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Transcript
Chapter 18
Sampling Distribution
Models
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Normal Model

When we talk about one data value and the
Normal model we used the notation:
N(μ, σ)
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 3
Scenario




Suppose everyone in AP Stats at Geneva High
School took a survey which entailed the question
“Do you believe in ghosts?”
Are the answers received going to be categorical
or quantitative?
Would everyone get the same proportion of
students that said “yes” and “no”?
Why or why not?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 4
Categorical Data

Deals with sample proportions
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 5
The Central Limit Theorem for Sample
Proportions




Rather than showing real repeated samples,
imagine what would happen if we were to actually
draw many samples.
Now imagine what would happen if we looked at
the sample proportions for these samples.
The histogram we’d get if we could see all the
proportions from all possible samples is called
the sampling distribution of the proportions.
What would the histogram of all the sample
proportions look like?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 6
Modeling the Distribution of
Sample Proportions (cont.)




We would expect the histogram of the sample
proportions to center at the true proportion, p, in
the population.
As far as the shape of the histogram goes, we
can simulate a bunch of random samples that we
didn’t really draw.
It turns out that the histogram is unimodal,
symmetric, and centered at p.
More specifically, it’s an amazing and fortunate
fact that a Normal model is just the right one for
the histogram of sample proportions.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 7
Modeling the Distribution of
Sample Proportions (cont.)



Modeling how sample proportions vary from
sample to sample is one of the most powerful
ideas we’ll see in this course.
A sampling distribution model for how a
sample proportion varies from sample to sample
allows us to quantify that variation and how likely
it is that we’d observe a sample proportion in any
particular interval.
To use a Normal model, we need to specify its
mean and standard deviation. We’ll put µ, the
mean of the Normal, at p.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 8
Modeling the Distribution of
Sample Proportions (cont.)

When working with proportions, knowing the
mean automatically gives us the standard
deviation as well—the standard deviation we will
use is
pq
n

So, the distribution of the sample proportions is
modeled with a probability model that is

pq 
N  p,

n 

Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 9
Modeling the Distribution of
Sample Proportions (cont.)

A picture of what we just discussed is as follows:
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 10
The Central Limit Theorem for Sample
Proportions (cont)

Because we have a Normal model, for example, we know
that 95% of Normally distributed values are within two
standard deviations of the mean.

So we should not be surprised if 95% of various polls gave
results that were near the mean but varied above and below
that by no more than two standard deviations.

This is what we mean by sampling error. It’s not really an
error at all, but just variability you’d expect to see from one
sample to another. A better term would be sampling
variability.
 A reasonable variance/error is ±2σ
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 11
How Good Is the Normal Model?


The Normal model gets better as a good model
for the distribution of sample proportions as the
sample size gets bigger.
Just how big of a sample do we need? This will
soon be revealed…
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 12
Assumptions and Conditions


Most models are useful only when specific
assumptions are true.
There are two assumptions in the case of the
model for the distribution of sample proportions:
1. The Independence Assumption: The sampled
values must be independent of each other.
2. The Sample Size Assumption: The sample
size, n, must be large enough.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 13
Assumptions and Conditions (cont.)



Assumptions are hard—often impossible—to
check. That’s why we assume them.
Still, we need to check whether the assumptions
are reasonable by checking conditions that
provide information about the assumptions.
The corresponding conditions to check before
using the Normal to model the distribution of
sample proportions are the Randomization
Condition, the 10% Condition and the
Success/Failure Condition.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 14
Assumptions and Conditions (cont.)
1. Randomization Condition: The sample should
be a simple random sample of the population.
2. 10% Condition: the sample size, n, must be no
larger than 10% of the population.
3. Success/Failure Condition: The sample size
has to be big enough so that both np (number of
successes) and nq (number of failures) are at
least 10.
…So, we need a large enough sample that is not
too large.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 15
A Sampling Distribution Model
for a Proportion


A proportion is no longer just a computation from
a set of data.
 It is now a random variable quantity that has a
probability distribution.
 This distribution is called the sampling
distribution model for proportions.
Even though we depend on sampling distribution
models, we never actually get to see them.
 We never actually take repeated samples from
the same population and make a histogram. We
only imagine or simulate them.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 16
A Sampling Distribution Model
for a Proportion (cont.)

