



But Which Normal?The CLT says that the sampling distribution of any mean or proportion is approx-imately Normal. But which Normal model? We know that any Normal is specifiedby its mean and standard deviation. For proportions, the sampling distribution iscentred at the population proportion. For means, it’s centred at the populationmean. What else would we expect?

What about the standard deviations, though? We noticed in our dice simula-tion that the histograms got narrower as we averaged more and more dicetogether. This shouldn’t be surprising. Means vary less than the individual obser-vations. Think about it for a minute. Which would be more surprising, having oneperson in your Statistics class who is over 205 cm tall or having the mean of 100students taking the course be over 205 cm? The first event is fairly rare.16 You mayhave seen somebody this tall in one of your classes sometime. But finding a classof 100 whose mean height is over 205 cm tall just won’t happen. Why? Becausemeans have smaller standard deviations than individuals.

How much smaller? Well, we have good news and bad news. The good newsis that the standard deviation of y falls as the sample size grows. The bad news isthat it doesn’t drop as fast as we might like. It only goes down by the square root ofthe sample size. Why? The Math Box will show you that the Normal model for thesampling distribution of the mean has a standard deviation equal to

where σ is the standard deviation of the population. To emphasize that this is astandard deviation parameter of the sampling distribution model for the samplemean, , we write SD ( ) or � ( ).yyy

SD1y2 =

s

2n

Activity: The StandardDeviation of Means. Experimentto see how the variability of themean changes with the samplesize.

Activity: The SamplingDistribution of the Mean. TheCLT tells us what to expect. Inthis activity you can work withthe CLT or simulate it if youprefer.

16 If students are a random sample of adults, fewer than 1 out of 10 000 should be taller than205 cm. Why might university students not really be a random sample with respect to height?Even if they’re not a perfectly random sample, a university student over 205 cm tall is still rare.

THE SAMPLING DISTRIBUTION MODEL FOR A MEAN

When a simple random sample is drawn from any (much bigger) populationwith mean and standard deviation �, its sample mean, , has a sampling

distribution with the same mean but whose standard deviation is

(and we write ). No matter what population the random

sample comes from, the shape of the sampling distribution is approximatelyNormal as long as the sample size is large enough. The larger the sample used,the more closely the Normal model approximates the sampling distributionfor the mean.

= SD1y2 =

s

1ns1y2

s

1nm

ym

Why is ? We know that is a sum divided by n:

As we saw in Chapter 16, when a random variable is divided by a constantits variance is divided by the square of the constant:

Var(y) 5Var(y1 1 y2 1 y3 1c1 yn)

n2

y 5y1 1 y2 1 y3 1c1 yn

n

ySD1y2 =

s

1n

(continued)



We draw the y's randomly from a big population, ensuring they are inde-pendent. For independent random variables, variances add:

All n of the y’s were drawn from our population, so they all have the samevariance, σ2:

The standard deviation of is the square root of this variance:

In a very similar manner, we can also prove that the expected value of thesample mean, E( ), is equal to m, which shows that the mean of the measure-ments, , in a simple random sample is an unbiased estimator of the populationmean, m (in the absence of non-sampling sources of bias).

yy

SD(y) 5 Ås2

n5

s

!n

y

Var(y) 5s2 1 s2 1 s2 1c1 s2

n2 5ns2

n2 5s2

n

Var(y) 5Var(y1) 1 Var(y2) 1 Var(y3) 1c1 Var(yn)

n2

Using the CLT for means

Recap: A university physical education department asked a random sample of 200 female students toself-report their heights and weights, but the percentage of students with body mass indexes over 25seemed suspiciously low. One possible explanation may be that the respondents “shaded” their weightsdown a bit. The CDC reports that the mean weight of 18-year-old women is 65.4 kg, with a standarddeviation of 23.45 kg, but these 200 randomly selected women reported a mean weight of only 63.7 kg.

Question: Based on the Central Limit Theorem and the 68–95–99.7 Rule, does the mean weight in thissample seem exceptionally low, or might this just be random sample-to-sample variation?

The conditions check out okay:

Ç Randomization Condition: The women were a random sample and their weightscan be assumed to be independent.

FOR EXAMPLE

We now have two closely related sampling distribution models that we canuse when the appropriate assumptions and conditions are met. Which one we usedepends on which kind of data we have:

u When we have categorical data, we calculate a sample proportion, ; thesampling distribution of this random variable has a Normal model with a mean at the true proportion (“Greek letter”) p, and a standard deviation of

. We’ll use this model in Chapters 19 through 22.

u When we have quantitative data, we calculate a sample mean, ; the samplingdistribution of this random variable has a Normal model with a mean at the true mean, , and a standard deviation of . We’ll use this model inChapters 23, 24, and 25.

The means of these models are easy to remember, so all you need to be carefulabout is the standard deviations. Remember that these are standard deviations ofthe statistics and . They both have a square root of n in the denominator. Thattells us that the larger the sample, the less either statistic will vary. The only differ-ence is in the numerator. If you just start by writing SD( ) for quantitative dataand SD( ) for categorical data, you’ll be able to remember which formula to use.pN

y

ypN

SD1y2 =

s

1nm

y

SD1pN2 = Apqn =

2pq

1n

pN

(continued)



For Example (continued)

Ç 10% Condition: They sampled fewer than 10% of all women at the college.Ç Large Enough Sample Condition: The distribution of university women’s weights

is likely to be unimodal and reasonably symmetric, so the CLT applies to meansof even small samples; 200 values is plenty.

The sampling model for sample means is approximately Normal with E( ) 65.4 and

SD( ) . The expected distribution of sample means is=

s

2n =

23.45

2200= 1.66y

=y

The 68–95–99.7 Rule suggests that although the reported mean weight of 63.7 kgis somewhat lower than expected, it does not appear to be unusual. Such variabil-ity is not all that extraordinary for samples of this size.

60.42 62.08 63.74 67.06 68.72 70.3865.40Sample Means

68%

95%

99.7%

63.7

Working with the Sampling Distribution Model for the Mean

Asking the probability that the total weight ofa sample of 10 men exceeds 1000 kg is equiva-lent to asking the probability that their meanweight is greater than 100 kg.

Plan State what we want to know.

The mean weight of adult men in Canada is 82 kg with a standard deviation of 16 kg.17

Question: An elevator in our building has a weight limit of 1000 kg. What’s the probability that if10 men get on the elevator, they will overload its weight limit?

STEP–BY–STEP EXAMPLE

Ç Independence Assumption: It’s reason-able to think that the weights of 10 ran-domly sampled men will be independent ofeach other. (But there could beexceptions—for example, if they were allfrom the same family or if the elevatorwere in a building with a diet clinic!)

Model Think about the assumptions andcheck the conditions.

