Consider the following setup. We have a $p$-dimensional parameter vector $\theta$ that specifies the model completely and a maximum-likelihood estimator $\hat{\theta}$. The Fisher information in $\theta$ is denoted $I(\theta)$.
What is usually referred to as the Wald statistic is
$$(\hat{\theta} - \theta)^T I(\hat{\theta}) (\hat{\theta} - \theta)$$
where $I(\hat{\theta})$ is the Fisher information evaluated in the maximum-likelihood estimator. Under regularity conditions the Wald statistic follows
asymptotically a $\chi^2$-distribution with $p$-degrees of freedom when $\theta$ is the true parameter. The Wald statistic can be used to test a simple hypothesis $H_0 : \theta = \theta_0$ on the entire parameter vector.
With $\Sigma(\theta) = I(\theta)^{-1}$ the inverse Fisher information the Wald test statistic of the hypothesis $H_0 : \theta_1 = \theta_{0,1}$ is
$$\frac{(\hat{\theta}_1 - \theta_{0,1})^2}{\Sigma(\hat{\theta})_{ii}}.$$
Its asymptotic distribution is a $\chi^2$-distribution with 1 degrees of freedom.
For the normal model where $\theta = (\mu, \sigma^2)$ is the vector of the mean and the variance parameters, the Wald test statistic of testing if $\mu = \mu_0$ is
$$\frac{n(\hat{\mu} - \mu_0)^2}{\hat{\sigma}^2}$$
with $n$ the sample size.
Here $\hat{\sigma}^2$ is the maximum-likelihood estimator of $\sigma^2$ (where you divide by $n$). The $t$-test statistic is
$$\frac{\sqrt{n}(\hat{\mu} - \mu_0)}{s}$$
where $s^2$ is the unbiased estimator of the variance (where you divide by the $n-1$). The Wald test statistic is almost but not exactly equal to the square of the $t$-test statistic, but they are asymptotically equivalent when $n \to \infty$. The squared $t$-test statistic has an exact $F(1, n-1)$-distribution, which converges to the $\chi^2$-distribution with 1 degrees of freedom for $n \to \infty$.
The same story holds regarding the $F$-test in one-way ANOVA.
