In a study conducted for the effectiveness of drug A on a certain disease, 60 randomly selected patients with this disease were given medication and 36 patients recovered. Calculate the confidence interval for the efficacy of this drug for a 99% confidence level.
Solution
Step 1 Specify the confidence level $(1-\alpha)$
Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.
Step 2 Given information
Given that sample size $n =60$, observed value of $X$ is $X=36$.
The estimate of the proportion is $\hat{p} =\frac{X}{n} =\frac{36}{60}=0.6$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval for population proportion is
$$ \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned} $$
where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ 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 $Z_{\alpha/2} = Z_{0.005} = 2.58$.
Step 5 Compute the margin of error
The margin of error for proportions is
$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 2.58 \sqrt{\frac{0.6*(1-0.6)}{60}}\\ & =0.163. \end{aligned} $$
Step 6 Determine the confidence interval
$99$% confidence interval estimate for population proportion is
$$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.6 - 0.163 & \leq p \leq 0.6 + 0.163\\ 0.4368 & \leq p \leq 0.7632. \end{aligned} $$
$99$% confidence interval estimate for population proportion $p$ is $(0.4368,0.7632)$.
Thus the $99$% confidence interval estimate for the efficacy of this drug is $(0.4368,0.7632)$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
