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!
Best Answer
Some people call both "the Wald test".
See, for example, Wikipedia on the Wald test. Note that if you square the Z statistic you get the chi-square statistic; similarly the square of the two tailed Z critical value is the chi square critical value.
So they're not really doing different things.
I wouldn't say one really generalizes the other in the case of a single parameter (aside from the Z-version arguably has the advantage of making a one-tailed test possible), but the Wald chi-square approach readily extends to multiple parameters, so in that sense you could say it was more general.