Statement of no difference between population proportions: $H_0: p_1 = p_2$

Statement that there is a difference between population proportions ($H_a: p_1 > p_2$, $H_a: p_1 < p_2$, or $H_a: p_1 \neq p_2$).

$p_1$ and $p_2$ represent the population proportions of two different groups being compared.

A weighted average of the sample proportions, used when the null hypothesis assumes equal population proportions.

A procedure for measuring the strength of evidence against a null hypothesis.

Ensures the samples are representative of the populations, avoiding bias and allowing for generalization of results.

Ensures that the observations within each sample are independent of one another. Often checked using the 10% condition.

To ensure that the sampling distribution of the difference in sample proportions is approximately normal, allowing us to use the z-distribution for inference.

The probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true.

Rejecting the null hypothesis means there is sufficient evidence to support the alternative hypothesis. The observed difference is statistically significant.

$\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2}$, where $x_1$ and $x_2$ are the number of successes, and $n_1$ and $n_2$ are the sample sizes.

$z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)(\frac{1}{n_1} + \frac{1}{n_2})}}$

Calculate: $n_1\hat{p}_c$, $n_1(1-\hat{p}_c)$, $n_2\hat{p}_c$, and $n_2(1-\hat{p}_c)$.

All must be greater than or equal to 10 to satisfy the Normal condition.

$z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}$