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By von der Linden W., Dose V., von Toussaint U.

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1. exp The hypothesis implies on average μ = N q2 green balls. 1 ng = 12) deviates from this value by n = μ − ng . Now the probability for a deviation from the mean as large or even larger than the observed deviation n∗ is computed: μ− n∗ P := N P (ng |N, q1 ) + ng =0 P (ng |N, q1 ). 1. This quantity is called the ‘P value’. g. 5%) then the data are said to be significant, because the deviations can hardly be caused by chance, and the hypothesis is rejected. 2 as a function of exp the number of green balls ng in the sample.

Next we compute the mean N , based on the same approximations: 1 N = Z Nmax NN −L N=nmax ≈ nmax 1 ≈ Z Nmax N −L+1 dN N=nmax 1 1+ . L−2 This result is only sensible for sample sizes L > 2. For samples of size 1 and 2 the posterior probability depends on the chosen cutoff value Nmax and neither the norm nor the first moment exists. e. the probability for N being less than a given threshold. g. 90% of the probability mass, 90% of the probability mass is in the interval nmax N L−1 . 1 must hold. Therefore 1 I90% = [nmax , nmax 10 L−1 ].

The likelihood function in this example is the binomial distribution. The remaining task is to clarify what H and in particular H really means, given the background information I. This is a crucial step in serious hypothesis testing, as we will discuss in Part IV [p. 255]. In the coin example it corresponds to assigning values to the prior PDF P (q|N, A, I). The present background information is encoded as P (q|A, I) = δ(q − 1/2) for A = H 1(0 ≤ q ≤ 1) for A = H . That means for us a coin is only fair if the probability q is precisely 1/2.

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