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

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%.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 500, \pi = (445)/(500) = 0.89`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

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

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.89  + 2.575\sqrt{(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%.