29
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ @sed so the t-test is not a Wald test? $\endgroup$
    – guest
    Commented Sep 13, 2014 at 22:01
  • 1
    $\begingroup$ when n is large, the t-test is essentially identical to the wald test. $\endgroup$
    – marsei
    Commented Sep 13, 2014 at 22:20
  • $\begingroup$ @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? $\endgroup$
    – guest
    Commented Sep 13, 2014 at 22:33

2 Answers 2

32
$\begingroup$

As Wasserman defines the Wald test, the statistic used in the t-test is the Wald-statistic defined there (aside, perhaps, from a scale factor in the estimate of the standard error that asymptotically goes to $1$):

$$W=\frac{\hat{\theta}-\theta_0}{\hat{\text{se}}(\hat{\theta})}$$

However, the Wald test uses an asymptotic argument to compare that statistic with a standard normal distribution (asymptotically such a scale factor disappears). [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).

$\endgroup$
2
$\begingroup$

@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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.