Is Student's t test a Wald test? Is Student's t test a Wald test?
I've read the description of Wald tests from Wasserman's All of Statistics.
It seems to me that the Wald test includes t-tests. Is that correct? If not, what makes a t-test not a Wald test?
 A: As Wasserman defines the Wald test, the statistic used in the t-test is certainly the Wald-statistic defined there:
$$W=\frac{\hat{\theta}-\theta_0}{\hat{\text{se}}(\hat{\theta})}$$
However, the Wald test uses an asymptotic argument to compare that statistic with a standard normal distribution. [The Wald test when dealing with a single parameter can be cast either as a Z-test or a chi-square; in the section being discussed, Wasserman is talking about the Z-form; if you square it, you would have the chi-squared form.]
The t-test relies on an exact small-sample argument to compare the test statistic with a t-distribution.
So, to answer your title question, strictly speaking, no the t-test  is not a Wald test.
Note, though, that they're asymptotically equivalent (i.e. as the sample size, $n\to\infty$, they will reject the same cases); certainly some people - if a bit loosely - call a test based on a t-statistic a Wald-test, whether the statistic is compared with the asymptotic normal distribution or the small-sample result (t-distribution).
A: @Glen_b has provided an excellent answer to the topic. I want to add that, in the t-test, the distribution is the t-distribution. For example, you'd need to know the degree of freedom for your statistics. However, the wald-test relies on the chi-square distribution (square of standard normal). Of course, as the degree of freedom goes to infinity, they're both asymptotically equivalent. 
One would prefer only the wald-test for a sufficient large sample.
