Module 22: Proportions: One Sample
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Transcript of Module 22: Proportions: One Sample
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Module 22: Proportions: One Sample
This module presents confidence intervals and tests of hypotheses for proportions for the situation with one random sample from a population.
Reviewed 06 June 05 /MODULE 22
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Proportions
P = Population parameter, the proportion of population with characteristic
x = Number in sample with the characteristic
n = Total number in sample
p = x/n, the sample estimate of the proportion with the characteristic
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Hypothesis tests and confidence intervals are based on the normal approximation to the binomial distribution. For the hypothesis:
H0: P = P0 vs. H1: P P0
The options for the test statistic are:
where Q0 = 1- P0
0
(1 )
p Pz
p p
n
0
0 0
p Pz
P Q
n
0
0 0(1 )
p Pz
P P
n
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Example: AJPH, Nov. 1977;67:1033 - 1036
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Table 1 on the previous page indicates that 85.39% of the 1,109 males kept their appointments. Suppose that you had worked in this area previously and the percent of appointments kept before had always been 90% or higher. A question this raises is whether this observed rate is sufficiently different from 90% for you to think your population was different. That is, you would like to test the hypothesis:
H0: PM = 0.90 vs. H1: PM 0.90
One Sample Hypothesis test
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Hypothesis test for males keeping appointments
1. The hypothesis: H0: PM = 0.90 vs. H1: PM 0.90
2. The assumptions: Independence
3. The level: = 0.05
4. The test statistic:
5. The rejection region: Reject H0: PM = 0.90 if z is not between 1.96
(1 )Mp P
zp p
n
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6. The result:
M947
0.85 P 0.901,109
xp
n
0.85 0.90 =
(1 ) (0.85)(1 0.85)
1,109
Mp Pz
p p
n
0.85 0.90 0.05 0.05 = =
0.0107(0.85)(0.15) 0.000115
1,109
= - 4.66
7. The conclusion: Reject H0: PM = 0.90 since z is not
between ± 1.96
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Confidence Interval for P
(1 ) (1 )1.96 1.96 0.95
p p p pC p P p
n n
Confidence Interval for P, with = 0.05
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Appointment keeping behavior for males
9470.85
1,109
xp
n
(1 ) (1 )1.96 1.96 0.95
p p p pC p P p
n n
0.85 (0.15) 0.85 (0.15)0.85 1.96 0.85 1.96 0.95
1,109 1,109C P
(0.85)(0.15) 0.000115 0.0107
1,109
[0.85 1.96(0.0107) 0.85 1.96(0.0107)] 0.95
[0.85 0.021 0.85 0.021] 0.95
[0.829 0.871] 0.95
Note:
C P
C P
C P
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Social Class I, Asthma or Wheezy bronchitis n = 191 p = 0.047 or 4.7%
1. The hypothesis: H0: P = 0.10 vs. H1: P
0.10
2. The assumptions: Independence
3. The -level : = 0.05
4. The test statistic:
5. The rejection region: Reject H0: P = 0.10, if z is not between ±1.96
0.10
(1 )
pz
p pn
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6. The result:
7. The conclusion: Reject H0: P = 0.10, since z is not between ± 1.96
0.047 0.100 0.053
0.047(1 0.047) 0.047(0.953)191 191
z
0.053 0.053 0.053
0.01530.0448 0.0002191
3.46z
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(1 ) (1 )1.96 1.96 0.95
0.047 1.96(0.0153) 0.047 1.96(0.0153) 0.95
0.047 0.03 0.047 0.03 0.95
0.017 0.077 0.95
1.7% 7.7% 0.95
p p p pC p P p
n n
C P
C P
C P
C P
.
2.2% 8.8% 0.95
vs
C P
Confidence Interval for P with = 0.05
n = 191, p = 0.047, p(1-p) = 0.047(0.953) = 0.0448