Can you post the data/info you have?
You can likely plug those values right into the formula.
Hi, I have a 3x2 contingency table, which contains the observed and expected allele frequencies from a hypothetical population and I'd like to test whether the deviations from HWE are statistically significant. However I do not have access to the population size (though we assume it is 'large') - I only have proportions (i.e. all expected/observed values are <1).
Can Chi-Squared or Fisher's exact handles this? As I believe Chi-squared requires cell counts to be >5 and Fisher's exact is based on a discrete distribution so uses integer values only. If not, is there an alternate test that can be used? What would you do in this scenario? Thanks
Can you post the data/info you have?
You can likely plug those values right into the formula.
Stop cowardice, ban guns!
Acather96 (05-04-2017)
Hi, thanks:
Expected: 0.7225, 0.255, 0.0225
Observed; 0.725, 0.25, 0.025
I know its obviously isn't significant, but I need to do a formal test and give a P-value.
Cheers
You really can't do any formal testing unless you know the raw counts or the total number sampled.
Could the Expected and Observed be the other way round because usually the Expected values are "simpler" than the Observed values. If that is the case, you can perhaps make some progress towards finding a likely value for the total number.
Acather96 (05-04-2017)
No, they are the correct way round. Unfortunately there is no information relating to the total number sampled, merely that the population is large and evolved for 100 generations.
OK. So you have a large population and every single member was seen and classified?
Acather96 (05-04-2017)
Yes, this is an idealized/theoretical population and we can assume that all members are sampled accurately.
A commonly used test would Chi squared on the raw counts. However, the p value depends on the number sampled. In this case the number needed to show a significant difference is quite large. If the total number of units sampled and classified is 10 000, then the chi square test gives p = 0.15 (just multiply your proportions by 10 000 and do a chi square test). If you have sampled and classified 20 000 units, then the p value is 0.02. So a formal test and a p value and the significance all depend and how many were counted.
No other test will do better than the Chi squared test.
What do you plan to do with the results?
Stop cowardice, ban guns!
Tweet |