I have a theoretical problem about hypothesis testing with no "preferable clue" and punctual hypothesis: to make it more understandable I'll try to formulate it as an example.
Let's play the following game: I have developed a software that samples a number from a Gaussian distribution having variance 1. The mean of the distribution can either be $1$ or $1 + \epsilon$ with $\epsilon>0$ fixed and known. You can sample from the software $N$ realizations, then you have to guess which is the real mean I implemented.
Which is the minimum $N$ in order to have a 99% confidence about your answer?
The idea is to construct a confidence interval that shrinks as $N \to \infty$. However in this case there's no "null hypothesis to be proven wrong" and, however, the alternative hypothesis is just a point (either $\mathbb{E}[X]=1$ or $\mathbb{E}[X] =1+\epsilon$ depending on what we choose as null hypothesis).
How would you solve such a problem?
You can use Hoeffding's inequality to guess the correct answer with high probability pretty quickly. It's a pretty standard fact about the Gaussian distribution that if $Z$ has a Gaussian distribution with variance $1$, then $$ \mathbb{P}(|Z-\mathbb{E}[Z]| \geq t) \leq 2\exp(-t^2/2) $$ for all $t>0$.
Let $X_1, \dots, X_N$ be $N$ independent samples from your mystery distribution and let $\bar{X} = \frac{1}{N}(X_1 + \dots + X_N)$. The mean of $\bar{X}$ will be either $1$ or $1+\epsilon$. So we can apply Hoeffding's inequality with a margin of error of (say) $\epsilon/3$ and get that $$ \mathbb{P}(|\bar{X} - \mathbb{E}[\bar{X}]| \geq \epsilon/3) \leq 2\exp \left( \frac{-(1/2)(N\epsilon/3)^2}{N(1/4)} \right) = 2\exp\left( -\frac29 N\epsilon^2 \right). $$ (Note: it's possible that the constant $2/9$ is not quite correct, but there is some constant in there that makes this inequality true. I put $2/9$ for the sake of concreteness but I may have used an incorrect number for the sub-Gaussian norm of the normal distribution).
Let's say we want the above probability to be at least $99\%$. Then just by rearranging it suffices to take $$ N > \frac{(9/2)\log(.01/2)}{\epsilon^2} \approx \frac{23.84}{\epsilon^2}. $$
The conclusion of all this is the following: if you take $N$ independent samples of your mystery distribution (where $N$ satisfies the above inequality) and take the average, there is a $>99\%$ chance that the average will be within $\epsilon/3$ of the true mean of the distribution. Since the two options for the true mean are $1$ or $1+\epsilon$, then you have a $>99\%$ chance of guessing the true mean correctly because the sample mean will be closer to one than the other.