Question

Suppose you want to investigate the proportion of women police officers. How large should a sample be if the margin of error is .05 for a 92% confidence interval

Answer 1

A sample of 307 is needed.

Explanation:

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)`.

The margin of error is of:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

In this question, we have that:

We dont know the true proportion of women police officers, so we use
`\pi = 0.5`, which is when the largest sample size would be needed.

92% confidence level

So
`\alpha = 0.08`, z is the value of Z that has a pvalue of
`1 - (0.08)/(2) = 0.96`, so
`Z = 1.75`.

How large should a sample be if the margin of error is .05 for a 92% confidence interval

We need a sample of n, and n is found when
`M = 0.05`. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.05 = 1.75\sqrt{(0.5*0.5)/(n)}`

`0.05√(n) = 1.75*0.5`

`√(n) = 17.5`

`(√(n))^2 = (17.5)^2`

`n = 306.25`

Rounding up

A sample of 307 is needed.