Misleading survey responses:In a survey of 1002 people , 701 said thatthey voted in a recent presidential election (based on data fromICR Research Group). Voting records show that 61% of eligible voteractually did vote. a. Find a 99% confidence interval estimate ofthe proportion of people who say that they voted. b. Are the survey results consistent with theactual voter turnout fo 61%? Why or why not? Misleading survey responses:In a survey of 1002 people , 701 said thatthey voted in a recent presidential election (based on data fromICR Research Group). Voting records show that 61% of eligible voteractually did vote. a. Find a 99% confidence interval estimate ofthe proportion of people who say that they voted. b. Are the survey results consistent with theactual voter turnout fo 61%? Why or why not? a. Find a 99% confidence interval estimate ofthe proportion of people who say that they voted. b. Are the survey results consistent with theactual voter turnout fo 61%? Why or why not?
Explanation / Answer
From the given information that n = 1002 and 701 said that they voted in a recent presidentialelection. That is x = 701 Proportion of the persons said that they are voted p(hat) =701/1002 = 0.6996 ~ 0.7 H0 : p=0.5 against Ha : P (not equal)0.5 99 % confidence interval for proportion of people (0.662297, 0.736905) The result is consistent with the 61% of the proportion ofpersons that are voted. Since the upper interval is in the above interval.