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Monte Carlo Error using modelXY

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  • axelgans
    Hello dear NMRPipe users, I m using the modelXY program provided by NMRpipe to fit T1 relaxation curves. It seems to work fine but when I compare to the
    Message 1 of 2 , Mar 1 5:25 AM
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      Hello dear NMRPipe users,

      I'm using the modelXY program provided by NMRpipe to fit T1 relaxation curves. It seems to work fine but when I compare to the results from other fitting programs (like minuit or origin) I get similar values except that the Monte Carlo error from NMRPipe is systematically about twice the value I get from the other programs. Does anybody know why this is the case ? Does the DATA MONTE_CARLO_ERROR given by NMRPipe correspond to a +/- value or do I have to divide it by two ?
      Thank you in advance,
      Axel.
    • Frank Delaglio
      By default, the modelXY program reports an error range which accounts for 90% of the monte carlo trials (argument -cf 0.9). It may well be that the other
      Message 2 of 2 , Mar 4 12:05 PM
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        By default, the modelXY program reports an error range which accounts
        for 90% of the monte carlo trials (argument -cf 0.9).

        It may well be that the other software is reporting a range which
        accounts for something like +/- 1 standard deviations, which roughly
        speaking, would probably give values about 1/2 of the modelXY defaults.

        You can reduce the modelXY -cf argument accordingly if you wish.


        Quoting axelgans <axel.gansmuller@...>:

        > Hello dear NMRPipe users,
        >
        > I'm using the modelXY program provided by NMRpipe to fit T1
        > relaxation curves. It seems to work fine but when I compare to the
        > results from other fitting programs (like minuit or origin) I get
        > similar values except that the Monte Carlo error from NMRPipe is
        > systematically about twice the value I get from the other programs.
        > Does anybody know why this is the case ? Does the DATA
        > MONTE_CARLO_ERROR given by NMRPipe correspond to a +/- value or do I
        > have to divide it by two ?
        > Thank you in advance,
        > Axel.
        >
        >
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