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Two-Sample Tests for Proportions

Learn how to compare proportions from two independent populations

Lesson Objectives

By the end of this lesson, you will be able to:

Formulate hypotheses for comparing two population proportions
Calculate the pooled proportion under the null hypothesis
Check conditions for valid two-proportion z-tests
Conduct and interpret two-proportion z-tests
Apply two-proportion tests to real-world scenarios

1. When Do We Compare Two Proportions?

Many research questions involve comparing success rates, percentages, or proportions between two groups:

Medical Research: Is the cure rate higher for Drug A than Drug B?
Marketing: Do more customers prefer Product X or Product Y?
Education: Is the pass rate different for online vs. in-person classes?
Political Polling: Do men and women vote differently on an issue?
Quality Control: Is the defect rate the same for two manufacturing plants?

Definition: Two-Sample Proportion Test

A two-proportion z-test is used to test whether two population proportions (p₁ and p₂) are equal based on data from two independent random samples.

2. Hypotheses for Two-Proportion Tests

We compare two population proportions: p₁ and p₂.

Test Type	Null Hypothesis (H₀)	Alternative Hypothesis (Hₐ)
Two-tailed	H₀: p₁ = p₂ or H₀: p₁ - p₂ = 0	Hₐ: p₁ ≠ p₂ or Hₐ: p₁ - p₂ ≠ 0
Right-tailed	H₀: p₁ = p₂	Hₐ: p₁ > p₂ or Hₐ: p₁ - p₂ > 0
Left-tailed	H₀: p₁ = p₂	Hₐ: p₁ < p₂ or Hₐ: p₁ - p₂ < 0

Example: Setting Up Hypotheses

Scenario: A researcher wants to test if the proportion of college graduates differs between two cities.

H₀: p₁ = p₂ (the proportions are equal)
Hₐ: p₁ ≠ p₂ (the proportions are different) — two-tailed test

3. Conditions for Two-Proportion z-Test

Before conducting the test, verify these conditions:

Independence:
- The two samples are independent of each other
- Both samples are random samples from their populations
- Each sample size is less than 10% of its population (if sampling without replacement)
Success-Failure Condition: For both samples, we need enough successes and failures:
- n₁p̂₁ ≥ 10 and n₁(1 - p̂₁) ≥ 10
- n₂p̂₂ ≥ 10 and n₂(1 - p̂₂) ≥ 10

Important: If the success-failure condition is not met, the normal approximation may not be valid. You would need to use exact methods (like Fisher's exact test) instead.

4. The Pooled Proportion

Under the null hypothesis (H₀: p₁ = p₂), we assume the two populations have the same proportion. We estimate this common proportion by combining (pooling) the data from both samples.

Pooled Proportion

p̄ = (x₁ + x₂) / (n₁ + n₂)

where:
x₁ = number of successes in sample 1
x₂ = number of successes in sample 2
n₁ = size of sample 1
n₂ = size of sample 2

Alternatively, if you're given sample proportions:

p̄ = (n₁p̂₁ + n₂p̂₂) / (n₁ + n₂)

Why Pool? The pooled proportion p̄ is our best estimate of the common population proportion under the assumption that H₀ is true (i.e., p₁ = p₂). We use this pooled proportion in the test statistic formula.

5. Test Statistic for Two Proportions

Test Statistic for Two-Proportion z-Test

z = (p̂₁ - p̂₂) / √[p̄(1 - p̄)(1/n₁ + 1/n₂)]

where:
p̂₁, p̂₂ = sample proportions
p̄ = pooled proportion
n₁, n₂ = sample sizes

This test statistic follows a standard normal distribution (z-distribution) when the null hypothesis is true and conditions are met.

Decision Rules

Two-tailed: Reject H₀ if |z| > z* (e.g., z* = 1.96 for α = 0.05)
Right-tailed: Reject H₀ if z > z* (e.g., z* = 1.645 for α = 0.05)
Left-tailed: Reject H₀ if z < -z* (e.g., z* = -1.645 for α = 0.05)

Or use the p-value approach: Reject H₀ if p-value < α.

6. Complete Example: Drug Efficacy Study

Example 1: Comparing Drug Cure Rates

Research Question: Is Drug A more effective than Drug B at curing a disease?

Data:

Drug A: 120 out of 200 patients cured (p̂₁ = 0.60)
Drug B: 90 out of 180 patients cured (p̂₂ = 0.50)

Significance level: α = 0.05

Step 1: State hypotheses

H₀: p₁ = p₂ (cure rates are equal)
Hₐ: p₁ > p₂ (Drug A has a higher cure rate) — right-tailed test

Step 2: Check conditions

Independent random samples from two groups
n₁p̂₁ = 200(0.60) = 120 ≥ 10, n₁(1-p̂₁) = 200(0.40) = 80 ≥ 10
n₂p̂₂ = 180(0.50) = 90 ≥ 10, n₂(1-p̂₂) = 180(0.50) = 90 ≥ 10
All conditions satisfied!

