Difference between two independent proportions

#1
Hello,

I am not a statistician, however, for my latest small research, I would like to test whether there is a difference between two independent groups of samples in terms of proportions (one with a condition and one without a condition).

I have researched this problem and find out that different researchers did use two different approaches:
a) z-test to test the difference between two independent proportions
b) chi-quadrate

Now I would like to kindly ask you personal opinion/experience which one is more accurtae in this matter and why ?

Looking forward to hearing (reading) your precious thoughts. Thank you very much !

Cheers,
Zuzana.
 
#3
Actually, I am at the stage of defining a sample size for my study. This is why I have been evaluating these different approaches. But my expected proportions (also for the purpose of power analysis) are 50% to 50% in the intervention group and 70% to 30% in the control group. Does this influence the choice of the test ? (but I would expect a sample size of approx. 300 samples - 150 in each group). Thx for your thoughts.
 

hlsmith

Omega Contributor
#4
Yeah, the typical approach for this question would be to use the chi-square test. Though, if any expected cell frequency is <5, the Fisher Exact test gets used. I am sure the z-test is fine to use, when samples get larger things normalize.


It might also come down to your field. I am in medicine and we hardly ever use the Z over chi-square.


I am not familiar with the chi-quadrate. What is that in particular? Do you have a link. So is it just a phrasing that says you are comparing two rates from a 2x2 table?
 
#5
Hi, following up on this discussion. I'd like to ask, WHY, for example in medicine, is chi-square typically used instead of z? I am learning Statistics at a beginner level and came across this problem of a fair coin: https://frank.itlab.us/datamodel/node42.html
I wondered why you would not just use the z-test? I cannot figure out at this point when to use the chi-square instead of the z-test if working with (1 or 2) proportions. Help is appreciated; thanks.