Let p be the success probability of a Bernoulli random variable Y,i.e. ,p = Pr(Y = 1).It can be shown that ,the fraction of successes in a sample,is asymptotically distributed N(p,
.Using the estimator of the variance of
,
,construct a 95% confidence interval for p.Show that the margin for sampling error simplifies to 1/
if you used 2 instead of 1.96 assuming,conservatively,that the standard error is at its maximum.Construct a table indicating the sample size needed to generate a margin of sampling error of 1%,2%,5% and 10%.What do you notice about the increase in sample size needed to halve the margin of error? (The margin of sampling error is 1.96×SE(
))
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