# Solved (Free): A survey was conducted on a random sample of 1,000 Baltimore residents. Residents were asked whether they have health insurance. 650 individuals surveyed

#### ByDr. Raju Chaudhari

Apr 5, 2021

A survey was conducted on a random sample of 1,000 Baltimore residents. Residents were asked whether they have health insurance. 650 individuals surveyed said they do have health insurance, and 350 said they do not have health insurance. Find the 95% CI for the proportion of Baltimore residents with health insurance.

#### Solution

Given that sample size $n =1000$, observed value of $X$ (number of individuals havinf health insurance) is $X=650$.

##### 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 $n =1000$, observed value of $X$ is $X=650$.

The estimate of the proportion is $\hat{p} =\frac{X}{n} =\frac{650}{1000}=0.65$.

##### 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 MARKDOWN_HASHd57c9a47c98e9677cf7bc95478054db0MARKDOWNHASH 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{0.65*(1-0.65)}{1000}}\\ & =0.03. \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\\ 0.65 - 0.03 & \leq p \leq 0.65 + 0.03\\ 0.6204 & \leq p \leq 0.6796. \end{aligned}

$95$% confidence interval estimate for population proportion $p$ is $(0.6204,0.6796)$.

Thus the $95$% confidence interval estimate for proportion of Baltimore residents with health insurance is $(0.6204,0.6796)$.