Question

A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him/her. If the candidate wants a 3% margin of error at a 90% confidence level, what size of sample is needed

Answer 1

A sample size of 752 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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

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

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a p-value of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

If the candidate wants a 3% margin of error at a 90% confidence level, what size of sample is needed?

We have no estimate of the proportion, so we use
`\pi = 0.5`.

The sample size is n for which M = 0.03. So

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

`0.03 = 1.645\sqrt{(0.5*0.5)/(n)}`

`0.03√(n) = 1.645*0.5`

`√(n) = (1.645*0.5)/(0.03)`

`(√(n))^2 = ((1.645*0.5)/(0.03))^2`

`n = 751.67`

Rounding up:

A sample size of 752 is needed.