# 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?

• The Wald test statistic is almost but not exactly equal to the square of the t-test statistic - see the accepted answer stats.stackexchange.com/questions/60438/… – marsei Sep 13 '14 at 20:54
• @sed so the t-test is not a Wald test? – guest Sep 13 '14 at 22:01
• when n is large, the t-test is essentially identical to the wald test. – marsei Sep 13 '14 at 22:20
• @sed what are the "essential" elements of the tests that compare identical? Are you saying the t-test is the wald test when n is large? What aspects are not identical when n is large? – guest Sep 13 '14 at 22:33

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]

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).

@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.