Question

Liam wants to estimate the percentage of people who lease a car. He surveys 240 individuals and finds that 54 lease a car. Find the margin of error for the confidence interval for the population proportion with a 95% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

Answer 1

Answer:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The confidence interval is 95% and the significance level is $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ and the critical value would be:

$z_{\alpha/2}= 1.96$

And the margin of error would be:

$ME = 1.96 \sqrt{\frac{0.225 (1-0.225)}{240}}= 0.0528$

Step-by-step explanation:

We have the following info given:

$n =240$ the sample size selected

$X=54$ the number of people who lease a car

$\hat p =\frac{54}{240}= 0.225$ the estimated proportion of people who lease a car

The margin of error is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The confidence interval is 95% and the significance level is $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ and the critical value would be:

$z_{\alpha/2}= 1.96$

And the margin of error would be:

$ME = 1.96 \sqrt{\frac{0.225 (1-0.225)}{240}}= 0.0528$