Hello Mike Here is Emmnauel's answer (he is the author of the programm):
Let "sum" be the sum of actual difference in timing at turns of series S1 against S2. The test is the following H0: sum = 0 H1: sum > 0 The idea of the test is to built all the possible outcomes (ie all the different "sum") by randomization and then to test what is the probability of the observed (or more extreme) outcome. So what we get is P( >= sum under H0). The rejection probability is built as 1-P( >= sum under H0) and gives the confidence level at which the null can be rejected. It can be seen as the risk of acception H0 while H0 is false Let's look at your results (see below). The first line is the following test H0 : S1 leads S2 by k=0 which we wrote as H0 : k<1. Under H0 the probability to observe a sum at least greated 21 is 0.2%, such that we can reject H0 with a confidence level of 99.8%. The test can be performed in a sequential way, starting for the maximum lead to be tested. In your case for 4 leads, H0 : S1 leads S2 by k=0, we have P( >= sum under H0) = 96.1%. Meaning that risk of accepting H0 while H0 is true is below 5%.... Éric. 2013/4/23 Mike <[email protected]> > Thank you for your answer! > > However, I don't understand why there is a lead of 3 periods at a standard > 5% level. > > Isn't it 100% - 5% = 95%? > > That means in return that i accept the hypothesis as long as the rejection > prob. higher than 95% is? > > > > > -- > View this message in context: > http://mailinglists.scilab.org/Interpretation-of-Banerji-test-result-tp4026503p4026587.html > Sent from the Scilab users - Mailing Lists Archives mailing list archive > at Nabble.com. > _______________________________________________ > users mailing list > [email protected] > http://lists.scilab.org/mailman/listinfo/users >
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