A poll taken this year asked 1012 adults whether they were fans of a particular
ID: 3206026 • Letter: A
Question
A poll taken this year asked 1012 adults whether they were fans of a particular spoil and 75% said they were. Last year, 80% of a similar-size sample had reported being fans of the sport. Complete parts a through e below. Find the margin of error for the poll taken this year if one wants 90% confidence in the estimate of the percentage of adults who are fans of the sport. ME = (Round to three decimal places as needed.) Explain what that margin of error means. There is a 90% chance that the true proportion of adults who are fans of the sport is within plusminus ME of this sample proportion. In 9C% of samples of adults, the proportion who are fans of the sport will be within plusminus ME of the sample proportion. One is 90% confident that this sample proportion is within plusminus ME of the true proportion of adults who are fans of the sport. One is 90% plusminus ME confident that the true proportion of adults who are fans of the sport is equal to this sample proportion. If one wanted to be 80% confident instead of 90% confident, would the margin of error be larger or smaller? To be less confident, the interval needs to contain the true proportion more often, so the margin of error would be larger. To be less confident, the interval needs to contain the true proportion less often, so the margin of error would be smaller. To be less confident, the interval needs to contain the true proportion more often, so the margin of error would be smaller. To be less confident, the interval needs to contain the true proportion less often, so the margin of error would be larger. Find the margin of error for the poll taken this year if one wants 80% confidence in the estimate of the percent of adults who are fans of the sport. ME = (Round to three decimal places as needed.) In general, if all other aspects of the situation remain the same, will smaller margins of error produce greater or less confidence in the interval? Greater confidence Less confidenceExplanation / Answer
n=1012
p=.75
q=1-0.75
= 0.25
mean = n*p = 1012*0.75 = 759
variance = n*p*q = 1012*0.75*0.25
= 189.75
SD = sqrt(variance) = 13.77
Standard Error(S.E) = SD/sqrt(1012)
= 0.43
crirtica value at 90% confidence interval is 1.64
Margin of Error (ME) = critical value * Standard Error
= 1.64 * 0.43
= 0.71
b) One is 90% confident that the sample proportion is within +/- ME of the true portions of the adult who are fans of the sport.
c) B
d) For 80% confidence interval the critical value is 1.28
ME = 1.28 * 0.43
= 0.55
e) Less Confidence