An insurance company checks police records on 500 randomly selected auto accidents and notes that teenagers were at the wheel in 80 of them. Construct a 95% confidence interval estimate of the proportion of auto accidents that involve teenage drivers.

#### 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 `$n =500$`

, observed value of $X$ is `$X=80$`

.

The estimate of the proportion is `$\hat{p} =\frac{X}{n} =\frac{80}{500}=0.16$`

.

##### 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

` $$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.16*(1-0.16)}{500}}\\ & =0.032. \end{aligned} $$ `

##### Step 6 Determine the confidence interval

`$95$%`

confidence interval estimate for population proportion is

` $$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.16 - 0.032 & \leq p \leq 0.16 + 0.032\\ 0.1279 & \leq p \leq 0.1921. \end{aligned} $$ `

Thus, `$95$%`

confidence interval estimate for population proportion $p$ is `$(0.1279,0.1921)$`

.

#### Further Reading

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