Question

In a randomly selected sample of 500 Phoenix residents, 445 supported mandatory sick leave for food handlers. Legislators want to be very confident that voters will support this issue before drafting a bill. What is the 99% confidence interval for the percentage of Phoenix residents who support mandatory sick leave for food handlers?

Answer 1

Answer:

The 99% confidence interval for the percentage of Phoenix residents who support mandatory sick leave for food handlers is between 85.40% and 92.60%.

Step-by-step explanation:

Confidence interval for the proportion:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 500, \pi = \frac{445}{500} = 0.89$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.89 - 2.575\sqrt{\frac{0.89*0.11}{500}} = 0.8540$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.89 + 2.575\sqrt{\frac{0.89*0.11}{500}} = 0.9260$

For the percentage:

Multiply the proportion by 100.

0.8540*100 = 85.40%

0.9260*100 = 92.60%

The 99% confidence interval for the percentage of Phoenix residents who support mandatory sick leave for food handlers is between 85.40% and 92.60%.