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.
>
>