Question

insurance companies are interested in knowing the population proportion of drivers who always buckle up before riding in a car. they randomly survey 420 drivers and find that 284 claim to always buckle up. construct a 99% confidence interval for the population proportion that claim to always buckle up. < select an answer < do not round between steps. round answers to at least 4 decimal places. > next question

Explanation:

Step1: Calculate sample proportion

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 284$ (number of successes) and $n=420$ (sample size).
$\hat{p}=\frac{284}{420}\approx0.6762$

Step2: Find $z -$ value

For a $99\%$ confidence interval, the significance level $\alpha=1 - 0.99=0.01$. Then $\frac{\alpha}{2}=0.005$. The $z -$ value $z_{\frac{\alpha}{2}}$ is the $z$ - score such that the area to the right of it is $0.005$. From the standard normal table, $z_{\frac{\alpha}{2}} = 2.576$

Step3: Calculate margin of error

Margin of error $E=z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.6762$, $n = 420$ and $z_{\frac{\alpha}{2}}=2.576$
$E=2.576\sqrt{\frac{0.6762\times(1 - 0.6762)}{420}}$
First calculate $(1-\hat{p})=1 - 0.6762=0.3238$
$\hat{p}(1 - \hat{p})=0.6762\times0.3238 = 0.2190$
$\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2190}{420}\approx0.0005214$
$\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\approx\sqrt{0.0005214}\approx0.0228$
$E=2.576\times0.0228\approx0.0587$

Step4: Calculate confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-E$\hat{p}-E=0.6762- 0.0587=0.6175$
$\hat{p}+E=0.6762 + 0.0587=0.7349$

Answer:

$0.6175