17 Author’s estimate, using Canadian Community Health Survey data and other sources.


About Variation 489

Ç Randomization Condition: I’ll assumethat the 10 men getting on the elevatorare a random sample from the population.

Ç 10% Condition: 10 men is surely less than10% of the population of possible elevatorriders.

Ç Large Enough Sample Condition: I believethe distribution of population weights isonly mildly skewed, so my sample of 10 menseems large enough.

The mean for all weights is = 82 kg and thestandard deviation is � 16 kg. Since the con-ditions are satisfied, the CLT says that thesampling distribution of has a Normal modelwith mean 82 and standard deviation

SD( ) =

s

2n=

16

210L 5.06y

y

s

mState the parameters and the samplingmodel.

Note that if the sample were larger we’dbe less concerned about the shape of thedistribution of all weights.

P( ) P( )1 � 0.9998 � 0.0002

=z 7 3.56=y 7 100

z =

y - m

SD(y)=

100 - 825.06

= 3.56

Plot Make a picture. Sketch the modeland shade the area we’re interested in.Here the mean weight of 100 kg appearsto be far out on the right tail of the curve.

Mechanics Use the standard deviationas a ruler to find the z-score of the cutoffmean weight. We see that an average of100 kg is more than three standard devia-tions above the mean.

Find the resulting probability from a tableof Normal probabilities such as Table Z, acomputer program, or a calculator.

77 82 87 92 977267y

The chance that a random collection of 10 menwill exceed the elevator’s weight limit is only0.0002. So, if they are a random sample, it isquite unlikely that 10 people will exceed thetotal weight allowed on the elevator.

Conclusion Interpret your result in theproper context, being careful to relate it tothe original question.

About VariationMeans vary less than individual data values. That makes sense. If the same test isgiven to many sections of a large course and the class average is, say, 80%, somestudents may score 95% because individual scores vary a lot. But we’d be shocked(and pleased!) if the average score of the students in any section was 95%. Averages



are much less variable. Not only do group averages vary less than individualvalues, but common sense suggests that averages should be more consistent for

larger groups. The fact that SD( ) has n in the denominator shows that the

variability of sample means decreases as the sample size increases. There’s a catch,though. The standard deviation of the sampling distribution declines only withthe square root of the sample size and not, for example, with 1/n. The mean of a

random sample of four has half the standard deviation of an individual

data value. To cut the standard deviation in half again, we’d need a sample of 16,and a sample of 64 to halve it once more.

If only we had a much larger sample, we could get the standard deviation ofthe sampling distribution really under control so that the sample mean could tellus still more about the unknown population mean, but larger samples cost moreand take longer to survey. And while we’re gathering all that extra data, the pop-ulation itself may change, or a news story may alter opinions. There are practicallimits to most sample sizes. As we shall see, that nasty square root limits howmuch we can make a sample tell about the population. This is an example ofsomething that’s known as the Law of Diminishing Returns.

a 1

14=

12b

=

s

1ny

A Billion Dollar Misunderstanding? In the late 1990s, the Bill and MelindaGates Foundation began funding an effort to encourage the breakup of large schoolsinto smaller schools. Why? It had been noticed that smaller schools were more com-mon among the best-performing schools than one would expect. In time, theAnnenberg Foundation, the Carnegie Corporation, the Center for CollaborativeEducation, the Center for School Change, Harvard’s Change Leadership Group, theOpen Society Institute, Pew Charitable Trusts, and the U.S. Department ofEducation’s Smaller Learning Communities Program all supported the effort. Wellover a billion dollars was spent to make schools smaller.

But was it all based on a misunderstanding of sampling distributions? StatisticiansHoward Wainer and Harris Zwerling18 looked at the mean test scores of schools inPennsylvania. They found that indeed 12% of the top-scoring 50 schools were fromthe smallest 3% of Pennsylvania schools—substantially more than the 3% we’dnaively expect. But then they looked at the bottom 50. There they found that 18%were small schools! The explanation? Mean test scores are, well, means. We arelooking at a rough real-world simulation in which each school is a trial. Even if allPennsylvania schools were equivalent, we’d expect their mean scores to vary. How much? The CLT tells us that means of test scores vary according to . Smaller

schools have (by definition) smaller n’s, so the sampling distributions of their meanscores naturally have larger standard deviations. It’s natural, then, that small schoolshave both higher and lower mean scores.

On October 26, 2005, the Seattle Times reported:

The Gates Foundation announced last week it is moving away from itsemphasis on converting large high schools into smaller ones and insteadgiving grants to specially selected school districts with a track record ofacademic improvement and effective leadership. Education leaders at theFoundation said they concluded that improving classroom instruction andmobilizing the resources of an entire district were more important first stepsto improving high schools than breaking down the size.

s

1n

18 Wainer, H. and Zwerling, H., 2006, “Legal and empirical evidence that smaller schools donot improve student achievement.” The Phi Delta Kappan 87:300–303. Discussed in HowardWainer, 2007, “The most dangerous equation.” American Scientist pp. 249–256; also atwww.americanscientist.org.


Sampling Distribution Models 491

19 Don’t forget to adjust for inflation.20 But add gender to the mix of influences on height, and if the average height differencebetween males and females is very big, we can lose the Normality, possibly ending up withbi-modality. What went wrong? One variable had a big effect on the final result.

The Real World and the Model WorldBe careful. We have been slipping smoothly between the real world, in which wedraw random samples of data, and a magical mathematical model world, inwhich we describe how the sample means and proportions we observe in the realworld might behave if we could see the results from every random sample that wemight have drawn. Now we have two distributions to deal with. The first is thereal world distribution of the sample, which we might display with a histogram(for quantitative data) or with a bar chart or table (for categorical data). The sec-ond is the math world sampling distribution of the statistic, which we may be ableto model with a Normal model based on the Central Limit Theorem. Don’t con-fuse the two.

For example, don’t mistakenly think the CLT says that the data are Normally dis-tributed as long as the sample is large enough. In fact, as samples get larger, weexpect the distribution of the data to look more and more like the population fromwhich they are drawn—skewed, bimodal, whatever—but not necessarily Normal.You can collect a sample of CEO salaries for the next 1000 years,19 but the histogramwill never look Normal. It will be skewed to the right. The Central Limit Theoremdoesn’t talk about the distribution of the data from the sample. It talks about thesample means and sample proportions of many different random samples drawn fromthe same population. Of course, the CLT does require that the sample be bigenough when the population shape is not unimodal and symmetric, but the factthat, even then, a Normal model is useful is still a very surprising and powerfulresult.

