Question

insurance companies are interested in knowing the population proportion of drivers who always buckle up before riding in a car. they randomly survey 402 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. do not round between steps. round answers to at least 4 decimal places.

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 284$ and $n=402$.
$\hat{p}=\frac{284}{402}\approx0.7065$

Step2: Find $z$-score

For a $99\%$ confidence interval, the significance level $\alpha=1 - 0.99=0.01$, and $\alpha/2=0.005$. The $z$-score $z_{\alpha/2}=z_{0.005}$. From the standard normal table, $z_{0.005} = 2.576$

Step3: Calculate margin of error

The margin of error $E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.7065$, $n = 402$, and $z_{\alpha/2}=2.576$
$E=2.576\sqrt{\frac{0.7065\times(1 - 0.7065)}{402}}$
First calculate $0.7065\times(1 - 0.7065)=0.7065\times0.2935 = 0.2074$
Then $\sqrt{\frac{0.2074}{402}}\approx\sqrt{0.000516}= 0.0227$
$E=2.576\times0.0227\approx0.0585$

Step4: Construct confidence interval

The confidence interval is $\hat{p}-E$0.7065-0.0585 < p<0.7065 + 0.0585$
$0.6480

Answer:

$0.6480 < p < 0.7650$