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## Main Question or Discussion Point

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:

Reduced Chi-Sqr =

Adj. R-Square =

y0 = 1,60364 (Standard Error = 0,0012)

A1 =

t1 =

A2 =

t2 =

Reduced Chi-Sqr =

Adj. R-Square =

y0 = 1,05106 (Standard Error = 8,31296E-4)

A1 =

t1 =

A2 =

t2 =

Reduced Chi-Sqr =

Adj. R-Square =

y0 = 0,60035 (Standard Error = 0,00425)

A1 =

t1 =

A2 =

t2 =

Reduced Chi-Sqr =

Adj. R-Square =

y0 = 0,16076 (Standard Error = 8,19131E-4)

A1 =

t1 =

A2 =

t2 =

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..

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