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## Pearson Chi-squared test of independence

version 1.0.0 (1.09 KB) by
Pearson Chi-Square Test for Independence: calculates the chi-square, p-value and the test decision,

10 Downloads

Updated 17 Jul 2020

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Example;

x = [61, 54, 32, 26, 47, 39, 13, 26, 12, 59; 7, 14, 36, 42, 21, 29, 55, 42, 56, 9];
[h, chi, p] = chi2ind(x, 0.05)

h =

logical

1

chi =

173.0768

p =

0

### Cite As

Ryan Ward (2021). Pearson Chi-squared test of independence (https://www.mathworks.com/matlabcentral/fileexchange/78339-pearson-chi-squared-test-of-independence), MATLAB Central File Exchange. Retrieved .

### Comments and Ratings (3)

Gerrik Labra

Thank you Marius. I used this for a neuroengineering project to find significant transfer of entropy, and no matter what I put in the p value is always zero. Please update your code.

Marius Marinescu

Hello,
I think the degrees of freedom is incorrectly computed.
They must be
df=r*c-(r+c-1); % =(r-1)*(c-1);
where r is the numbers of rows and c the number of columns. You have the following formula
df = size(chi, 1)-1*size(chi,2);
where size(chi) is always one since chi in your code is scalar.

Marius Marinescu

##### MATLAB Release Compatibility
Created with R2018a
Compatible with R2018b and later releases
##### Platform Compatibility
Windows macOS Linux

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