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?

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

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 500, \pi = \frac{445}{500} = 0.89[/tex]

99% confidence level

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

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.89 - 2.575\sqrt{\frac{0.89*0.11}{500}} = 0.8540[/tex]

The upper limit of this interval is:

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

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%.