I see also nice things, data, generic and cocktail have very very similar
numbers!
> o The mm2 cut is much less efficient on the signal than it used to
> be (54.1% -> 42.3% for the single cut)
this is due to an overall positive shift in the distribution (then the
cut is harder...)
>
> o The overall efficiency for signal is at 30.1% which is not quite
> what we give in table 12 from the fit. I am not quite sure about
> the difference, but think that this is probably due to the fact
> that in this table the normalization sample does not have a p* cut
> applied.
yes, but if you divide by 1-0.820 (eff of the lepton cuts) and multiply
by 0.866 (kaon cut) you get 0.321 that is a bit far from the 0.345 from
the fit. I am assuming no correlation between kaon cut and lepton cut and
this is not completely true. Urs, may you redo them with the same
normalization as in the fit and adding the kaon cut, just as a crosscheck?
> o The signal efficiency on the depleted sample is quite high.
>
the situation is actually better. Multiply by the kaon cut eff
0.230*(1-0.866) = 0.03
a bit lower than the fit one (0.045). Again we should redo it with the
kaon cut in.
Daniele
> For this table I had to run anaRecoil. I find that the default fit is
> different from the default fit obtained with the output of anaQA, when
> both are run without PidKilling (for Kaons):
>
> k00 CF default, ursl-121002, no Kaon PK 0.018483 +- 0.00254913(stat) +- 0.00142799(MC stat)
> k09 CF default again 0.018483 +- 0.00254913(stat) +- 0.00142799(MC stat)
> k21 CF anaRecoil 0.019723 +- 0.00264981(stat) +- 0.00144295(MC stat)
>
> I don't yet know why that is (or could be). There is no difference
> between k00 and k09, both are with anaQA and they should be identical,
> it's just a paranoia check that we have no random sequence hidden
> somewhere.
>
> Cheers,
> --U.
>
>
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