Hypothesis testing for Pareto distributions I wish to to some simple hypothesis testing of the form provided by T-Tests and ANOVA. However, my data is not normally distributed (it follows a Pareto distribution).
My understanding is that T-Tests make the assumption that the data is normally distributed and hence I won't be able to use them - is that correct? Is there something else I can do?
EDIT Here is some more info about my problem.
I'm trying to do some quality analysis on software defects, and am having trouble knowing where to start. One basic question I want to answer is:

Does software produced in department X have more defects than department Y?

As some background, we group changes to software as "patches", in which case the question becomes

Does the average patch from department X have more defects than department Y?

Here is a histogram of bugs / patch, N = 3700.

There is a philosophical issue of what it means to have "more" defects that I don't have a great answer to. The obvious choice for one of my limited knowledge is to compare the mean defects in each group, but as others have pointed out that's not clearly the best choice. The measure linked to by Procrastinator ($P(X<Y)$) seems like it captures my intuition well.
 A: Interesting problem. This looks like an independent probability problem rather than a statistics problem, i.e., not a problem yielding a statistic like a $t$-statistic. For example, if the probability of having zero defects is 80%, then the probability of having one or more defects is 20%. If we assume that the probability of having any defect is unrelated to whether or not there is a defect in the "object" already, then yes, the probabilities chain, and look like a power function, because it is a power function. This relates to the probability of having only one defect ($p$) as $\sum _{i=1}^{\infty } p^i=\frac{p}{1-p}$. So for our example, we ask how did we get to having a 20% total defect rate and the answer is when single failure rate is $\frac{p}{1-p}=0.2$ or $p=\frac{1}{6}$.
Now the second part of the question is when is a failure rate significantly different from another failure rate. I think that this means that once we have the single failure rate from one experiment, $p1$, and another single failure rate from another experiment, $p2$, we can do a two-sample binomial probability test as outlined here. That is, I think so, but, we may need to be careful with what numbers we are substituting in. For example, the total number I do not think includes the multiple failure objects, just the number with no failures and those with single failures for the two-sample test. So now comes the tricky part, the calculated single failure rate from including multiple failures as part of the calculation may differ from the observed number of objects with only a single failure. It may be better to use the calculated single failure rate, rather than the observed rate, and it is even possible to calculate binomial probabilities for non-integer numbers of observations by using a continuous form of the binomial distribution. However, for approximate answers and for a large number of observations, that may be overkill, and nearest integer results are likely good enough for most purposes.
