Question 1200392: You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estimation for the population proportion. You would like to be 98% confident that you estimate is within 1.5% of the true population proportion. How large of a sample size is required?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! at 98% two tailed confidence interval, the critical z-score is plus or minus 2.326347877.
the standard error is equal to the standard deviation divided by the square root of the sample size.
you want the margin of error to be maximum of .015.
your critical z-score formula becomes z = .015 / s which becomes 2.326347877 = .015 / s, where s is the standard error.
solve for s to get s = .015 / 2.326347877 = .0064478749.
the formula for s is s = sqrt(p * q / n)
s is maximum when p = .5.
you get s = sqrt(.5*.5/n)
since s = .0064478749, then you get .0064478749 = sqrt(.25/n) = sqrt(.25) / sqrt(n).
solve for sqrt(n) to get sqrt(n) = sqrt(.25) / .0064478749 = 77.54492924.
solve for n to get n = that squared = 6013.21605.
when n = that, maximum s will be sqrt(.25/6013.21605) = .0064478749.
this indicates that the sample size shuld be greater than 6013.21605.
nearest integer greater than 6013.21605 = 6014.
regardless of what the mean proportin is, the margin of error should be less tha .015 when the sample size is 6014 or more.
to test this out, we'll look at some possible mean proportiong to see if this is true.
at p = .5, you get s = sqrt(.5*.5/6014) = .0064474546.
you get 2.326347877 = (x-m) / .0064474546
solve for (x-m) to get (x-m) = 2.326347877 * .0064474546 = .014999 which is less than .015.
when p = .9, you get s = sqrt(.9*.1)/6014) = .0038604720.
your z-score formule becomes 2.326347877 = (x-m) / .008604720.
solve for (x-m) to get (x-m) = .0089994134 which is less than .015.
you should be guaranteed that the magin of error will be less than .015 when the sample size is 6014 or greater.
i had to do this many times to make sure i did it correctly.
i'm reasonably certain that i finally got it right.
my best guess is a sample size of 6014 or greater will satisfy the requirement that the margin of error is less than .015 regardless of the value of p.
|
|
|
