I've read about two ways to test the significance of just one predictor when fitting a linear model with least squares.

Suppose I fit the following linear model using least squares using standard assumptions. $$y = X \beta + \varepsilon \tag{1}$$ where $\beta = (\beta_1, \cdots, \beta_i, \cdots, \beta_p)^T$.

The first method I've seen is the most basic way, where we compute the test statistic $$t = \frac{\hat{\beta}_i}{se(\hat{\beta}_i)} . \tag{2}$$ Then assuming that we've already shown that $\hat{\beta}_i \sim N(X\beta, \sigma^2 I)$ under the assumption that $X$ is non-stochastic, the test statistic $t$ follows a $T$ distribution with $\nu$ degrees of freedom, $$t \sim \frac{Z}{\sqrt{\chi^2/\nu}} = T_\nu .$$

The second approach I've seen is a likelihood ratio testing approach, where $\Omega$ is a linear model with $p$ parameters and $\omega$ is a linear model with $q$ parameters formed by imposing linear constraints on the model $\Omega$. The likelihood ratio test then gives us the following test statistic with an F distribution. $$f_{general} = \frac{(RSS_\omega - RSS_\Omega)/(p-q)}{RSS_\Omega / (n-p)} \sim F_{p-q,n-p}$$.

If we let $\Omega$ be the model in Equation (1), and let $\omega$ be that model but with $\beta_i$ removed (or imposing the constraint that $\beta_i=0$) for some particular $i$, then we get $$f = \frac{(RSS_\omega - RSS_\Omega)}{RSS_\Omega / (n-p)} \sim F_{1,n-p} . \tag{3}$$

We know that the square of a T-distributed random variable is an F-distribution random variable, $$T^2_\nu = F_{1,\nu}.$$

So what's the difference between testing with the test statistic in Equation (2) versus with the test statistic in Equation (3)?

I think that the Neyman-Pearson Lemma tells us that testing with Equation (3) would give us the most powerful test for $H_0: \beta_i=0$, but can the same be said of Equation (2)?

Also, I don't see why it can't be that $t^2 \neq f$, so is it possible for (2) to fail to reject $H_0$ but then for (3) to actually reject $H_0$ for the same model and data?


1 Answer 1


(2) and (3) are equivalent. They Reject or not at the same time. And yes the value of the t statistic squared will be the value of the F stat (because we have 1 numerator df). All you are doing is comparing the t and F statistics to different Tables where the alpha % tail value of the t table is squared to give the corresponding value on the F table. Look at the tabled values.


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