Let p be the success probability of a Bernoulli random variable Y, i.e., p = Pr(Y = 1). It can be shown that $\hat { p }$ , the fraction of successes in a sample, is asymptotically distributed N(p, $\frac { p ( 1 - p ) } { n }$ Using the estimator of the variance of $\hat { p }$ , $\frac { \hat { p } ( 1 - \hat { p } ) } { n }$ , construct a 95% confidence interval for p. Show that the margin for sampling error simplifies to 1/ $\sqrt { n }$ 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( $\hat { p }$ ))

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The Answer of Let p be the success probability of a Bernoulli random variable Y, i.e., p = Pr(Y = 1). It can be shown that $\hat { p }$ , the fraction of successes in a sample, is asymptotically distributed N(p, $\frac { p ( 1 - p ) } { n }$ Using the estimator of the variance of $\hat { p }$ , $\frac { \hat { p } ( 1 - \hat { p } ) } { n }$ , construct a 95% confidence interval for p. Show that the margin for sampling error simplifies to 1/ $\sqrt { n }$ 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( $\hat { p }$ ))

Let P Be the Success Probability of a Bernoulli Random p^\hat { p }p^​

Let P Be the Success Probability of a Bernoulli Random $\hat { p }$