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Let P Be the Success Probability of a Bernoulli Random

question 63

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

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