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.

#### Solution

#### 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 `$(0.2689,0.4584)$`

.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators