Question

1. insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. they randomly survey 415 drivers and find that 315 claim to always buckle up. construct a 99% confidence interval for the population proportion that claim to always buckle up.

Explanation:

Step1: Calculate sample proportion $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 315$ (number of successes) and $n=415$ (sample size).
$\hat{p}=\frac{315}{415}\approx0.759$

Step2: Find $z -$ value for 99% confidence level

For a 99% confidence level, 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 the margin of error $E$

The formula for the margin of error for a proportion is $E=z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.759$, $n = 415$ and $z_{\frac{\alpha}{2}}=2.576$
First, calculate $(1-\hat{p})=1 - 0.759 = 0.241$
Then $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.759\times0.241}{415}}\approx\sqrt{\frac{0.183}{415}}\approx\sqrt{0.000441}\approx0.021$
$E=2.576\times0.021\approx0.054$

Step4: Construct the confidence interval

The confidence interval for the population proportion $p$ is given by $\hat{p}-ESubstitute $\hat{p}=0.759$ and $E = 0.054$
$0.759-0.054

Answer:

The 99% confidence interval is $(0.705,0.813)$