Mean test of log-normal distibutions I have two data sets, which are assumed to be log-normally distributed. How can I test whether their means are statistically different from each other?
I guess I cannot use the 2-sample t-test, because data is not normal. Are there any easy solutions for the issue?
 A: If you can assume equality of the $\sigma$ parameters (the population standard deviation of the logs), then a test for equality of $\mu$ (the population mean of the logs) will be the same as a test for equality of the mean of the lognormal.
That is, under that assumption, you can take logs and do a two sample equal-variance t-test (a Welch test wouldn't work for that though).
Alternatively, under the same assumption of equality of $\sigma$ parameters, a Wilcoxon-Mann-Whitney test will also be a test of equality of means (against a location shift in the logs or equivalently a scale shift in the original variables). You don't have to take logs to do this test - it works the same either way.
A: Take the log of your data, and run standard ANOVA.
If you have the data sets $x_i$ and $y_i$ both from lognormal distribution, it means that $\ln x_i$ and $\ln y_i$ are from normal distribution. 
As @whuber noted, this would be a test on medians of $x,y$, but if you assume that $\sigma_x=\sigma_y$, then equal medians will imply equal means. On the other hand if you don't assume the equal variances, then you may end up with the same means and different variances. In this regard having the same mean is less informative. So, it's better to analyze medians.
A: What about doing a likelihood ratio test where under the null, the location and scale parameters of the lognormal are equal, and under the alternative they are distinct? In that case data from sample 1 would have $E(X) = \exp(\mu_1+.5\sigma_1^2)$ and data from sample 2 would have $E(Y) = \exp(\mu_2+.5\sigma_2^2)$. Under the null, set $\mu_1 = \mu_2$ and $\sigma_1 = \sigma_2$ then you would have equal means and the difference in the -2log(likelihood) values under each hypothesis would have a Chi-square distribution. 
