/ questioin about statistics - fisher exact test / chi squared

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jussy on 28 Feb 2013
Very random topic for UKC but anyway!

Am just writing up some data and have a few stats related queries as haven't done them in ages...

I am using Fishers exact test with 2 x 2 tables.

So one of my results is:
The 2 variables in the table are:
1 - pt is obese or non-obese
2 - pt has low expression of tumour marker or high expression

Results are:
Low expression - 50 pts are not obese and 12 pts are obese
High expression - 40 pts are not obese and 23 pts are obese.

so the p value is 0.026.
so does ths mean the people with a low expression are MORE likely to be non-obese than those with a high expression? or how do you describe it?

Thanks


kathrync - on 28 Feb 2013
In reply to jussy:

All that the statistic itself tells you is that there is a significant relationship between obesity and expression of your tumour marker. The p-value tells you nothing about how that relationship is manifest just that it exists.

The nature of the relationship must be inferred from the data, however even then you can't say anything about causality. You can say that obese patients are more likely to express the tumour marker than non-obese patients, but you have to be careful not to imply that having the tumour marker causes obesity or indeed that being obese causes the tumour marker to be expressed.
lithos on 01 Mar 2013
In reply to jussy:

why a fisher with n that large usually use a chi sq test of association(/independence)

as kathryn states the test tells you if there is an association or if they are independent variables, thats all,
Mark Torrance on 01 Mar 2013
> Results are:
> Low expression - 50 pts are not obese and 12 pts are obese
> High expression - 40 pts are not obese and 23 pts are obese.
>
> so the p value is 0.026.
> so does ths mean the people with a low expression are MORE likely to be non-obese than those with a high expression? or how do you describe it?

Yes.

What you have found is that there is a very low probability that in the population from which your sample was drawn, whether or not someone is obese is unrelated to tumour marker. So you reject the possibility of no association. Which means there must be an association.

And the association is as you've described (low expression more rather than less likely to be non-obese). The probability of the actual pattern (in the population) being the opposite of what you observed in your sample is even less than p = .026.

Whether or not you can infer cause depends on theory, study design and so forth. It's nothing to do with the fact that chi square is a "test of association".

jussy on 01 Mar 2013
In reply to Mark Torrance:

Thanks for the replies everybody.


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