Question 1191258: You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately p = 37%. You would like to be 99% confident that your estimate is within 4% of the true population proportion.
How large of a sample size is required?
Found 2 solutions by Boreal, math_tutor2020:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! error=half-interval, which is z(0.995)*sqrt(0.37*0.63/n)
0.04=2.576*sqrt(0.2331/n)
square both sides
0.0016=6.636*0.2331/n
=6.636*0.2331/0.0016
=966.74=967
967 units
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
The margin of error for the confidence interval of a population proportion p is
E = z*sqrt(phat(1-phat)/n)
where z is the critical value, phat is the sample proportion, and n is the sample size.
The term "phat" is often written as "p hat" or "p-hat", but I'm deciding to use a shorter version for the sake of compactness.
phat estimates the population proportion p.
Solving for n gets us
E = z*sqrt(phat(1-phat)/n)
E/z = sqrt(phat(1-phat)/n)
(E/z)^2 = phat(1-phat)/n
n*(E/z)^2 = phat(1-phat)
n = phat(1-phat)*(z/E)^2
Since we estimate p = 0.37, this will be plugged into each phat.
At 99% confidence, the z critical value is roughly z = 2.576 (use a calculator or table to find this value).
The desired error we want is E = 0.04
With all that in mind, we can then compute the minimum sample size n
n = phat(1-phat)*(z/E)^2
n = 0.37(1-0.37)*(2.576/0.04)^2
n = 966.749616
We round up to n = 967
We always round up regardless how close or far the decimal value is to the nearest largest integer.
For example, if we got the result n = 966.00001, we would still round up to n = 967
This rounding up is to ensure we clear the hurdle needed to form the min sample size.
Answer: 967
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