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.