If the candidate only wants a 10% margin of error at a 95% confidence level, what size of sample is needed?

Answer 1

Assuming that this sample is for estimating a proportion, otherwise we are missing standard deviation.

Sample size >= (z-critical / margin of error) ^2 * p * (1-p), where p = population proportion.

Since p is not give, we assume it is 0.5. the z-critical for 95% confidence level is 1.96.

Sample size >= (1.96/0.1)^2 * 0.25 = 96.04, so required sample size is 97.