How to find uncertainties in Intercept and Slope of a fitted line
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Dinithi Siriwardana Pathiranage
2022 年 6 月 29 日
回答済み: William Rose
2022 年 6 月 30 日
Hi everyone,
I am trying to calculate the uncertainties in the intercept and the slope of a fitted line seperately. Here is my code to fitting a linear line. Can anyone help me to find the uncertainties using polyfit function or any other function?
Thank you.
for bands = 1:7
% Fit line to data using polyfit
c(bands,:) = polyfit(Sun_Zenith_AA_ASD,ratio(:,bands),1);
% Evaluate fit equation using polyval
y_est(bands,:) = polyval(c(bands,:),Sun_Zenith_AA_ASD);
% Add trend line to plot
p_1= plot(Sun_Zenith_AA_ASD,y_est(bands,:),'b--','LineWidth',2)
hold on
% calculating r-sq and RMSE
mdl{:,bands} = fitlm(Sun_Zenith_AA_ASD,ratio(:,bands));
r_sq(bands,:) = round((mdl{:,bands}.Rsquared.Adjusted),4);
RMSE(bands,:) = round((mdl{:,bands}.RMSE),4);
end
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William Rose
2022 年 6 月 30 日
The uncertainties in the fitted parameters are available in the model structure which is created when you call fitlm(). Use coefCI(mdl) to get the 95% confidence intervals for the estimated a0 and estimated a1. Here is what I mean.
Sun_Zenith_AA_ASD=[1:20]'; %x-values to demonstrate
%Next: generate 7 sets of y values: y=a0 + a1*x + noise
a0=70:-10:10; %intercept values
a1=-3:3; %slope values
ratio=a0.*ones(20,1)+a1.*Sun_Zenith_AA_ASD + 2*randn(20,7); %y-values
%Next: Fit columns 1 and 2 of y-data.
%Column 1: a0=70, a1=-3. Column 2: a0=60, a1=-2. Etc.
for i=1:7
mdl=fitlm(Sun_Zenith_AA_ASD,ratio(:,i)); %fit the linear model
a0est=mdl.Coefficients{1,1};
a1est=mdl.Coefficients{2,1};
confInt=coefCI(mdl);
%Display results on console.
fprintf('Model %d: Est.intercept=%.2f, 95%% C.I.=%.2f,%.2f\n',...
i,a0est,confInt(1,:));
fprintf(' Est. slope =%.2f, 95%% C.I.=%.2f,%.2f\n',...
a1est,confInt(2,:));
end
Try it. Good luck.
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