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)
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’})
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?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)
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’})
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? 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)
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

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