Still, sampling distribution models are important
because
 they act as a bridge from the real world of data
to the imaginary world of the statistic and
 enable us to say something about the
population when all we have is data from the
real world.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 17
The Sampling Distribution Model
for a Proportion (cont.)
 Provided that the sampled values are
independent and the sample size is large
enough, the sampling distribution of p̂
(sample proportion of success) is modeled
by a Normal model with
Mean:  ( p̂)  p
 Standard deviation: SD( p̂) 

Copyright © 2010, 2007, 2004 Pearson Education, Inc.
pq
n
Slide 18 - 18
Example p. 434 #15

Based on past experience, a bank
believes that 7% of the people
who receive loans will not make
payments on time The bank has
recently approved 200 loans.
 a) What are the mean and
standard deviation of the
proportion of clients in this
group who may not make timely
payments?
 b) What assumptions underlie
you model? Are the conditions
met? Explain.
 c) What’s the probability that
over 10% of these clients will
not make timely payments?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 19
Ch 18 Homework

p. 432 #3, 5, 7, 9, 16, 22
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 20
What About Quantitative Data?




Proportions summarize categorical variables.
The Normal sampling distribution model looks like
it will be very useful.
Can we do something similar with quantitative
data?
We can indeed. Even more remarkable, not only
can we use all of the same concepts, but almost
the same model.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 21
Quantitative Data

Deals with sample means
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 22
Simulating the Sampling Distribution of a Mean


Like any statistic computed from a random
sample, a sample mean also has a sampling
distribution.
We can use simulation to get a sense as to what
the sampling distribution of the sample mean
might look like…
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 23
Means – The “Average” of One Die

Let’s start with a simulation of 10,000 tosses of a
die. A histogram of the results is:
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 24
Means – Averaging More Dice

Looking at the average of
two dice after a simulation
of 10,000 tosses:
Copyright © 2010, 2007, 2004 Pearson Education, Inc.

The average of three dice
after a simulation of
10,000 tosses looks like:
Slide 18 - 25
Means – Averaging Still More Dice

The average of 5 dice
after a simulation of
10,000 tosses looks like:
Copyright © 2010, 2007, 2004 Pearson Education, Inc.

The average of 20 dice
after a simulation of
10,000 tosses looks like:
Slide 18 - 26
Means – What the Simulations Show


As the sample size (number of dice) gets larger,
each sample average is more likely to be closer
to the population mean.
 So, we see the shape continuing to tighten
around 3.5
And, it probably does not shock you that the
sampling distribution of a mean becomes Normal.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 27
The Fundamental Theorem of Statistics


The sampling distribution of any mean becomes
more nearly Normal as the sample size grows.
 All we need is for the observations to be
independent and collected with randomization.
 We don’t even care about the shape of the
population distribution!
The Fundamental Theorem of Statistics is called
the Central Limit Theorem (CLT).
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 28
The Fundamental Theorem of Statistics (cont.)


The CLT is surprising and a bit weird:
 Not only does the histogram of the sample
means get closer and closer to the Normal
model as the sample size grows, but this is
true regardless of the shape of the population
distribution.
 The more skewed, the larger the sample size
we need.
The CLT works better (and faster) the closer the
population model is to a Normal itself. It also
works better for larger samples.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 29
Assumptions and Conditions

The CLT requires essentially the same
assumptions we saw for modeling proportions:
 Independence Assumption: The sampled
values must be independent of each other.
 Sample Size Assumption: The sample size
must be sufficiently large.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 31
Assumptions and Conditions (cont.)

We can’t check these directly, but we can think about
whether the Independence Assumption is plausible.
We can also check some related conditions:

Randomization Condition: The data values must
be sampled randomly.