The CLT in NatureThe CLT we have given here is adequate for our purposes, assuming independentobservations (as is common in statistical studies), but actually there have beenmany generalizations of this theorem over the years. It turns out that even if theindividual random variables exhibit some dependence (though not too great), ordo not all come from the same distribution (as long as each has but a small effecton the total), Normality will still arise if enough variables are averaged (orsummed). Perhaps this more general CLT effect helps explain all those variables innature that possess an approximately Normal frequency distribution. Are heightsin a genetically homogenous population nearly Normal because the final resultcomes from summing (or averaging) many small genetic and environmental influ-ences? Possibly.20 Was your height a sum of such random influences, somewhatlike summing the random values in a sample?

Many other variables are formed by multiplying many small influencestogether rather than adding them. But guess what? Take the log (converting multi-plication to addition), and you may just get Normality!

Sampling Distribution ModelsLet’s summarize what we’ve learned about sampling distributions. At the heart isthe idea that the statistic itself is a random quantity. We can’t know what our statisticwill be because it comes from a random sample. It’s just one instance of somethingthat happened for our particular random sample. A different random sample



would have given a different result. This sample-to-sample variability is whatgenerates the sampling distribution. The sampling distribution shows us the dis-tribution of possible values that the statistic could have had.

We could simulate that distribution by pretending to take lots of samples.While a sampling distribution can take on pretty much any form, depending on theparticular statistic and study design, fortunately, for the mean or the proportion,the CLT tells us that we can model the sampling distribution with a Normal model.

Two basic truths about sampling distributions are:

1. Sampling distributions arise because samples vary. Each random sample willcontain different cases, and therefore a different value of the statistic.

2. Although we can simulate a sampling distribution, the Central LimitTheorem saves us the trouble for means and proportions.

Here’s a picture showing the process going into the sampling distributionmodel:

Simulation: The CLT forReal Data. Why settle for apicture when you can see it inaction?

FIGURE 18.5We start with a population model,which can have any shape. It can evenbe bimodal or skewed (as this one is).We label the mean of this model and its standard deviation, .

We draw one real sample (solid line)of size n and show its histogram andsummary statistics. We imagine (orsimulate) drawing many other samples(dotted lines), which have their ownhistograms and summary statistics.

sm

We (imagine) gathering all the means into a histogram.

The CLT tells us we can model the shape of this histogramwith a Normal model. The mean of this Normal is , andthe standard deviation

is .SD1y2 =

s

2n

m

–3sn –2

sn –1

sn +1

sn +2

sn +3

sn

n

m

s

s1 s2 s3

y2y1 y3

m

s

JUST CHECKING4. Human gestation times have a mean of about 266 days, with astandard deviation of about 16 days. If we record the gestation times of a

sample of 100 women, do we know that a histogram of the times will be wellmodelled by a Normal model?

5. Suppose we look at the average gestation times for a sample of 100 women. Ifwe imagined all the possible random samples of 100 women we could take andlooked at the histogram of all the sample means, what shape would it have?

6. Where would the centre of that histogram be?

7. What would be the standard deviation of that histogram?

18_deve_ch18.qxd 2/3/11 6:04 PM Page 492

*Finite Population Correction 493

The View from Here . . .We’ve seen that when it comes to understanding the data’s story, our biggestnemesis is variability. Whether we’re looking at information collected in a surveyor the results of an experiment, we have only a narrow glimpse of the larger worldwe wish we knew about. That glimpse is subject to error, the variability that natu-rally arises from sample to sample and group to group. What we’ve now learnedis that this unavoidable variability is predictable and quantifiable. The CentralLimit Theorem quantifies sampling error.

In this chapter our discussion has been purely theoretical. We’ve imaginedthat we know the population parameters—proportions, means, and standarddeviations—and we’ve learned to describe the behaviour of statistics observed inmany, many samples. But that’s not the way the real world works. In practice,we’ll never know the population parameters, and we’ll have data from just a sin-gle study. We poll 1000 voters, but we want to know whom the entire electoratewill choose. We’ve seen a new medicine lower blood pressure more than aplacebo, but we need to know whether that difference is statistically significant.21

We now must combine our ability to design studies and summarize the data withthe CLT’s insights about sampling variability in order to use the sample statisticswe observe to make inferences about the population parameters we wish weknew.

That’s the path that lies ahead. In Chapters 19 to 22, we’ll learn how to makeinferences about proportions. In Chapters 23 to 25, we’ll turn our attention tomeans. The Central Limit Theorem will be with us all the way.

21 We first encountered “statistically significant” in Chapter 13. The term refers to results sounusual that they’re probably not merely sample-to-sample variation. You’ll be seeing—and saying—those words a lot from here on.

*Finite Population CorrectionWhen drawing a simple random sample from a finite population of size N, theactual variance of the sample proportion and sample mean are given by the for-mulas:

The latter multiplicative part, appended to our earlier formulas, is

called the finite population correction. This factor is very close to one if N is muchbigger than n, in which case these formulas reduce to our earlier formulas forSD’s. Also note that they give you the correct result of SD = 0 if you sample theentire population, since then the sample mean (or proportion) must equal the truemean (or proportion), and hence has no variability at all.

The limitation of our earlier formulas and derivations is that we assumedindependent observations. But when you sample (without replacement) from thepopulation, you change it. Imagine sampling five cards from a deck of 52. Everytime you draw, say, a face card, you reduce the number of face cards left and thechance that the next card will be a face card. Suppose instead that you draw fivecards from 10 decks shuffled together. Every time you draw a card now, theremaining deck composition changes little, and the draws are nearly independent.

Our earlier formulas are good enough most of the time, but what if we fail tocorrect for finiteness when we should (if we violate that 10% Condition)? The result

AN - nn - 1

SD(p̂) 5 Åpqn

3 ÅN 2 nN 2 1

and SD(y2) 5s

!n3 Å

N 2 nN 2 1



is that we overestimate the variability of our sample, and our results will actuallybe more precise or accurate than suggested by our calculations.22 Not a bad way toerr; we call this a conservative error (sorry, but there is no place for liberal thinking inStatistics). But what is this worth if we may lose the Normality? Fear not—moregeneral versions of the CLT (relaxing the Independence Assumption) assure us ofNormality even if we take a very big bite out of the population.

22 And the same goes for two sample procedures to be discussed in later chapters.23 For proportions, of course, there is a rule: the Success/Failure Condition. That works forproportions because the standard deviation of a proportion is linked to its mean.

