# 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.

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!

• FYI I used this great post as an inspiration: stats.stackexchange.com/questions/15664/… – Drew75 Nov 3 '13 at 6:50
• Thanks, I had to fix the data and as you can see the N for female is smaller now. The data is log normal so that is why I thought to take the log. However, my goal is just to show that female's perform better along this measure from an anecdotal point of view. I feel like because of the sample size I will never be able to conduct the proper test. Am I wrong? – CJ12 Nov 3 '13 at 16:19
• If you have a bimodal distribution, the central limit theorem will converge to something that is most likely meaningless. – Mikhail Dec 25 '15 at 16:10

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.

• Thanks, I added in the coef. of variation, and the data changed slightly but they are not equal. The distribution of the data is pretty close to log normal so I am wondering the best way to test these means against each other. I feel like the sample size will limit me, but can I get around this? – CJ12 Nov 3 '13 at 16:20
• Why are the data now quite different to the data my answer related to? – Glen_b Nov 3 '13 at 21:54
• Forgot I was only analyzing in a given time range, sorry about this. – CJ12 Nov 3 '13 at 22:30
• My comments no longer apply to the new data. – Glen_b Nov 3 '13 at 23:28
• Should I repost? – CJ12 Nov 3 '13 at 23:45

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.

• Although this is a nice idea and well-explained, it would help to see some justification for why comparing medians is appropriate when a comparison of means is desired. – whuber Nov 7 '13 at 17:52