For this problem, the p-value is bounded by 1⁄36 from below, that is, p-value > 1⁄36. The supremum of (1/36)(1+34p0) is 35⁄36 and the infimum is 1⁄36. Therefore, I’m not taking the supremum, actually the cartoon took the infimum, when you take the infimum you are assuming the neutrino detector measures without errors and this is a problem. The p-value, for this example, is a number between 1⁄36 and 35⁄36.
I did not understand “the big problem” with my argument…
For this problem, the p-value is bounded by 1⁄36 from below, that is, p-value > 1⁄36. The supremum of (1/36)(1+34p0) is 35⁄36 and the infimum is 1⁄36. Therefore, I’m not taking the supremum, actually the cartoon took the infimum, when you take the infimum you are assuming the neutrino detector measures without errors and this is a problem. The p-value, for this example, is a number between 1⁄36 and 35⁄36.
I did not understand “the big problem” with my argument…