SOLUTION: In a survey of 212 people at the local track and field championship, 72% favored the home team winning. Find the margin of error for the survey, and give and interval that is likel

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Question 193875: In a survey of 212 people at the local track and field championship, 72% favored the home team winning. Find the margin of error for the survey, and give and interval that is likely to contain the exact percent of all people who favor the home team winning.
I got the Margin of Error: 6.86% but I have no idea on how to get the interval.
Thanks
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
In a survey of 212 people at the local track and field championship, 72% favored the home team winning. Find the margin of error for the survey, and give and interval that is likely to contain the exact percent of all people who favor the home team winning.
I got the Margin of Error: 6.86% but I have no idea on how to get the interval.
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To generate a confidence interval you need a sample-proportion (p-hat)
and a margin of error (E).
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Your p-hat is 72% because that is the proportion of the sample you are
considering.
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C.I. : p-hat - E < p < p-hat + E
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Cheers,
Stan H.