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

1. ## 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

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

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