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The problem reads as follows:

Suppose there are $n$ boxes with $k$ coins in each. The authorities suspect that there is one fake coin in each box. To check it, they randomly select one coin from each box and test it. Suppose that $r$ coins were found to be fake during the test. Is the hypothesis true at significance level $\alpha$? Suppose $k=100, n=10, r=1, \alpha=0.01$.

Okay, so it seems that the null hypothesis here is the presence of a fake coin (what is the alternative here, by the way?)

As far as I understand, in order to do hypothesis testing, we need to choose the test statistic and the way we calculate the p-value. It seems reasonable to me to choose the number of fake coins $r$ as test statistic and calculate the p-value as $$ p(t)=P(r\leq{t}|H_0). $$

My question is: is my problem interpretation and formulation correct? If so, this leads to suspicious numerical results and thus I suspect some kind of mistake here.

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  • $\begingroup$ "true at significance level α" is a very unfortunate wording (since it doesn't make any sense). But your interpretation looks right. What is suspicious about your numerical results? $\endgroup$
    – conjugateprior
    Commented Mar 26, 2016 at 22:00
  • $\begingroup$ @conjugateprior This is absolutely correct because, AFAIK, you can reject the null hypothesis at a given significance level but not accept it. However, that's how the problem is formulated. As for the numerical results, it is easy to calculate that the probability of even not finding a single fake coin is $P(r\leq{0}|H_0)=\prod_{i=1}^{n}\frac{k-1}{k}=\left(\frac{99}{100}\right)^{10} \approxeq 0.9$, so there is no point in even performing the test since the null won't be rejected anyway. $\endgroup$
    – Vossler
    Commented Mar 26, 2016 at 22:29
  • $\begingroup$ Or perhaps you just performed the test, and the next task is to explain it in hypothesis testing terms. $\endgroup$
    – conjugateprior
    Commented Mar 26, 2016 at 22:42
  • $\begingroup$ @conjugateprior Do you mean that this kind of result may be what the exercise was made for? That may be the case, but I'm still not completely sure that this is the case so I'm not closing the question yet...Thank you! $\endgroup$
    – Vossler
    Commented Mar 26, 2016 at 22:48
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    $\begingroup$ On an earlier question: the alternative hypothesis is not specified here, because the textbook is thinking in Fisherian rather than Neyman/Pearson mode. p values don't need to know about alternatives because they are explicitly constructed under the assumption that the null is true. Concepts like power, on the other hand, do require thinking of alternatives. $\endgroup$
    – conjugateprior
    Commented Mar 26, 2016 at 22:59

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Just for the sake of completeness: I found out that my understanding was correct and I was overthinking the problem.

That's what happens when the statements are ambiguous, though.

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