Sampling Distribution of a Sample Proportion
Lecture 26
Section 8.4
Robb T. Koether
Hampden-Sydney College
Mon, Mar 1, 2010
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Outline
1
The Central Limit Theorem for Proportions
2
Applications
3
Assignment
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Outline
1
The Central Limit Theorem for Proportions
2
Applications
3
Assignment
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The Central Limit Theorem for Proportions
Theorem (The Central Limit Theorem for Proportions)
For any population, the sampling distribution of p̂ has the following
mean and standard deviation:
µp̂ = p
r
σp̂ =
p(1 − p)
.
n
Furthermore, the sampling distribution of p̂ is approximately
normal, provided n is large enough.
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The Central Limit Theorem for Proportions
The Sample Size
n is large enough if
np ≥ 5 and n(1 − p) ≥ 5.
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Outline
1
The Central Limit Theorem for Proportions
2
Applications
3
Assignment
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Applications
Suppose that 60% of all high-school students own a cell phone.
If we survey 150 high-school students, how likely is it that we will
find that at least 65% of them own a cell phone?
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Applications
If p = 0.60 and our sample size is n = 150, then p̂ is normal with
mean µp̂ = 0.60 and
r
σp̂ =
(0.60)(0.40) √
= 0.0016 = 0.04.
150
Want to know the probability that p̂ ≥ 0.65.
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Applications
10
8
6
4
2
0.55
0.60
0.65
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0.70
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Applications
10
8
6
4
2
0.55
0.60
0.65
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0.70
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Applications
The probability that p̂ is greater than 0.65 is
normalcdf(.65,E99,.60,.04) = 0.1056.
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Applications
What if our sample size were 600 instead of 150?
Then p̂ is normal with mean µp̂ = 0.60 and
r
σp̂ =
(0.60)(0.40) √
= 0.0004 = 0.02.
600
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Applications
20
15
10
5
0.55
0.60
0.65
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0.70
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Applications
20
15
10
5
0.55
0.60
0.65
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0.70
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Applications
The probability that p̂ is greater than 0.65 is
normalcdf(.65,E99,.60,.02) = 0.0062.
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Applications
Suppose that the true proportion of Americans who strongly
disapprove of President Obama’s performance is 40%.
Can we use a sample of 1500 Americans to disprove the
hypothesis that the rate is 45%?
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Applications
Which is the correct question to ask?
I
I
Given that we observed p̂ = 0.40 in a sample, what is the
probability that p = 0.45?
If p really is 0.45, what is the probability that we would observe
p̂ = 0.40 (or more extreme)?
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Applications
Which is the correct question to ask?
I
I
Given that we observed p̂ = 0.40 in a sample, what is the
probability that p = 0.45?
If p really is 0.45, what is the probability that we would observe
p̂ = 0.40 (or more extreme)?
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Hypothesis Testing
Let us test the hypotheses
H0 : 45% of all Americans strongly disapprove of President
Obama’s performance.
H1 : Less than 45% of all Americans strongly disapprove of
President Obama’s performance.
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Hypothesis Testing
Using the Central Limit Theorem, the null hypothesis predicts that
the distribution of p̂ is
I
I
I
Normal, with
Mean p, which is 0.45
q (hypothetically).
Standard deviation
(0.45)(0.55)
1500
= 0.0128.
That is, p̂ is N(0.45, 0.0128).
Find the probability that p̂ ≤ 0.40.
Design a decision rule so that α = 0.05.
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Homework Problem
Exercise 8.12, page 528.
Suppose that 60% of all students at a large university access course
information using the Internet.
(a) Sketch a picture of the distribution for the possible sample
proportions you could get based on a simple random sample of
100 students.
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Homework Problem
Exercise 8.12, page 528.
Suppose that 60% of all students at a large university access course
information using the Internet.
(a) Sketch a picture of the distribution for the possible sample
proportions you could get based on a simple random sample of
100 students.
8
6
4
2
0.50
0.55
0.60
0.65
0.70
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0.75
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Homework Problem
Exercise 8.12, page 528.
(b) Sketch a picture of the distribution for the possible sample
proportions you could get based on a simple random sample of
100 students.
(c) Use the 68–95–99.7 rule for normal distributions to complete the
following statements:
(i) There is a 68% chance that the sample proportion is between
_____ and _____.
(ii) There is a 95% chance that the sample proportion is between
_____ and _____.
(iii) It is almost certain that the sample proportion is between _____ and
_____.
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Homework Problem
Exercise 8.12, page 528.
