Section 6-4

Report
ELEMENTARY
STATISTICS
Section 6-5 Estimating a Population Proportion
EIGHTH
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman EDITION
MARIO F. TRIOLA
1
Assumptions
1. The sample is a simple random
sample.
2. The conditions for the binomial
distribution are satisfied
3. The normal distribution can be used
to approximate the distribution of sample
proportions because np  5 and nq  5
are both satisfied.
ˆ
ˆ
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
2
Notation for Proportions
p=
ˆp = nx
population proportion
sample proportion
of x successes in a sample of size n
(pronounced ‘p-hat’)
qˆ = 1 - pˆ = sample proportion
of x failures in a sample size of n
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
3
Definition
Point Estimate
ˆ
The sample proportion p is the best
point estimate of the population
proportion p.
(In other books population proportion
can be noted as )
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
4
Margin of Error of the Estimate of p
ME =
z
pˆ qˆ
n
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
5
Confidence Interval for
Population Proportion
pˆ - ME < p < pˆ + ME
where
ME = z 
pˆ qˆ
n
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
6
Confidence Interval for
Population Proportion
pˆ - E < p < pˆ + E
p = pˆ + E
(pˆ - E, pˆ + E)
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
7
Round-Off Rule for Confidence
Interval Estimates of p
Round the confidence
interval limits to
three significant digits.
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
8
Do people lie about voting?
• In a survey of 1002 people, 701 people said they
voted in a recent election. Voting records showed
that 61% of eligible voters actually did vote.
Using these results, find the following about the
people who “said” they voted:
• a) Find the point estimate
• b) Find the 95% confidence interval estimate of
the proportion
• c) Are the survey results are consistent with the
actual voter turnout of 61%?
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
9
Do people lie about voting?
• In a survey of 1002 people, 701 people said
they voted in a recent election. Voting records
showed that 61% of eligible voters actually did
vote. Using these results, find the following
about the people who said they voted:
• a) Find the point estimate
pˆ 
x
n

701
 0.6996  0.700
1002
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
10
Do people lie about voting?
• b) Find the 95% confidence interval estimate
of the proportion
E  z / 2
ˆˆ
pq
 1.96
(.70)(.30)
n
 .0284
1002
pˆ  E  p  pˆ  E
.7  .0284  p  .7  .0284
.671  p  .728
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
11
Do people lie about voting?
• c) Are the survey results consistent with the
actual voter turnout of 61%?
We are 95% confident that the true proportion
of the people who said they vote is in the
interval 67.1% <p<72.8% .
Because 61% does not fall inside the interval,
we can conclude our survey results are not
consistent with the actual voter turnout.
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
12
Determining Sample Size
ME = z 
pˆ qˆ
n
(solve for n by algebra)
n=
( z pˆ qˆ
2
)
ME2
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
13
Sample Size for Estimating Proportion p
ˆ
When an estimate of p is known:
n=
(
2 pq
)
z ˆ ˆ
E2
When no estimate of p is known:
n=
(
2 0.25
)
z
E2
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
14
Two formulas
for proportion sample size
n=
n=
( z )2 pˆ qˆ
ME2
( z (0.25)
2
)
ME2
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
15
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? A 1997 study indicates 16.9% of U.S. households
used e-mail.
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
16
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? A 1997 study indicates 16.9% of U.S. households
used e-mail.
ˆˆ
n = [z/2 ]2 p q
E2
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
17
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? A 1997 study indicates 16.9% of U.S. households
used e-mail.
ˆˆ
n = [z/2 ]2 p q
E2
= [1.645]2 (0.169)(0.831)
0.042
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
18
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? A 1997 study indicates 16.9% of U.S. households
used e-mail.
ˆˆ
n = [z/2 ]2 p q
E2
= [1.645]2 (0.169)(0.831)
0.042
= 237.51965
= 238 households
To be 90% confident that our
sample percentage is within
four percentage points of the
true percentage for all
households, we should
randomly select and survey
238 households.
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
19
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? There is no prior information suggesting a
possible value for the sample percentage.
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
20
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? There is no prior information suggesting a
possible value for the sample percentage.
n = [z/2 ]2 (0.25)
E2
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
21
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? There is no prior information suggesting a
possible value for the sample percentage.
n = [z/2 ]2 (0.25)
E2
= (1.645)2 (0.25)
0.042
= 422.81641
= 423 households
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
22
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? There is no prior information suggesting a
possible value for the sample percentage.
n = [z/2 ]2 (0.25)
E2
= (1.645)2 (0.25)
0.042
= 422.81641
= 423 households
With no prior information,
we need a larger sample to
achieve the same results
with 90% confidence and an
error of no more than 4%.
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
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Finding the Point Estimate and E
from a Confidence Interval
ˆ
(upper confidence interval limit) + (lower confidence interval limit)
Point estimate of p:
ˆ
p=
2
Margin of Error:
E = (upper confidence interval limit) - (lower confidence interval limit)
2
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
24
Finding the Point Estimate and E
from a Confidence Interval
Given the confidence interval .214< p < .678
ˆ
Find the point estimate of p:
ˆ
p = .678 + .214 = .446
2
Margin of Error:
E = .678 - .214 = .232
2
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
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