# equivalence between the likelihood ratio test and t-tests

The linked sites (link1, link2) demonstrate that the likelihood ratio tests and the corresponding one- and two-sample t-tests are equivalent. However, based on my understanding, the null distribution of the likelihood ratio test (i.e., chi-square distribution) is an approximation, whereas the t-distribution in the t-tests is not. Despite this difference, why are they equivalent?

1. With a likelihood ratio statistic, $$\Lambda$$, the chi-squared distribution is an asymptotic approximation to the distribution of $$-2\log(\Lambda)$$ under quite broad conditions.
2. Under the conditions where some t-test is exact, the corresponding likelihood ratio statistic might be shown to be monotonic in $$|t|$$ or $$t^2$$ (whichever is convenient), and therefore the two statistics should lead to equivalent tests, if the correct null distribution for that statistic is used; it won't be distributed as chi-squared in small samples.
Note in particular, that the square of a $$t$$-distributed random variable has an $$F$$ distribution, and that asymptotically as the denominator d.f. goes to infinity, the $$F$$ distribution goes to a scaled chi-squared. You may be able to derive an equivalent statistic to $$-2\log(\Lambda)$$ that has a small-sample $$F$$ distribution and in the limit as $$n$$ goes to infinity that $$F$$ will be proportional to a chi-squared with the same d.f. as the numerator d.f. of the F statistic (and indeed the same d.f. as the $$t$$).