WHAT CAN GO WRONG?u Don’t confuse the sampling distribution with the distribution of the sample. When you take a

sample, you always look at the distribution of the values, usually with a histogram,and you may calculate summary statistics. Examining the distribution of the samplelike this is wise. But that’s not the sampling distribution. The sampling distribution isan imaginary collection of the values that a statistic might have taken for all the ran-dom samples—the one you got and the ones that you didn’t get. We use the samplingdistribution model to make statements about how the statistic varies.

u Beware of observations that are not independent. The CLT depends crucially on theassumption of independence. If our elevator riders are related, are all from the sameschool (for example, an elementary school), or in some other way aren’t a randomsample, then the statements we try to make about the mean are going to be wrong.Unfortunately, this isn’t something you can check in your data. You have to thinkabout how the data were gathered. Good sampling practice and well-designed ran-domized experiments ensure independence.

u Watch out for small samples from skewed populations. The CLT assures us that the sam-pling distribution model is Normal if n is large enough. If the population is nearlyNormal, even small samples (like our 10 elevator riders) work. If the population isvery skewed, then n will have to be large before the Normal model will work well. Ifwe sampled 15 or even 20 CEOs and used to make a statement about the mean of allCEOs’ compensation, we’d likely get into trouble because the underlying data distri-bution is so skewed. Unfortunately, there’s no good rule of thumb.23 It just dependson how skewed the data distribution is. Always plot the data to check.

y

CONNECTIONSThe concept of a sampling distribution connects to almost everything we have done. The funda-mental connection is to the deliberate application of randomness in random sampling and random-ized comparative experiments. If we didn’t employ randomness to generate unbiased data, thenrepeating the data collection would just get the same data values again (with perhaps a few newmeasurement or recording errors). The distribution of statistic values arises directly because differ-ent random samples and randomized experiments would generate different statistic values.

The connection to the Normal distribution is obvious. We first introduced the Normal modelbefore because it was “nice.” As a unimodal, symmetric distribution with 99.7% of its area withinthree standard deviations of the mean, the Normal model is easy to work with. Now we see thatthe Normal holds a special place among distributions because we can use it to model the samplingdistributions of the mean and the proportion.

We used simulation to understand sampling distributions. In fact, some important sampling dis-tributions were discovered first by simulation.


WHAT HAVE WE LEARNED?

Way back in Chapter 1 we said that statistics is about variation. We know that no sample fully andexactly describes the population; sample proportions and means will vary from sample to sample.That’s sampling error (or, better, sampling variability). We know it will always be present—indeed, theworld would be a boring place if variability didn’t exist. You might think that sampling variability wouldprevent us from learning anything reliable about a population by looking at a sample, but that’s justnot so. The fortunate fact is that sampling variability is not just unavoidable—it’s predictable!

We’ve learned how the Central Limit Theorem describes the behaviour of sample proportions—shape,centre, and spread—as long as certain assumptions and conditions are met. The sample must beindependent, random, and large enough that we expect at least 10 successes and failures. Then:

u The sampling distribution (the imagined histogram of the proportions from all possible samples)is shaped like a Normal model.

u The mean of the sampling model is the true proportion in the population.

u The standard deviation of the sample proportions is .

And we’ve learned to describe the behaviour of sample means as well, based on this amazing resultknown as the Central Limit Theorem—the fundamental theorem of Statistics. Again the sample mustbe independent and random—no surprise there—and needs to be larger if our data come from a pop-ulation that’s not roughly unimodal and symmetric. Then:

u Regardless of the shape of the original population, the shape of the distribution of the means of allpossible samples can be described by a Normal model, provided the samples are large enough.

u The centre of the sampling model will be the true mean of the population from which we took thesample.

u The standard deviation of the sample means is the population’s standard deviation divided by the

square root of the sample size, .s

1n

Apqn

TermsSampling distribution model 475. Different random samples give different values for a statistic. The sampling distribution model

shows the behaviour of the statistic over all the possible samples for the same size n.

Sampling variability 476. The variability we expect to see from one random sample to another. It is sometimes called(sampling error) sampling error, but sampling variability is the better term.

Sampling distribution model 478. If assumptions of independence and random sampling are met, and we expect at least 10 for a proportion successes and 10 failures, then the sampling distribution of a proportion is modelled by a Normal

model with a mean equal to the true proportion value, p, and a standard deviation equal to .

Sampling distribution model 482. If assumptions of independence and random sampling are met, and the sample size is largefor a mean enough, the sampling distribution of the sample mean is modelled by a Normal model with a mean

equal to the population mean, , and a standard deviation equal to .

Central Limit Theorem 484. The Central Limit Theorem (CLT) states that the sampling distribution model of the samplemean (and proportion) from a random sample is approximately Normal for large n, regardless of thedistribution of the population, as long as the observations are independent.

Skillsu Understand that the variability of a statistic (as measured by the standard deviation of its sampling

distribution) depends on the size of the sample. Statistics based on larger samples are less variable.

u Understand that the Central Limit Theorem gives the sampling distribution model of the mean forsufficiently large samples regardless of the underlying population.

u Be able to demonstrate a sampling distribution by simulation.

s

1nm

Apq

n

What Have We Learned? 495



u Be able to use a sampling distribution model to make simple statements about the distribution ofa proportion or mean under repeated sampling.

u Be able to interpret a sampling distribution model as describing the values taken by a statistic inall possible realizations of a sample or randomized experiment under the same conditions.

EXERCISES

1. Send money. When it sends out its fundraising letter, aphilanthropic organization typically gets a return fromabout 5% of the people on their mailing list. To see whatthe response rate might be for future appeals, it did asimulation using samples of size 20, 50, 100, and 200. Foreach sample size, it simulated 1000 mailings with successrate p = 0.05 and constructed the histogram of the 1000sample proportions, shown below. Explain how these his-tograms demonstrate what the Central Limit Theoremsays about the sampling distribution model for sampleproportions. Be sure to talk about shape, centre, andspread.

2. Character recognition. An automatic character recog-nition device can successfully read about 85% of hand-written credit card applications. To estimate what mighthappen when this device reads a stack of applications,the company did a simulation using samples of size 20,50, 75, and 100. For each sample size, they simulated 1000samples with success rate p = 0.85 and constructed the

histogram of the 1000 sample proportions, shown here.Explain how these histograms demonstrate what theCentral Limit Theorem says about the sampling distribu-tion model for sample proportions. Be sure to talk aboutshape, centre, and spread.

3. Send money, again. The philanthropic organization inExercise 1 expects about a 5% success rate when it sendsfundraising letters to the people on their mailing list. InExercise 1 you looked at the histograms showing distri-butions of sample proportions from 1000 simulated mail-ings for samples of size 20, 50, 100, and 200. The samplestatistics from each simulation were as follows:

0.00Sample Proportions

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20 0.0497 0.047950 0.0516 0.0309

100 0.0497 0.0215200 0.0501 0.0152


Exercises 497

a) According to the Central Limit Theorem, what shouldthe theoretical mean and standard deviations be forthese sample sizes?

b) How close are those theoretical values to what wasobserved in these simulations?

c) Looking at the histograms in Exercise 1, at what sam-ple size would you be comfortable using the Normalmodel as an approximation for the sampling distribu-tion?

d) What does the Success/Failure Condition say aboutthe choice you made in part c)?

