SOLUTION: Here is another one I need help with. Thanks in Advance. 1. If I am surveying the constituents for my favorite candidate ... and I wish to be 95% confident of the estimate I dev

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Question 135397: Here is another one I need help with. Thanks in Advance.
1. If I am surveying the constituents for my favorite candidate ... and I wish to be 95% confident of the estimate I develop for the proportion of the constituents that favor my candidate and i want the error to be within 2% (0.02) of the actual population proportion

>> how many folks do there need to be in my sample ??

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
If I am surveying the constituents for my favorite candidate ... and I wish to be 95% confident of the estimate I develop for the proportion of the constituents that favor my candidate and i want the error to be within 2% (0.02) of the actual population proportion
>> how many folks do there need to be in my sample ??
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Formula: E = z*sqrt(pq/n)
sqrt(n) = z*(sqrt(pq)}/E
n = [z/E}^2*pq
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z-score for 95^ confidence = 1.96
Since p is not given, avoid bias by letting p=1/2=q
E is given as 0.02
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n = [1.96/0.02]^2(1/4)
n = 9604*(1/4)
n = 2401 (desired sample size)
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Cheers,
Stan H.