| T O P I C R E V I E W |
| yjiean |
Posted - 01/31/2013 : 8:36:20 PM
Origin Ver. and Service Release: OriginPro 8.6.0 Sr3
Operating System: Windows 7
I want to fit my data to a linear fit, passing through (0,0). However, while my individual data errors are large, the standard error of the slope is very small. How can I make the analysis more representative of the data points? Thank you.
x y y error
0 0 0
5 2.21338 5.64255
10 6.15661 13.48302
15 9.64017 18.24125
20 12.24915 20.91573
25 15.14843 23.54393
30 17.61415 24.67954
35 21.46274 25.05672
40 23.48769 24.4519
45 26.95351 24.99781
50 29.49941 25.83878
55 32.96064 27.49744
60 35.56967 27.34075
65 37.38313 27.35566
70 39.60066 29.00508
75 42.0113 30.43893
80 44.99208 30.84694
85 46.56707 31.17669
90 48.58238 31.41765
95 50.90368 31.07246
100 52.89489 31.57719
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| 4 L A T E S T R E P L I E S (Newest First) |
| Sam Fang |
Posted - 02/04/2013 : 05:03:54 AM
For calculated parameters, the terms "standard error" and "standard deviation" are same in fitting.
In linear fit, not y error magnitude itself but its relative magnitude to other points makes sense. You can try to scale each point's y error by the same factor, e.g. 10, and you will find the fitted result doesn't change.
If you want the parameter's standard error can reflect the change of y error, you can clear Use Reduced Chi-Sqr check box in Fit Options branch in Linear Fit's dialog.
Sam
OriginLab Technical Services |
| yjiean |
Posted - 02/02/2013 : 10:56:16 AM
What I actually mean is that, how do I get the standard deviation, instead of the standard error? |
| yjiean |
Posted - 02/01/2013 : 5:51:35 PM
Hi Hideo
Thank you for your response. I have been doing that thus far. The issue I have is that I get a slope of 0.55 and the standard error is 0.00366, for the data points given. However, from the individual data points, the error is about the same order of magnitude as the data points. So I should be expecting an error on the same magnitude as the obtained value?
Jie An
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| Hideo Fujii |
Posted - 02/01/2013 : 10:48:59 AM
Hi yjiean,
You can take the YErr column values for the "instrumental weighting" (reciprocal of the square of YErr),
as described at the following part of the Origin help:
http://www.originlab.com/www/helponline/Origin/en/UserGuide/Linear_Regression_Dialog.html#Fit_Options
In the instrumental weighting scheme, by definition, you cannot put zero(0) at the YErr.
You need to remove the row of {X=Y=YErr=0}, or put a very small value (say 1E-10) to its YErr.
Anyway, to let the fitted line pass thru (0, 0), you can turn ON the "Fix Intercept"
check box (under the Fit Options" option branch in the Linear Fit tool), and set "Fix Intercept at" to "0".
Does it make sense?
--Hideo Fujii
OriginLab |
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