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When performing a Kolmogorov-Smirnov test, we compute the KS distance between the empirical cumulative distribution function, and the theoretical one. Then, we compare this measure with the corresponding critical value, and reject the null hypothesis (the sample is drawn from the reference distribution) if the distance is larger than the critical value.

But if we look at the table of critical values, we can see that they increase as $\alpha$ gets smaller. My understanding was that $\alpha$ was the risk of making an error, so I would expect that the higher $\alpha$ is, the higher the risk, and so the higher the critical value should be.

Why does the critical value increase when $\alpha$ decreases?

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The level of significance α is supposed to the probability at which the null hypothesis $H_0$ can be rejected. But when checking the statistical tables for $t_\alpha$, $z_\alpha$, and $χ^2_\alpha$, the values increase with decrease in $\alpha$ leading to the conception that the acceptance region increases as the strictness increases.

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    $\begingroup$ Yes - though I might have said this as: $\alpha$ is the probability under the null hypothesis of being beyond the critical value, so a larger magnitude for the critical value corresponds to a smaller probability $\alpha$ of an observation being beyond it. $\endgroup$
    – Henry
    Sep 3, 2022 at 15:21
  • $\begingroup$ @Henry I am confused. If I am testing wheter my sample is drawn from a given reference distribution with $\alpha = 0.1$, can I say that roughly speaking, there is a $10 \%$ chance that my test is not reliable ? $\endgroup$
    – Kuifje
    Sep 5, 2022 at 9:14
  • $\begingroup$ Not quite: you can say that if your sample is indeed drawn from that distribution then there would be a $10\%$ chance your observation is in the rejection region (beyond the $\alpha=0.1$ critical value) but you cannot without more analysis say anything about what happens if your sample is drawn from a different distribution $\endgroup$
    – Henry
    Sep 5, 2022 at 9:18

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