Plus Four CI for difference between two population proportions
In this tutorial we will discuss some step by step numerical examples to estimate plus four confidence interval for the difference between two population proportions.
Example
For two independent samples A and B following information is available:
Sample A has 790 number of successes from a sample size of 805.
Sample B has 798 number of successes from a sample size of 811.
Construct a 95% confidence interval for the difference between the proportion of successes for sample A and B.
Solution
Here the confidence coefficient is at least 90% and both the sample sizes are at least 5 we use plus four confidence interval for the difference between the proportion of successes.
Step 1 Specify the confidence level $(1-\alpha)$
Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.
Step 2 Given information
Given that
| . | Sample A | Sample B |
|---|---|---|
| Sample size | 805 | 811 |
| No. of successes | 790 | 798 |
The estimate of sample proportion of success in Sample A based on plus four rule is
$\hat{p}_1=\dfrac{X1+1}{n1+2}=\dfrac{790+1}{802+2}=0.9838$
and the estimate of sample proportion of success in Sample A based on plus four rule is
$\hat{p}_2=\dfrac{X2+1}{n2+2}=\dfrac{798+1}{811+2}=0.9828$.
Step 3 Specify the formula
$100(1-\alpha)$% plus four confidence interval for difference in population proportions is
$$ \begin{aligned} \hat{p}_1-\hat{p}_2 - E \leq p \leq \hat{p}_1-\hat{p}_2 + E. \end{aligned} $$
where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1+2}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2+2}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.
Step 4 Determine the critical value
The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.
Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.
Step 5 Compute the margin of error
The margin of error for proportions is
$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1+2}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2+2}}\\ & = 1.96 \sqrt{\frac{0.9838*(1-0.9838)}{802+2}+\frac{0.9828*(1-0.9828)}{811+2}}\\ & =0.0123. \end{aligned} $$
Step 6 Determine the confidence interval
$95$% plus four confidence interval estimate for population proportion is
$$ \begin{aligned} \hat{p}_1-\hat{p}_2 - E & \leq p \leq \hat{p}_1-\hat{p}_2 + E\\ 0.9838-0.9828 - 0.0123 & \leq p \leq 0.9838-0.9828 + 0.0123\\ -0.0113 & \leq p \leq 0.0133. \end{aligned} $$
Thus, $95$% plus four confidence interval for the difference in population proportion of successes $p_1-p_2$ is $(-0.0113,0.0133)$.
