  The Wilcoxon Signed Rank test is typically called upon when testing whether a
symmetric distribution has a specified centre and the Gaussianity is in
question. As with all insurance policies it comes with a cost, even if small,
in terms of power versus a t-test, when the distribution is indeed Gaussian. In
this note we further show that even when the distribution tested is Gaussian
there need not be power loss at all, if the alternative is of a mixture type
rather than a shift. The signed rank test may turn out to be more powerful than
the t-test, and the supposedly conservative strategy, might actually be the
more powerful one. Drug testing and functional magnetic imaging are two such
scenarios.
  Wilcoxon' signed rank test will typically be called upon by a researcher when
testing for the location of a single population, using a small sample and
Gaussianity is dubious. As all insurance policies, it will come with a cost--
power. It is well known, that under a Gaussian setup, the signed rank test is
less powerful than, say, a t-test. The works of Pitman and others have
reassured us that this power loss is surprisingly small. In this note we argue
that the power loss might actually be smaller than typically assumed. In
particular, if the deviation from the null Gaussian distribution is of a
mixture type and not a shift type, the signed rank test is no longer dominated
by the t-test and can actually be more powerful.