10% Condition: When the sample is drawn without
replacement, the sample size, n, should be no
more than 10% of the population.

Large Enough Sample Condition: The CLT doesn’t
tell us how large a sample we need. For now, you
need to think about your sample size in the context
of what you know about the population.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 32
But Which Normal?



The CLT says that the sampling distribution of
any mean or proportion is approximately Normal.
But which Normal model?
 For proportions, the sampling distribution is
centered at the population proportion.
 For means, it’s centered at the population
mean.
But what about the standard deviations?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 33
But Which Normal? (cont.)

The Normal model for the sampling distribution of
the mean has a standard deviation equal to
SD y  

n
where σ is the population standard deviation.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 34
Central Limit Theorem

So the normal model dealing with sample means
is written like…
N ( ,
Copyright © 2010, 2007, 2004 Pearson Education, Inc.

n
)
Slide 18 - 36
About Variation



The standard deviation of the sampling
distribution declines only with the square root of
the sample size (the denominator contains the
square root of n).
Therefore, the variability decreases as the sample
size increases.
While we’d always like a larger sample, the
square root limits how much we can make a
sample tell about the population. (This is an
example of the Law of Diminishing Returns.)
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 37
Example p. 437 #49

A waiter believes the distribution of his tips has a
model 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 4 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.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 38

k
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 39
Homework

Chapter 18 Homework: p. 432 #31, 33, 37, 50,
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 40
The Real World and the Model World
Be careful! Now we have two distributions to deal
with.


The first is the real world distribution of the sample,
which we might display with a histogram.
The second is the math world sampling distribution
of the statistic, which we model with a Normal
model based on the Central Limit Theorem.
Don’t confuse the two!
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 41
Sampling Distribution Models


Always remember that the statistic itself is a
random quantity.
 We can’t know what our statistic will be
because it comes from a random sample.
Fortunately, for the mean and proportion, the CLT
tells us that we can model their sampling
distribution directly with a Normal model.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 42
Sampling Distribution Models (cont.)

There are two basic truths about sampling
distributions:
1. Sampling distributions arise because
samples vary. Each random sample will have
different cases and, so, a different value of
the statistic.
2. Although we can always simulate a sampling
distribution, the Central Limit Theorem saves
us the trouble for means and proportions.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 43
What Can Go Wrong?

Don’t confuse the sampling distribution with the
distribution of the sample.
 When you take a sample, you look at the
distribution of the values, usually with a
histogram, and you may calculate summary
statistics.
 The sampling distribution is an imaginary
collection of the values that a statistic might
have taken for all random samples—the one
you got and the ones you didn’t get.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 45
What Can Go Wrong? (cont.)


Beware of observations that are not independent.
 The CLT depends crucially on the assumption
of independence.
 You can’t check this with your data—you have
to think about how the data were gathered.
Watch out for small samples from skewed
populations.
 The more skewed the distribution, the larger
the sample size we need for the CLT to work.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 46
What have we learned?


Sample proportions and means will vary from
sample to sample—that’s sampling error
(sampling variability).
Sampling variability may be unavoidable, but it is
also predictable!
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 47
What have we learned? (cont.)


We’ve learned to describe the behavior of sample
proportions when our sample is random and large
enough to expect at least 10 successes and
failures.
We’ve also learned to describe the behavior of
sample means (thanks to the CLT!) when our
sample is random (and larger if our data come
from a population that’s not roughly unimodal and
symmetric).
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 48
Homework

Chapter 18 Homework: p. 432 #2(1), 4(3), 6(5),
8(7), 10(9), 16, 22, 30(29), 34, 38, 50, 52
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 49
Example p. 432 #1

See book
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 50
Example p. 432 #3

