[Scilab-users] Interpretation of Banerji test result

[Eric Dubois]

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 <autor52 at hotmail.com>

> 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?
>
>
>
>
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