Comparing means with unequal variance and very different sample sizes I am trying to compare the means of the same variable between men and women. This is the statistics:
     N        Mean        Variance    Coef. Var.     Gender    
   2000      26.12         10.89         0.13         Male        
     50      56.10         25.01         0.09        Female

Neither variable is normally distributed but taking the log makes it pretty darn close. What is the appropriate way to test the means between males and females? Should I use the log or not? Any additional advice using Stata would be helpful.
My initial reaction is that females fare better than men, but I want to be statistically rigorous.  
 A: The traditional test for comparing two sample means is the t-test. There are no assumptions about the sizes of the samples, so it is OK if they are different. 
However, you touch upon the normality assumption. Even if the population is not normally distributed, the Central Limit Theorem allows us to infer normality as the sample sizes increase. This means your test will be approximate, but the sample size for female is a little low.
Finally, the result of the t-test will be different for the original and log-ed data. Do you have a specific reason based on your data to use the logarithm? Perhaps there is another assumption you would like to test about the behavior of the log of your data? Do not take the log simply to create a normal curve if there is no deeper meaning, but for fun compare the difference between the two results anyway!
A: Taking logs and testing the mean on the log scale would normally not correspond to a difference in means on the original scale.
However:
[Edit: my comments apply to an earlier version of the data, and don't apply to the data that are presently in the question. As such, my comments really apply to the situation where the coefficient of variation in two close-to-lognormal samples are very similar, rather than to the case now at hand.]
The coefficient of variation is almost identical in the two samples, which does suggest that you might consider these as having a scale shift; if you think the logs look reasonably close to normal, then that would suggest lognormal distributions with common coefficient of variation. In that case a difference of means on the log-scale would actually indicate a scale-shift on the original scale (and hence that one of the means is a multiple of the other mean on the original scale).
That is, under an assumption of equal variance and normal distribution on the log-scale, a rejection of equality of means implies that the means on the original scale have a ratio that differs from 1. 
It seems like that would be a reasonable assumption.
There are other things you could do, though.
A: From the data you cannot infer that the variance between males and females is same, in fact the opposite is almost certainly true.  Also, since 50 is indeed a bit low, suppose you cannot assume normality.
Compare each female's value with the median of men's values.  If median female is neither better nor worse than a median male (null hypothesis), then each female will have 1/2 chance to be better than a median male.  The chance that K or less females are worse than a median male is $P(K) = 2^{-50} \sum_{m=0}^K {50 \choose K}$ .  Here we consider the error in the male's median to be negligible, since there are much more males than females, and the variance between males is smaller than the variance between females.
