Monday August 14th

[A-A-Abdelhamid]

From Friday: The acceptance for 400GeV and 0.1 ns smuons is 0.3928

Acceptance from the paper: 0.360

I tried to use "TEfficiency" to get the uncertainty :

Efficiency: 0.39297683957781
Uncertainty (Up): 0.43299669773397065
Uncertainty (Down): 0.39297683957781

Using this code:

histogram1 = TH1F()
histogram2= TH1F()

# Fill the histogram with single value
n = good_event #accepted events
d = 12135.  #unaccepted events

histogram1.Fill(0.5,n)  # Fill the center of the single bin with your value
histogram2.Fill(0.5,n+d)

# Create a TEfficiency object using the histogram
eff = TEfficiency(histogram1, histogram2)

efficiency = eff.GetEfficiency(1)
uncertainty_up = eff.GetEfficiencyErrorUp(1)  
uncertainty_down = eff.GetEfficiencyErrorLow(1)  

My understanding that after applying the acceptance we apply the efficiency to get the expected number of signals:

effi=(0.35621+0.35519) /2 #eff at log -1.5 ns and log -0.5 ns from the paper
final= n*effi
print(final)

output:

2794.3792000000003 #events

Is this so good so far? I tried to validate what I got but I got stuck

[A-A-Abdelhamid]

Is this the final number of expected signals after selection cuts (and efficiency)?

Table 14: Cutflow for SR-eμ for 2 representative signal points. For the τ̃ mass and lifetime points, the number of Monte Carlo events generated are: 30,000 for (200 GeV, 0.1 ns), and 104,000 for (300 GeV, 0.1 ns).

https://atlas.web.cern.ch/Atlas/GROUPS/PHYSICS/PAPERS/SUSY-2018-14/