What is a 95% confidence interval for this proportion and what is the sample siz

Question

What is a 95% confidence interval for this proportion and  what is the sample size required to estimate this proportion within 5%?

0.6065 0.594475 0.609351 0.59213 0.651685 0.660682 0.624394 0.54855 0.580872 0.576196 0.565428 0.574544 0.562025 0.600096 0.589394 0.614111 0.630723 0.591657 0.555374 0.598626 0.600915 0.617345 0.58968 0.628302 0.571802 0.610067 0.615338 0.609584 0.604889 0.572609 0.583361 0.60772 0.561712 0.698264 0.634606 0.597896 0.56449 0.592514 0.604167 0.591483 0.630357 0.606719 0.610114 0.606303 0.583529 0.581453 0.567802 0.621274 0.59841 0.585713

Explanation / Answer

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.5
ME = 0.05
n = ( 1.96 / 0.05 )^2 * 0.5*0.5
= 384.16 ~ 385

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