Monday, May 9, 2011

finding a sample size

I want a confidence interval of 95% with a population size of p+-.o5
where p+-z{[P(1-P)/n]^(1/2)} which can also be written as n=[z^2[P(1-P)]/m^2
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1 comment:

  1. mtoshachMay 9, 2011 at 1:47 PM

    with a margin of error of .05, a population size of n=384.16 (which is roughly 385) is necessary

    because z=1.96
    P=.5
    and m=.05
    thus

    n={1.96^2[.5(1-.5)]}/(.05^2)=384.16

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