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Re: Chi2 test

From: Nicholas Jankowski
Subject: Re: Chi2 test
Date: Thu, 21 Feb 2019 18:46:47 -0500

On Thu, Feb 21, 2019 at 6:34 PM Nicholas Jankowski <address@hidden> wrote:
do you have a 

On Thu, Feb 21, 2019 at 12:44 PM Javier Garcia Falaguera <address@hidden> wrote:
Many thanks for the support. I just find out that this function 
[pval_mu, chisq_mu, df_mu] = chisquare_test_homogeneity (d1, d2, c1)

Doesnt allow any of the values of c1 to be less than 0 (even 0.99 induces to error).

As commented in some mail before, the result is that we get a division by 0, lead to NaN error. It could be great to check it and change it at some point.

As I asked in an earlier email, do you have a good reference we can use for expected behavior, inputs/outputs, maybe a paper or another free open source implementation that we could use as a example... anything? 

a trivial test: 

>> [a b c]=chisquare_test_homogeneity([.1 .2 .3],[.2 .3 .1],[.7 .8 .9])
a =  1
b = 0
c =  3

no error, and c < 1

just to clarify, NaN is not necessarily an error. it's a result (unless you are getting an actual error message, in which case please report the error message). in this case a divide by zero is caused by the lines:

(x is your d1, y is your d2, c is your c1)

55: n_x   = sum (x * ones (1, df+1) < ones (l_x, 1) * c);
58: n_y   = sum (y * ones (1, df+1) < ones (l_y, 1) * c);
59: chisq = l_x * l_y * sum ((n_x/l_x - n_y/l_y).^2 ./ (n_x + n_y));

There is no error from putting in c<1 that I can see. the NaN comes from c being smaller than d1 and d2, making n_x and n_y zero, making chisq a divide by zero.

If this isn't what should happen, you should provide some reference that can describe a desired result.

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