Same table with just columns switched. TP FP FN TN with both P's in the same row and same for N's and your table on wikipedia page is FP TN TP FN See this video https://www.youtube.com/watch?v=Dsa9ly4OSBk&list=LLJnPqxdVeMk0dJ2h4MgGKuA&index=1
I edited aforesaid quote,I mistakenly wrote False +ve 2 times........ sorry for inconvenience. now this gives, \[ \matrix{ & H1:did not lie & H0:Lied \\+ve result & FN(type II error) & TP \\ -ve result &TN & FP(type I error)} \] PS: I didn't make table on Wikipedia page
I don't think you understand what TP, FP, FN, TN means. You can't have FN in the +ve row and FP in the -ve row. +ve row will have two positives (because it's the row of positve result). One of them would be FALSE and other will be TRUE. Similarly for -ve (Regardless of what Ho and H1 are)
\[ \matrix{ & H1:did not lie & H0:Lied \\+ve result & FN(type II error) & TP \\ -ve result &TN & FP(type I error)} \] Okay! if I accept it \[ \matrix{ & H1:did not lie & H0:Lied \\+ve result & FP & TP \\ -ve result &TN & FN} \] then FN gives type I error & FP gives type II error, which is not the definition says.
Because you're using different definition for +ve and -ve result. On this page http://en.m.wikipedia.org/wiki/False_positives_and_false_negatives "In statistical hypothesis testing the analogous concepts are known as type I and type II errors, where a positive result corresponds to rejecting the null hypothesis, and a negative result corresponds to not rejecting the null hypothesis" and you're doing the opposite. If you stick with the definitions given on that page, only then FP will be a Type I error and FN a Type II error.
Hi C2H6O, it takes long time........ but after understanding "Multiple comparisons problem" I realize you were correct
