Powerpoint sampling distribution

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
SAMPLING DISTRIBUTIONS
8-1

SAMPLING DISTRIBUTION OF THE SAMPLE
MEANS: OBJECTIVES
1. Describe the distribution of the sample mean:
normal population
2. Describe the distribution of the sample mean:
nonnormal population
8-2

DISTRIBUTION OF COINS
Histogram of
coins collected
over 6 months.
8-3

SIMULATION
Use the coin distribution to find samples and take
the mean of each sample.
Now take all the possible samples (through
simulation) and put the sample means ( ) in a
distribution.
This is called
The Sampling Distribution of the Mean
8-4
x

SIMULATION
Top :
Distribution of Coins
Middle:
One Sample
Bottom:
Distribution of sample
means
8-5

Statistics such as are random variables since
their value varies from sample to sample.
As such, they have probability distributions
associated with them.
8-6
x

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
is the probability distribution of all possible values of the
random variable computed from a sample of size n from a
population with mean μ and standard deviation σ.
8-7
x x

8-8
Illustrating Sampling Distributions
Step 1: Obtain a simple random sample of size n.

8-9
Step 2: Compute the sample mean.

8-10
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.

OBJECTIVE 1
 Describe the Distribution of the Sample Mean:
Normal Population
8-11

What role does n, the sample size, play in the standard deviation
of the distribution of the sample mean?
8-12

What role does n, the sample size, play in the
standard deviation of the distribution of the
sample mean?
8-13
As the size of the sample increases, the
standard deviation of the distribution of
the sample mean decreases.

8-14
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
will have mean and standard deviation
The standard deviation of the sampling distribution
of is called the standard error of the mean and
is denoted .
The Mean and Standard Deviation of the
Sampling Distribution of
x
x
σx
σx =
σ
n
.
µx = µ
x

If a random variable X is normally distributed, the distribution of
the sample mean is normally distributed.
8-15
The Shape of the Sampling
Distribution of If X is Normalx
x

OBJECTIVE 2
 Describe the Distribution of the Sample Mean: Nonnormal
Population
8-16

 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 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. 8-17
Key Points
σ
n

8-18
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?

8-19
of 3.2 minutes.
Solution: is approximately normally distributed
with mean = 11.4 and std. dev. = .3.2
35
= 0.5409
x

8-20
of 3.2 minutes.
(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?
Solution: is approximately normally distributed
with mean = 11.4 and std. dev. = .
Solution: Using StatCrunch with above, P = 0.23.
3.2
35
= 0.5409
x

SAMPLING DISTRIBUTION OF THE SAMPLE
PROPORTION: OBJECTIVES
1. Describe the sampling distribution of a sample proportion
2. Compute probabilities of a sample proportion
8-21

OBJECTIVE 1
 Describe the Sampling Distribution of a Sample Proportion
8-22

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
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.
8-23
ˆp ˆp =
x
n
ˆp

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.
8-24
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.
8-25
Parallel Example 1: Computing a Sample Proportion
Solution: ˆp =
349
1745
= 0.2

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.
Note: We are using simulations to create the
histograms on the following slides.
8-26
Parallel Example 2: Using Simulation to Describe the
Distribution of the Sample Proportion

 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.
8-27
Key Points

For a simple random sample of size n with population proportion
p:
• The shape of the sampling distribution of is approximately
normal provided np(1 – p) ≥ 10.
• The mean of the sampling distribution of is
• The standard deviation of the sampling distribution of is
8-28
ˆp
ˆp µˆp =p
ˆp
σˆp =
p(1−p)
n
ˆp

• 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 mean of the
sampling distribution of is p, and the standard deviation is
8-29
ˆp
σˆp =
p(1−p)
n
ˆp

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. 8-30
Parallel Example 3: Describing the Sampling Distribution of
the Sample Proportion

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 =
(Note: this is very close to the standard deviation of 0.072
found using simulation in Example 2.)
8-31
Solution
0.42(1− 0.42)
50
= 0.0698

OBJECTIVE 2
 Compute Probabilities of a Sample Proportion
8-32

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?
8-33
Parallel Example 4: Compute Probabilities of a Sample
Proportion

• n = 90 is less than 5% of the population size
• np(1 – p) = 90(.188)(1 – .188) ≈ 13.7 ≥ 10
• is approximately normal with mean=0.188
and standard deviation =
(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) Use StatCrunch with mean = .188, standard
deviation as calculated to be 0.0412, then the
probability is 0.0485
8-34
Solution
ˆp
(0.188)(1−0.188)
90
=0.0412

• 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?
8-35
Solution
ˆp
Using StatCrunch the probability = 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.
ˆp =
24
90
= 0.2667

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