- #1
Chlorophylle
- 4
- 0
Hello! :-)
I' have a problem...
I fitted experimental data with a second order exponential theoretical curve:
y = A1*exp(-x/t1)+ A2*exp(-x/t2) + y0
and I have these data for 4 repetitions of the same experiment:
first time:
Reduced Chi-Sqr = 6,81066E-4
Adj. R-Square = 0,41005
y0 = 1,60364 (Standard Error = 0,0012)
A1 = 0,03892 (Standard Error = 0,00349)
t1 = 1,80889 (Standard Error = 0,30923)
A2 = 0,29705 (Standard Error = 0,01466)
t2 = 0,05342 (Standard Error = 0,00445)second time:
Reduced Chi-Sqr = 9,52433E-4
Adj. R-Square = 0,34084
y0 = 1,05106 (Standard Error = 8,31296E-4)
A1 = 0,27818 (Standard Error = 0,02292)
t1 = 0,02764 (Standard Error = 0,00432)
A2 = 0,09086 (Standard Error = 0,01005)
t2 = 0,47223 (Standard Error = 0,06265)third time:
Reduced Chi-Sqr = 6,13655E-4
Adj. R-Square = 0,45827
y0 = 0,60035 (Standard Error = 0,00425)
A1 = 0,02735 (Standard Error = 0,0032)
t1 = 4,40428 (Standard Error = 1,76829)
A2 = 0,27926 (Standard Error = 0,01113)
t2 = 0,0895 (Standard Error = 0,00593)
forth time:
Reduced Chi-Sqr = 9,87583E-4
Adj. R-Square = 0,36993
y0 = 0,16076 (Standard Error = 8,19131E-4)
A1 = 0,28751 (Standard Error = 0,02368)
t1 = 0,02714 (Standard Error = 0,00429)
A2 = 0,10903 (Standard Error = 0,01115)
t2 = 0,42818 (Standard Error = 0,05112)
I discarded the second because of the big standard error of t1. But, this has the bigger adj. R squared.
I thought that the first and third are the best, because values A1, A2, t1, t2 are near.
Is this correct?
How can I evaluate them by the reduced chi-squared and adj. R squared. R squared is so low because of the noise..
I' have a problem...
I fitted experimental data with a second order exponential theoretical curve:
y = A1*exp(-x/t1)+ A2*exp(-x/t2) + y0
and I have these data for 4 repetitions of the same experiment:
first time:
Reduced Chi-Sqr = 6,81066E-4
Adj. R-Square = 0,41005
y0 = 1,60364 (Standard Error = 0,0012)
A1 = 0,03892 (Standard Error = 0,00349)
t1 = 1,80889 (Standard Error = 0,30923)
A2 = 0,29705 (Standard Error = 0,01466)
t2 = 0,05342 (Standard Error = 0,00445)second time:
Reduced Chi-Sqr = 9,52433E-4
Adj. R-Square = 0,34084
y0 = 1,05106 (Standard Error = 8,31296E-4)
A1 = 0,27818 (Standard Error = 0,02292)
t1 = 0,02764 (Standard Error = 0,00432)
A2 = 0,09086 (Standard Error = 0,01005)
t2 = 0,47223 (Standard Error = 0,06265)third time:
Reduced Chi-Sqr = 6,13655E-4
Adj. R-Square = 0,45827
y0 = 0,60035 (Standard Error = 0,00425)
A1 = 0,02735 (Standard Error = 0,0032)
t1 = 4,40428 (Standard Error = 1,76829)
A2 = 0,27926 (Standard Error = 0,01113)
t2 = 0,0895 (Standard Error = 0,00593)
forth time:
Reduced Chi-Sqr = 9,87583E-4
Adj. R-Square = 0,36993
y0 = 0,16076 (Standard Error = 8,19131E-4)
A1 = 0,28751 (Standard Error = 0,02368)
t1 = 0,02714 (Standard Error = 0,00429)
A2 = 0,10903 (Standard Error = 0,01115)
t2 = 0,42818 (Standard Error = 0,05112)
I discarded the second because of the big standard error of t1. But, this has the bigger adj. R squared.
I thought that the first and third are the best, because values A1, A2, t1, t2 are near.
Is this correct?
How can I evaluate them by the reduced chi-squared and adj. R squared. R squared is so low because of the noise..
Last edited: