Matlab function that can calculate pvalues from z-scores rho-values in a connectivity matrix
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Chiara Bulgarelli
le 17 Juil 2017
Commenté : turquoise_squid
le 6 Juil 2018
I am performing connectivity analyses and my output for each subject is a very big matrix (44x44) with rho values that I transformed in z scores. I then averaged these together among all my subjects. I work with Matlab and I would like to calculate the pvalue from the rho z-scores I have in each of the 1936 cells. I know that there are calculators of pvalues from rho-values available online, but of course I can't do it manually for each of my cell, so I am looking for a Matlab function that can help me doing that in my whole big matrix. Does anyone know anything about this? I haven't been able to find anything except for the classical 'corrcoef' function, but this is not helpful, my correlations have been already performed! Any suggestion is really appreciated! Thanks, Chiara
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turquoise_squid
le 6 Juil 2018
Dear Chiara, did you find an answer for you question? I can replicate the same finding that you had with the diagonals having the value 0.8413 instead of the expected p-value of 0 in the p-value matrix. Any help on this would be greatly appreciated.
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Star Strider
le 17 Juil 2017
Your z-scores imply that you are assuming your data are normally distributed.
Use the Statistics and Machine Learning Toolbox normcdf (link) function to calculate the probabilities.
You can also calculate it from the erfc function with:
P = @(z) erfc(-z/sqrt(2))/2; % Equivalent to ‘normcdf’
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Star Strider
le 19 Juil 2017
My pleasure.
The Fisher’s z-score is the correct statistic to use.
The matrix diagonal I was referring to are in the ‘P’ matrices returned by corr and corrcoef. Those should be zero, but are actually calculated correctly as 1. So if yours are also 1, the calculation is correct, although the interpretation implies that they should be 0.
You need to calculate the p-value individually for every element in your z matrix. The normcdf function will accept matrix arguments, and do this automatically.
I do not understand the p-value of 0.8413, or what the problem is with it. The normcdf function (without other arguments) assumes a mean of 0 and a standard deviation of 1, so your z-scores should work.
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