A clinical trial is run to evaluate the effectiveness of a new drug to prevent preterm delivery. A total of n=250 pregnant women agree to participate and are randomly assigned to receive either the new drug or a placebo and followed through the course of pregnancy. Among 125 women receiving the new drug, 24 deliver preterm and among 125 women receiving the placebo, 38 deliver preterm. Construct a 95 % confidence interval for the difference in proportions of women who deliver preterm.
Solution
Given information
| . | New Drug | Placebo |
|---|---|---|
| Sample size | $n_1=125$ | $n_2=125$ |
| Observed no. of Preterm delivery | $X_1=24$ | $X_2=38$ |
We wish to determine $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.
Step 1 Specify the confidence level
Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.
Step 2 Given information
Given that $X_1 = 24$, $X_2 = 38$, $n_1 = 125$, $n_2 = 125$.
The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{24}{125}=0.192$ and the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{38}{125}=0.304$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ 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 $E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$ 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 the difference $(p_1-p_2)$ is
$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\ & = 1.96 \sqrt{\frac{0.192*(1-0.192)}{125}+\frac{0.304*(1-0.304)}{125}}\\ &= 0.1062. \end{aligned} $$
Step 6 Determine the confidence interval
$95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is
$$ \begin{aligned} (\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\ (0.192-0.304) - 0.1062 & \leq (p_1-p_2) \leq (0.192-0.304) + 0.1062\\ -0.2182 & \leq (p_1-p_2) \leq -0.0058 \end{aligned} $$
Thus, $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(-0.2182,-0.0058)$.
Interpretation
We can be $95$% confident that the difference in proportions of women who deliver preterm $(p_1-p_2)$ is between $-0.2182$ and $-0.0058$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
