Question 1206930: A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him/her. If the candidate wants a 5% margin of error at a 90% confidence level, what size of sample is needed? Be sure to round accordingly.
The candidate would need to survey people in the community in order to be within a 5% margin of error at a 90% confidence level.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: 271
Explanation
The margin of error for a proportion is
E = z*sqrt(phat*(1-phat)/n)
Let's isolate n.
E = z*sqrt(phat*(1-phat)/n)
sqrt(phat*(1-phat)/n) = E/z
phat*(1-phat)/n = (E/z)^2
n/( phat*(1-phat) ) = (z/E)^2
n = phat*(1-phat)*(z/E)^2
This formula comes up a lot so it's best to either memorize it, or write it down in your notes. The same can be said about the original formula mentioned.
At 90% confidence, the z critical value is roughly z = 1.645
This is another thing to memorize or have on a reference sheet somewhere.
The error we want to have is E = 0.05 to represent being at most 5 percentage points off of the true proportion, ie a "5% margin of error".
We aren't told the value of phat, so the default is to go with phat = 0.5 which is the most conservative estimate.
It's right at the midpoint of phat = 0 and phat = 1.
Here are the input values
z = 1.645 (approximate)
E = 0.05
phat = 0.5
Then we'll have the following
n = phat*(1-phat)*(z/E)^2
n = 0.5*(1-0.5)*(1.645/0.05)^2
n = 270.6025
n = 271
Always make sure to round UP to the nearest whole number when it comes to minimum sample size problems.
Refer to part (c) of this question for more info.
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