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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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