R2 between histogram and (weibull) distribution

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R2 between histogram of parameter 'Weibull.Wh_Wp' and weibull distribution of parameter 'Weibull.Wh_Wp'.
[parmHat, parmCI]=wblfit(Weibull.Wh_Wp);
X=linspace(min(Weibull.Wh_Wp),max(Weibull.Wh_Wp));
plot(X,wblpdf(X,parmHat(1),parmHat(2)),'Color','k', 'LineWidth',1.3)
hold on
histogram(Weibull.Wh_Wp,'BinWidth',10,'FaceColor',[0.4660 0.6740 0.1880], 'EdgeColor', [0.45 0.45 0.45],'Normalization','pdf')
Below the figure of the parameter 'Weibull.Wh_Wp' with its histogram and fitted weibull distribution.

Accepted Answer

Star Strider
Star Strider on 14 Nov 2021
Try something like this —
wblfcn = @(p,x) wblpdf(x,p(1),p(2));
Weibull.Wh_Wp = wblrnd(150, 3, 1, 100);
X=linspace(min(Weibull.Wh_Wp),max(Weibull.Wh_Wp));
[parmHat,parmCI] = wblfit(Weibull.Wh_Wp);
plot(X,wblpdf(X,parmHat(1),parmHat(2)),'Color','k', 'LineWidth',1.3)
hold on
h1 = histogram(Weibull.Wh_Wp,'BinWidth',10,'FaceColor',[0.4660 0.6740 0.1880], 'EdgeColor', [0.45 0.45 0.45],'Normalization','pdf');
ctrs = h1.BinEdges(1:end-1) + diff(h1.BinEdges(1:2))/2;
y = h1.Values;
wblmdl = fitnlm(ctrs, h1.Values, wblfcn, parmHat)
wblmdl =
Nonlinear regression model: y ~ wblpdf(x,p1,p2) Estimated Coefficients: Estimate SE tStat pValue ________ _______ ______ __________ p1 151.42 9.9346 15.241 1.7925e-12 p2 2.6423 0.36199 7.2994 4.6607e-07 Number of observations: 22, Error degrees of freedom: 20 Root Mean Squared Error: 0.00251 R-Squared: 0.373, Adjusted R-Squared 0.341 F-statistic vs. zero model: 41.9, p-value = 7.05e-08
I’m not certain that an value on a distribution fit using maximum likelihood is statistically correct, however this will provide it if desired. See the documentation for fitnlm for details on it.
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  4 Comments
Dora de Jong
Dora de Jong on 14 Nov 2021
Thank you so much for explaining and taking the time. It has helped a lot.

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R2020a

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