About comparison to experimental reference values

[chastiell]

Hi again, probably this seems to be a simply question, but in last days i becomes a really strange one.

We all know that there are many kinds of constants in physics, some of them, are found experimentally with great accuracy in too expensive projects, but no matter how accurate can be measured such constant, they still have uncertainty. So imagine we made an experiment were any constant of this kind is measured , charge electron, charge to mass ratio, Newton gravitatory constant , ...

How to achieve an hypothesis proof on this constants considering that they have uncertainty ?

I has been searching this on Internet, but the nearest answer is given by the comparison of two gaussian variables with the t student distribution using:

t={|\bar{X_1}-\bar{X_2}|\over \sqrt{{\sigma_1^2\over n_1}+{\sigma_1^2\over n_2}}}
where
\sigma_i^2
are the estimators of variance of each variable

but in this form is assumed that the values n_i are known , thing that is not true for the experimental reference values , and this restrict us to always use a mean value, what if we determine this by linear fit, nonlinear fit ?

 

[Simon Bridge]

You are correct to use statistical methods to do the comparison ... the same as with any two statistical variables.
There are a wide range of methods depending on the kind of data you have and what you want to be able to say.

Rule of thumb:
Reference values should be treated like any other measurement.
If you just want to see if two measurements agree, then take the difference and propagate the errors - if they agree then the difference will be within some stated confidence value of zero.

 

[chastiell]

You are correct to use statistical methods to do the comparison ... the same as with any two statistical variables.
There are a wide range of methods depending on the kind of data you have and what you want to be able to say.

Rule of thumb:
Reference values should be treated like any other measurement.
If you just want to see if two measurements agree, then take the difference and propagate the errors - if they agree then the difference will be within some stated confidence value of zero.

Thanks for the answer it is really useful ! . After reading this, i thought the next question ... suppose that the t student test is used to do the comparison ... considering that experimental reference value is treated like any other measurent the number of degrees of freedom would be n or n-1 ?

 

[Simon Bridge]

How do you normally determine the degrees of freedom?

 

[chastiell]

number of degrees of freedom ndf :
ndf = n-r

n : number of data
r : number of restrictions

usually in this kind of test i use r = 1

 

[Simon Bridge]

Then you have your answer.