The "z" hypothesis test is derived from the fact the mean estimator $\hat\Theta$ is distributed normally. If we don't know the variance, we just estimate it ($\widehat{se}$).
The Wald test, derived from the fact that the fisher information of the MLE is distributed chi squared.
We basically get the same result with the two tests.
I just want to make sure that the Wald test is the generalization of the "z" test - or is there any difference?
Thanks!