Adjustment of medians in the Siegel-Tukey test vs. the Wilcoxon test I have two (unpaired) distributions. The distributions are far from being normal, so I'm using the Wilcoxon test to compare their medians: the Wilcoxon test does not reject the null-hypothesis (p-value = 0.5584). Then I'm comparing variances of these distributions using the Siegel-Tukey test, which turns out to reject the null-hypothesis (p-value = 0.03351) suggesting difference in variances.
However, since the literature says that difference in medians might have affected the results of the Siegel-Tukey test, I've decided to rerun the Siegel-Tukey test with median adjustment (despite the previous result of the Wilcoxon test). Surprisingly, after the median adjustment the Siegel-Tukey test can no longer reject the null hypothesis (p-value = 0.1144), i.e., we can no longer claim difference in medians.
Does this mean that I should have relied on the Wilcoxon tests and that adjusting the means was wrong in this case? Does this mean that the Siegel-Tukey test is not powerful enough to distinguish difference in the variances in the second case? 

 A: The Wilcoxon test is no more a test of medians than it is a test of means. 
If means exist, then under certain assumptions, it can be a test of means. Under the same assumptions, it's a test of medians.
More generally, under location shift alternatives, the Wilcoxon test tests whether the median of pairwise cross-population differences is different from zero. More generally still, it's a test of whether $P(X<Y)\neq \frac{1}{2}$.

Then I'm comparing variances of these distributions using the Siegel-Tukey test, which turns out to reject the null-hypothesis (p-value = 0.03351) suggesting difference in variances.

It suggests a difference in spread (in some sense). The population variance needn't exist; this test would still work.
It's difficult to guess what it is about the samples that lead to the outcomes you describe (certainly not without the samples, but even then it may be tricky), but you'd never expect two different ways of treating the samples to yield identical p-values. [Note also that adjusting for the difference in median no longer means all sample arrangements are equally likely; I don't think the test is exactly distribution free any more.]
It may be that there's just enough location shift in the unadjusted data to tip the Siegel-Tukey over the 5% critical value. Or it may be that the significance level of the median-adjusted version of the test is affected by that adjustment just enough to push the otherwise significant difference a little way the other side of your p-value.
Or it might be something else. 
