I'm analysing paired data for equivalency and it's not normally distributed, i.e., the difference of the paired result is not normally distributed due to, amongst other things, outliers. If it were normally distributed I would use a Two One Sided T-test (TOST).

My questions are:

  1. Can two one-sided Wilcoxon tests with suitably defined limits ("mu" in R terminology) be used just like the Two One Sided T-test approach?
  2. If so, is it possible to establish a single p-value? Or is the larger p-value for the limit specified used?
  3. Are there any published references for doing this? (I have searched but couldn't find anything immediately available, although my search may not have been thorough enough.)

Thanks in advance!

  • $\begingroup$ You can collect some thought here: stats.stackexchange.com/questions/49782/… $\endgroup$
    – Michael M
    Commented Jan 22, 2014 at 12:09
  • $\begingroup$ TOST stands for two one-sided tests - it is a generally applicable procedure. $\endgroup$
    – Aniko
    Commented Jan 22, 2014 at 13:49
  • $\begingroup$ Well... did you find a solution? i have the same problem. Have a two dependent group but differences ara not normally distribuited. I think the comment above doesn't apply to our situation as that article talks about independent two groups. How did you manage your data? thanks. $\endgroup$ Commented Nov 20, 2023 at 17:33

1 Answer 1


The short answer is yes, you can do it, since the TOST methodology is not restricted to t-tests. The p-value is the larger of the two p-values. A quick Google search led me to a methodological article (Meier U. Nonparametric equivalence testing with respect to the median difference. Pharm Stat. 2010 Apr-Jun;9(2):142-50) describing this procedure in detail.

  • $\begingroup$ Thank you for this answer - it directly address the what I asked and adds to my understanding. Thanks also for the reference, obviously I wasn't using the right search terms, i.e., TOST, etc., versus distributions. This is also very useful and adds to my knowledge. $\endgroup$
    – bitcyber
    Commented Jan 24, 2014 at 11:55

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