Plus four Confidence Interval for difference between two population proportions
Let $X_1$ be the observed number of individuals possessing certain attributes (number of successes) in arandom sample of size $n_1$ from a large population with population proportion $p_1$ and let $X_2$ be the observed number of individuals possessing certain attributes (number of successes) in a random sample of size $n_2$ from a large population with population proportion $p_2$. The two sample are independent.
Formula
$100(1-\alpha)$% plus four confidence interval for the difference $(p_1-p_2)$ is
$(\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E$
where,
$\hat{p_1}=\dfrac{X_1+1}{n_1+2}$and$\hat{p_2}=\dfrac{X_2+1}{n_2+2}$are the estimate of proportions from two samples using plus four rule,$E = Z_{\alpha/2} \sqrt{\dfrac{\hat{p}_1*(1-\hat{p}_1)}{n_1+2}+\dfrac{\hat{p}_2*(1-\hat{p}_2)}{n_2+2}}$is the margin of error,$1-\alpha$is the confidence coefficient,$Z_{\alpha/2}$is the critical value of $Z$.
