If a candidate only wants a 2% margin of error at a 90% confidence level, what size of sample is needed?

Answer 1

So, assuming you are talking about a candidate for office based on the wording of the question, we may apply our knowledge from a Bernoulli distribution. In a Bernoulli, we know that:

:

CI = sample probability ± MOE

MOE = Critical Value × sqrt (sample probability × (1-sample probability) / n)

We know that this critical value is 1.645 and that MOE ≤ .02. Let's plug those in:

.02 ≥ 1.645 × sqrt (sample probability × (1-sample probability) / n)

sqrt (sample probability × (1-sample probability) / n) ≤ .02/1.645

We know that variance is maximized in a Bernoulli when it is at 50%. Thus, any other probabilities would lead to a smaller variance. So let's use 50% for a worst-case scenario:

sqrt (.5 × (.5) / n) ≤ .02/1.645

sqrt (.25 / n) ≤ .02/1.645

.25 / n ≤ (.02/1.645)^2

.25 / [(.02/1.645)^2] ≤ n

1691.265 ≤ n

Thus, there must be 1,692 people or more to ensure that there is a 2% margin of error in a 90% confidence interval.