In a random sample of 500 high school juniors in a Western state, 360 of them said that they intended to continue their education at an in-state college or university. Find a 95% confidence interval for the corresponding true proportion.


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 size $n =500$, observed value of $X$ is $X=360$.

The estimate of the proportion is $\hat{p} =\frac{X}{n} =\frac{360}{500}=0.72$.

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 $Z_{\alpha/2} = Z_{0.025} = 1.96$.

Step 5 Compute the margin of error

The margin of error for proportions is

$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.72*(1-0.72)}{500}}\\ & =0.039. \end{aligned} $$

Step 6 Determine the confidence interval

$95$% confidence interval estimate for population proportion is

$$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.72 - 0.039 & \leq p \leq 0.72 + 0.039\\ 0.6806 & \leq p \leq 0.7594. \end{aligned} $$
Thus, $95$% confidence interval estimate for population proportion $p$ is $(0.6806,0.7594)$.

Further Reading