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L Let pp Be the Success Probability of a Bernoulli Random Variable

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L Let pp be the success probability of a Bernoulli random variable YY , i.e., p=Pr(Y=1)p = \operatorname { Pr } ( Y = 1 ) . It can be shown that p^\hat { p } , the fraction of successes in a sample, is asymptotically distributed N(p,p(1p)n)N \left( p , \frac { p ( 1 - p ) } { n } \right) . Using the estimator of the variance of p^,p^(1p^)n\hat { p } , \frac { \hat { p } ( 1 - \hat { p } ) } { n } , construct a 95%95 \% confidence interval for pp . Show that the margin for sampling error simplifies to 1/n1 / \sqrt { n } if you used 2 instead of 1.961.96 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%1 \% , 2 \% , 5 \% and 10%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(p^)1.96 \times S E ( \hat { p } ) .


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