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Max Eastwood
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Value of Information: 8 examples
Max Eastwood4y*10

(0.05 * 0.8) / ((0.05 * 0.8) + (0.95 * 0.2)) = 0.1739

I'm wondering why 0.05 (alpha) was used in that formula? True positive and false negative rates depends on statistical power (1-beta) and beta, and in case of beta 0.2, rate of "Melatonin is working" in case of negative result is 0.457 (not a 0.1739)

"Melatonin is working" branch (prior P(W) = 0.8) have 2 possibilities 
True positive, P("W"|W) = 1-b = 0.8
False negative, P("~W"|W) = b = 0.2

"Melatonin is not working" branch (prior P(~W) = 0.2) have 2 possibilities 
False positive, P("W"|~W) = a = 0.05
True negative, , P("~W"|~W) = 1-a = 0.95

P(W|"~W") = P("~W"|W) * P(W) / (P("~W"|W) * P(W) + P("~W"|~W) * P(~W)) = 

(0.2 * 0.8) / ((0.2 * 0.8) + (0.95 * 0.2)) = 0.457, not 0.1739 (~3 fold difference)

I'm a bit confused because i'm getting different results, but maybe i'm wrong and someone can correct me?

I'm planning to make blind experiment with melatonin, but want to learn more stats and better understand VOI, before i start

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