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\subsection{Choosing Sample Size}  you can find the sample size needed for a given confidence interval. $ME=z*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ and you can solve for n.\\  with some algebra $n=\frac{(z*)^2\hat{p}(1-\hat{p})}{ME^2}$  Special case of 95 percent is $n=\frac{1}{m^2}$