I am trying to test the null E[X]=0, against the local alternative E[X]>0, for a random variable X, subject to mild to medium skew and kurtosis of the random variable. Following suggestions by Wilcox in ‘Introduction to Robust Estimation and Hypothesis Testing’, I have looked at tests based on the trimmed mean, the median, as well as the M-estimator of location (Wilcox’ “one-step” procedure). These robust tests do outperform the standard t-test, in terms of power, when testing with a distribution that is non-skewed, but leptokurtotic.

However, when testing with a distribution that is skewed, these one-sided tests are either far too liberal or far too conservative under the null hypothesis, depending on whether the distribution is left- or right-skewed, respectively. For example, with 1000 observations, the test based on the median will actually reject ~40% of the time, at the nominal 5% level. The reason for this is obvious: for skewed distributions, the median and the mean are rather different. However, in my application, I really need to test the mean, not the median, not the trimmed mean.

Is there a more robust version of the t-test that actually tests for the mean, but is impervious to skew and kurtosis?Ideally the procedure would work well in the no-skew, high-kurtosis case as well. The ‘one-step’ test is almost good enough, with the ‘bend’ parameter set relatively high, but it is less powerful than the trimmed mean tests when there is no skew, and has some troubles maintaining the nominal level of rejects under skew.

background:the reason I really care about the mean, and not the median, is that the test would be used in a financial application. For example, if you wanted to test whether a portfolio had positive expected log returns, the mean is actually appropriate because if you invest in the portfolio, you will experience all the returns (which is the mean times the number of samples), instead of n duplicates of the median. That is, I really care about the sum of n draws from the R.V. X.

**Answer**

Why are you looking at non-parametric tests? Are the assumptions of the t-test violated? Namely, ordinal or non-normal data and inconstant variances? Of course, if your sample is large enough you can justify the parametric t-test with its greater power despite the lack of normality in the sample. Likewise if your concern is unequal variances, there are corrections to the parametric test that yield accurate p-values (the Welch correction).

Otherwise, comparing your results to the t-test is not a good way to go about this, because the t-test results are biased when the assumptions are not met. The Mann-Whitney U is an appropriate non-parametric alternative, if that’s what you really need. You only lose power if you are using the non-parametric test when you could justifiably use the t-test (because the assumptions are met).

And, just for some more background, go here: Student’s t Test for Independent Samples.

**Attribution***Source : Link , Question Author : shabbychef , Answer Author : MarianD*