Download Sampling Distribution Proportion

Sampling Distribution of
a Sample Proportion
Lecture 37
Section 8.2
Tue, Mar 30, 2004
Parameters and Statistics

The purpose of a statistic is to
estimate a population parameter.
A sample mean is used to estimate the
population mean.
 A sample proportion is used to estimate
the population proportion.
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Example
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See Example 8.1, p. 464.
The Census Bureau surveys 3000
employees and asks them, “Have the
job skills demanded by your job
increased over the past few years?”
57% replied, “Yes.”
That is a sample proportion.
What is the population proportion?
Some Questions
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What if the survey were repeated?
Would the survey results again be
57%?
Would the sample proportion be close
to 57%?
Might it be 99%?
Might it be 1%?
Some Questions
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We hope that the sample proportion is
close to the population proportion.
How close can we expect it to be?
Would it be worth it to collect a larger
sample?
If the sample were larger, would we
expect it to be closer?
 How much closer?

The Sampling Distribution
of a Statistic

Sampling Distribution of a Statistic –
The distribution of values of the
statistic over all possible samples of
the size n from that population.
The Sample Proportion
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Let p be the population proportion.
Then p is a fixed value (for a given
population).
Let p^ (“p-hat”) be the sample
proportion.
Then p^ is a random variable; it takes
on a new value every time a sample is
collected.
Example
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Suppose that this class is 1/3
freshmen.
Suppose that we take a sample of 2
students, selected with replacement.
Find the sampling distribution of p^.
Example
1/3
1/3
F
P(FF) = 1/9
N
P(FN) = 2/9
F
P(NF) = 2/9
N
P(NN) = 4/9
2/3
2/3
1/3
N
F
2/3
Example

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Let X = number of freshmen in the
sample.
The probability distribution of X is
x
0
1
2
P(X = x)
1/9
4/9
4/9
Example
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Let p^ = proportion of freshmen in
the sample.
The sampling distribution of p^ is
x
0
1/2
1
P(p^ = x)
1/9
4/9
4/9
Simulating Sampling with
the TI-83

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Use the TI-83 to simulate sampling 2
people (with replacement) from a
population in which 1/3 are freshmen.
Use the function randBin(n, p).
n = sample size (n = 2).
 p = proportion of freshmen (p = 1/3).
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The function will report the number of
freshmen in the sample.
Example
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For example, randBin(30, 1/3) = 6.
This represents a sample proportion of
6 out of 30, or 6/30 = 0.20.
If we press ENTER several more times,
we get 11, 10, 11, 9, and 13.
These represent sample proportions of
11/30, 10/30, 11/30, 9/30, and 13/30.
Example
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The expression
randBin(n, p, k)
will take k samples of size n and put
the results in a list.
For example, randBin(30, 1/3, 100)
produces the list
{13, 11, 11, 9, 10, 11, 10, 8, 9, …}.
The Histogram
15
10
5
0.1
0.2
0.3
0.4
0.5
0.6
p^
Larger Sample Size
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Now we will select samples of size 120
instead of size 30.
randBin(120, 1/3, 100) produces
{38, 47, 33, 49, 34, 47, 41, 37, …}
The Histogram
25
20
15
10
5
0.1
0.2
0.3
0.4
0.5
0.6
p^
Observations and
Conclusions
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Observation: The values of p^ are
clustered around p.
Conclusion: p^ is probably close to p.
Observation: As the sample size
increases, the clustering is tighter.
Conclusion: Larger samples give more
reliable estimates.
One More Observation
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Observation: The distribution of p^
appears to be approximately normal.
The Histogram
15
10
5
0.1
0.2
0.3
0.4
0.5
0.6
p^
The Histogram
15
10
5
0.1
0.2
0.3
0.4
0.5
0.6
p^
One More Conclusion
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Conclusion: We can use the normal
distribution to calculate how close we
can expect p^ to be.
However, we must know  and  for
the distribution of p^.
The Sampling Distribution
of p^
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It turns out that the sampling
distribution of p^ has
Mean p.
 Variance p(1 – p)/n.
 Standard deviation (p(1 – p)/n).
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