SOLUTION: Margin of Error for Proportions 1. A survey was given to a random sample of 125 voters in the United States to ask about their preference for a presidential candidate. Of those su

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Question 1175298: Margin of Error for Proportions
1. A survey was given to a random sample of 125 voters in the United States to ask about their preference for a presidential candidate. Of those surveyed, 76% of the people said they preferred Candidate A. At the 95% confidence level, what is the margin of error for this survey expressed as a percentage to the nearest tenth? (Do not write \pm±).
Confidence Intervals for Proportions
2. A survey was given to a random sample of 60 voters in the United States to ask about their preference for a presidential candidate. Of those surveyed, 27 respondents said that they preferred Candidate A. Determine a 95% confidence interval for the proportion of people who prefer Candidate A, rounding values to the nearest thousandth. (,)
Answer by ewatrrr(24785) About Me  (Show Source):
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Hi

ME = z%2Asqrt%28%28p%281-p%29%29%2Fn%29
ME = 1.96%2Asqrt%28%28.76%28.24%29%29%2F125%29 = .075
CI: (.685 , .835 )

2) mean= 27/60= .45
ME = 1.96sqrt%28%28.45%29%28.55%29%2F60%29  = .126 
CI: (.324 , .576) 
   
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