How to use chi2gof within CUPID
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Two examples of usage of the Matlab's "Chi-square goodness-of-fit test" (chi2gof) function are the following:
First (comparing two frequency distributions):
Population = [996, 749, 370, 53, 9, 3, 1, 0];
Sample = [647, 486, 100, 22, 0, 0, 0, 0];
Population2 = [996, 749, 370, sum(Population(4:8))];
Sample2 = [647, 486, 100, sum(Sample(4:8))];
x = [];
for i = 1:length(Sample2)
x = [x,i*ones(1,Sample2(i))];
end
edges = .5+(0:length(Sample2));
[h,p,k] = chi2gof(x,'Expected',Population2,'Edges',edges)
Second (fit a distribution to data):
bins = 0:5;
obsCounts = [6 16 10 12 4 2];
n = sum(obsCounts);
pd = fitdist(bins','Poisson','Frequency',obsCounts');
expCounts = n * pdf(pd,bins);
[h,p,st] = chi2gof(bins,'Ctrs',bins,...
'Frequency',obsCounts, ...
'Expected',expCounts,...
'NParams',1)
addpath('.../Cupid-master')
% (1) create a "truncated dataset"
pd = makedist('Weibull','a',3,'b',5);
t = truncate(pd,3,inf);
data_trunc = random(t,10000,1);
% (2) fit a distribution (in this case the "Weibull2") to the "truncated test"
fittedDist = TruncatedXlow(Weibull2(2,2),3);
% (3) estimate the Weibull parameters by maximum likelihood, allowing for the truncation.
fittedDist.EstML(data_trunc);
% (4) plot both the "truncated test" (through the histogram) and the "fitting distribution"
% (in this case the "Weibull2" with Weibull's parameters estimated by maximum likelihood)
figure
xgrid = linspace(0,100,1000)';
histogram(data_trunc,100,'Normalization','pdf','facecolor','blue')
line(xgrid,fittedDist.PDF(xgrid),'Linewidth',2,'color','red')
xlim([2.5 6])
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Risposta accettata
Jeff Miller
il 23 Giu 2023
Yes, that is correct. The successive bin probabilities are the differences of the successive CDF values, and the expected number is the total N times the bin probability--just as you have computed it.
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