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When performing linear regression with one variable, one can compute Fisher's test with value $F$, and derive Student's test $T=\sqrt{F}$.

When there is more than one variable, the relationship $T=\sqrt{F}$ no longer holds. Is there a relationship between $F$ and the different Student tests $T_i$ (one for each variable)? Perhaps a system of equations?

EDIT :

I am referring to the F test in Excel's linear regression report:

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Note : crossposted here with no answer.

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    $\begingroup$ What "Fisher's test" are you referring to? $\endgroup$
    – utobi
    Commented Oct 11, 2022 at 13:00
  • $\begingroup$ @utobi I am reffering to the F test which is run by Excel when performing linear regression. $\endgroup$
    – Kuifje
    Commented Oct 11, 2022 at 15:27
  • $\begingroup$ and by "more than one variable" do you mean more than one predictor? $\endgroup$
    – utobi
    Commented Oct 11, 2022 at 17:07
  • $\begingroup$ Yes, by variable I mean explanatory variable / predictor. $\endgroup$
    – Kuifje
    Commented Oct 12, 2022 at 7:57
  • $\begingroup$ the relation between $t$ and $F$ is always true no matter what is the number of predictors, as long as you test a single predictor. If you want to test jointly more than one predictor, you have to switch to an $F$ or a Wald test. $\endgroup$
    – utobi
    Commented Oct 12, 2022 at 8:22

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I'm not sure that this is what you mean, but the $T^2$ Hotelling test is a generalization of the $t$-test for multivariate normal means.

Suppose that $X \sim N(\mu, \Sigma)$, where the number of dimensions is $p$. The number of observations is $n$. The test statistic is $$T = n (\bar{X} - \mu)' \hat{\Sigma}^{-1} (\bar{X} - \mu)$$, which using SVD decomposition for $\hat{\Sigma}$ is, suppose that $S$ is a diagonal matrix of the eigen-values and $U$ is the eigenvectors of $\hat{\Sigma}$. You obtain

$$ T= n (U(\bar{X} -\mu))' S^{-1} U(\bar{X} - \mu),$$ which can be viewed as sum of squared $t$-statistics.

The statistic is related to $F$ distribution by $\frac{n-p}{p(n-1)}T \sim F_{p, n-p}$.

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