(b) Sketch a picture of the distribution for the possible sample
proportions you could get based on a simple random sample of
100 students.
(c) Use the 68–95–99.7 rule for normal distributions to complete the
following statements:
(i) There is a 68% chance that the sample proportion is between 0.551
and 0.649.
(ii) There is a 95% chance that the sample proportion is between
_____ and _____.
(iii) It is almost certain that the sample proportion is between _____ and
_____.
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Homework Problem
Exercise 8.12, page 528.
(b) Sketch a picture of the distribution for the possible sample
proportions you could get based on a simple random sample of
100 students.
(c) Use the 68–95–99.7 rule for normal distributions to complete the
following statements:
(i) There is a 68% chance that the sample proportion is between 0.551
and 0.649.
(ii) There is a 95% chance that the sample proportion is between 0.502
and 0.698.
(iii) It is almost certain that the sample proportion is between _____ and
_____.
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Homework Problem
Exercise 8.12, page 528.
(b) Use the 68–95–99.7 rule for normal distributions to complete the
following statements:
(i) There is a 68% chance that the sample proportion is between 0.551
and 0.649.
(ii) There is a 95% chance that the sample proportion is between 0.502
and 0.698.
(iii) It is almost certain that the sample proportion is between 0.453 and
0.747.
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Homework Problem
Exercise 8.12, page 528.
(c) Would it be likely to observe a sample proportion of 0.50, based on
a simple random sample of size 100, if the population proportion
were 0.60? Explain.
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Homework Problem
Exercise 8.12, page 528.
(c) Would it be likely to observe a sample proportion of 0.50, based on
a simple random sample of size 100, if the population proportion
were 0.60? Explain.
(The question should ask how likely it is to observe a sample
proportion at least as low as 0.50.)
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Homework Problem
Exercise 8.12, page 528.
(c) Would it be likely to observe a sample proportion of 0.50, based on
a simple random sample of size 100, if the population proportion
were 0.60? Explain.
(The question should ask how likely it is to observe a sample
proportion at least as low as 0.50.)
No. A sample proportion of 0.50 is more than 2 standard
deviations from the mean.
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Homework Problem
Exercise 8.12, page 528.
(d) Sketch a picture of the distribution for the possible sample
proportions you could get based on a simple random sample of
400 students.
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Homework Problem
Exercise 8.12, page 528.
(d) Sketch a picture of the distribution for the possible sample
proportions you could get based on a simple random sample of
400 students.
15
10
5
0.50
0.55
0.60
0.65
0.70
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0.75
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Homework Problem
Exercise 8.12, page 528.
(i) How does this picture differ from the one in part (a)?
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Homework Problem
Exercise 8.12, page 528.
)
(i) How does this picture differ from the one in part (a)?
It is only half as wide.
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Homework Problem
Exercise 8.12, page 528.
)
(i) How does this picture differ from the one in part (a)?
It is only half as wide.
How will the increased sample size affect the range of values you
gave in (i)–(iii) of part (b)
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Homework Problem
Exercise 8.12, page 528.
)
(i) How does this picture differ from the one in part (a)?
It is only half as wide.
How will the increased sample size affect the range of values you
gave in (i)–(iii) of part (b)
They will each be only half as wide.
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Outline
1
The Central Limit Theorem for Proportions
2
Applications
3
Assignment
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Assignment
Homework
i)
Read Sections 8.1 - 8.2, pages 499 - 508.
Exercises 7 - 14, page 526.
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Assignment
Homework Answers
8. (a) 0.0264.
(b) Yes. The probability of that is 3.117 × 10−6 .
10. (a) The sampling distribution for the sample proportion of men is
normal with mean 0.49 and standard deviation 0.04999.
(b) 0.0548.
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Assignment
Homework Answers
12. (a) Draw a normal curve with mean 0.60 and standard deviation
0.04899.
(b)
(i) 0.55101 and 0.64899.
(ii) 0.50202 and 0.69798.
(iii) 0.45303 and 0.74697.
(c) No. According to part (b)(ii), a proportion as low as 0.50 has only
about a 2 21 % chance of occurring.
(d) Sketch a normal curve with mean 0.60 and standard deviation
0.02449.
(i) It is narrower.
(ii) They will each be half as wide as before.
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Assignment
Homework Answers
14. (a) It is normal with mean 0.50 and standard deviation 0.02887.
(b) p̂ = 0.56. p-value of 0.56 is 0.0188.
(c) Reject H0 . A majority of shoppers favor longer shopping hours.
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