section 9.2 inferences about two proportions
DESCRIPTION
Section 9.2 Inferences About Two Proportions. Objective Compare the proportions of two populations using two samples from each population. Hypothesis Tests and Confidence Intervals of two proportions use the z -distribution. p 1 First population proportion n 1 First sample size - PowerPoint PPT PresentationTRANSCRIPT
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Objective
Compare the proportions of two populations using two samples from each population.
Hypothesis Tests and Confidence Intervals of two proportions use the z-distribution
Section 9.2Inferences About Two Proportions
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Notation
p1 First population proportion
n1 First sample size
x1Number of successes in first sample
p1 First sample proportion
First Population
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p2 Second population proportion
n2 Second sample size
x2Number of successes in second sample
p2 Second sample proportion
Second Population
Notation
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The pooled sample proportion p
=p n1 + n
2
x1 + x
2
Definition
q = 1 – p
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(1) Have two independent random samples
(2) For each sample:The number of successes is at least 5 The number of failures is at least 5
Requirements
Both requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval
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Tests for Two Proportions
The goal is to compare the two proportions
H0 : p1 = p
2
H1 : p1 p
2
Two tailed
H0 : p1 = p
2
H1 : p1 < p
2
Left tailed
H0 : p1 = p
2
H1 : p1 > p
2
Right tailed
Note: We only test the relation between p1 and p2
(not the actual numerical values)
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Finding the Test Statistic
+
z =( p
1 – p
2 ) – ( p
1 – p
2 )^ ^
n1
pqn2
pq
Note: p1 – p
2 =0 according to H0
This equation is an altered form of the test statistic for a single proportion (see Ch. 8-3)
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Test Statistic
Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics)
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Steps for Performing a Hypothesis Test on Two Proportions
• Write what we know
• State H0 and H1
• Draw a diagram
• Calculate the sample and pooled proportions
• Find the Test Statistic
• Find the Critical Value(s)
• State the Initial Conclusion and Final Conclusion
Note: Same process as in Chapter 8
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The table below lists results from a simple random sample of front-seat occupants involved in car crashes.
Use a 0.05 significance level to test the claim that the fatality rate of occupants is lower for those in cars equipped with airbags.
Example 1
What we know: x1 = 41 x2 = 52 α = 0.05
n1 = 11541 n2 = 9853 Claim: p1 < p2
p1 : Proportion of fatalities with airbagsp2 : Proportion of fatalities with no airbags
Claimp1 < p2
Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements
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H0 : p1 = p2
H1 : p1 < p2
Example 1
Left-TailedH1 = Claim
Pooled Proportion
Given: x1 = 41 x2 = 52 α = 0.05
n1 = 11541 n2 = 9853 Claim: p1 < p2
Sample Proportions
z = –1.9116 –zα = –1.645
z-dist.
Test Statistic
Critical Value
Diagram
Initial Conclusion: Since z is in the critical region, reject H0
Final Conclusion: We Accept the claim the fatality rate of occupants is lower for those who wear seatbelts
(Using StatCrunch)
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H0 : p1 = p2
H1 : p1 < p2
Example 1
Left-TailedH1 = Claim
Given: x1 = 41 x2 = 52 α = 0.05
n1 = 11541 n2 = 9853 Claim: p1 < p2
z = –1.9116 –zα = –1.645
z-dist.Diagram
Initial Conclusion: Since P-value is less than α (with α = 0.05), reject H0
Final Conclusion: We Accept the claim the fatality rate of occupants is lower for those who wear seatbelts
Stat → Proportions → Two sample → With summary
Null: prop. diff.=Alternative
Sample 1: Number of successes: .Number of observations:
Sample 2: Number of successes: .Number of observations:
● Hypothesis Test
P-value = 0.028
41
5211541
9853
0<
Using StatCrunch
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Confidence Interval Estimate
We can observe how the two proportions relate by looking at the Confidence Interval Estimate of p1–p2
CI = ( (p1–p2) – E, (p1–p2) + E )
Where
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Use the same sample data in Example 1 to construct a 90% Confidence Interval Estimate of the difference between the two population proportions (p1–p2)
Example 2
x1 = 41 x2 = 52 p1 = 0.003553 n1 = 11541 n2 = 9853 p2 = 0.005278
CI = (-0.003232, -0.000218 )
Note: CI negative implies p1–p2 is negative. This implies p1<p2
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Use the same sample data in Example 1 to construct a 90% Confidence Interval Estimate of the difference between the two population proportions (p1–p2)
Example 2
x1 = 41 x2 = 52 p1 = 0.003553 n1 = 11541 n2 = 9853 p2 = 0.005278
Note: CI negative implies p1–p2 is negative. This implies p1<p2
CI = (-0.003232, -0.000218 )
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Stat → Proportions → Two sample → With summary
Level
Sample 1: Number of successes: .Number of observations:
Sample 2: Number of successes: .Number of observations:
● Confidence Interval41
5211541
9853
0.9
Using StatCrunch
Use the same sample data in Example 1 to construct a 90% Confidence Interval Estimate of the difference between the two population proportions (p1–p2)
Example 2
Note: CI negative implies p1–p2 is negative. This implies p1<p2
CI = (-0.003232, -0.000218 )
x1 = 41 x2 = 52 n1 = 11541 n2 = 9853
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If a confidence interval limits does not contain 0, it implies there is a significant difference between the two proportions (i.e. p1 ≠ p2).
