= Solution
Write $Q=\sum_i(X_i-\bar X)^2$ and $Q_0=\sum_i(X_i-\mu_0)^2=Q+n(\bar X-\mu_0)^2$. The normal-sample likelihood is
$$
L(\mu,\sigma^2)=(2\pi\sigma^2)^{-n/2}\exp\left[-\frac{\sum_i(X_i-\mu)^2}{2\sigma^2}\right].
$$
Without the mean restriction, its maximum has $\widehat\mu=\bar X$ and $\widehat\sigma^2=Q/n$. Under the null, the maximizing variance is $Q_0/n$. Substituting them gives the <generalized likelihood-ratio test> statistic
$$
\Lambda=\frac{\sup_{H_0}L}{\sup L}=\left(\frac Q{Q_0}\right)^{n/2}=\boxed{\left(1+\frac{T^2}{n-1}\right)^{-n/2}},
$$
where $S^2=Q/(n-1)$ and $T=\sqrt n(\bar X-\mu_0)/S$. Assume $n\geq2$; $Q>0$ almost surely for positive population variance.
The ratio decreases strictly as $|T|$ increases, so rejecting for small $\Lambda$ is equivalent to rejecting for large $|T|$. Under $H_0$, $T$ has <Student's t-distribution> with $n-1$ degrees of freedom. Therefore the exact size-$\alpha$ rule is
$$
\boxed{|T|>t_{n-1,\,1-\alpha/2}}.
$$
This is the two-sided <Student's t-test>. Equivalently reject when $\Lambda<(1+t_{n-1,1-\alpha/2}^2/(n-1))^{-n/2}$. No large-sample chi-squared approximation is needed for this <normal-mean likelihood ratio with unknown variance>.
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