How can I obtain the coefficient values, confidence intervals by the Fit

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Mohamed Lamine ASKOURA 2014-4-3

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https://ww2.mathworks.cn/matlabcentral/answers/124366-how-can-i-obtain-the-coefficient-values-confidence-intervals-by-the-fit

关闭： MATLAB Answer Bot 2021-8-20

hello, I'm new to Matlab, I have this equation:

   c.*(((a+1./sqrt((1/b).^2+X.^2))./((1/b).^2+X.^2).*exp(-a.*sqrt((1/b).^2+X.^2)))+5.33267*(a+1./(sqrt((5.33267./b).^2+X.^2)))./((5.33267./b).^2+X.^2).*exp(a.*sqrt((5.33267./b).^2+X.^2)))./(4.*pi.*b)

I have to fit to find the two parameters a and b. The X and Y are:

X= [1.5 2 3 4 4.5 6 7.5 8 9 10 10.5 12 13.5 14 15 16 16.5 18 19.5 20 21 22 22.5 24 26 28 30 ];
Y=[2.029 0.796 0.456 0.24 0.204 0.122 0.077 0.061 0.053 0.042 0.035 0.027 0.021 0.02 0.017 0.016 0.015 0.015 0.013 0.012 0.01 0.009 0.009 0.008 0.007 0.0067 0.006];

knowing that a must be between [0,01732-0.95]; and b must be between [1.0001-3.1].

Thank you in advance for your help

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Answer 1

Star Strider 2014-4-3

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I replaced: a = P(1), b = P(2), c = P(3) but otherwise did not change your function:

f = @(P,X)  P(3).*(((P(1)+1./sqrt((1/P(2)).^2+X.^2))./((1/P(2)).^2+X.^2).*exp(-P(1).*sqrt((1/P(2)).^2+X.^2)))+5.33267*(P(1)+1./(sqrt((5.33267./P(2)).^2+X.^2)))./((5.33267./P(2)).^2+X.^2).*exp(P(1).*sqrt((5.33267./P(2)).^2+X.^2)))./(4.*pi.*P(2));
X = [1.5 2 3 4 4.5 6 7.5 8 9 10 10.5 12 13.5 14 15 16 16.5 18 19.5 20 21 22 22.5 24 26 28 30 ];
Y = [2.029 0.796 0.456 0.24 0.204 0.122 0.077 0.061 0.053 0.042 0.035 0.027 0.021 0.02 0.017 0.016 0.015 0.015 0.013 0.012 0.01 0.009 0.009 0.008 0.007 0.0067 0.006];
P0 = [0.5;  2.0;  1.0];
[B,resnorm,residual,exitflag,output,lambda,J] = lsqcurvefit(f, P0, X, Y, [0.01732;  1.0001;  -Inf], [0.95;  3.1;  Inf]);
Bci = nlparci(B,residual,'jacobian',J);
[Ypred,delta] = nlpredci(f,X,B,residual,'Jacobian',J);
figure(1)
plot(X, Y, 'xb', 'LineWidth',1.5)
hold on
% plot(X, Ypred, '-r')
errorbar(X, Ypred, delta, '-r')
plot(X, Ypred, '-r')
hold off
legend('Data', 'Predicted ±95% CI', 'Location', 'NorthEast')
grid
sprm = ['a'; 'b'; 'c'];
fprintf(1,'Parameter Estimates and 95%%Confidence Intervals:\n')
for k1 = 1:3
    fprintf(1,'\t%s = %13.5E\t(%13.5E  %13.5E)\n', sprm(k1), B(k1), Bci(k1,:))
end

You did not specify what ‘c’ is, so I added it as P(3), a parameter to be estimated. I used the Optimization Toolbox function lsqcurvefit to fit your data, because the Statistics Toolbox function nlinfit does not allow for constrained parameter estimates. The other functions, nlparci and nlpredci are Statistics Toolbox functions.

Unfortunately, neither a nor b are statistically significant:

Parameter Estimates and 95% Confidence Intervals:
  a =   1.73200E-02  ( -2.26964E-01    2.61604E-01)
  b =   3.08837E+00  ( -1.26825E+00    7.44499E+00)
  c =   9.51825E+01  (  1.60626E+01    1.74302E+02)