Chapter 3

Report
Chapter
8
Sampling
Distributions
© 2010 Pearson Prentice©Hall.
rights reserved
2010AllPearson
Prentice Hall. All rights reserved
8-1
Section 8.1
Distribution of the Sample Mean
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
8-2
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 .

© 2010 Pearson Prentice Hall. All rights reserved
8-3
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 .

© 2010 Pearson Prentice Hall. All rights reserved
8-4
Illustrating Sampling Distributions
Step 1: Obtain a simple random
sample of size n.
© 2010 Pearson Prentice Hall. All rights reserved
8-5
Illustrating Sampling Distributions
Step 1: Obtain a simple random
sample of size n.
Step 2: Compute the sample
mean.
© 2010 Pearson Prentice Hall. All rights reserved
8-6
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-7
Objective 1
• Describe the Distribution of the Sample MeanSamples from Normal Populations
© 2010 Pearson Prentice Hall. All rights reserved
8-8
Illustrate concept using the sampling
distribution applet by simulating sampling from
a normal population.
© 2010 Pearson Prentice Hall. All rights reserved
8-9
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-10
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
2.471
Note: x =2.479 for this sample
© 2010 Pearson Prentice Hall. All rights reserved
8-11
Sample Means for Samples of Size n=5
© 2010 Pearson Prentice Hall. All rights reserved
8-12
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-13
© 2010 Pearson Prentice Hall. All rights reserved
8-14
What role does n, the sample size, play in the
standard deviation of the distribution of the
sample mean?
© 2010 Pearson Prentice Hall. All rights reserved
8-15
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-16
Parallel
ParallelExample
Example2:2:The
TheImpact
ImpactofofSample
Sample
Size on Sampling
VariabilityVariability
Size on Sampling
• 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)
© 2010 Pearson Prentice Hall. All rights reserved
8-17
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-18
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 .
© 2010 Pearson Prentice Hall. All rights reserved
8-19
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.

© 2010 Pearson Prentice Hall. All rights reserved
8-20
IQ scores are normally distributed with mean 100 and
standard deviation 15. Suppose we obtain a simple
random sample of size n = 20 from this population.
Therefore, the distribution of the sample mean is also
normally distributed. The mean of the sampling
distribution of x will be
 x    100
and the standard deviation of the sample mean will be
x 

n

15
20
© 2010 Pearson Prentice Hall. All rights reserved
8-21
15
15
20
100
© 2010 Pearson Prentice Hall. All rights reserved
8-22
Parallel
theof Sample
ParallelExample
Example3:2:Describing
The Impact
Distribution
of on
theSampling
Sample Mean
Size
Variability
IQ scores are approximately normally distributed with
mean 100 and standard deviation 15.
What is the probability that in a simple random sample of
10 people, we obtain a sample mean of at least 105?
© 2010 Pearson Prentice Hall. All rights reserved
8-23
• The mean of the sampling distribution of the
sample mean
a) equals the median of the population
b) equals the mean of the population
c) equals 0
d) cannot be determined
e) Not sure
© 2010 Pearson Prentice Hall. All rights reserved
8-24
• If X is a normal random variable, the
sampling distribution of the sample mean
a) has a shape that is skewed right
b) has a shape that is skewed left
c) has a shape that is normal
d) has a shape that cannot be determined
e) Not sure
© 2010 Pearson Prentice Hall. All rights reserved
8-25
Suppose a random variable X is normally
distributed with mean 100 and standard
deviation 15. What is the probability that a
random sample of 20 individuals from this
population results in a sample mean of at least
105?
© 2010 Pearson Prentice Hall. All rights reserved
8-26
Objective 2
• Describe the Distribution of the Sample MeanSamples from a Population That is Not Normal
© 2010 Pearson Prentice Hall. All rights reserved
8-27
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
© 2010 Pearson Prentice Hall. All rights reserved
8-28
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-29
© 2010 Pearson Prentice Hall. All rights reserved
8-30
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 n regardless of the
sample size.
• The Central Limit Theorem: the shape of the
 sample mean becomes
distribution of the
approximately normal as the sample size n
increases, regardless of the shape of the
population.
© 2010 Pearson Prentice Hall. All rights reserved
8-31
© 2010 Pearson Prentice Hall. All rights reserved
8-32
Parallel Example 5: Using the Central Limit Theorem
Suppose that the mean time for an oil change at a “10minute 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?
© 2010 Pearson Prentice Hall. All rights reserved
8-33
Parallel Example 5: Using the Central Limit Theorem
Suppose that the mean time for an oil change at a “10minute 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?

