jamovi

Hi!
I am currently using Jamovi for analysing some data and I have a doubt.
I need to interpret the Mann-Whitney U output, especially the effect size available in the software (rank biserial correlation).
I have read the paper by Kerby but I haven't been able to understand which formula, whithin the ones proposed by the author, has been used to perform the effect size.

Could you give me some clarification about that and about the interpretation of the rbc value?

Thanks in advance,
Lydia

here's the code which performs the calculation:

https://github.com/jamovi/jmv/blob/master/R/ttestis.b.R#L254-L283

hopefully that answers some questions for you.

cheers

jonathon

Lydia_L wrote: I have read the paper by Kerby but I haven't been able to understand which formula, whithin the ones proposed by the author, has been used to perform the effect size.

Hi, Lydia.

Dave S. Kerby (2014) reports:
"Hans Wendt (1972) presented a third formula, one based on U. Wendt was motivated to develop his formula because he observed in published research a" neglect of correlation in favor of significance statistics "(p. 463). His goal was to derive an easy-to-use formula that would promote the reporting of effect sizes with the Mann-Whitney U test. The Wendt formula computes the rank-biserial correlation from U and from the sample size (n) of the two groups: r = 1 - (2U) / (n1 * n2)."
The above is the formula for effect size (Rank biserial correlation) for Mann-Whitney U test in jamovi.

Cheers,
Maurizio

Hey,

Thanks for all the help @MAgojam

Does the Wilcoxon test in Jamovi use the same formula to output the rank biserial correlation? Would anyone be in position to clarify how to interpret the output value. How good or bad is a value of .5?

Also, if I want to compare a range of values that are either -1 or 1 to zero, would the Wilcoxon test work as a good substitute to the one sample sign test. To my understanding, the one sample sign test has very low power!?

Hi, @dama.
As you computed your variables (-1, 1), you already have ready-made vectors representing a dominance matrix like in a signed test, whose average is the effect size (Cliff's Delta).
I am attaching a jamovi file and a screenshot illustrating the steps taken, which may help you.
I wrote you a little function in Rj which summarizes what could be useful to you for your variables which probably look like the one called Score_sign in the attached file.
Sorry if it's a bit of a fruit salad, but I don't have much time to spare.
For the magnitude of the effect size (Cliff's delta) you might be interested in reading this:
https://citeseerx.ist.psu.edu/viewdoc/d ... 1&type=pdf

Cheers,
Maurizio

Hey,

I am super grateful for your help @MAgojam. Many thanks for the Rj function. Very kind of you.

I am sorry for bothering you ones again. If I understand your beautiful code correctly, cliffs delta is also the rank biserial correlation. Is that correct?

When doing the calculations, I get it correct as you have described it for the Mann Whitney U test, however, when doing it for Wilcox, I am not getting a "correct" result.

For Mann Whitney U, the rank biserial correlation seems to be the same as Cliffs Delta.

The formula for cliffs delta and rank biserial correlation for Mann Whitney U is [1-(2U)/(n1xn2).

Is the corresponding formula for Wilcox [(Sum of favorable ranks/sum of ranks)-(Sum of unfavorable ranks/sum of ranks)?

When using the stated formula for Wilcox I get -.429, as indicated in the picture at "mean".

(34/63)-(7/63)=-.429

If I remove the sum of zeros from the sum of ranks, I get the rank biserial correlation. I had 22 zeros.

(34/41)-(7/41)=-.659

Once again, I deeply appreciate your help @MAgojam

Hi, @dama.
I realize I give you answers in this post, which also refer to another post "One sample sign test" and this could be considered "Off-Topic".
Those who come here, specifically for the post, may not find themselves with the suggestions as an answer to your questions.
I would close this post and if you agree, to learn more, I would continue with "Send private message" as a method offered by the Forum.

Cheers,
Maurizio

Hey, I’ve sent you a private message. Unclear if it has gone through.

In any case, many thanks for your help.