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I know statistical test usually requires "enough" samples to estimate the uncertainties. But what if I have only one sample in each group, and also I "know" which distribution this one sample comes from, is it reasonable to construct a statistical test?

To be more specific, Let's say we have a case-control study and we want to know if there is some significant improvement in "case" than in "control".

The data we have is only one sample, $x$ for the case group, and $y$ for the control group, so $x$ and $y$ are both scalars.

The other information we also have is that $X\sim exp(\lambda)$ and $Y\sim exp(\mu)$, where $\lambda$ and $\mu$ are both some known parameters.

My question is to test my hypothesis that "case is greater than control", can I just construct a test statistic $T = X-Y$? And I will be able to find the distribution of $T$ by this post. Does this make sense?

My confusion mainly comes from how to write up $H_0$ and $H_1$. It seems like I am testing $H_0:x=y$ but obviously this looks very weird... Because usually we will test $H_0: \lambda = \mu$...

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  • $\begingroup$ If you know the distribution, you know everything. You do not need statistics, probability calculus is enough. $\endgroup$ – cure Oct 15 '20 at 22:28
  • $\begingroup$ Could you please be more specific how will you compute the probability for the hypothesis I am trying to test? $\endgroup$ – Jeffrey Oct 15 '20 at 22:35
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    $\begingroup$ I answered this question in the affirmative for an instructive special case at stats.stackexchange.com/a/1836/919. $\endgroup$ – whuber Oct 16 '20 at 14:58
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    $\begingroup$ @whuber Thanks for the very interesting answer! But I am still confused that in my question the two distributions are fully known. I guess in this case maybe there is no way to do any statistical test? $\endgroup$ – Jeffrey Oct 16 '20 at 19:11
  • $\begingroup$ Jeffrey, although the distribution families are known, the distributions are not. Your situation is identical. $\endgroup$ – whuber Oct 16 '20 at 21:12