= Solution
<Wilks theorem> states that, under an interior null parameter and the usual identifiability, smoothness and nonsingular-information regularity conditions, $-2\log\Lambda$ for nested models converges under the null to a chi-square distribution with degrees of freedom equal to the difference of parameter <dimensions>. Boundary or nonidentifiable nulls need not have this distribution.
Write $\bar X=n^{-1}\sum X_i$, $Q=\sum(X_i-\bar X)^2$ and $S=\sum X_i^2=Q+n\bar X^2$. The unrestricted maximum-likelihood estimates are $\widehat\mu=\bar X$ and $\widehat\sigma^2=Q/n$. Under the null the <variance> estimate is $S/n$. Substituting both maxima into the Gaussian likelihood gives
$$
\boxed{\Lambda=(Q/S)^{n/2},\qquad -2\log\Lambda=n\log\left(1+\frac{n\bar X^2}{Q}\right)\ \overset{H_0}{\approx}\ \chi_1^2.}
$$
This is for $n\ge2$, with positive true <variance>; $Q>0$ almost surely. The lost parameter <dimension> is one, even though <variance> is estimated in both models.
The two-sided <Student t-test> uses $T=\sqrt n\bar X/s$, where $s^2=Q/(n-1)$, and rejects when $|T|>t_{n-1,1-\alpha/2}$. Under the null $T$ has exactly the <Student t-distribution> with $n-1$ degrees of freedom. The likelihood-ratio statistic is
$$
-2\log\Lambda=n\log(1+T^2/(n-1))=T^2+o_P(1)
$$
under the null, so its chi-square approximation agrees asymptotically with the squared asymptotic standard-normal t statistic. In fact the expression is strictly increasing in $T^2$: with exact finite-sample calibration the tests have \b[identical rejection regions for every $n$], not just approximately for large samples.
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