Test statistic for Wald test - Normal distribution If given X a random variable: $X1,...,Xn ~ N \sim(\mu, \sigma^2)$ ..
I want to perform a Wald test for:
$\mathrm{H}_\mathrm{0}: \mu = \mathrm{\mu}_\mathrm{0}$
$\mathrm{H}_\mathrm{1}: \mu \neq \mathrm{\mu}_\mathrm{0}$
What would be the test statistic here?
Generaly, it is $T.S = |W| = (\hat{\theta} - \mathrm{\theta}_\mathrm{0})/ \hat{se}$
is it simply $\hat{\mu}$ in this case? And in order to find what is it equal to, I need to find the MLE for $\mu$, correct? and that is by finding Fisher Information $I(\mu)$.
Any help would be much appreciated. Thanks!
 A: Definition: The Wald test statistic is defined by $$W = n(\hat\theta - \theta_0)'I(\hat\theta)^{-1}(\hat\theta - \theta_0),$$ where $\hat\theta$ is the MLE of $\theta$, $\theta_0$ the fixed value under $H_0$, and $I(\hat\theta)$ is the Fisher information evaluated at the estimate $\hat\theta$.
Remark: It ca be shown that $W$ is asymptotically $\chi^2_p$-distributed, where $p$ are the number of components of $\theta$. The square root of a $\chi^2_1$-distributed random variable is standard normally distributed. Hence in the case $p=1$, $\sqrt W$ can be expressed as $$\sqrt W = \sqrt n\frac{\sqrt{(\hat\theta - \theta_0)^2}}{\sqrt{I(\hat\theta)}} = \frac{\hat\theta - \theta_0}{\mathit{SE}(\hat\theta)},$$ where $\mathit{Se}(\hat\theta) = \frac{\sqrt{I(\hat\theta)}}{\sqrt n}$ is the standard error of $\hat\theta$.
Example: Let $X_1, X_2, \dots, X_n$ be an iid sample from a $\mathcal N(\theta,1)$ distribution. Note that $p = 1$ in this case. The MLE of $\theta$ is given by $$\hat\theta = \frac 1n\sum_{i=1}^n X_i.$$ The Fisher information in the given setting is $I(\theta) = 1$. Hence the Wald statistic for testing the null hypothesis $H_0: \theta = \theta_0$ is given by $$W = n\left(\frac 1n\sum_{i=1}^nX_i - \theta_0\right)^2.$$ Considering the square root $\sqrt W$ yields the usual test statistic.
