comparing two proportions - testing problems

#1
hi guys..

base data: only average proportions and sample sizes,
issue: comparing specifical mortality in 8clusters with county
chosen test: chi-square of proportions via contingency tables
alpha=0,05

problem:
cluster1 cluster8 county
proportion in % 0,413 0,241 0,287
n 5104 64 000 557 613
p-value 0,095 0,038

so cluster8 unequal, cluster1 equal, despite the average proportions..

any better test? or just take it as it is? browsed google for two hours, no idea..
just repeating - base data - I have only average proportions and sample sizes, for each cluster and county..
thanks..

P.S. tried even z-test but p-values are quite all around zero..

m
 

hlsmith

Omega Contributor
#2
The formatting of the text in your post is confusing, but I think I followed. You have to select your test before collecting data - if you start trying all tests you are biasing things. Also, if you are making two comparisons, you may want to adjust your alpha level, to address false discovery concerns.


Yes, it is likely what it is and you have to take it.
 
#3
Thanks for reply, well text format looks different after pressing enter ;).. thing is, that there are big differences between sample sizes.. I chose chi-square for testing with those results.. so main question is if I chose right test, or if there is another solution for testing proportions with big difference between sample sizes.. for example if n1=5000, n2=50000, n3=5000000, etc. Thanks
 
#5
Chi sq should be fine. The clusters are not a subset of the community are they.
clusters are subsets, so finally I made comparison clusterX vs (community-clusterX) - is that right?

and is there any test where can I compare subset with community straight?

anyway - another problem is that all mortality rates are from population data, so not a random sampling.. but somehow I have to decide (statistically) which villagecluster is most problematic comparing to county rate..
so what to do? some testing or just compare the differences between rates?

thanks
m
 
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