1
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
2
Notation
p1 First population proportion
n1 First sample size
x1Number of successes in first sample
p1 First sample proportion
First Population
3
p2 Second population proportion
n2 Second sample size
x2Number of successes in second sample
p2 Second sample proportion
Second Population
Notation
4
The pooled sample proportion p
=p n1 + n
2
x1 + x
2
Definition
q = 1 – p
5
(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
6
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)
7
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)
8
Test Statistic
Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics)
9
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
10
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
11
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
13
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
14
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
15
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
17
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
18
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
13
Example 3
19
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.
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