SOLUTION: In a random sample of 1,500 voters showed that 52% favor candidate A over candidate B in the
primary elections. Give a 99% confidence interval for the proportion of voters t
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primary elections. Give a 99% confidence interval for the proportion of voters t
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Question 1131561: In a random sample of 1,500 voters showed that 52% favor candidate A over candidate B in the
primary elections. Give a 99% confidence interval for the proportion of voters that prefer candidate A.
You can put this solution on YOUR website! half interval is z* sqrt(p*(1-p))/n=2.576*sqrt(.52*.48/1500)=0.0332
the point estimate is 0.52
the interval is to two decimal places (0.49, 0.55), to three it is (0.487, 0.553)
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