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Objectives
1. Describe the distribution of the sample mean:
samples from normal populations
2. Describe the distribution of the sample mean:
samples from a population that is not normal
© 2010 Pearson Prentice Hall. All rights reserved
8-1
Statistics such as x are random variables
since their value varies from sample to
sample. As such, they have probability
distributions associated with them. In this
 we focus on the shape, center and
chapter
spread of statistics such as x .

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8-2
The sampling distribution of a statistic is a
probability distribution for all possible values
of the statistic computed from a sample of size
n.
The sampling distribution of the sample mean
x is the probability distribution of all possible
values of the random variable x computed
from a sample of size n from a population with
mean  and standard deviation .

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8-3
Illustrating Sampling Distributions
Step 1: Obtain a simple random sample of size n.
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8-4
Illustrating Sampling Distributions
Step 1: Obtain a simple random sample of size n.
Step 2: Compute the sample mean.
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8-5
Illustrating Sampling Distributions
Step 1: Obtain a simple random sample of size n.
Step 2: Compute the sample mean.
Step 3: Assuming we are sampling from a finite
population, repeat Steps 1 and 2 until all
simple random samples of size n have
been obtained.
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8-6
Objective 1
• Describe the Distribution of the Sample MeanSamples from Normal Populations
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8-7
Parallel Example 1: Sampling Distribution of the Sample
Mean-Normal Population
The weights of pennies minted after 1982 are
approximately normally distributed with mean
2.46 grams and standard deviation 0.02 grams.
Approximate the sampling distribution of the
sample mean by obtaining 200 simple random
samples of size n = 5 from this population.
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8-8
The data on the following slide represent the
sample means for the 200 simple random samples
of size n = 5.
For example, the first sample of n = 5 had the
following data:
2.493
2.466 2.473 2.492
Note:
x=2.479 for this sample
2.471
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8-9
Sample Means for Samples of Size n =5
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8-10
The mean of the 200 sample means is 2.46, the
same as the mean of the population.
The standard deviation of the sample means is
0.0086, which is smaller than the standard
deviation of the population.
The next slide shows the histogram of the
sample means.
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8-11
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8-12
What role does n, the sample size, play in the
standard deviation of the distribution of the
sample mean?
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8-13
What role does n, the sample size, play in the
standard deviation of the distribution of the
sample mean?
As the size of the sample gets larger, we do not
expect as much spread in the sample means
since larger observations will offset smaller
observations.
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8-14
Parallel Example 2: The Impact of Sample Size on Sampling
Variability
• Approximate the sampling distribution of the
sample mean by obtaining 200 simple random
samples of size n = 20 from the population of
weights of pennies minted after 1982 (=2.46
grams and =0.02 grams)
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8-15
The mean of the 200 sample means for n =20 is still 2.46, but the
standard deviation is now 0.0045 (0.0086 for n = 5).
As expected, there is less variability in the distribution
of the sample mean with n =20 than with n =5.
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8-16
The Mean and Standard Deviation of the
Sampling Distribution of x
Suppose that a simple random sample of size n is
drawn from a large population with mean  and
standard deviation . The sampling
 distribution of
x willhave mean x   and standard deviation
.
x 
n
The standard deviation of the sampling distribution of
x is called
the standard error of the mean and is

denoted  x .
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8-17
The Shape of the Sampling
Distribution of x If X is Normal
If a random variable X is normally distributed,
the distribution of the sample mean x is

normally distributed.

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8-18
Parallel Example 3: Describing the Distribution of the
Sample Mean
The weights of pennies minted after 1982 are
approximately normally distributed with mean
2.46 grams and standard deviation 0.02 grams.
What is the probability that in a simple random
sample of 10 pennies minted after 1982, we
obtain a sample mean of at least 2.465 grams?
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8-19
Solution
•
x
0.02
is normally distributed with x =2.46 and  x 
 0.0063 .
10
2.465  2.46
 0.79 .
• Z
0.0063
• P(Z>0.79)=1-0.7852


=0.2148.
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8-20
Objective 2
• Describe the Distribution of the Sample MeanSamples from a Population That is Not Normal
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8-21
Parallel Example 4: Sampling from a Population that is Not
Normal
The following table and histogram
give the probability distribution
for rolling a fair die:
Face on Die
Relative Frequency
1
0.1667
2
0.1667
3
0.1667
4
0.1667
5
0.1667
6
0.1667
=3.5, =1.708
Note that the population distribution is NOT normal
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8-22
Estimate the sampling distribution of x by obtaining
200 simple random samples of size n=4 and
calculating the sample mean for each of the 200
samples. Repeat for n = 10 and 30.
distribution of the sample
Histograms of the sampling
mean for each sample size are given on the next slide.
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8-23
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8-24
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8-25
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8-26
Key Points from Example 4
• The mean of the sampling distribution is equal to
the mean of the parent population and the
standard deviation of the sampling distribution of

the sample mean is
regardless of the sample
n
size.
• The Central Limit Theorem: the shape of the
distribution 
of the sample mean becomes
approximately normal as the sample size n
increases, regardless of the shape of the
population.
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8-27
Parallel Example 5: Using the Central Limit Theorem
Suppose that the mean time for an oil change at a “10-minute
oil change joint” is 11.4 minutes with a standard deviation
of 3.2 minutes.
(a) If a random sample of n = 35 oil changes is selected,
describe the sampling distribution of the sample mean.
(b) If a random sample of n = 35 oil changes is selected, what
is the probability the mean oil change time is less than 11
minutes?
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8-28
Parallel Example 5: Using the Central Limit Theorem
Suppose that the mean time for an oil change at a “10-minute
oil change joint” is 11.4 minutes with a standard deviation
of 3.2 minutes.
(a) If a random sample of n = 35 oil changes is selected,
describe the sampling distribution of the sample mean.
Solution: x is approximately normally distributed
with mean=11.4 and std. dev. = 3.2  0.5409.
35
(b) If a random sample of n = 35 oil changes is selected, what
is the probability the mean oil change time is less than 11

