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Mann-Whitney W in JASP, isn't it usually a U?

Hello!

I've conducted Bayesian Mann-Whitney tests in JASP, but the statistic in my tables is labelled W, not U. I'm used to it being called a Mann-Whitney U test. In case my paper's reviewers question me on this - what's the reason for the different naming convention?


Thank you!

Comments

  • Hi

    I believe that Wilcoxon first defined the test which was further adapted by Mann-Whitney - hence the analysis is often termed the Wilcoxon- Mann-Whitney test. I think R reports U as W which are essentially the same value. I am ready to stand corrected though.


    Mark

  • Hi Eniseg2,

    As DrMark pointed out, the history of the test is a little complicated, leading to different naming conventions of both the test itself (Wikipedia already lists 4 names) and the test statistic (U or W). For the test statistics, this is further explained in the R helpfile for the wilcox.test function:

    The literature is not unanimous about the definitions of the Wilcoxon rank sum and Mann-Whitney tests. The two most common definitions correspond to the sum of the ranks of the first sample with the minimum value subtracted or not: R subtracts and S-PLUS does not, giving a value which is larger by m(m+1)/2 for a first sample of size m. (It seems Wilcoxon's original paper used the unadjusted sum of the ranks but subsequent tables subtracted the minimum.)

    So R (and JASP) subtracts a constant from the sum of the pooled ranks of the first group. Subtracting or not has no implications for the corresponding effect sizes, p-value, and Bayes factor though, since that is just a matter of convention.

    Kind regards,

    Johnny

  • Thank you very much! :) Now I just have to ask a quick question with regard to the interpretation.

    I had specified effects, e.g. "Give me the B10 in favour of Group 1 scoring higher that Group 2". If my BF10 is very low, and in turn the BF01 is high, does this mean the evidence supports the opposite of my spec? I.e. that instead of Group1 scoring higher than Group2, as I expected, Group2 is now scoring higher? And BF10=1 means the evidence goes towards the null model, i.e. the groups score equally?

  • No, not necessarily. If BF0+ is high (so evidence for H0 versus the directional alternative hypothesis that the effect is positive) this can happen because the effect is absent or because the effect is actually negative. With small samples, a high BF0+ usually indicates the effect in the sample is negative.

    E.J.

  • Thank you. Our sample is around 85 people. How do I know how to interpret the evidence for H0 vs. the directional alternative hypothesis? If BF0+ is high, and if it is just 1?

  • Hi Eniseg,


    For help on interpretation, the following figure from one of our guidelines papers can be helpful (scroll down to the third figure and caption: https://www.researchgate.net/publication/330569075_The_JASP_Guidelines_for_Conducting_and_Reporting_a_Bayesian_Analysis/figures

    You can also find the full paper with general guidelines on Bayesian analysis here: https://psyarxiv.com/yqxfr

    First, a Bayes factor of 1 is inconclusive: it cannot distinguish between the two hypotheses (since it is relative, this can mean they both predict the data equally well, or equally poorly).

    Second, the Bayes factor is a continuous metric of the two hypotheses' predictive qualities - some scholars have proposed some guidelines for categorizing certain Bayes factor values, which you can find in the figure I linked, but it remains a bit of subjective practice. In short, a Bayes factor of 5 in favor of H+ (i.e., the positive directional alt. hypothesis) means the data are 5 times more likely under H+ than under H0. It is then up to you (and your audience) to decide if this consitutes enough evidence to base a theory/paper on.

    Cheers,

    Johnny

  • Brilliant, thank you! :) I had read the paper and understood it the way you explained. I just wanted to make sure I have properly understood it before I make a fool of myself!

    Thank you, Johnny & EJ.

    Best,

    Anna

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