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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!

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  • $\begingroup$ This is a good question but it may need to be contextualized because at the broad level at which it is stated, the "z hypothesis" is in conflict with the statement "if we don't know the variance, we just estimate it". Imagine that you are trying to estimate the unknown parameter in a binomial random variable. You take a sample. You calculate a sample mean. Now to test the hypothesis that the population sample equals a particular value, you calculate the z-statistic as the ratio of sample estimate minus hypothesized value of population parameter over the standard error. $\endgroup$ – ColorStatistics Dec 26 '18 at 13:34
  • $\begingroup$ This standard error in the denominator of the z-statistic is computed using the hypothesized population parameter. Therefore, once you decided on the "z hypothesis" you want to test, which means that you decided that you want to test whether your population parameter equals a particular value, you've automatically decided the standard error that will be used as the denominator of the z-statistic (employing the Central Limit Theorem). No separate estimation of the standard deviation need take place. $\endgroup$ – ColorStatistics Dec 26 '18 at 13:34
  • $\begingroup$ In the context of same problem, a Wald statistic is computed as z squared, where the denominator in the z-statistic uses the ML estimate of the population parameter, as opposed to the hypothesized (in the null) value of the population parameter. In other words, the Wald statistic employs the standard error evaluated at the ML estimate, whereas the z-statistic employs the null standard error. In this sense, the z-test is a more generic test than the Wald test. Only if null in z-test is that population parameter equals ML estimate, then exactly same hypothesis is tested by the z and Wald tests. $\endgroup$ – ColorStatistics Dec 26 '18 at 13:34
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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.

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