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Thread: Z-test vs. t-test and calculating std error

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    Z-test vs. t-test and calculating std error




    Hello,

    I have two questions concerning proportions.

    First, say I have one sample of 1000 units and 10 percent have the characteristic A while 8 percent have B - how would I test whether these two proportions are significantly different, is A actually more common than B? I understand that I can compute a z-test, but how would I calculate the standard error for this test?

    What I can find (say, Agresti & Finlay) either seems to describe comparing a proportion in one sample to some population parameter my, or comparing proportions from two different samples - but in my case, I'm dealing with one sample only.

    And more generally, is it correctly understood that when comparing proportions I would use a z-test, not a t-test (if I'm wrong, how would I compute the standard error in a t-test?).

    For any help, thanks in advance,
    deenha

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    Hope the following steps help you a bit.
    Each of your groups with one of the two characteristics you say follows a binomial distribution
    X~b(n,px)
    Y~(b(m,py)
    n=m=1000

    The proportions of whom you want to test the difference (large sample)
    px(estimate)=X/n ~ N(px,px*qx/n)
    Same with py(estimate)

    If the samples are independent, compute the difference px(estimate)-py(estimate)
    then the statistical function and make it a N(0,1)
    The standard error you are searching equals the addition of the two Normals

    H0 : px=py
    Under the null hypothesis px=py=p, change the numbers of your proportions in the Z
    function you created above (careful not the estimates) and you will come up with your Z-test
    Keep up
    Last edited by pateras34; 06-18-2009 at 09:12 AM.

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