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2 Sample Proportion Test Statistics Meaning

2 Sample Proportion Test Formula:

\[ z = \frac{p_1 - p_2}{\sqrt{\bar{p}(1 - \bar{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] \[ \text{where } \bar{p} = \frac{x_1 + x_2}{n_1 + n_2} \]

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1. What is the 2 Sample Proportion Test?

The 2 Sample Proportion Test is a statistical hypothesis test used to determine whether two population proportions are significantly different from each other. It calculates a z-score that measures the standardized difference between two sample proportions.

2. How Does the Calculator Work?

The calculator uses the 2 sample proportion test formula:

\[ z = \frac{p_1 - p_2}{\sqrt{\bar{p}(1 - \bar{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] \[ \text{where } \bar{p} = \frac{x_1 + x_2}{n_1 + n_2} \]

Where:

Explanation: The z-score measures how many standard errors the difference between the two proportions is away from zero, indicating whether the difference is statistically significant.

3. Importance of Z-Score Calculation

Details: The z-score is crucial for hypothesis testing in comparing proportions from two independent samples. It helps determine if observed differences are due to chance or represent true population differences.

4. Using the Calculator

Tips: Enter the number of successes and sample sizes for both groups. Ensure success counts are between 0 and sample sizes, and sample sizes are positive integers.

5. Frequently Asked Questions (FAQ)

Q1: What does the z-score represent?
A: The z-score represents how many standard errors the difference between proportions is from zero. Higher absolute values indicate stronger evidence against the null hypothesis.

Q2: When is this test appropriate?
A: This test is appropriate when comparing proportions from two independent samples with sufficiently large sample sizes (typically n > 30 for each group).

Q3: How do I interpret the z-score?
A: Typically, |z| > 1.96 suggests statistical significance at α = 0.05 level, |z| > 2.58 at α = 0.01 level.

Q4: What are the assumptions of this test?
A: The test assumes independent samples, binomial distribution, and sufficiently large sample sizes (np > 5 and n(1-p) > 5 for both samples).

Q5: When should I use this instead of chi-square test?
A: The z-test for proportions is essentially equivalent to chi-square test for 2x2 contingency tables. The z-test provides directionality (which proportion is larger) while chi-square only tests for association.

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