How to calculate R^2 using 1 - (SSR/SST)? For normal fit distribution.
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Hello, I have used the fitlm function to find R^2 (see below), to see how good of a fit the normal distribution is to the actual data. The answer is 0.9172.
How can I manually calculate R^2?
R^2 = 1 - (SSR/SST) or in other words 1 - ((sum(predicted - actual)^2) / ((sum(actual - mean of actual)^2)). I am having a hard time getting the correct answer.
Table = readtable("practice3.xlsx");
actual_values = Table.values;
actual_values = sort(actual_values);
normalfit = fitdist(actual_values,'Normal'); % fit the normal distribution to the data
cdfplot(actual_values); % Plot the empirical CDF
x = 0:2310;
hold on
plot(x, cdf(normalfit, x), 'Color', 'r') % plot the normal distribution
hold off
grid on
nonExceedanceProb = sum(actual_values'<=actual_values,2)/numel(actual_values);
Table.nonExceedanceProb=nonExceedanceProb;
mdl=fitlm(cdf(normalfit, actual_values),Table.nonExceedanceProb);
mdl.Rsquared.Ordinary % R^2
mdl.SSR
mdl.SST
% How can I manually calculate R^2 (or SSR and SST)?
% SSR = sum(((predicted data - actual data).^2))
% TSS = sum((actual data - mean(actual data)).^2)
% Rsquared = 1 - SSR/TSS
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Accepted Answer
Torsten
on 15 Feb 2023
Edited: Torsten
on 15 Feb 2023
In my opinion, it does not make sense to fit a linear function to the value pairs (cdf(normalfit, actual_values),Table.nonExceedanceProb) as you do above.
In principle, the blue points below should lie on the red line. This would mean that the empirical cdf is perfectly reproduced by the normal distribution.
So if you really want to compare the two distributions, you should consider the distance of the blue points (achieved quality of fit) to the red line (perfect fit).
Table = readtable("practice3.xlsx");
actual_values = Table.values;
actual_values = sort(actual_values);
normalfit = fitdist(actual_values,'Normal'); % fit the normal distribution to the data
nonExceedanceProb = sum(actual_values'<=actual_values,2)/numel(actual_values);
hold on
plot(nonExceedanceProb,cdf(normalfit, actual_values),'o')
plot([0 1],[0 1])
xlabel('P(empirical)')
ylabel('P(normal)')
hold off
grid on
12 Comments
Torsten
on 21 Feb 2023
corr(yi,fi) is the pearson correlation coeffcient - I don't know why he wanted to square it.
Anyway: congratulations that you finished your assignment successfully.
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