4. Character recognition, again. The automatic charac-ter recognition device discussed in Exercise 2 successfullyreads about 85% of handwritten credit card applications.In Exercise 2 you looked at the histograms showing dis-tributions of sample proportions from 1000 simulatedsamples of size 20, 50, 75, and 100. The sample statisticsfrom each simulation were as follows:

a) According to the Central Limit Theorem, what shouldthe theoretical mean and standard deviations be forthese sample sizes?

b) How close are those theoretical values to what wasobserved in these simulations?

c) Looking at the histograms in Exercise 2, at what sam-ple size would you be comfortable using the Normalmodel as an approximation for the samplingdistribution?

d) What does the Success/Failure Condition say aboutthe choice you made in part c)?

5. Coin tosses. In a large class of introductory Statisticsstudents, the professor has each person toss a coin 16times and calculate the proportion of his or her tossesthat were heads. The students then report theirresults, and the professor plots a histogram of theseproportions.a) What shape would you expect this histogram to be?

Why?b) Where do you expect the histogram to be centred?c) How much variability would you expect among these

proportions?d) Explain why a Normal model should not be used here.

6. M&M’s. The candy company claims that 10% of theM&M’s it produces are green. Suppose that the candiesare packaged at random in small bags containing about50 M&M’s. A class of elementary school students learningabout percents opens several bags, counts the variouscolours of the candies, and calculates the proportion thatare green.a) If we plot a histogram showing the proportions of

green candies in the various bags, what shape wouldyou expect it to have?

b) Can that histogram be approximated by a Normalmodel? Explain.

c) Where should the centre of the histogram be?d) What should the standard deviation of the proportion

be?

7. More coins. Suppose the class in Exercise 5 repeats thecoin-tossing experiment.a) The students toss the coins 25 times each. Use the

68–95–99.7 Rule to describe the sampling distributionmodel.

b) Confirm that you can use a Normal model here.c) They increase the number of tosses to 64 each. Draw

and label the appropriate sampling distributionmodel. Check the appropriate conditions to justifyyour model.

d) Explain how the sampling distribution model changesas the number of tosses increases.

8. Bigger bag. Suppose the class in Exercise 6 buys biggerbags of candy, with 200 M&M’s each. Again the studentscalculate the proportion of green candies they find.a) Explain why it’s appropriate to use a Normal model

to describe the distribution of the proportion of greenM&M’s they might expect.

b) Use the 68–95–99.7 Rule to describe how this propor-tion might vary from bag to bag.

c) How would this model change if the bags containedeven more candies?

9. Just (un)lucky? One of the students in the introduc-tory Statistics class in Exercise 7 claims to have tossed hercoin 200 times and found only 42% heads. What do youthink of this claim? Explain.

10. Too many green ones? In a really large bag of M&M’s,the students in Exercise 8 found 500 candies, and 12% ofthem were green. Is this an unusually large proportion ofgreen M&M’s? Explain.

11. Speeding. Provincial police believe that 70% of thedrivers travelling on a particular section of the Trans-Canada Highway exceed the speed limit. They plan toset up a radar trap and check the speeds of 80 cars.a) Using the 68–95–99.7 Rule, draw and label the distri-

bution of the proportion of these cars the police willobserve speeding.

b) Do you think the appropriate conditions necessary foryour analysis are met? Explain.

12. Smoking pot. The Canadian Drug Use and MonitoringSurvey estimates that 33% of 15- to 24-year-olds usedcannabis in 2008. Using the 68–95–99.7 Rule, describe thesampling distribution model for the proportion ofcannabis users among a randomly selected group of fifty15- to 24-year-olds. Be sure to discuss your assumptionsand conditions.

13. Vision. It is generally believed that nearsightednessaffects about 12% of all children. A school district has reg-istered 170 incoming kindergarten children.a) Can you apply the Central Limit Theorem to describe

the sampling distribution model for the sample pro-portion of children who are nearsighted? Check theconditions and discuss any assumptions you need tomake.

n mean st. dev.

20 0.8481 0.080350 0.8507 0.050975 0.8481 0.0406

100 0.8488 0.0354



b) Sketch and clearly label the sampling model, based onthe 68–95–99.7 Rule.

c) How many of the incoming students might the schoolexpect to be nearsighted? Explain.

14. Mortgages. In early 2009, rising unemployment andthe house-price slump in Alberta led to a mortgage delin-quency rate of 0.54% of all big bank mortgages (rising forthe nineteenth straight month), the highest delinquencyrate in Canada. Suppose a large bank holds 1731mortgages.a) Can you apply the Central Limit Theorem to describe

the sampling distribution model for the sample pro-portion of delinquencies? Check the conditions anddiscuss any assumptions you need to make.

b) Sketch and clearly label the sampling model, based onthe 68–95–99.7 Rule.

c) How many of these homeowners might the bankexpect to be delinquent on their mortgages? Explain.

15. Loans. Based on past experience, a bank believes that7% of the people who receive loans will not make pay-ments on time. The bank has recently approved 200 loans.a) What are the mean and standard deviation of the pro-

portion of clients in this group who may not maketimely payments?

b) What assumptions underlie your model? Are the con-ditions met? Explain.

c) What is the probability that over 10% of these clientswill not make timely payments?

16. Contacts. Assume that 30% of students at a universitywear contact lenses.a) We randomly pick 100 students. Let represent the

proportion of students in this sample who wear con-tacts. What is the appropriate model for the distribu-tion of ? Specify the name of the distribution, themean, and the standard deviation. Be sure to verifythat the conditions are met.

b) What is the approximate probability that more thanone third of this sample wear contacts?

17. Big families. According to Statistics Canada, 8.8% ofCanadian families have five or more members. If we takea random sample of 400 families, use the 68-95-99.7 Ruleto describe the sampling distribution model for the per-centage of those 400 families that will be of size five orbigger. Are the appropriate conditions met?

18. Binge drinking. A national study found that 44% ofuniversity students engage in binge drinking (five drinksat a sitting for men, four for women). Use the 68–95–99.7Rule to describe the sampling distribution model for theproportion of students in a randomly selected group of200 who engage in binge drinking. Do you think theappropriate conditions are met?

19. Big families, again. It was mentioned in Exercise 17that 8.8% of Canadian families have five or more mem-bers. Suppose we take a random sample of 625 rural farmfamilies and find that 16% of those families have five ormore members. Is this a very unusual result? Would youconclude that farm families differ from other Canadianfamilies? Why?

pN

pN

20. Binge sample. After hearing of the national result that44% of students engage in binge drinking, a professorsurveyed a random sample of 244 students at his univer-sity and found that 96 of them admitted to binge drinkingin the past week. Should he be surprised at this result?Explain.

21. Polling. Just before a Quebec referendum on sover-eignty, a local newspaper polls 400 voters in an attempt topredict whether it will succeed. Suppose that sovereigntyhas the support of 48% of the voters. What is the proba-bility that the newspaper’s sample will lead them to pre-dict victory? Be sure to verify that the assumptions andconditions necessary for your analysis are met.

