I know that ME = z(a/2)*SE = z(a/2) * (sd/sqrt(n)). This implies, n = z(a/)^2 * sd / ME^2. For level of significance 0.05, I am 95% confident about my results. Z(a/2) = z(0.05/2) = 1.96. I choose 0.5 as standard deviation, and a margin of error (confidence interval) of +/- 5%, then sample size is calculated as:
1.96^2 * 0.5 / 0.05^2
=768.32
=738 (approximately)