Assuming we use the same sample for both intervals, how would the margin of erro

Question

Assuming we use the same sample for both intervals, how would the margin of error of a 99% confidence interval compare with the margin of error of a 90% confidence interval? It would be smaller because it omits only 1% of the possible samples instead of 10%. It would be the same because the sample is the same. It would be larger because higher confidence requires a larger margin of error. We can't tell because the margin of error is random. We can't tell because it depends on the size of the population. Suppose we didn't actually know the true proportion of all Ohio State students who bought football tickets. We could then take a sample to try and estimate it. From our sample, we find that 124 of 200 people actually bought tickets. Form a 90% confidence interval for the true proportion of students who bought tickets using the sample described in this problem. when forming this confidence interval, do not round your answer until the very end of the problem. Choose the answer that is closest to what you calculate. 0.55 plusminus 0.056 0.55 plusminus 0.067 0.62 plusminus 0.056 0.62 plusminus 0.067 In the previous problem where you formed a confidence interval for the true proportion of students who bought football tickets, what would happen to our interval if we instead used a 95% level of confidence? Assume everything else would remain the same (sample size, sample proportion). It would be narrower. It would be the same because the sample is the same. It would be wider. It is impossible to tell without seeing the actual data.

Explanation / Answer

7) option c is correct

8)here p=124/200=0.62 ; n=200

std error =(p(1-p)/n)1/2 =0.034

for 90% CI; z=1.64485

hence margin of error =z*std error =0.056

therefore option c is correct

9)option C is correct

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