An opinion poll during the presidential campaign sampled n = 500 potential voter

Question

An opinion poll during the presidential campaign sampled n = 500 potential voters in June. A primary purpose of the poll was to obtain an estimate of the proportion of potential voters who favor each candidate. Assume a planning value for the population proportion of 0.50 and a 95% confidence level. What was the "expected" margin of error for the June poll? a 0.054 b 0.049 c 0.044 d 0.038 As the November election gets closer, better precision and smaller margins of error are desired. Assume the following margin of error is requested for the survey to be conducted during the presidential campaign. Compute the recommended sample size The requested margin of error is plusminus 3 percentage points. a 1085 b 1068 c 1020 d 1005

Explanation / Answer

22. Here z=1.96 and p=0.50

So SE(p)=sqrt(p(1-p)/n)=sqrt(0.5*0.5/500)=0.044 so answer is c

23. Here p=0.5 and E=0.03

So n=z^2*p*(1-p)/E^2=1.96^2*0.5*0.5/0.0009=1068 so b is answer

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