Step 3: Calculate pooled proportion

p̄ = (x₁ + x₂) / (n₁ + n₂)
p̄ = (120 + 90) / (200 + 180)
p̄ = 210 / 380
p̄ ≈ 0.5526

Step 4: Calculate test statistic

z = (p̂₁ - p̂₂) / √[p̄(1 - p̄)(1/n₁ + 1/n₂)]
z = (0.60 - 0.50) / √[0.5526(0.4474)(1/200 + 1/180)]
z = 0.10 / √[0.2472 × 0.01056]
z = 0.10 / √0.00261
z = 0.10 / 0.0511
z ≈ 1.96

Step 5: Find critical value and p-value

For α = 0.05 (right-tailed): z-critical = 1.645
For z = 1.96: p-value ≈ 0.025

Step 6: Make decision

Since z = 1.96 > 1.645 (or p-value = 0.025 < 0.05), we reject H₀.

Step 7: Conclusion

There is sufficient evidence at the 0.05 significance level to conclude that Drug A has a higher cure rate than Drug B. The difference in cure rates (60% vs. 50%) is statistically significant.

7. Example: Two-Tailed Test

Example 2: Gender Differences in Support

Research Question: Is there a difference in support for a policy between men and women?

Data:

Men: 156 out of 300 support the policy (p̂₁ = 0.52)
Women: 198 out of 350 support the policy (p̂₂ = 0.566)

Significance level: α = 0.05, two-tailed test

Step 1: Hypotheses

H₀: p₁ = p₂ (no gender difference)
Hₐ: p₁ ≠ p₂ (there is a gender difference)

Step 2: Check conditions

All success-failure conditions met (verify yourself for practice!)

Step 3: Calculate pooled proportion

p̄ = (156 + 198) / (300 + 350) = 354 / 650 ≈ 0.545

Step 4: Calculate test statistic

z = (0.52 - 0.566) / √[0.545(0.455)(1/300 + 1/350)]
z = -0.046 / √[0.248 × 0.00619]
z = -0.046 / 0.0391
z ≈ -1.18

Step 5: Decision

For α = 0.05 (two-tailed): z-critical = ±1.96
Since |z| = 1.18 < 1.96, we fail to reject H₀.

Step 6: Conclusion

There is insufficient evidence at the 0.05 significance level to conclude that men and women differ in their support for the policy. The observed difference (52% vs. 56.6%) could reasonably be due to sampling variability.

8. Confidence Interval for Difference in Proportions

In addition to hypothesis testing, you can construct a confidence interval for the difference p₁ - p₂. Note that for confidence intervals, we do NOT use the pooled proportion!

Confidence Interval for p₁ - p₂

(p̂₁ - p̂₂) ± z* × √[(p̂₁(1-p̂₁)/n₁) + (p̂₂(1-p̂₂)/n₂)]

where z* is the critical value for the desired confidence level
(e.g., z* = 1.96 for 95% confidence)

Key Difference:

Hypothesis test: Uses pooled proportion p̄
Confidence interval: Uses individual sample proportions p̂₁ and p̂₂

Example: CI for Drug Cure Rate Difference

Using the drug data: p̂₁ = 0.60, p̂₂ = 0.50, n₁ = 200, n₂ = 180

95% CI = (0.60 - 0.50) ± 1.96√[(0.60×0.40/200) + (0.50×0.50/180)]
= 0.10 ± 1.96√[0.0012 + 0.00139]
= 0.10 ± 1.96√0.00259
= 0.10 ± 1.96(0.0509)
= 0.10 ± 0.0998
= (0.0002, 0.1998) or (0.02%, 19.98%)

Interpretation: We are 95% confident that Drug A's cure rate is between 0.02% and 19.98% higher than Drug B's cure rate. Since the interval does NOT contain 0, we can conclude Drug A is significantly better.

Check Your Understanding

Question 1: Why do we use a pooled proportion in hypothesis testing but not in confidence intervals?

Answer: In hypothesis testing, we assume H₀ is true (p₁ = p₂), so we pool the data to get a single estimate. In confidence intervals, we're estimating the actual difference p₁ - p₂, so we use the individual sample proportions to preserve any real difference.

Question 2: A study has n₁ = 50 with 8 successes and n₂ = 60 with 10 successes. Can you conduct a two-proportion z-test?

Answer: No! The success-failure condition is violated. For sample 1: n₁p̂₁ = 8 < 10. For sample 2: n₂p̂₂ = 10, which is barely acceptable, but n₁p̂₁ fails. You would need a larger sample size or use an exact test instead.

Question 3: If a 95% confidence interval for p₁ - p₂ is (-0.05, 0.15), what can you conclude about a two-tailed test at α = 0.05?

Answer: Fail to reject H₀. Since the confidence interval contains 0, there is no significant difference between p₁ and p₂ at the 0.05 level. The confidence interval and hypothesis test always agree when using the same α.

Key Takeaways

Two-proportion z-tests compare proportions from two independent samples
Pooled proportion: p̄ = (x₁ + x₂) / (n₁ + n₂) — used in hypothesis testing
Test statistic: z = (p̂₁ - p̂₂) / √[p̄(1 - p̄)(1/n₁ + 1/n₂)]
Check success-failure condition: At least 10 successes and 10 failures in each sample
Confidence intervals use individual p̂₁ and p̂₂, NOT the pooled proportion
Use z-distribution (not t-distribution) for proportion tests
Applications: Medical trials, quality control, survey comparisons, A/B testing

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