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

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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

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

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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.