Plus Four CI for difference between two population proportions
The plus four method produces more accurate confidence interval for difference between two population proportions. We assume that we had four additional observations. One success and one failure in each of the two samples.
The plus four method when the confidence coefficient is at least 90% and both the sample sizes are at least 5.
Let $X_1$ be the observed number of individuals possessing certain attributes (number of successes) in a random 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 $p_2$. Assume that the two sample are independent.
Then the plus four estimates of the two proportions are $\hat{p_1}=\dfrac{X_1+1}{n_1+2}$ and $\hat{p_2}=\dfrac{X_2+1}{n_2+2}$.
Let $C=1-\alpha$ be the confidence coefficient. Our objective is to construct a $100(1-\alpha)$% plus four confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.
The standard error of the estimated difference $\hat{p}_1-\hat{p}_2$ is
$$ \begin{aligned} SE =\sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1+2}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2+2}} \end{aligned} $$
The margin of error for the estimated difference $(\hat{p}_1-\hat{p}_2)$ is
$$ \begin{aligned} E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1+2}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2+2}} \end{aligned} $$
where $Z_{\alpha/2}$ is the value from normal statistical table.
Then $100(1-\alpha)$% confidence interval for the difference $(p_1-p_2)$ is
$$ \begin{aligned} (\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E. \end{aligned} $$
Assumptions
a. The sample proportions are from the two random samples that are independent.
b. For each of the two samples the sample sizes are at least 5.
Step by Step Procedure
Step by step procedure to estimate the confidence interval for difference between two population proportions is as follows:
Step 1 Specify the confidence level $(1-\alpha)$
Step 2 Given information
Specify the given information, sample sizes $n_1$ and $n_2$. The observed number of successes $X_1$ and $X_2$.
The plus four estimates of population proportions are $\hat{p}_1 =\dfrac{X_1+1}{n_1+2}$ and $\hat{p}_2 = \dfrac{X_2+1}{n_2+2}$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval for the difference $(p_1-p_2)$ is
$$ \begin{aligned} (\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E. \end{aligned} $$
where $E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1+2}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2+2}}$.
Step 4 Determine the critical value
Determine the critical value $Z_{\alpha/2}$ from the normal statistical table that corresponds to the desired confidence level.
Step 5 Compute the margin of error
The margin of error for the difference $(p_1-p_2)$ is
$$ \begin{aligned} E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1+2}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2+2}}. \end{aligned} $$
Step 6 Determine the confidence interval
Thus, $100(1-\alpha)$% confidence interval estimate for the difference $(p_1-p_2)$ is $(\hat{p}_1 -\hat{p}_2) \pm E$ or $\big((\hat{p}_1-\hat{p}_2) -E, (\hat{p}_1-\hat{p}_2) +E\big)$.