22. Seeds. Information on a packet of seeds claims that thegermination rate is 92%. What is the probability thatmore than 95% of the 160 seeds in the packet will germi-nate? Be sure to discuss your assumptions and check theconditions that support your model.

23. Apples. When a truckload of apples arrives at a packingplant, a random sample of 150 is selected and examinedfor bruises, discolouration, and other defects. The wholetruckload will be rejected if more than 5% of the sample isunsatisfactory. Suppose that 8% of the apples on the truckdo not meet the desired standard. What is the probabilitythat the shipment will be accepted anyway?

24. Airline no-shows. About 12% of ticketed passengersdo not show up for their flights, so airlines like to over-book for flights. If an airplane has sold tickets to 400 peo-ple for the next flight, how many seats, at the least, wouldyou hope to be available on the plane in order to be verysure of having enough seats for all who show up?Comment on the assumptions and conditions that supportyour model, and explain what “very sure” means to you.

25. Nonsmokers. You are planning to open a new “PekingDuck” specialty restaurant in Beijing, and need to planthe seating arrangement. Unlike in Canada, you’ll need toplan for smokers as well as nonsmokers. While somenonsmokers do not mind being seated in the smokingsection of a restaurant, you estimate that about 60% of thecustomers will demand a smoke-free area. Your restau-rant will have 120 seats. How many seats should be in thenonsmoking area in order to be very sure of havingenough seating there? Comment on the assumptions andconditions that support your model, and explain what“very sure” means to you.

26. Meals. A restauranteur anticipates serving about 180people on a Friday evening, and believes that about 20%of the patrons will order the chef’s steak special. Howmany of those meals should he plan on serving in orderto be pretty sure of having enough steaks on hand tomeet customer demand? Justify your answer, includingan explanation of what “pretty sure” means to you.

27. Sampling. A sample is chosen randomly from a popu-lation that can be described by a Normal model.a) What is the sampling distribution model for the sam-

ple mean? Describe shape, centre, and spread.b) If we choose a larger sample, what is the effect on this

sampling distribution model?


Exercises 499

28. Sampling, part II. A sample is chosen randomly froma population that was strongly skewed to the left.a) Describe the sampling distribution model for the sam-

ple mean if the sample size is small.b) If we make the sample larger, what happens to the

sampling distribution model’s shape, centre, andspread?

c) As we make the sample larger, what happens to theexpected distribution of the data in the sample?

29. Waist size. A study measured the waist size of 250 men,finding a mean of 36.33 inches and a standard deviation of4.02 inches. Here is a histogram of these measurements

a) Describe the histogram of waist size.b) To explore how the mean might vary from sample

to sample, the researchers simulated by drawingmany samples of size 2, 5, 10, and 20, with replace-ment, from the 250 measurements. Here are his-tograms of the sample means for each simulation.Explain how these histograms demonstrate what theCentral Limit Theorem says about the sampling distri-bution model for sample means.

Waist Size (inches)30 40 50

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30. CEO compensation. In Chapter 4 we saw the distribu-tion of the total compensation of the chief executive offi-cers (CEOs) of the 800 largest U.S. companies (the Fortune800). The average compensation (in thousands of dollars)is 10 307.31 and the standard deviation is 17 964.62. Here isa histogram of their annual compensations:

a) Describe the histogram of total compensation. Aresearch organization simulated sample means bydrawing samples of 30, 50, 100, and 200, with replace-ment, from the 800 CEOs. The histograms show thedistributions of means for many samples of each size.

0Compensation in $1000

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b) Explain how these histograms demonstrate what theCentral Limit Theorem says about the sampling distri-bution model for sample means. Be sure to talk aboutshape, centre, and spread.

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c) Comment on the “rule of thumb” that “With a samplesize of at least 30, the sampling distribution of the meanis nearly Normal”24?

31. Waist size revisited. Researchers measured the waistsizes of 250 men in a study on body fat. The true meanand standard deviation of the waist sizes for the 250 menare 36.33 inches and 4.019 inches, respectively. In Exercise29, you looked at the histograms of simulations that drewsamples of sizes 2, 5, 10, and 20 (with replacement). Thesummary statistics for these simulations were as follows:

a) According to the Central Limit Theorem, what shouldthe theoretical mean and standard deviation be foreach of these sample sizes?

b) How close are the theoretical values to what wasobserved in the simulation?

c) Looking at the histograms in Exercise 29, at whatsample size would you be comfortable using theNormal model as an approximation for the samplingdistribution?

d) What about the shape of the distribution of waist sizeexplains your choice of sample size in part c)?

32. CEOs revisited. In Exercise 30 you looked at the annualcompensation for 800 CEOs, for which the true mean andstandard deviation were (in thousands of dollars) 10 307.31and 17 964.62, respectively. A simulation drew samples ofsizes 30, 50, 100, and 200 (with replacement) from the totalannual compensations of the Fortune 800 CEOs. The sum-mary statistics for these simulations were as follows:

a) According to the Central Limit Theorem, what shouldthe theoretical mean and standard deviation be foreach of these sample sizes?

b) How close are the theoretical values to what wasobserved from the simulation?

c) Looking at the histograms in Exercise 30, at whatsample size would you be comfortable using theNormal model as an approximation for the samplingdistribution?

d) What about the shape of the distribution of totalcompensation explains your answer in part c)?

33. GPAs. A university’s data about incoming high-schoolgraduates indicates that the mean of their high schoolaverages was 73, with a standard deviation of 7; the dis-tribution was roughly mound-shaped and only slightlyskewed. The students are randomly assigned to first-yearwriting seminars in groups of 25. What might the meanhigh school average of one of these seminar groups be?Describe the appropriate sampling distribution model—shape, centre, and spread—with attention to assumptionsand conditions. Make a sketch using the 68–95–99.7 Rule.

34. Home values. Assessment records indicate that thevalue of homes in a small city is skewed right, with amean of $280 000 and standard deviation of $90 000. Tocheck the accuracy of the assessment data, officials planto conduct a detailed appraisal of 100 homes selected atrandom. Using the 68–95–99.7 Rule, draw and label anappropriate sampling model for the mean value of thehomes selected.

