# Confusion in hypothesis testing for the linear model

Consider the usual normal linear model $y=X\beta + \epsilon$ where $X$ is $N\times (p+1)$ and $\epsilon \sim N(0,\sigma^2I_N)$ I'm reading up on hypothesis testing in this model: $H_0:$ $\beta_j=0$,$H_1:$ $\beta_j\neq0$. A statistic of interest for this test is $\frac{\hat{\beta}_j}{\hat{\sigma}\sqrt{v_j}}$ where $v_j=(X^TX)^{-1}_{jj}$.

What I fail to understand is the link between this kind of hypothesis testing and the usual definition. The framework for hypothesis testing is usually the following: we have a sample $X_1,\ldots, X_n$ where the $X_i$ are i.i.d and follow the probability distribution $P_{\theta}$ where $\theta$ is unknown. We consider a partition of the parameter space and some $R\subset \mathbb R^n$ (the rejection region).

Here, what is the sample ? What are the probability distributions $P_{\theta}$ ? What is $R$ ?

A sample is $\{y_i, x_{i1}, x_{i2}, \ldots, x_{ip} \}$ for $i \in \{ 1, 2, \ldots , N\}$. Or simply matrix $X$ and vector $y$.
We assume that $\varepsilon \sim N(0, \sigma^2 I_N)$ which translates to $y \sim N(X\beta, \sigma^2 I_N)$. So family of $N$-dimensional normal distributions with mean vector $X\beta$ and covariance matrix $\sigma^2 I_N$ plays role of $P_\theta$ and $\beta$ plays a role of $\theta$.
Statistic from your question follows $t$ distribution with $N-p-1$ degrees of freedom, so rejection region is $(-\infty, -t^\star) \cup (t^\star, +\infty)$, where $t^\star$ is $(1-\alpha/2)$- quantile of $t$ distribution with $N-p-1$ degrees of freedom and $\alpha$ is your significance level (typically $\alpha=0.05$).
• In my answer rejection region is subset of values that test statistic can possibly take. So, to be consistent with Casella and Bergrer is sholud write that rejection region is set of $X$'s and $y$'s such that $\hat\beta_j / \hat \sigma \sqrt{v_j}$ lies in $(-\infty, -t^\star) \cup (t^\star, +\infty)$. – Łukasz Deryło Jul 4 '17 at 8:25