Question

A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 5% margin of error at a 99% confidence level, what size of sample is needed

Answer 1

A sample size of 664 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:

`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`.

If the candidate only wants a 5% margin of error at a 99% confidence level, what size of sample is needed

A sample size of n is needed.

n is found when M = 0.05.

We have no estimate, so we use
`\pi = 0.05`. Then

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

`0.05 = 2.575\sqrt{(0.5*0.5)/(n)}`

`0.05√(n) = 2.575*0.5`

`√(n) = (2.575*0.5)/(0.05)`

`(√(n))^(2) = ((2.575*0.5)/(0.05))^(2)`

`n = 663.1`

Rounding up

A sample size of 664 is needed.