Question

A researcher would like to estimate the population proportion of adults living in a certain town who have at least a high school education. No more information is available about its value. How large of a sample size is needed to estimate it to within 0.15 of the true value with 99% confidence?

Answer 1

We would need a sample of size at least 74.

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:

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

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`.

How large of a sample size is needed to estimate it to within 0.15 of the true value with 99% confidence?

We need a sample of size at least n

n is found when
`M = 0.15`

We don't know the proportion, so we estimate
`\pi = 0.5`, which is the case for which we are going to need the largest sample size.

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

`0.15 = 2.575\sqrt{(0.5*0.5)/(n)}`

`0.15√(n) = 2.575*0.5`

`√(n) = (2.575*0.5)/(0.15)`

`(√(n))^(2) = ((2.575*0.5)/(0.15))^(2)`

`n = 73.67`

Rounding up

We would need a sample of size at least 74.