35. The trial of the pyx. In 1150, it was recognized inEngland that coins should have a standard weight ofprecious metal as the basis for their value. A guinea,for example, was supposed to contain 128 grains ofgold. (There are 12.7 grains in a gram.) In the “trialof the pyx,” coins minted under contract to theCrown were weighed and compared to standard coins(which were kept in a wooden box called the pyx).Coins were allowed to deviate by no more than 0.28grains—roughly equivalent to specifying that the stan-dard deviation should be no greater than 0.09 grains(although they didn’t know what a standard deviationwas in 1150). In fact, the trial was performed by weigh-ing 100 coins at a time and requiring the sum to deviateby no more than 100 × 28 = 28 or 28 grains—equivalentto the sum having a standard deviation of about ninegrains.a) In effect, the trial of the pyx required that the mean

weight of the sample of 100 coins have a standarddeviation of 0.09 grains. Explain what was wrong withperforming the trial in this manner.

b) What should the limit have been on the standard devi-ation of the mean?

Note: Because of this error, the Crown was exposed tobeing cheated by private mints that could mint coins withgreater variation and then, after their coins passed thetrial, select out the heaviest ones and recast them at theproper weight, retaining the excess gold for themselves.The error persisted for over 600 years, until sampling dis-tributions became better understood.

36. Safe cities. Allstate Insurance Company identified the 10 safest and 10 least-safe cities from among the 200 largest cities in the United States, based on the meannumber of years drivers went between automobileaccidents. The cities on both lists were all smaller thanthe 10 largest cities. Using facts about the samplingdistribution model of the mean, explain why this is notsurprising.24 Actually, this rule of thumb is generally qualified with the condi-

tion that the population not be severely skewed or have extremeoutliers.

n mean st. dev.

2 36.314 2.8555 36.314 1.805

10 36.341 1.27620 36.339 0.895

n mean st. dev.

30 10 251.73 3359.6450 10 343.93 2483.84

100 10 329.94 1779.18200 10 340.37 1230.79


Exercises 501

Exercises 37–54 require the use of Normal tables or technology.

37. Pregnancy. Assume that the duration of human preg-nancies can be described by a Normal model with mean266 days and standard deviation 16 days.a) What percentage of pregnancies should last between

270 and 280 days?b) At least how many days should the longest 25% of all

pregnancies last?c) Suppose a certain obstetrician is currently providing

prenatal care to 60 pregnant women. Let representthe mean length of their pregnancies. According to theCentral Limit Theorem, what is the distribution of thissample mean, ? Specify the model, mean, and stan-dard deviation.

d) What is the probability that the mean duration ofthese patients’ pregnancies will be less than 260 days?

38. Calgary in January. The average January temperaturein Calgary varies from year to year. These averageJanuary temperatures themselves have an average, overmany past years, of –7.9 degrees Celsius, with a standarddeviation of 5.0 degrees. Assume a Normal model applies.a) During what percentage of years does Calgary have a

January average above 0 degrees?b) In the 20% coldest Januaries, the average January tem-

perature reaches no higher than how many degrees?c) Suppose you will be in Calgary for the next four

years. Describe the sampling distribution model forthe average of the next four January average tempera-tures. Assume climatic conditions are not changing.

d) What is the probability that the average of the nextfour January average temperatures will exceed �4degrees? If this happens, would you consider this evi-dence of climatic warming (at least locally, inCalgary), or just Normal variation?

39. Pregnant again. The duration of human pregnanciesmay not actually follow a Normal model, as described inExercise 37.a) Explain why it may be somewhat skewed to the left.b) If the correct model is in fact skewed, does that change

your answers to parts a), b), and c) of Exercise 37?Explain why or why not for each.

40. At work. Some business analysts estimate that thelength of time people work at a job has a mean of 6.2 yearsand a standard deviation of 4.5 years.a) Explain why you suspect this distribution may be

skewed to the right.b) Explain why you could estimate the probability that

100 people selected at random had worked for theiremployers an average of 10 years or more, but youcould not estimate the probability that an individualhad done so.

41. Dice and dollars. You roll a die, winning nothing if thenumber of spots is odd, $1 for a two or a four, and $10 fora six.a) Find the expected value and standard deviation of

your prospective winnings.b) You play twice. Find the mean and standard deviation

of your total winnings.c) You play 40 times. What is the probability that you

win at least $100?

y

y

42. New game. You pay $10 and roll a die. If you get asix, you win $50. If not, you get to roll again. If youget a six this time, you get your $10 back.a) Create a probability model for this game.b) Find the expected value and standard deviation of

your prospective winnings.c) You play this game five times. Find the expected value

and standard deviation of your average winnings.d) 100 people play this game. What is the probability that

the person running the game makes a profit?

43. Canadian families. Statistics Canada reported thefollowing distribution for family size in Canada.

a) Find the mean and standard deviation of family size(treat “6 or more” as “6”).

b) If we select a random sample of 40 families, wouldyou expect their sizes to follow a Normal model?Explain.

c) Consider the mean sizes of random samples of 40families. Describe the sampling model for these means(shape, centre, and spread).

44. Museum membership. A museum offers several levelsof membership, as shown in the table.

a) Find the mean and standard deviation of the dona-tions.

b) During the museum’s annual membership drive, ithopes to sign up 50 new members each day. Wouldyou expect the distribution of the donations for a dayto follow a Normal model? Explain.

c) Consider the mean donation of the 50 new memberseach day. Describe the sampling model for thesemeans (shape, centre, and spread).

45. Canadian families, again. In a certain rural area, thereare 36 farm families whose average size is 3.0. What is theprobability that the average family size will be at least 3.0 ifwe were to draw a random sample of 36 families from theentire Canadian population, as described in exercise 43?Does this calculation tell you anything about farm families?

46. Joining the museum. One of the museum’s phonevolunteers sets a personal goal of getting an average

Size Percent of Families

6 or more 2.85 6.04 15.03 15.92 33.61 26.8

Member Amount of Percent ofCategory Donation ($) Members

Individual 50 41Family 100 37Sponsor 250 14Patron 500 7Benefactor 1000 1



donation of at least $100 from the new members sheenrols during the membership drive. If she gets 80 newmembers and they can be considered a random sample ofall the museum’s members, what is the probability thatshe can achieve her goal?

47. Pollution. Carbon monoxide (CO) emissions for a cer-tain kind of truck vary with mean 2.9 L/km and standarddeviation 0.4 L/km. A company has 80 of these trucks inits fleet. Let represent the mean CO level for the com-pany’s fleet.a) What is the approximate model for the distribution of

? Explain.b) Estimate the probability that is between 3.0 and

3.1 L/km.c) There is only a 5% chance that the fleet’s mean CO

level is greater than what value?

48. Potato chips. The weight of potato chips in a medium-size bag is stated to be 300 g. The amount that thepackaging machine puts in these bags is believed tohave a Normal model with mean 306 g and standard deviation 3.6 g.a) What fraction of all bags sold are underweight?b) Some of the chips are sold in “bargain packs” of three

bags. What is the probability that none of the three isunderweight?

c) What is the probability that the mean weight of thethree bags is below the stated amount?

d) What is the probability that the mean weight per bagof a 24-bag case of potato chips is below 300 g?

