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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採用された回答
Jeff Miller
2023 年 6 月 23 日
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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