Question 663359: suppose that we conducted a survey in which we asked 800 people the following question: "in the event of an emergency, would you seek medical assistance at the emergency room of our hospital?" If 180 of the respondents indicated that they would seek care at the ER, construct a 95% confidence interval for the true proportion.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! suppose that we conducted a survey in which we asked 800 people the following question: "in the event of an emergency, would you seek medical assistance at the emergency room of our hospital?" If 180 of the respondents indicated that they would seek care at the ER, construct a 95% confidence interval for the true proportion.
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p-hat = 180/800 = 0.225
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ME = 1.96*sqrt[0.225*0.775/800] = 0.0289
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95% CI: 0.225-0.0289 <= p <= 0.225+0.0289
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Cheers,
Stan H.
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