# Testing $A\mu + b = 0$ for a normal distribution

Say $$\{x_i\}_{i=1}^m$$ are i.i.d observations from a multivariate normal distribution $$N(\mu, \Sigma$$).

How could I test the hypothesis $$H_0:A \mu + b=0$$ against $$H_0:A \mu + b\neq0$$, such that A is some invertible $$2\times 2$$ matrix and b is a 2D vector?

Under $$H_0$$, the quadratic form $$(A\bar x+b)^T (A\Sigma A^T)^{-1}(A\bar x+b)$$ is chi-square with 2 degrees of freedom. If you substitute $$\Sigma$$ by its estimate $$\hat\Sigma=\frac1{m-1} \sum_{i=1}^m (x_i-\bar x)(x_i-\bar x)^T$$, then $$(A\bar x+b)^T (A\hat\Sigma A^T)^{-1}(A\bar x+b)$$ is instead Hotelling T-squared distributed with $$2$$ and $$m-1$$ degrees of freedom.
• Can you explain why we are using the quadratic form to get inferences of $H_0$? We simply reject $H_0$ if $T>F$ right? Sep 11 '20 at 2:54