Reduce the inputs for the chi2gof function (follow-up question of: Using chi2gof to test two distributions)

This is a follow-up question of Using chi2gof to test two distributions.

@Allie had the following inputs, i.e. the binned observational data (x), the binned model data (y), and the bin edges (bins):

x= [41 22 11 10 9 5 2 3 2];
y= [38.052 24.2655 15.4665 9.8595 6.2895 4.011 2.562 1.6275 2.8665];
bins=[0:9:81];

and asked how to use "chi2gof to test if two distributions come from a common distribution (null hypothesis) or if they do not come from a common distribution (alternative hypothesis)".

The accepted answer comes from @Jeff Miller:

bins=[0:9:81];
xvals = bins(1:end-1)+4.5;   % Here are some fake data values that belong in each bin.
xcounts= [41 22 11 10 9 5 2 3 2];  % These are the counts of the data values in each bin.
y= [38.052 24.2655 15.4665 9.8595 6.2895 4.011 2.562 1.6275 2.8665];
[h,p,stat]=chi2gof(xvals,'Edges',bins,'Expected',y,'Frequency',xcounts,'EMin',1)

h = 0

p = 0.7905

stat = struct with fields:
    chi2stat: 4.6864
          df: 8
       edges: [0 9 18 27 36 45 54 63 72 81]
           O: [41 22 11 10 9 5 2 3 2]
           E: [38.0520 24.2655 15.4665 9.8595 6.2895 4.0110 2.5620 1.6275 2.8665]

However, by re-doing all the calculations, i.e.

% inputs
alpha = 0.05;
x= [41 22 11 10 9 5 2 3 2];                                           % <-- observed data
y= [38.052 24.2655 15.4665 9.8595 6.2895 4.011 2.562 1.6275 2.8665];  % <-- expected data

% Chi-square goodness-of-fit test
chi2 = sum(((x-y).^2)./y);       % <-- chi-square value
df = length(x)-1;                % <-- degrees of freedom
p = chi2cdf(chi2,df,'upper');    % <-- p-value calculated from the chi-square distribution with "df" degrees of freedom
if p>alpha                       % <-- accept or reject the null hypothesis, based on the p-value
    h = 0;                       % <-- disp('we fail to reject the null hypothesis')
else
    h = 1;                       % <-- disp('we reject the null hypothesis')
end
table([chi2; df; p; h],'RowNames',{'chi-square','df','p-value','h'},'VariableNames',{'Stats'})

ans = 4x1 table
                   Stats 
                  _______

    chi-square     4.6864
    df                  8
    p-value       0.79051
    h                   0

I have realized that "xvals" and the "bins" are not necessary for the Chi-square goodness-of-fit test.

Then, is it possible to still use the chi2gof function, but just employing "x" and "y" (and "alpha") as the only inputs?

Answer 1

[Arnav]

Hi @Sim,

You are correct in seeing that xvals and bins are not necessary for the chi-square test if you have the actual bin counts x, the expected bin counts y, and the significance level alpha.

Going through the chi2gof Documentation, the first argument i.e., x is mandatory. Because of this we need to provide a xvals even if we do not use the binning of the function. So, any dummy vector of length equal to the number of bins will work as xvals.

You can find the documentation here: https://www.mathworks.com/help/stats/chi2gof.html#btv1j1v-3

The following code snippet gets the results of chi-square test by only using x, y and alpha with chi2gof. This works the same as the custom code you provided.

% inputs
alpha = 0.05;
x = [41 22 11 10 9 5 2 3 2];
y = [38.052 24.2655 15.4665 9.8595 6.2895 4.011 2.562 1.6275 2.8665];

% Chi-square goodness-of-fit test
[h, p_value, stats]=chi2gof(1:length(x), 'Expected', y, 'Frequency', x, 'EMin', 0, 'Alpha', alpha)

h = 0

p_value = 0.7905

stats = struct with fields:
    chi2stat: 4.6864
          df: 8
       edges: [1.0000 1.8889 2.7778 3.6667 4.5556 5.4444 6.3333 7.2222 8.1111 9.0000]
           O: [41 22 11 10 9 5 2 3 2]
           E: [38.0520 24.2655 15.4665 9.8595 6.2895 4.0110 2.5620 1.6275 2.8665]

[Sim]

Thanks a lot @Arnav, very kind! :-)