Solved (Free): Of 99 adults selected randomly from one town, 63 have health insurance. Find a 95% confidence interval for the true proportion

ByDr. Raju Chaudhari

Mar 14, 2021

Of 99 adults selected randomly from one town, 63 have health insurance. Find a 95% confidence interval for the true proportion of all adults in the town who do not have health insurance.

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 $n =99$. The number of adults who have health insurance is 63. So the number of adults who do not have health insurance is $X=99 -63 = 36$.

The estimate of the proportion is $\hat{p} =\frac{X}{n} =\frac{36}{99}=0.364$.

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{0.364*(1-0.364)}{99}}\\ & =0.095. \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\\ 0.364 - 0.095 & \leq p \leq 0.364 + 0.095\\ 0.2689 & \leq p \leq 0.4584. \end{aligned}
Thus, $95$% confidence interval estimate for population proportion $p$ is $(0.2689,0.4584)$.