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AnaBelchior
Portugal
3 Posts |
 Posted - 10/23/2012 : 06:48:51 AM
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Dear all, I intend to perform a linear regression (linear-quadratic)using an independent variable, two independent variablea and a weighting variable. I tried no use the non-linear curve fit and specify my equation but the results were not the expected. Then I tried to use tje linear fit, but the main problem is that I don't know where to put such information. Does anyone performed this before?
Many thanks. Best Regards, Ana |
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Kathy_Wang
China
159 Posts |
Posted - 10/25/2012 : 12:01:01 AM
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Hi,
Your question is still not clear to us. Could you please share us with the equation you're using and a sample of data which you want to fit? Also, please tell us what results you are expecting. This would be really helpful for us to understand your problem better.
In order to send us files, please click the Send File to Tech Support button and follow the instructions.
Kathy
Originlab |
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AnaBelchior
Portugal
3 Posts |
Posted - 10/29/2012 : 06:55:44 AM
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Hi,
Thanks for your email.
I need to determine the values of a and b (equation below) using the folowing equation:
SF=exp(-a*D-B*D^2)
and values:
D SF
0 1
0.005 0.69
0.01 0.67
0.05 0.56
0.1 0.54
0.5 0.35
1 0.29
1.5 0.27
2 0.15
Thank you a lot,
Ana Belchior
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Kathy_Wang
China
159 Posts |
Posted - 10/30/2012 : 01:43:07 AM
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Your equation is not a linear equation, so you have to use the non-linear curve fit tool with a user defined fitting function (which is what you had done already).
The problem is that the data points doesn't seem to follow the equation well, to improve the fit, one thing you could try is to take natural logarithm for your fitting function on both side, so it would be:
ln(SF)=-a*D-b*D^2
and take ln(SF) as the new dependent variable.
By doing this, you could get a bigger adjusted R square value (about 0.615, while for the previous one only 0.28), which indicates the fitting is improved.
If the fitting is still not good enough for you, you may need to modify the fitting equation to make it follow the data trend better.
Kathy
Originlab |
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Topic |
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Linear regression