49. Tips. A waiter believes the distribution of his tips has amodel that is slightly skewed to the right, with a mean of$9.60 and a standard deviation of $5.40.a) Explain why you cannot determine the probability

that a given party will tip him at least $20.b) Can you estimate the probability that the next four

parties will tip an average of at least $15? Explain.c) Is it likely that his 10 parties today will tip an average

of at least $15? Explain.

50. Groceries. Grocery store receipts show that customerpurchases have a skewed distribution with a mean of $32and a standard deviation of $20.a) Explain why you cannot determine the probability

that the next customer will spend at least $40.b) Can you estimate the probability that the next 10 cus-

tomers will spend an average of at least $40? Explain.

c) Is it likely that the next 50 customers will spend anaverage of at least $40? Explain.

51. More tips. The waiter in Exercise 49 usually waits onabout 40 parties over a weekend of work.a) Estimate the probability that he will earn at least

$500 in tips.b) How much does he earn on the best 10% of such

weekends?

52. More groceries. Suppose the store in Exercise 50 had312 customers today.a) Estimate the probability that the store’s revenues were

at least $10 000.

yy

y

b) If, in a typical day, the store serves 312 customers,how much does the store take in on the worst 10% ofsuch days?

53. Big stats course. STA220H is a large introductoryStatistics course at the University of Toronto, with 720students enrolled. On the term test, the average score was72, with a standard deviation of 16. There are 24 tutorials,with 30 students in each. Xiu Xiu teaches one, Sabriteaches another. Assume that tutorial groups are formedquite randomly.a) We select a student at random from Sabri’s tutorial.

Find the probability that this student’s test score is atleast 80. Assume that scores are Normally distributed.

b) We select one student at random from each tutorial.Find the probability that Sabri’s student outperformedXiu Xiu’s student by at least seven marks on the test.(Hint: Work with the difference between the twoscores.) Assume that scores are Normally distributed.

c) What is the probability that the average for Sabri’stutorial is at least 80? Do you need to assume thatscores are Normally distributed?

d) What is the probability that the average test score inSabri’s tutorial is at least seven marks higher than theaverage test score for Xiu Xiu’s tutorial? (Hint: Workwith the difference between the two tutorial averages.)Do you need to assume that scores are Normallydistributed?

54. Milk. Although most of us buy milk by the litre, farm-ers measure daily production in kilograms or pounds.Ayrshire cows average 47 pounds of milk a day, with astandard deviation of six pounds. For Jersey cows, themean daily production is 43 pounds, with a standarddeviation of 5 pounds. Assume that Normal modelsdescribe milk production for these breeds.a) We select an Ayrshire at random. What is the probabil-

ity that she averages more than 50 pounds of milk aday?

b) What is the probability that a randomly selectedAyrshire gives more milk than a randomly selectedJersey? (Hint: Work with the difference in milk pro-duction between the two cows.)

c) A farmer has 20 Jerseys. What is the probability thatthe average production for this small herd exceeds45 pounds of milk a day?

d) A neighbouring farmer has 10 Ayrshires. What is theprobability that his herd average is at least fivepounds higher than the average for the Jersey herd?(Hint: Work with the difference of the two means.)

For the exercises below, go to www.onlinestatbook.com/stat_sim/sampling_dist/index.html where there is a Javaapplet illustrating sampling distributions. Click on the Beginbutton. According to the questions below, select first the desiredpopulation shape, at the top. Next, where you see Mean andN=5, change these according to the question (click on the smallarrows). Below you can also change None and N=5 if the ques-tion asks you to compare two different statistics or two differentsample sizes. After these are set, you are ready to sample andlearn. Always start by choosing Sample: Animated, whichproduces a single sample and calculates the chosen statistic.Then click on Sample: 1,000 several times to add more samples,

18_deve_ch18.qxd 2/3/11 6:04 PM Page 502

Exercises 503

until the picture of the sampling distribution stabilizes nicely.Note that measures of centre and spread appear to the side of alldistributions.

55. Mean or median? Choose the (default) Normal popu-lation at the top. What is the centre of this population?Choose Mean and N==25, and below choose Median andN==25. Draw samples and examine the two sampling dis-tributions. Which of these sampling distributions are cen-tred at the population centre (unbiased)? Which statisticwould you prefer to use as an estimator of the true popu-lation centre—if unknown to you—for this type ofpopulation? Explain.

56. Effect of sample size. Choose the (default) Normalpopulation. Choose Mean at both lower levels, butchoose N==5 at one level and N==25 at the other. Predict themean and standard deviation of each of the two samplingdistributions you generate. Now generate the samplingdistributions and confirm your predictions.

57. Central Limit Theorem. Choose the skewed popula-tion at the top, and N==5 and N==25 at the two levelsbelow, along with the Mean in both cases. Predict theshapes, means, and standard deviations of the two sam-pling distributions you generate. Generate the samplingdistributions, and compare with your predictions.

58. Central Limit Theorem again. Use your computermouse to create an odd-looking population at the top.Choose Mean and N==25 at the first level below. Generatethe sampling distribution. What phenomenon does itsshape reflect? Try again to create another odd populationthat you think might lead to a very different samplingdistribution below. Did you succeed?

59. Central Limit Theorem, finis. Compare the samplingdistribution of the sample mean when sampling from theuniform population versus sampling from the skewedpopulation (both provided at the top). Start with a smallsample size and gradually increase it, first for the uni-form population, then for the skewed population. Abouthow big does the sample size have to be to produce anearly Normal sampling distribution for the mean, foreach type of population? Why the difference?

60. Another sampling distribution. Use the Normalpopulation at the top. Every statistic has its own sam-pling distribution. Let’s examine the sampling distribu-tion of the sample variance. Choose N==5 and Varianceat one of the levels below. The sample variance calcu-lated uses the denominator n rather than the n � 1 weare accustomed to using. Is the centre (mean) of thesampling distribution of this sample variance equal tothe actual population variance (printed out at the topnext to the population)? If so, we say that this statistic isan unbiased estimator of the population variance. Trychanging the statistic to Var (U). This statistic uses n � 1as its denominator. Which version of the sample vari-ance do you prefer? Explain. (We’ll learn more aboutthis denominator, which we call the degrees of freedom, inChapter 23.)

JUST CHECKINGAnswers

1. A Normal model (approximately).

2. At the actual proportion of all students who are infavour.

3.

4. No, this is a histogram of individuals. It may or maynot be Normal, but we can’t tell from the informationprovided.

5. A Normal model (approximately).

6. 266 days.

7. days.16

2100= 1.6

SD1Np2 = A10.5210.52

100= 0.05

Go to MathXL at www.mathxl.com or MyStatLab at www.mystatlab.com. You can practiseexercises for this chapter as often as you want. The guided solutions will help you find answers step by step. You’ll find a personalized study plan available to you too!


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