Solution:
Z 
11  11 .4
0 .5409
  0 .74 ,
P(Z<-0.74)=0.23.
© 2010 Pearson Prentice Hall. All rights reserved
8-34
• If a random variable X has a skewed right
distribution, then the shape of the distribution
of the sample mean for a sample of size n =
50 for X is
a) Approximately normal
b) Very skewed right
c) Somewhat skewed left
d) Uniform
e) Not sure
© 2010 Pearson Prentice Hall. All rights reserved
8-35
• If a random variable X has a standard
deviation σ = 20, then the standard error
of the mean for a sample of size n = 100 is
a) 2
b) 5
c) 20
d) 100
e) Not sure
© 2010 Pearson Prentice Hall. All rights reserved
8-36
Suppose X is a random variable whose distribution is
known to be skewed right with mean 50 and
standard deviation 10. What is the probability that a
random sample of size n = 40 results in a sample
mean less than 48?
© 2010 Pearson Prentice Hall. All rights reserved
8-37
Section 8.2
Distribution of the Sample Proportion
Objectives
1. Describe the sampling distribution of a
sample proportion
2. Compute probabilities of a sample proportion
© 2010 Pearson Prentice Hall. All rights reserved
8-38
Objective 1
• Describe the Sampling Distribution of a
Sample Proportion
© 2010 Pearson Prentice Hall. All rights reserved
8-39
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, pˆ
denoted (read “p-hat”) is given by
pˆ 
x
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-40
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-41
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
1745
 0 .2
© 2010 Pearson Prentice Hall. All rights reserved
8-42
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-43
© 2010 Pearson Prentice Hall. All rights reserved
8-44
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 sample 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.
© 2010 Pearson Prentice Hall. All rights reserved
8-45
Sampling Distribution of pˆ
For a simple random sample of size n with
population proportion p:

• The shape of the sampling distribution of pˆ is
approximately normal provided np(1-p)≥10.
• The mean of the sampling distribution of pˆ is
 pˆ  p .

• The standard deviation of the sampling
distribution
of pˆ is
 pˆ 

p(1  p)

n

© 2010 Pearson Prentice Hall. All rights reserved
8-46
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
mean of the sampling distribution of pˆ is p, and
the standard deviation is
 pˆ 
p(1  p)
n

© 2010 Pearson Prentice Hall. All rights reserved
8-47
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-48
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.)

© 2010 Pearson Prentice Hall. All rights reserved
8-49
• The shape of the sampling distribution of p
hat is
a) Approximately normal provided np(1 – p) > 10.
b) Approximately normal regardless of the sample
size.
c) Impossible to determine unless we know p.
d) Not sure
© 2010 Pearson Prentice Hall. All rights reserved
8-50
• True or False: The mean of the sampling
distribution of p hat is p.
© 2010 Pearson Prentice Hall. All rights reserved
8-51
Objective 2
• Compute Probabilities of a Sample Proportion
© 2010 Pearson Prentice Hall. All rights reserved
8-52
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 611 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?
© 2010 Pearson Prentice Hall. All rights reserved
8-53
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
 is the probability that at least 19%
6-11 years, what
are overweight?
Z 
0.19  0.188
0.0412
 0.0485 ,
P(Z>0.05)=1-0.5199=0.4801
© 2010 Pearson Prentice Hall. All rights reserved
8-54

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?
pˆ 
24
90
 0.2667 ,
Z 
0.2667  0.188
0.0412
 1.91
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.
© 2010 Pearson Prentice Hall. All rights reserved
8-55
According to the Centers for Disease Control and
Prevention, 18.8% of school-aged children, aged 6-11
years, were overweight in 2004.
What is the probability that a random sample of n = 100
children in an impoverished school district results in 26 or
more who are overweight?
© 2010 Pearson Prentice Hall. All rights reserved
8-56

similar documents