The American Cancer Society wants to estimate the proportion of high school seniors who smokes regularly. In a random sample of 1,000 seniors, 200 smoke regularly. Construct 90% confidence interval. Interpret your confidence interval.

Solution

Given that sample size $n = 1000$, observed $X = 200$.

Thus the sample proportion is
$\hat{p}=\frac{X}{n}=\frac{200}{1000}=0.2$.

The procedure for $0.9$ % confidence interval for the proportion of high school seniors who smoke regularly is as follows:

Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.

Step 2 Given information

Given that sample size $n =1000$, observed data $X=200$.

The estimate of the proportion of is $\hat{p} =\frac{X}{n} =\frac{200}{1000}=0.2$.

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 $Z_{\alpha/2}$.

Z-critical value
Z-critical value

Thus $Z_{\alpha/2} = Z_{0.05} = 1.64$.

Step 5 Compute the margin of error

The margin of error for proportions is

$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.64 \sqrt{\frac{0.2*(1-0.2)}{1000}}\\ & =0.021. \end{aligned} $$

Step 6 Determine the confidence interval

$90$% confidence interval estimate for population proportion is

$$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.2 - 0.021 & \leq p \leq 0.2 + 0.021\\ 0.1793 & \leq p \leq 0.2207. \end{aligned} $$
Thus, $90$% confidence interval estimate for population proportion $p$ is $(0.1793,0.2207)$.

Further Reading