I have two independent groups and they seems to form a Pareto distribution when the histograms are looked at. Is there a particular hypothesis test that is specifically made to compare the means of such distributions?
I should also mention that the distributions are highly unbalanced (one group has 1000 instances and the other has 35,000 instances).
I want to check if the mean of one group is greater than the mean of the other group.
With such huge sample sizes, you will certainly reject the null; it may be better to focus on measuring the size of the difference than testing for it.
In the case that the lower limit of both Pareto distributions ($x_m$ Wikipedia's parameterization) is the same, the test is straightforward -- it amounts to testing equality of the $\alpha$ parameter.
The mean, when it exists, is $\frac{\alpha x_m}{\alpha-1}$, which increases as $\alpha$ decreases toward 1.
Note that the log of a Pareto random variable is a shifted exponential whose lower limit (shift parameter) is $\log(x_m)$, and $\alpha$ is the rate parameter. So let's proceed this way -- by working with the logs.
In the case that $x_m$ is common to both groups but unknown, you could estimate its log by the smallest observation in either group, subtracting it from all other log-values (and then discarding that observation from whichever sample it occurred in).
You could then compare the means of the resulting exponentials via an F-test, as here (the ratio of the means of the shifted logs $\bar{X}/\bar{Y}$ should be $\sim F_{2n_x,2n_y}$), or you could do a straight likelihood ratio test. They should give very similar results (at considerably smaller sample sizes than yours they can differ more).
Pareto might not have a finite mean, so this comparison simply might not be reasonable.
You could make a simulation by bootstrapping from your samples and get a pretty good idea of if one group is larger than the other.
