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8/14/2019 Lecture 26 - Sampling Distribution Proportion.pdf
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Outline
1 Computing the Sampling Distribution ofp
2 The Central Limit Theorem for Proportions
3 Applications
4 Assignment
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Sampling Distributions
Definition (Sampling Distribution of a Statistic)
Thesampling distribution of a statisticis the distribution of values of
that statistic over all possible samples of a given size nfrom the
population.
We may sample with or without replacement.
For our purposes, it will be simpler to sample with replacement.
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The Sample Proportion
We will work out the sampling distribution for pfor sample sizes of1, 2, and 3.
Then I will show you the sampling distribution for pfor samplesizes of 4, 5, and 10.
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Example
Suppose that 45%of all people approve of President Obamasperformance.
Suppose that we select one person at random.
We may diagram the 2 possibilities.
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Example
Y
N
0.55
0.45
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Example
Y
N
0.55
0.45
NY
NN
1
0
(0.55)(0.45) = 0.2475
(0.55)2= 0.3025
Y
N0.55
0.45
Y
N0.55
0.45
YN 1 (0.45)(0.55) = 0.2475
YY 2 (0.45)2= 0.2025
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Example
Now we take a sample of 3 people, sampling with replacement.
Find the sampling distribution of p.
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Example
Y
N
0.55
0.45
NYY
NNY
2
1
(0.55)(0.45)2= 0.1114
(0.55)2
(0.45) = 0.1114
Y
N0.55
0.45
YNY 2 (0.45)(0.55)(0.45) = 0.1114
YYY 3 (0.45)3= 0.0911Y
N0.55
0.45
Y
N0.55
0.45
Y
N0.55
0.45
Y
N0.55
0.45
Y
N0.55
0.45
NYN
NNN
1
0
(0.55)(0.45)(0.55) = 0.1361
(0.55)3= 0.1664
YNN 1 (0.45)(0.55)2= 0.1361
YYN 2 (0.45)2(0.55) = 0.1114
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Example
Letpbe the sample proportion of people who strong disapproveof President Obamas performance.
The sampling distribution of pis
p P(p)0 0.1664
1/3 0.40842/3 0.3341
1 0.0911
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S S
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Samples of Sizen= 4
If we sample 4 people, then the sampling distribution of the
sample proportion is
p P(p)
0 0.09151/4 0.29952/4 0.36753/4 0.2005
1 0.0410
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S l f Si 5
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Samples of Sizen= 5
If we sample 5 people, then the sampling distribution of the
sample proportion is
p P(p)
0 0.05031/5 0.20592/5 0.33693/5 0.27574/5 0.1128
1 0.0185
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S l f Si 6
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Samples of Sizen= 6
If we sample 6 people, then the sampling distribution of the
sample proportion is
p P(p)
0 0.0277
1/6 0.13592/6 0.27803/6 0.30324/6 0.18615/6 0.0609
1 0.0083
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S l f Si 10
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Samples of Sizen= 10
If we sample 10 people, then the sampling distribution of the
sample proportion is
p P(p)
0.00 0.00250.10 0.0207
0.20 0.07630.30 0.16650.40 0.23840.50 0.23400.60 0.15960.70 0.07460.80 0.02290.90 0.00421.00 0.0003
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Outline
1 Computing the Sampling Distribution ofp
2 The Central Limit Theorem for Proportions
3 Applications
4 Assignment
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The Central Limit Theorem for Proportions
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The Central Limit Theorem for Proportions
Theorem (The Central Limit Theorem for Proportions)
For any population and any sample size, the sampling distribution
ofp has the following mean and standard deviation:
p = p
p =
p(1 p)
n .
Furthermore, the sampling distribution ofp is approximatelynormal, provided n is large enough.
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The Central Limit Theorem for Proportions
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The Central Limit Theorem for Proportions
The Sample Size
nis large enough if
np 5 andn(1 p) 5.
Ifnis small, then we have to work out the distribution by hand.
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Outline
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Outline
1 Computing the Sampling Distribution ofp
2 The Central Limit Theorem for Proportions
3 Applications
4 Assignment
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Applications
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Applications
Suppose that 60% of all high-school students own a cell phone.
If we survey 3 high-school students, how likely is it that we will findthat at least 2 of them own a cell phone?
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Applications
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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 willfind that at least 65% of them own a cell phone?
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Applications
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Applications
Ifp=0.60 and our sample size isn=150, thenpis normal withmeanp=0.60 and
p=
(0.60)(0.40)
150 =
0.0016= 0.04.
We want to know the probability that p 0.65.
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Applications
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Applications
0.55 0.60 0.65 0.70
2
4
6
8
10
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Applications
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Applications
0.55 0.60 0.65 0.70
2
4
6
8
10
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Applications
The probability that pis 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?
Thenpis normal with meanp=0.60 and
p=
(0.60)(0.40)
600 =
0.0004= 0.02.
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Applications
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Applications
0.55 0.60 0.65 0.70
5
10
15
20
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Applications
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pp
0.55 0.60 0.65 0.70
5
10
15
20
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Guessing on a Test
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g
Example (Guessing on a Test)
A student takes a math placement test with 25 multiple-choice
questions.
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Guessing on a Test
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Example (Guessing on a Test)
A student takes a math placement test with 25 multiple-choice
questions.
Each question has 5 choices.If a student guesses at each answer, what will his score most
likely be?
If a student scores 10 out of 25, can we be sure that he did not
guess at all 25 answers?
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Guessing on a Test
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Example (Guessing on a Test)If he guesses at all 25 answers, then the probability of getting any
one answer correct isp=0.20.
The proportion that he actually gets correct isp.
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Example (Guessing on a Test)If he guesses at all 25 answers, then the probability of getting any
one answer correct isp=0.20.
The proportion that he actually gets correct isp.
The distribution ofpis normal with mean 0.20 and standard
deviation
(0.20)(0.80)
25 =0.04.
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Guessing on a Test
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Example (Guessing on a Test)If he guesses at all 25 answers, then the probability of getting any
one answer correct isp=0.20.
The proportion that he actually gets correct isp.
The distribution ofpis normal with mean 0.20 and standard
deviation
(0.20)(0.80)
25 =0.04.
If we allow three standard deviations each way, then pis almostcertainly at least 0.08 and at most 0.32.
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Guessing on a Test
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Example (Guessing on a Test)If he guesses at all 25 answers, then the probability of getting any
one answer correct isp=0.20.
The proportion that he actually gets correct isp.
The distribution ofpis normal with mean 0.20 and standard
deviation
(0.20)(0.80)
25 =0.04.
If we allow three standard deviations each way, then pis almostcertainly at least 0.08 and at most 0.32.
That represents from 2 to 8 correct answers.
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Outline
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1 Computing the Sampling Distribution ofp
2 The Central Limit Theorem for Proportions
3 Applications
4
Assignment
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Assignment
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Homework
Read Sections 8.1 - 8.2, pages 499 - 508.
Exercises 7 - 14, page 526.
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