Question : what is statistical error of AN, AL extracted from phi-symmetric detector

Model of spin dependent yield:

     y=1.+sign*Al*P+sign*An*P*cos(phi);

where 'sign' is spin direction + or - 

Fig 1: Example of distributions for + & -, total N0=Np+Nm=500 events, number of bins nb=12

Fit function:  Double_t fitval =  par[0]*sin(phi) + par[1]*cos(phi) +par[2];

Extraction of epsL=Al*P= (C0p-C0m)/N0*nb, sig(epsL)=1/sqrt(N0)

Extraction of epsN=An*P= (Ccp-Ccm)/N0*nb, sig(epsN)=1/sqrt(N0*2) 

Conclusion: FOM for AN is half of FOM for AL. 

Input of the program:
  simuAN_err(float An=0.1, float P=0.5, float Al=-0.2, int n0=500)

Output of the program:
yield: AL*P : assumed= -0.10   measured=-0.068 +/-0.045  diff=-0.032  diff/sig=-0.7
fit:   AL*P : assumed= -0.10   measured=-0.054 +/-0.044  diff=-0.046  diff/sig=-1.0
fit:   AN*P : assumed=  0.05   measured=-0.073 +/-0.062  diff=0.123  diff/sig=2.0
 tot eve=500

Here's the link to the theory paper http://prl.aps.org/abstract/PRL/v103/i17/e172001

Fig 2.   p+p -> W -> mu 

Fig 3. The same simu and reco but with N0=50,000 events

yield: AL*P : assumed= -0.10   measured=-0.094 +/-0.004  diff=-0.006  diff/sig=-1.4
fit:   AL*P : assumed= -0.10   measured=-0.094 +/-0.004  diff=-0.006  diff/sig=-1.4
fit:   AN*P : assumed=  0.05   measured=0.053 +/-0.006  diff=-0.003  diff/sig=-0.4
 tot eve=50000