Standard error of regression curve

Hello,

I have two vectors, named as x and y. I would like to estimate the standard error of a regressed curve of these data points. The regression model is defined as ft in the below code. Can you please let me know if there is a way to estimate the standard error of b and m coefficients of this fit type?

x=[2;4;6;9;12;15;18]
y=[188;198;294;433;661;875;1097]
ft = fittype( 'b*x^m');
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
fitresult= fit( x, y, ft, opts );

Answer 1

[Adam Danz]

Follow this answer provided by MathWorks support [here]. Since the degrees of freedom are not defined in that answer, here's a more complete answer:

% Your data and their fit
x=[2;4;6;9;12;15;18]
y=[188;198;294;433;661;875;1097]
ft = fittype( 'b*x^m');
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
[fitresult, gof] = fit( x, y, ft, opts ); % <-- also include 2nd output!

% Compute standard error of the coefficients.
alpha = 0.95; % Choose an alpha level

ci = confint(fitresult, alpha) 
df = gof.dfe; 
t = tinv((1+alpha)/2, df); 
se = (ci(2,:)-ci(1,:)) ./ (2*t) % Standard Error

% Results in same order as coefficients listed in fitresult.
% se =
%     10.0399163997427        0.0994320537151288

[Hakan Süleyman]

Thank you so much for your answer Adam! That's exactly what I was looking for.

Short before submitting my question here I saw the post that you provided in the link. I just wasn't sure if that solution works only for polynomial regression models or also for other types of models, as I am not familiar with these functions yet.

[Adam Danz]

@Hakan Süleyman you have a good point and now I'm second guessing that approach for nonlinear models.

Unforunately I have to shut down for the day but this discussion linked below might be what you're looking for, particularly Richard Willey's answer.

https://www.mathworks.com/matlabcentral/answers/34234-how-to-obtain-std-of-coefficients-from-curve-fitting#answer_42952

I'd be interested if you find a more suitable approach and I may look into it later, myself.

[Hakan Süleyman]

The solution provided by MathWorks support (that you shared in your first comment) returns a very logical result. When I also tried the function given in Richard Willey's answer (in the link you provided in your latter comment), it gives exactly the same SE estimates:

x=[2;4;6;9;12;15;18]
y=[188;198;294;433;661;875;1097]
ft = fittype('b*x^m');
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
[fitresult, gof] = fit( x, y, ft, opts ); 
coeff=coeffvalues(fitresult);

myFit = NonLinearModel.fit(x,y, 'y ~ b0*x1^b1', [coeff(1), coeff(2)])
myFit.Coefficients.SE

% ans =
% 
%    10.0400
%     0.0994

It seems both ways work for custom fit types. I will only need to be sure for the mathematical part. Thanks a lot!