Hypothesis Testing with Two Samples: Two-Sample z-Test for Proportions

Chapter 8: Hypothesis Testing with Two Samples

Section 8.4: Testing the Difference Between Proportions

This section focuses on statistical methods for comparing two population proportions using hypothesis testing. The primary tool discussed is the two-sample z-test for proportions, which allows researchers to determine if there is a significant difference between the proportions of two independent groups.

Objectives

• Understand how to perform a two-sample z-test for the difference between two population proportions.

• Identify the necessary conditions for conducting the test.

• Interpret the results of the test in the context of real-world examples.

Two-Sample z-Test for Proportions

The two-sample z-test for proportions is used to test the difference between two population proportions. This test is appropriate when the following conditions are met:

• Random Sampling: Both samples must be randomly selected from their respective populations.

• Independence: The samples must be independent of each other.

• Sample Size: The samples must be large enough so that the sampling distribution of the difference in sample proportions is approximately normal. Specifically, the counts of successes and failures in both samples should each be at least 5.

Test Statistic and Formulas

• Weighted Estimate of Proportion: The pooled (weighted) estimate of the population proportion is calculated as: $\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}$ where $x_1$ and $x_2$ are the number of successes in samples 1 and 2, and $n_1$ and $n_2$ are the sample sizes.

• Standard Error: $SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}$

• Test Statistic (z): $z = \frac{\hat{p}_1 - \hat{p}_2}{SE}$ where $\hat{p}_1 = \frac{x_1}{n_1}$ and $\hat{p}_2 = \frac{x_2}{n_2}$ are the sample proportions.

Hypotheses

• Null Hypothesis ($H_0$): $p_1 = p_2$ (There is no difference between the population proportions.)

• Alternative Hypothesis ($H_a$): $p_1 \neq p_2$, $p_1 > p_2$, or $p_1 < p_2$ (Depending on the research question, the test can be two-tailed or one-tailed.)

Decision Rule

• Determine the level of significance ($\alpha$), typically 0.05 or 0.01.

• Find the critical value(s) from the standard normal distribution corresponding to $\alpha$.

• Reject $H_0$ if the calculated $z$ falls in the rejection region; otherwise, fail to reject $H_0$.

Example 1: Comparing Seat Belt Use in Two States

Scenario: A study compares the proportion of drivers who wear seat belts in Idaho and Wisconsin. Out of 200 drivers in Idaho, 87.5% wear seat belts; out of 250 drivers in Wisconsin, 92.0% wear seat belts. At the 10% significance level, can we reject the claim that the proportions are the same?

• Step 1: State the hypotheses:

  • $H_0: p_1 = p_2$

  • $H_a: p_1 \neq p_2$

• Step 2: Check conditions: Samples are random, independent, and large enough.

• Step 3: Calculate the pooled proportion and standard error.

• Step 4: Compute the test statistic $z$ and compare to critical values ($\pm 1.645$ for $\alpha = 0.10$).

• Step 5: Conclusion: If $z$ is not in the rejection region, fail to reject $H_0$. There is not enough evidence to conclude a difference in proportions.

Example 2: Effectiveness of PTSD Medication

Scenario: In a clinical trial, 28 out of 42 patients taking medication no longer meet PTSD criteria, while 12 out of 37 on placebo no longer meet criteria. At the 1% significance level, can we support the claim that the medication is more effective?

• Step 1: State the hypotheses:

  • $H_0: p_1 = p_2$

  • $H_a: p_1 > p_2$

• Step 2: Check conditions: Samples are random, independent, and large enough.

• Step 3: Calculate the pooled proportion and standard error.

• Step 4: Compute the test statistic $z$ and compare to the critical value ($z_{0.01} = 2.33$ for a right-tailed test).

• Step 5: Conclusion: If $z$ is in the rejection region, reject $H_0$. There is enough evidence to support the claim that the medication is more effective.

Summary Table: Two-Sample z-Test for Proportions

Step	Description	Formula
1. State Hypotheses	Set up $H_0$ and $H_a$	$H_0: p_1 = p_2$ $H_a: p_1 \neq p_2$, $p_1 > p_2$, or $p_1 < p_2$
2. Check Conditions	Random, independent, large samples	Each of $n_1\hat{p}_1$, $n_1(1-\hat{p}_1)$, $n_2\hat{p}_2$, $n_2(1-\hat{p}_2)$ at least 5
3. Calculate Pooled Proportion	Weighted estimate	$\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}$
4. Compute Standard Error	Standard error of difference	$SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}$
5. Find Test Statistic	Standardized difference	$z = \frac{\hat{p}_1 - \hat{p}_2}{SE}$
6. Decision	Compare $z$ to critical value(s)	Reject or fail to reject $H_0$

Additional info: The two-sample z-test for proportions is widely used in medical, social, and behavioral sciences to compare rates, proportions, or probabilities between two groups. It is important to ensure that the assumptions of the test are met to draw valid conclusions.