# What could the null hypotesis be in the case of a $F-$test for two $\chi^2$s coming from two fits of the same data?

I report one of the definiton of Fisher's $F$

Suppose that $\chi^2_A$ has the chi-square distribution with $\nu_A$ degrees of freedom and $\chi^2_B$ has the chi-square distribution with $\nu_B$ degrees of freedom, where $\nu_A,\nu_B∈(0,∞)$. Assume also that $\chi^2_A$ and $\chi^2_B$ are independent. The distribution of $$\mathrm{F}=\frac{\chi^2_A/\nu_A}{\chi^2_B/\nu_B}$$

is the $\mathrm{F}$ distribution with $\nu_A$ degrees of freedom in the numerator and $\nu_B$ degrees of freedom in the denominator.

Consider the following situation: I have a set of data and I try to use Least Square Method with two different functions, say a line and a parabola, to fit these data.

Then I calculate $\chi^2$ in both cases and in both cases I get a value smaller than the critical one, so I do not refuse any of the two fits.

(In order to choose between the two of course I can see where $\chi^2$ is smaller, but that can be a fluctuation and let's not consider that).

I heard that a more suitable way to understand if the two fits are both good is to use $F$, as defined above, to do a $F-$test.

Anyway I do not understand what the null hypotesis of the test would be in that case, neither I found anything on textbooks.

So how can one use $F$ to understand if two $\chi^2$ from different fits of the same data can be considered both "good"? In particular what the null hypotesis for that test would be?

My guess (surely wrong and confused) would be that, if the two $\chi^2$ are both "good" then $F \approx 1$ but "the two $\chi^2$ are both good" does not seem a real null hypotesis, since $F$ is defined as $\chi^2$s ratio indipendently from the fact that both the $\chi^2$s come from fits of the same set of data or not. As far as I understand the two $\chi^2$s could be any (as long as indipendent) and also from different sets of data.

• Could you validate my answer since I provided (for the line and parabola you mentioned) a test statistic and a code comparing the F-distribution with the sampling distribution of the test statistic? Sep 8, 2016 at 12:35

• Thanks for the answer, I read your useful answer and the pdf you linked there. If I may, I still don't understand two main things. Firstly in the example you made and in the pdf $F$ is defined as $$F=\frac{(\chi^2_A-\chi^2_B)/(\nu_A+\nu_B)}{\chi^2_B/\nu_B}$$ And not just $$F=\frac{\chi^2_A/\nu_A}{\chi^2_B/\nu_B}$$ Why is this necessary? And can it be defined as the form in my question in some cases (maybe when the models are not nested?)? Sep 6, 2016 at 17:52
• Secondly, still concerning the null hypotesis for the test, from your answer: "Under the null hypothesis that model 2 does not provide a significantly better fit than model 1, $F$ will have an $F$ distribution". Here is what I do not get: how can this null hypotesis make that ratio (however it is defined, as in my question or in your answer) follow an $F$ distribution (and don't follow it otherwise), what is the "key" in the hypotesis that transform that ratio in a ratio that follows a $F$ distribution? Or would it follow a $F$ distribution in any case? Sep 6, 2016 at 17:54
• Thanks for feedback. 1) My version is necessary because one or both of the two $\chi^2$ could not follow a $\chi^2$ distribution. Suppose you fit a parabola relationship with a line... 2) I'm not sure about that but you could investigate using the provided code. Sep 6, 2016 at 20:33