The philanthropic organization in Exercise 1 expects
about a 5% success rate when they send fundraising
letters to the people on their mailing list. In Exercise
1 you looked at the histograms showing distributions
of sample proportions from 1000 simulated mailings
for samples of size 20, 50, 100, and 200. The sample
statistics from each simulation were as follows:
 a) According to the CLT,
n
Mean
St. Dev
0.0497
0.0479
what should the theoretical 20
50
0.0516
0.0309
mean and standard
100
0.0497
0.0215
deviation be for these
200
0.0501
0.0152
sample sizes?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 51
Example p. 432 #3

The philanthropic organization in Exercise 1
expects about a 5% success rate when they send
fundraising letters to the people on their mailing
list. In Exercise 1 you looked at the histograms
showing distributions of sample proportions from
1000 simulated mailings for samples of size 20,
50, 100, and 200. The sample statistics from
each simulation were as follows:
n
Mean
St. Dev
 b) How close are those
0.0497
0.0479
theoretical values to what 20
50
0.0516
0.0309
was observed in these
100
0.0497
0.0215
simulations?
200
0.0501
0.0152
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 52
Example p. 432 #3

The philanthropic organization in Exercise 1 expects
about a 5% success rate when they send fundraising
letters to the people on their mailing list. In Exercise
1 you looked at the histograms showing distributions
of sample proportions from 1000 simulated mailings
for samples of size 20, 50, 100, and 200. The sample
statistics from each simulation were as follows:
 c) Looking at the histograms
n
Mean
St. Dev
in Exercise 1, at what sample 20
0.0497
0.0479
size would you be comfortable 50
0.0516
0.0309
100
0.0497
0.0215
using the Normal model
200
0.0501
0.0152
as an approximation for the
sampling distribution?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 53
Example p. 432 #3

The philanthropic organization in Exercise 1
expects about a 5% success rate when they send
fundraising letters to the people on their mailing
list. In Exercise 1 you looked at the histograms
showing distributions of sample proportions from
1000 simulated mailings for samples of size 20,
50, 100, and 200. The sample statistics from
each simulation were as follows:
n
Mean
St. Dev
 d) What does the
0.0497
0.0479
Success/Failure Condition 20
0.0516
0.0309
say about the choice you 50
100
0.0497
0.0215
made in part c?
200
0.0501
0.0152
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 54
Example p. 433 #5

In a large class of introductory Statistics students,
the professor has each person toss a coin 16
times and calculate the proportion of his or her
tosses that were heads. The students then report
their results, and the professor plots a histogram
of these several proportions.
 d) Explain why a Normal model should not be
used here.
 a) What shape would you expect this histogram
to be? Why?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 55
Example p. 433 #5

In a large class of introductory Statistics students,
the professor has each person toss a coin 16
times and calculate the proportion of his or her
tosses that were heads. The students then report
their results, and the professor plots a histogram
of these several proportions.
 b) Where do you expect the histogram to be
centered?
 c) How much variability would you expect
among these proportions?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 56
Example p. 433 #7

Suppose the class in Exercise 5 repeats the cointossing experiment.
 a) The students toss the coins 25 times each. Use
the 68-95-99.7 Rule to describe the sampling
distribution.
 b) Confirm that you can use the Normal model
here.
 c) They increase the number of tosses to 64 each.
Draw and label the appropriate sampling distribution
model. Check the appropriate conditions to justify
your model.
 d) Explain how the sampling distribution model
changes as the number of tosses increases.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 57
Example p. 433 #9

One of the students in the introductory Statistics
class in Exercise 7 claims to have tossed her coin
200 times and found only 42% heads. What do
you think of this claim? Explain.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 58
Example p. 434 #21

Just before a referendum on a school budget, a
local newspaper polls 400 random voters in an
attempt to predict whether the budget will pass.
Suppose that the budget actually has the support
of 52% of the voters. What’s the probability the
newspaper’s sample will lead them to predict
defeat (less than 50%)? Be sure to verify that the
assumptions and conditions necessary for our
analysis are met.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 59
Example p. 434 #29

See book
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 60
Example p. 432 #31

