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

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) (Show Source): You can put this solution on YOUR website!


Hi

ME = 
ME =  = .075
CI: (.685 , .835 )

2) mean= 27/60= .45
ME =   = .126 
CI: (.324 , .576) 
   
Wish You the Best in your Studies.

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