minutes?
1111.4
 0.74 , P(Z<-0.74)=0.23.
Solution: Z 
0.5409
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8-29
Section
8.2
Distribution of the
Sample Proportion
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Objectives
1. Describe the sampling distribution of a
sample proportion
2. Compute probabilities of a sample proportion
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8-31
Objective 1
• Describe the Sampling Distribution of a
Sample Proportion
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8-32
Point Estimate of a Population
Proportion
Suppose that a random sample of size n is obtained
from a population in which each individual either
does or does not have a certain characteristic. The
sample proportion, denoted (read “p-hat”) is
given by pˆ
x
pˆ 
n
where x is the number of individuals in the sample
with the specified characteristic. The sample

proportion is a statistic that estimates the

population proportion, p.
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8-33
Parallel Example 1: Computing a Sample Proportion
In a Quinnipiac University Poll conducted in May
of 2008, 1,745 registered voters nationwide were
asked whether they approved of the way George
W. Bush is handling the economy. 349 responded
“yes”. Obtain a point estimate for the proportion
of registered voters who approve of the way
George W. Bush is handling the economy.
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8-34
Parallel Example 1: Computing a Sample Proportion
In a Quinnipiac University Poll conducted in May
of 2008, 1,745 registered voters nationwide were
asked whether they approved of the way George
W. Bush is handling the economy. 349 responded
“yes”. Obtain a point estimate for the proportion
of registered voters who approve of the way
George W. Bush is handling the economy.
Solution:
pˆ 
349
 0.2
1745
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8-35
Parallel Example 2: Using Simulation to Describe the
Distribution of the Sample Proportion
According to a Time poll conducted in June of
2008, 42% of registered voters believed that
gay and lesbian couples should be allowed to
marry.
Describe the sampling distribution of the sample
proportion for samples of size n=10, 50, 100.
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8-36
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8-37
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8-38
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8-39
Key Points from Example 2
• Shape: As the size of the sample, n, increases,
the shape of the sampling distribution of the
sample proportion becomes approximately
normal.
• Center: The mean of the sampling distribution
of the sample proportion equals the population
proportion, p.
• Spread: The standard deviation of the
sampling distribution of the sample proportion
decreases as the sample size, n, increases.
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8-40
Sampling Distribution of pˆ
For a simple random sample of size n with
population proportion p:

ˆ is
• The shape of the sampling distribution
of p
approximately normal provided np(1-p)≥10.
ˆ is
• The mean of the sampling distribution of p
 pˆ  p.

• The standard deviation of the sampling
distribution
ˆ is
of p
p(1 p)

 pˆ 
n


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8-41
Sampling Distribution of pˆ
• The model on the previous slide requires that the
sampled values are independent. When sampling
from finite populations, this assumption is verified

by checking that the sample size n is no more than
5% of the population size N (n ≤ 0.05N).
• Regardless of whether np(1-p) ≥10 or not, the
ˆ is p, and
mean of the sampling distribution of p
the standard deviation is
 pˆ 
p(1 p)
n

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8-42
Parallel Example 3: Describing the Sampling Distribution of
the Sample Proportion
According to a Time poll conducted in June of
2008, 42% of registered voters believed that
gay and lesbian couples should be allowed to
marry. Suppose that we obtain a simple
random sample of 50 voters and determine
which believe that gay and lesbian couples
should be allowed to marry. Describe the
sampling distribution of the sample proportion
for registered voters who believe that gay and
lesbian couples should be allowed to marry.
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8-43
Solution
The sample of n=50 is smaller than 5% of the
population size (all registered voters in the
U.S.).
Also, np(1-p)=50(0.42)(0.58)=12.18≥10.
The sampling distribution of the sample
proportion is therefore approximately normal
with mean=0.42 and standard deviation=
0.42(1 0.42)
 0.0698.
50
(Note: this is very close to the standard deviation of 0.072
found using simulation in Example 2.)

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8-44
Objective 2
• Compute Probabilities of a Sample Proportion
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8-45
Parallel Example 4: Compute Probabilities of a Sample
Proportion
According to the Centers for Disease Control and
Prevention, 18.8% of school-aged children, aged
6-11 years, were overweight in 2004.
(a) In a random sample of 90 school-aged children, aged
6-11 years, what is the probability that at least 19%
are overweight?
(b) Suppose a random sample of 90 school-aged
children, aged 6-11 years, results in 24 overweight
children. What might you conclude?
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8-46
Solution
•
•
•
n=90 is less than 5% of the population size
np(1-p)=90(.188)(1-.188)≈13.7≥10
pˆ is approximately normal with mean=0.188 and
standard deviation = (0.188)(1 0.188)  0.0412
90

(a) In a random sample of 90 school-aged children, aged
6-11 years, what
 is the probability that at least 19%
are overweight?
0.19  0.188
Z
 0.0485 , P(Z>0.05)=1-0.5199=0.4801
0.0412
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8-47


Solution
• pˆ is approximately normal with mean=0.188 and
standard deviation = 0.0412
(b) Suppose a random sample of 90 school-aged
children, aged 6-11 years, results in 24 overweight
children. What might you conclude?
24
0.2667  0.188
pˆ 
 0.2667,
Z
1.91
90
0.0412
P(Z>1.91)=1-0.9719=0.028.
We would only expect to see about 3 samples in 100

resulting in a sample
proportion of 0.2667 or more.
This is an unusual sample if the true population
proportion is 0.188.
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8-48
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