# Solved (Free): In a study conducted for the effectiveness of drug A on a certain disease, 60 randomly selected patients with this disease were given medication

#### ByDr. Raju Chaudhari

Apr 16, 2021

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 $(0.4368,0.7632)$.