Error in the intercept
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I think this might be a dumb question, but how do you find the error of the intercept for a straigh line graph that I have fittded using polyfit? and polyval. I got the error for the slope by doing something like this.
[p2,s2] = polyfit(A2,B2,1);
[f2,delta] = polyval(p2,x,s2);
deltaf2=s2.normr/sqrt(s2.df);
C2=deltaf2^2*inv(s2.R)*inv(s2.R)';
deltap2=sqrt(diag(C2));
ok
댓글 수: 3
Ganessen Moothooveeren
2013년 3월 14일
you used this to find error in slope but which variable is the error in slope??..is it deltaf2?? [p2,s2] = polyfit(A2,B2,1); [f2,delta] = polyval(p2,x,s2); deltaf2=s2.normr/sqrt(s2.df); C2=deltaf2^2*inv(s2.R)*inv(s2.R)'; deltap2=sqrt(diag(C2));
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Adam Parry
2012년 7월 4일
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Star Strider
2012년 7월 4일
I believe both 'nlinfit' and 'lsqcurvefit' can give you the information you need to give to 'nlparci' to calculate the parameter confidence intervals. The principal difference between 'nlinfit' and 'lsqcurvefit' is that 'lsqcurvefit' allows parameter constraints. Both will give you either the covariance matrix or the jacobian as well as the other results that 'nlparci' can use as arguments.
You don't have to alter 'nlparci' to get the standard errors, since 'nlinfit' gives you the ability to calculate those from the 'COVB' matrix it returns. You already calculated the standard errors as 'deltap2' from the covariance matrix you calculated as 'C2' in your original code. I used your results to calculate the 'CI95' matrix.
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Star Strider
2012년 7월 3일
Not dumb at all. The problem is that if you want confidence limits on the estimated parameters, the 'polyfit' and 'polyval' functions won't get you there.
If you have the Statistics or Optimization Toolboxes, you can fit your model with 'lsqcurvefit' or 'nlinfit' respectively, then use 'nlparci' to get the confidence limits on the parameters. (Use 'nlpredci' to get confidence limits on the fitted data.)
If you don't have access the these, 'lscov' will likely give you what you need to calculate the confidence intervals yourself.
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