1
$\begingroup$

Let us suppose I have a value measured from experiment and given by $$V_{\text{exp}} \pm \sigma_{V_{\text{exp}}}$$ and a theoretical value given as $$V_{\text{the}} \pm \sigma_{V_{\text{the}}}$$

Is there a statistical way to measure how well $V_{\text{the}}$ matches with the $V_{\text{exp}}$.

In other words, what is the right way to tell that $V_{\text{the}}$ is a valid theory (or not) for the given experimental result?

It seems to be that we should take the difference,

$$ (V_{\text{exp}} \pm \sigma_{V_{\text{exp}}})- (V_{\text{the}} \pm \sigma_{V_{\text{the}}})$$

and that is $$(V_{\text{exp}} - V_{\text{the}}) \pm \sqrt{\sigma_{V_{\text{exp}}}^2 + \sigma_{V_{\text{the}}}^2} \equiv \Delta V \pm \sigma_{\Delta V}$$

If $$\Delta V - \sigma_{\Delta V} < 0 < \Delta V + \sigma_{\Delta V}$$ we say that the theory is valid I guess. But is a there a measure of how valid...like at which $\sigma$ ?

I guess it is $$\frac{\sigma_{\Delta V}}{\Delta V}$$, but I am not sure. Any help would be appreciated.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

Since you are only dealing with mean values and their standard errors, you can rephrase the task as comparing the (assumed) norm distribution of the sample mean and the distribution of the theoretical mean (and its variation, given a certain sample size).

Assuming normal distributions, this task can be achieved by the KS-test which allows to compare two distributions.

$\endgroup$
3
  • $\begingroup$ The KS test is not the solution to this problem. $\endgroup$
    – whuber
    Commented Mar 27, 2022 at 18:28
  • $\begingroup$ @whuber why can't we use the KS test? If only the mean and the standard deviations matter, then we can compute the KS test for the two corresponding normal distributions in order to see if the experimental distribution matches the theoretical normal distribution. Note that the original question omits higher moments so assuming a normal distribution looks good to me. I am curious to learn why this approach is not sound :) $\endgroup$
    – Ggjj11
    Commented Mar 27, 2022 at 20:06
  • 2
    $\begingroup$ The KS test doesn't work that way: you don't get to construct two distributions from a few parameters, estimated or not. It only applies to actual iid samples, which is not the case here. $\endgroup$
    – whuber
    Commented Mar 28, 2022 at 1:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.