Thus, we can interpret a relation between the two proportions from the confidence interval.
In general:
•If p1 = p2 then the CI should contain 0
•If p1 > p2 then the CI should be mostly positive
•If p1 > p2 then the CI should be mostly negative
Interpreting Confidence Intervals
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Drug Clinical TrialChantix is a drug used as an aid to stop smoking. The number of subjects experiencing insomnia for each of two treatment groups in a clinical trial of the drug Chantix are given below:
(a)Use a 0.01 significance level to test the claim proportions of subjects experiencing insomnia is the same for both groups.
(b)Find the 99% confidence level estimate of the difference of the two proportions. Does it support the result of the test?
What we know: x1 = 41x2 = 52α = 0.01
n1 = 129 n2 = 9853Claim: p1= p2
Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements
Number in group
Number experiencing insomnia
Chantix Treatment
129
19
Placebo
805
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Example 3
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H0 : p1 = p2
H1 : p1 ≠ p2
Two-TailedH0 = Claim
Pooled Proportion
Given: x1 = 19 x2 = 13 α = 0.01
n1 = 129 n2 = 805 Claim: p1= p2
Sample Proportions
z = 7.602
zα/2 = 2.576
z-dist.
Test Statistic
Critical Value
Diagram
Initial Conclusion: Since z is in the critical region, reject H0
Final Conclusion: We Reject the claim the proportions of the subjects experiencing insomnia is the same in both groups.
(Using StatCrunch)
-zα/2 = -2.576
Example 3a
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Using StatCrunchStat → Proportions → Two sample → With summary
Null: prop. diff.=Alternative
Sample 1: Number of successes: .Number of observations:
Sample 2: Number of successes: .Number of observations:
● Hypothesis Test
P-value < 0.0001
19
13129
805
0≠
H0 : p1 = p2
H1 : p1 ≠ p2
Two-TailedH0 = Claim
Given: x1 = 19 x2 = 13 α = 0.01
n1 = 129 n2 = 805 Claim: p1= p2
z-dist.Diagram
Initial Conclusion: Since the P-value is less than α (0.01), reject H0
Final Conclusion: We Reject the claim the proportions of the subjects experiencing insomnia is the same in both groups.
i.e. the P-value is very small
Example 3a
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Example 3b
x1 = 19 x2 = 13 p1 = 0.14729 n1 = 129 n2 = 805p2 = 0.01615
Note: CI does not contain 0 implies p1 and p2 have significant difference.
Use the same sample data in Example 3 to construct a 99% Confidence Interval Estimate of the difference between the two population proportions (p1–p2)
CI = (0.0500, 0.2123 )
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Stat → Proportions → Two sample → With summary
Level
Sample 1: Number of successes: .Number of observations:
Sample 2: Number of successes: .Number of observations:
● Confidence Interval19
13129
805
0.9
Using StatCrunch
Use the same sample data in Example 3 to construct a 99% Confidence Interval Estimate of the difference between the two population proportions (p1–p2)
CI = (0.0500, 0.2123 )
x1 = 19 x2 = 13 n1 = 129 n2 = 805
Example 3b
Note: CI does not contain 0 implies p1 and p2 have significant difference.