Sampling Distribution Proportion
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Transcript 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.
Example
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
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
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
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
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
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
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
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).
The function will report the number of
freshmen in the sample.
Example
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
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
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
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
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
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^
It turns out that the sampling
distribution of p^ has
Mean p.
Variance p(1 – p)/n.
Standard deviation (p(1 – p)/n).