Transcript Lecture 28 - Sampling Distribution Mean

Sampling Distribution
of a Sample Mean
Lecture 28
Section 8.4
Mon, Mar 20, 2006
Sampling Distribution of the
Sample Mean

Sampling Distribution of the Sample Mean–
The distribution of sample means over all
possible samples of the size n from that
population.
With or Without Replacement?
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
If the sample size is small in relation to the
population size (< 5%), then it does not matter
whether we sample with or without replacement.
The calculations are simpler if we sample with
replacement.
In any case, we are not going to worry about it.
Example
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Suppose a population consists of the numbers
{1, 2, 3}.
Using samples of size n = 1, 2, or 3, find the
sampling distribution ofx.
Draw a tree diagram showing all possibilities.
The Tree Diagram

n=1
1
mean = 1
2
mean = 2
3
mean = 3
The Sampling Distribution
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The sampling distribution ofx is
x
1
2
P(x)
1/3
1/3
3
1/3
The parameters are
=2
 2 = 2/3 = 0.6667

The Tree Diagram
mean
1
2
3
1
1
2
1.5
3
2
1
1.5
2
2
3
2.5
1
2
2
2.5
3
3
The Sampling Distribution
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The sampling distribution ofx is
x
P( x)
1
1/9
1.5
2/9
2
3/9
2.5
2/9
3
1/9
The parameters are
=2
 2 = 2/6 = 0.3333

The Tree Diagram
1
1
2
3
1
2
2
3
1
3
2
3
1
2
3
1
4/3
5/3
1
2
3
1
2
3
1
2
3
1
2
3
4/3
5/3
2
5/3
2
7/3
4/3
5/3
2
1
2
3
1
2
3
1
2
3
2
7/3
8/3
5/3
2
7/3
2
7/3
8/3
1
2
3
7/3
8/3
3
5/3
2
7/3
The Sampling Distribution
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The sampling distribution ofx is
The parameters are
=2
 2 = 2/9 = 0.2222

x
P(x)
1
1/27
1.333
3/27
1.667
6/27
2
7/27
2.333
6/27
2.667
3/27
3
1/27
Sampling Distributions
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
Run the program
Central Limit Theorem for Means.exe.
Use n = 30 and population = {1, 2, 3}; generate
100 samples.
100 Samples of Size n = 30
 = 0.75
 = 0.079
Observations and Conclusions
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Observation #1: The values ofx are clustered
around .
Conclusion #1:x is probably close to .
Larger Sample Size
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Now we will select 100 samples of size 120
instead of size 30.
Run the program
Central Limit Theorem for Means.exe.
Pay attention to the spread (standard deviation)
of the distribution.
100 Samples of Size n = 120
 = 0.75
 = 0.0395
Observations and Conclusions
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Observation #2: As the sample size increases,
the clustering is tighter.
Conclusion #2A: Larger samples give more
reliable estimates.
Conclusion #2B: For sample sizes that are large
enough, we can make very good estimates of
the value of .
Larger Sample Size
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Now we will select 10000 samples of size 120
instead of only 100 samples.
Run the program
Central Limit Theorem for Means.exe.
Pay attention to the shape of the distribution.
10,000 Samples of Size n = 120
 = 0.75
 = 0.0395
10,000 Samples of Size n = 120
More Observations and Conclusions
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Observation #3: The distribution ofx appears
to be approximately normal.
One More Conclusion
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Conclusion #3: We can use the normal
distribution to calculate just how close to  we
can expectx to be.
However, we must know the values of  and 
for the distribution ofx.
That is, we have to quantify the sampling
distribution ofx.
The Central Limit Theorem
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Begin with a population that has mean  and
standard deviation .
For sample size n, the sampling distribution of
the sample mean is approximately normal with
Mean of x  
Variance of x 
2
n
Standard deviation of x 

n
The Central Limit Theorem
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The approximation gets better and better as the
sample size gets larger and larger.
That is, the sampling distribution “morphs”
from the distribution of the original population
to the normal distribution.
For many populations, the distribution is almost
exactly normal when n  10.
For almost all populations, if n  30, then the
distribution is almost exactly normal.
The Central Limit Theorem
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Therefore, if the original population is exactly
normal, then the sampling distribution of the
sample mean is exactly normal for any sample
size.
This is all summarized on pages 536 – 537.
Example
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Based on past data, a student’s average score on
individual homework problems is 7.3 points out
of 10, with a standard deviation of about 2.9
points.
This counts only those homework problems that
were attempted.
Example
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Over the course of the semester, I sample
(grade) approximately 100 homework problems
per student.
Thus, n = 100.
What is the sampling distribution of the
homework average (for the typical student)?
Example
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Based on the Central Limit Theorem for Means,
the sampling distribution
Is normally distributed (n  30)
 Has a mean of 7.3 points.
 Has a standard deviation of 2.9/100 = 0.29 points.
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On a 100-point scale, that would be
An average of 73.
 A standard deviation 2.9.
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Example
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What is the probability that a student’s
homework average (100 sampled problems) will
be within 5 points of his true average for all
problems?
Example
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For the typical student, that would be a
homework average between 68 and 78.
normalcdf(68, 78, 73, 2.9) = 0.9153, or about
92%
Point of Fact
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Since the sample size (n = 100) is a sizable
fraction of the population size (N = 400) and we
are sampling without replacement, we should
take into account the “finite population
correction factor” of (N – n)/(N – 1) for the
variance ofx.
x 
N n  


N 1  n 
Point of Fact
For n = 100 and N = 400, this factor is 0.8671.
 Thus, in fact, the typical standard deviation is
only about 0.251, or 2.51 points out of 100.
 Recompute:
normalcdf(68, 78, 73, 2.51) = 0.9536, or 95.36%.
