9.3 Two-Groups: Proportion

The hypothesis test for comparing the proportion between two groups is written as follows: $$H_0: \hat{p}_1-\hat{p}_2=0\\ H_a: \hat{p}_1-\hat{p}_2 \neq 0$$ We first need to calculate the pooled estimate, i.e., the total number of success over the total number of observations. You can think of this as a weighted average. Let this weighted average be $\hat{p}^*$.: $$\hat{p}^*=\frac{p_1+p_2}{n_1+n_2}$$ Then, the standard error can be calculated as follows: $$S.E. = \sqrt{\hat{p}^* \cdot (1-\hat{p}^*) \cdot \left( \frac{1}{n_1}+\frac{1}{n_2} \right)}$$ The resulting test statistic is written as $$z = \frac{\hat{p}_1-\hat{p}_2}{S.E.}$$ Consider the hypothesis about the proportion of gun owners differentiated by male vs. female. The data is in gssgun. A summary of the data is in the following table:

owngun	male	female
yes	232	208
no	335	501

The pooled proportion is calculated as follows: $$\hat{p}^*=\frac{232+208}{440+836}=0.345$$ Given those numbers, the standard error is calculated as: $$S.E. = \sqrt{0.345 \cdot (1-0.345) \cdot \left( \frac{1}{440}+\frac{1}{836} \right)}=0.028$$ The test statistic is: $$z = \frac{0.41-0.29}{0.028}=4.286$$ Thus, we reject the hypothesis that the proportion of gun owners among males and females is the same.