Excluding outliers with MAD
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- Posts: 16
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Excluding outliers with MAD
Hi, is there a way to make a filter for the exclusion of outliers using the median absolute deviation (MAD) proposed by Leys et al., 2013 (https://www.sciencedirect.com/science/a ... 3113000668)
i.e. Median ± 2.5 * MAD.
Thanks in advance
i.e. Median ± 2.5 * MAD.
Thanks in advance
Re: Excluding outliers with MAD
Hey @Vicente_inefo,
if you need it, my suggestion is the example with this screenshot: Cheers,
Maurizio
UPDATE 2022-09-28
Sorry for the hint attached with the broken formula.
The attachment with the correct suggestion is in the post below, as a response to @anagrammarian, who drew my attention to the breakup.
Maurizio
if you need it, my suggestion is the example with this screenshot: Cheers,
Maurizio
UPDATE 2022-09-28
Sorry for the hint attached with the broken formula.
The attachment with the correct suggestion is in the post below, as a response to @anagrammarian, who drew my attention to the breakup.
Maurizio
Last edited by MAgojam on Wed Sep 28, 2022 5:27 pm, edited 1 time in total.
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Re: Excluding outliers with MAD
Thanks for the suggestion @MAgojam, I think it could be useful. I was thinking on calculate robust MAD using JASP and then calculate upper and lower limits for the MAD method for detecting outliers and applying it in JAMOVI
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Re: Excluding outliers with MAD
Maurizio,
I had a q re your formula. The original paper says M - 3 or 2.5* MAD and M + 3 or 2.5*MAD as the cutoffs. For MAD, they have an additional multiplier b which is 1.4826 i.e. MAD = b* MED(ABS( var - MED(var))) where b = 1.4826. Is there a reason why you omitted b in your formula? I noticed significantly more outliers obviously without that and adding 1.4826 and using 2.5 seemed to make sense.
Let me know what you think and thanks for introducing me to MAD.
-manoj
I had a q re your formula. The original paper says M - 3 or 2.5* MAD and M + 3 or 2.5*MAD as the cutoffs. For MAD, they have an additional multiplier b which is 1.4826 i.e. MAD = b* MED(ABS( var - MED(var))) where b = 1.4826. Is there a reason why you omitted b in your formula? I noticed significantly more outliers obviously without that and adding 1.4826 and using 2.5 seemed to make sense.
Let me know what you think and thanks for introducing me to MAD.
-manoj
Re: Excluding outliers with MAD
The reason is that I did not double check how I should have done and forgot to add the scale constant as defined by (Huber, 1981).anagrammarian wrote:Is there a reason why you omitted b in your formula?
I apologize to everyone who has passed by the post and trusted my suggestion.
Thank you so much manoj, for getting my attention and replying with the correct suggestion. Cheers,
Maurizio
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Re: Excluding outliers with MAD
I could confirm that the result of the proposed formula matches with the MAD result reported in JASP.
Thanks to all the participants.
Thanks to all the participants.
Re: Excluding outliers with MAD
should we add a MAD() function to our formula list?
jonathon
jonathon
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Re: Excluding outliers with MAD
It sounds good to me!!
Re: Excluding outliers with MAD
if someone can create an issue here, describing in detail how it's calculated, i'll see about adding it.
https://github.com/jamovi/jamovi/issues
cheers
jonathon
https://github.com/jamovi/jamovi/issues
cheers
jonathon
Re: Excluding outliers with MAD
If it is of interest and can help, Rj is there:
Maurizio
Cheers,Maurizio