Question

insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. they randomly survey 382 drivers and find that 288 claim to always buckle up. construct a 90% confidence interval for the population proportion that claim to always buckle up. do not round between steps. round answers to at least 4 decimal places. question help: message instructor submit question

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 288$ (number of successes) and $n=382$ (sample size).
$\hat{p}=\frac{288}{382}\approx0.7539$

Step2: Find $z -$ value

For a $90\%$ confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z -$ value $z_{\alpha/2}=z_{0.05}$. From the standard normal table, $z_{0.05} = 1.645$

Step3: Calculate the margin of error

The margin of error $E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.7539$, $n = 382$, and $z_{\alpha/2}=1.645$
$E=1.645\sqrt{\frac{0.7539\times(1 - 0.7539)}{382}}$
First, calculate $(1-\hat{p})=1 - 0.7539 = 0.2461$
$\hat{p}(1-\hat{p})=0.7539\times0.2461\approx0.1856$
$\frac{\hat{p}(1-\hat{p})}{n}=\frac{0.1856}{382}\approx0.000486$
$\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\approx\sqrt{0.000486}\approx0.0220$
$E=1.645\times0.0220\approx0.0362$

Step4: Calculate the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-ESubstitute $\hat{p}=0.7539$ and $E = 0.0362$
$\hat{p}-E=0.7539-0.0362 = 0.7177$
$\hat{p}+E=0.7539+0.0362=0.7901$

Answer:

$0.7177 < p < 0.7901$