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Chapter
8
Sampling
Distributions
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Section
8.1
Distribution of the
Sample Mean
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Describe the distribution of the sample mean:
normal population
2. Describe the distribution of the sample mean:
nonnormal population
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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 chapter we focus
on the
shape, center and spread of statistics such
as x .
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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 μ
and standard deviation σ.
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• Let’s consider a population consisting of
{4, 5, 9} - ages of someone’s children
If two ages are randomly selected with
replacement from the population, identify the
sampling distribution of the sample mean.
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Sample Distribution of Mean
x
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Sample
Sample Mean
Probability
4,4
4.0
1/9
4,5
4.5
1/9
4,9
6.5
1/9
5,4
4.5
1/9
5,5
5.0
1/9
5,9
7.0
1/9
9,4
6.5
1/9
9,5
7.0
1/9
9,9
9
1/9
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Sampling Distribution of
Mean(condensed)
x
Sample Mean
Probability
4.0
1/9
4.5
2/9
5.0
1/9
6.5
2/9
7.0
2/9
9.0
1/9
Using this table, let’s calculate the mean of the
sample mean:
4(1/9)+4.5(2/9)+5(1/9)+…+9(1/9)=6.0
The mean of the population {4,5,9} is also 6.0!
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Because the mean of the sample means is equal
to the mean of the population, we conclude that
the values of the sample mean do target the
value of the population mean.
The mean of the sample mean is equal to the
population mean µ
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Illustrating Sampling Distributions
Step 1: Obtain a simple random sample of size n.
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Illustrating Sampling Distributions
Step 1: Obtain a simple random sample of size n.
Step 2: Compute the sample mean.
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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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Objective 1
• Describe the Distribution of the Sample Mean:
Normal Population
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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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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
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2.471
Sample Means for Samples of Size n = 5
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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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What role does n, the sample size, play in the
standard deviation of the distribution of the
sample mean?
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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 increases, the standard
deviation of the distribution of the sample mean
decreases.
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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 (μ = 2.46
grams and σ = 0.02 grams)
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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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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 μ and
standard deviation σ. 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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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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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?
This is not an individual value!
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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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Objective 2
• Describe the Distribution of the Sample Mean:
Nonnormal Population
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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
μ = 3.5, σ = 1.708
Note that the population distribution is NOT normal
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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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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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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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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
3.2
with mean = 11.4 and std. dev. =
 0.5409 .
35
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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
3.2
with mean = 11.4 and std. dev. =
 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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Designing Elevators
When designing elevators, an obviously important
consideration is the weight capacity. An Ohio college
studetn died when he tried to escape from a dormitory
elevator that was overloaded with 24 passengers. The
elevator was rated for a capacity of 16 passengers with a
total weight of 2500 lb. Weight of adults are changing over

time(table below). WE assume
a worst-case scenario in
which all of the passengers are males. If an elevator is
loaded to a capacity of 2500 lb with 16 males, the mean
weight of a passenger is 156.25 lb.
µ

Males
Females
182.9 lb
165.0 lb
40.8 lb
45.6 lb
The distribution is Normal for both males and females.
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(a) Find the probablity that 1 randomly selected adult male has
a weight greater than 156.25
(b) Find the probability that a sample of 16 randomly selected
adult males has a maximum capacity of 156.25 lb
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(a) Approach used for an individual Value z  0.65
(b) Sample has a mean:
40.8
 x  10.2  x 
 10.2
16
z  2.61
• Now use Table V, to find the probabilities…
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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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Objective 1
• Describe the Sampling Distribution of a
Sample Proportion
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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 pˆ (read “p-hat”) is
given by
x
pˆ 
n
where x is the number
of
individuals
in
the
sample

with the specified characteristic. The sample
proportion 
pˆ is a statistic that estimates the
population proportion, p.
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Parallel Example 1: Computing a Sample Proportion
In a Quinnipiac University Poll conducted in May
of 2008, 1745 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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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.
349
 0.2
Solution: p̂ 
1745
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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.
( p  0.42)
Note: We are using simulations to create the histograms on the
following slides.
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• Note: The mean of the sample proportions
equals the population proportion
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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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Sampling Distribution of p̂
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

8-52
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Sampling Distribution of p̂
• 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 
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p(1 p)
n

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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 voters 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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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)
50
 0.0698
(Note: this is very close to the standard deviation of 0.072 found
using simulation in Example 2.)
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Objective 2
• Compute Probabilities of a Sample Proportion
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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 611 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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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)
90

 0.0412
(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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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
p̂ 
 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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