Researchers measured the Waist Sizes of 250 men in
a study on body fat. The true mean and standard
deviation of the Waist Sizes for the 250 men are
36.33 in and 4.019 in, respectively. In Exercise 29
you looked at the histograms of simulations that drew
samples of sizes 2, 5, 10, and 20 (with replacement).
The summary statistics for these simulations were as
follows:
n
Mean
St. Dev
 a) According to the CLT,
2
36.314
2.855
36.314
1.805
what should the theoretical 5
10
36.341
1.276
mean and standard
20
36.339
0.895
deviation be for each of
these sample sizes?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 61
Example p. 432 #31

Researchers measured the Waist Sizes of 250
men in a study on body fat. The true mean and
standard deviation of the Waist Sizes for the 250
men are 36.33 in and 4.019 in, respectively. In
Exercise 29 you looked at the histograms of
simulations that drew samples of sizes 2, 5, 10,
and 20 (with replacement). The summary
statistics for these simulations were as follows:
n
Mean
St. Dev
 b) How close are those
2
36.314
2.855
theoretical values to what 5
36.314
1.805
10
36.341
1.276
was observed in these
20
36.339
0.895
simulations?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 62
Example p. 432 #31

Researchers measured the Waist Sizes of 250 men in
a study on body fat. The true mean and standard
deviation of the Waist Sizes for the 250 men are
36.33 in and 4.019 in, respectively. In Exercise 29
you looked at the histograms of simulations that drew
samples of sizes 2, 5, 10, and 20 (with replacement).
The summary statistics for these simulations were as
follows:
n
Mean
St. Dev
 c) Looking at the histograms
2
36.314
2.855
in Exercise 29, at what
5
36.314
1.805
sample size would you be
36.341
1.276
comfortable using the Normal 10
20
36.339
0.895
model as an approximation
for the sampling distribution?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 63
Example p. 432 #31

Researchers measured the Waist Sizes of 250
men in a study on body fat. The true mean and
standard deviation of the Waist Sizes for the 250
men are 36.33 in and 4.019 in, respectively. In
Exercise 29 you looked at the histograms of
simulations that drew samples of sizes 2, 5, 10,
and 20 (with replacement). The summary
statistics for these simulations were as follows:
Mean
St. Dev
 d) What about the shape of n
36.314
2.855
the distribution of Waist Size 2
5
36.314
1.805
explains your choice of
10
36.341
1.276
sample size in part c?
20
36.339
0.895
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 64
Example p. 436 #33

A college’s data about the incoming freshman
indicates that the mean of their high school GPAs
was 3.4, with a standard deviation of 0.35; the
distribution was roughly mound-shaped and only
slightly skewed. The students are randomly
assigned to freshman writing seminars in groups
of 25. What might the mean GPA of one of these
seminar groups be? Describe the appropriate
sampling distribution model—shape, center,
spread—with attention to assumptions and
conditions. Make a sketch using the 68-95-99.7
Rule.
Slide 18 - 65
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Example p. 437 #49

A waiter believes the distribution of his tips has a
model 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 4 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.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 66

k
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 67
Example p. 438 #51

A waiter believes the distribution of his tips has a
model that is slightly skewed to the right, with a
mean of $9.60 and a standard deviation of $5.40.
The waiter usually waits on about 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?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Slide 18 - 68
Homework

Chapter 18 Homework: p. 432 #2(1), 4(3), 6(5),
8(7), 10(9), 16, 22, 30(29), 32(31), 34, 38, 50(49),
52(51)
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Practice 1

“Groovy” M&M’s are supposed to make up 30% of
the candies sold. In a large bag of 250 M&M’s,
what is the probability that we get at least 25%
groovy candies?
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
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Practice 2

At birth, babies average 7.8 pounds, with a
standard deviation of 2.1 pounds. A random
sample of 34 babies born to mothers living near a
large factory that may be polluting the air and
water shows a mean birth weight of only 7.2
pounds. Is that unusually low? Explain.
Copyright © 2010, 2007, 2004 Pearson Education, Inc.
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