An insurance company checks police records on 1200 randomly selected auto accidents and notes that teenagers were at the wheel in 180 of them. Construct a 90% confidence interval estimate of the proportion of auto accidents that involve teenage drivers. Show all work.


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 $n =1200$, observed value of $X$ is $X=180$.

The estimate of the proportion is $\hat{p} =\frac{X}{n} =\frac{180}{1200}=0.15$.

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 0.1
Z-critical 0.1

Thus $Z_{\alpha/2} = Z_{0.05} = 1.645$.

Step 5 Compute the margin of error

The margin of error for proportions is
$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.645 \sqrt{\frac{0.15*(1-0.15)}{1200}}\\ & =0.017. \end{aligned} $$

Step 6 Determine the confidence interval

$90$% confidence interval estimate for population proportion is

$$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.15 - 0.017 & \leq p \leq 0.15 + 0.017\\ 0.133 & \leq p \leq 0.167. \end{aligned} $$
Thus, $90$% confidence interval estimate for population proportion $p$ is $(0.133,0.167)$.

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