Is it possible to put inequality to null hypothesis? Based on a sample of i.i.d. Normal random variables $X_1, . . . , X_n$ with mean µ and variance $σ^2$, propose a test with asymptotic level 5% for the hypotheses:
$$ H_0: µ > σ$$
$$ H_1: µ \leq σ$$
What is the p-value of your test if the sample has size n = 100, the sample average is 2.41 and the sample variance is 5.20 ? If the sample size is n = 100, the sample average is 3.28 and the sample variance is 15.95 ? In the latter case, do you reject H0 at level 5% ? At level 10% ?
 A: As $n$ is great enuogh, you can use an asymptotic test:
$$\Lambda =-2 \log \lambda (\mathbf{x})\sim \chi_{(r)}^2$$
Where
$\lambda (\mathbf{x})$ is the generalized likelihood ratio and $r$ is the number of parameters specified in $\mathcal{H}_0$.
As $\mathcal{H}_0 $ is concerned you can consider $\mu=\sigma$. The hypothesis system does not change because of the definition of the size $\alpha$,
$$\alpha=\sup_{\theta \in \Theta_0}\mathbb{P}[ \lambda (\mathbf{x})<k] $$
A: The Tommik answer is likely the intended one for this textbook-like problem.  I may as well explain my (likely nonstandard) approach from my earlier comment:
Assume $\sigma, \mu$ are unknown, and $\sigma>0$.  You can reformulate the problem (as in a comment by Henry) as:
$$ H_0: \mu \geq \sigma, \quad H_1: \mu < \sigma $$
then devolop a test that accepts $H_0$ if
$$ M_n > S_n(1-c)$$
where $M_n, S_n^2$ are the sample mean and variance, and $c$ is a to-be-determined constant chosen to ensure $P[M_n \leq S_n(1-c)] \leq \alpha$ whenever $H_0$ holds (where $\alpha$ is either $0.05$ or $0.1$). Then, assuming $H_0$ holds:
\begin{align}
P[M_n \leq S_n(1-c)] &= P\left[\frac{\sqrt{n}(M_n-\mu)}{\sigma} \leq \frac{\sqrt{n}(S_n(1-c) - \mu)}{\sigma}\right]\\
&=P\left[\frac{\sqrt{n}(M_n-\mu)}{\sigma} \leq (1-c)\sqrt{\frac{n}{n-1}}\sqrt{\frac{(n-1)S_n^2}{\sigma^2}} - \frac{\mu}{\sigma}\sqrt{n}\right]\\
&\leq P\left[\frac{\sqrt{n}(M_n-\mu)}{\sigma} \leq (1-c)\sqrt{\frac{n}{n-1}}\sqrt{\frac{(n-1)S_n^2}{\sigma^2}} - \sqrt{n}\right]
\end{align}
where the final inequality holds because we assume $\mu \geq \sigma$ (and in fact this inequality holds with equality if $\mu = \sigma$).
It is known that (amazingly) $M_n$ and $S_n$ are independent and so the following variables $G$ and $X_{n-1}$ are independent:
\begin{align}
G &= \frac{\sqrt{n}(M_n-\mu)}{\sigma} \sim N(0,1)\\
X_{n-1} &= \sqrt{\frac{(n-1)S_n^2}{\sigma^2}} \sim \mbox{chi($n-1$) variable}
\end{align}
So we get (using independence of $G$ and $X_{n-1}$):
\begin{align}
P[M_n \leq S_n(1-c)] &\leq P\left[G \leq X_{n-1}(1-c)\sqrt{\frac{n}{n-1}} - \sqrt{n}\right]\\
&= \int_0^{\infty} F_G\left(x(1-c)\sqrt{\frac{n}{n-1}} - \sqrt{n}\right)f_{X_{n-1}}(x)dx
\end{align}
where equality holds if we assume $\mu = \sigma$, and where $F_G(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-t^2/2}dt$ is the CDF of a standard Gaussian, and $f_{X_{n-1}}(x)$ is the PDF of the chi($n-1$) variable. So we search for the smallest value $c>0$ for which
$$ \int_0^{\infty} F_G\left(x(1-c)\sqrt{\frac{n}{n-1}} - \sqrt{n}\right)f_{X_{n-1}}(x)dx \leq \alpha $$
For a given $c$ value, you can numerically integrate to find the left-hand-side.  Then play around with $c$ to yield a result close to the desired $\alpha$.(As I mentioned in my comment, this is a complicated integral to evaluate.)
