| 23 |
|
The data yields in the |
| 24 |
|
four regions are summarized in Table~\ref{tab:datayield}. |
| 25 |
|
The prediction of the ABCD method is is given by $A\times C/B$ and |
| 26 |
< |
is 1.5 events. (see Table~\ref{tab:datayield}. |
| 26 |
> |
is 1.5 $\pm$ 0.9 events (statistical uncertainty only, assuming |
| 27 |
> |
Gaussian errors). (see Table~\ref{tab:datayield}). |
| 28 |
|
|
| 29 |
|
\begin{table}[hbt] |
| 30 |
|
\begin{center} |
| 60 |
|
%estimate of the $t\bar{t}$ contribution. The result |
| 61 |
|
%of this exercise is {\color{red} xx} events. |
| 62 |
|
|
| 63 |
+ |
\clearpage |
| 64 |
+ |
|
| 65 |
|
\subsection{Background estimate from the $P_T(\ell\ell)$ method} |
| 66 |
|
\label{sec:victoryres} |
| 67 |
|
|
| 68 |
+ |
We first use the $P_T(\ell \ell)$ method to predict the number of events |
| 69 |
+ |
in a control region defined by $125<{\rm SumJetPt}<300$~GeV and |
| 70 |
+ |
\met/$\sqrt{\rm SumJetPt} > 8.5$. We find 6 events satisfying the |
| 71 |
+ |
corresponding selection with the \met/$\sqrt{\rm SumJetPt}$ cut replaced |
| 72 |
+ |
by a $P_T(\ell\ell)/\sqrt{\rm SumJetPt}$ cut. The predicted yield |
| 73 |
+ |
is then given by $N_A = K \cdot K_C \cdot N_{A'} = 10.4 \pm 4.2$ |
| 74 |
+ |
(statistical uncertainty only, assuming Gaussian errors), |
| 75 |
+ |
where we have taken $K = 1.73$ and $K_C = 1$. This yield is in good |
| 76 |
+ |
agreement with the observed yield of 11 events, as shown in |
| 77 |
+ |
Table~\ref{tab:victory_control} and displayed in Fig.~\ref{fig:victory} (left). |
| 78 |
+ |
{\color{ref} \bf Perform DY estimate for this control region}. |
| 79 |
+ |
|
| 80 |
+ |
Encouraged by the good agreement between predicted and observed yields |
| 81 |
+ |
in the control region, we proceed to perform the $P_T(\ell \ell)$ method |
| 82 |
+ |
in the signal region ${\rm SumJetPt}>300$~GeV. |
| 83 |
|
The number of data events in region $D'$, which is defined in |
| 84 |
|
Section~\ref{sec:othBG} to be the same as region $D$ but with the |
| 85 |
|
$\met/\sqrt{\rm SumJetPt}$ requirement |
| 86 |
< |
replaced by a $P_T(\ell\ell)/\sqrt{\rm SumJetPt}$ requirement |
| 86 |
> |
replaced by a $P_T(\ell\ell)/\sqrt{\rm SumJetPt}$ requirement, |
| 87 |
|
is $N_{D'}=2$. Thus the BG prediction is |
| 88 |
< |
$N_D = K \cdot K_C \cdot N_{D'} = 1.5$ |
| 89 |
< |
where $K=1.5 \pm xx$ as derived in Sec.~\ref{sec:victory} and |
| 90 |
< |
$K_C = 1$. |
| 91 |
< |
Note that if we were to subtract off from region $D'$ |
| 92 |
< |
the {\color{red} 0.8 $\pm$ 0.8} DY events estimated from |
| 93 |
< |
Section~\ref{sec:othBG}, the background |
| 94 |
< |
prediction would change to $N_D=1.8 \pm xx$ events. |
| 95 |
< |
|
| 96 |
< |
%%%TO BE REPLACED |
| 79 |
< |
%{\color{red}As mentioned previously, for the 11/pb analysis |
| 80 |
< |
%we use the $K$ factor from data and take $K=1$. |
| 81 |
< |
%This will change for the full dataset. We will also pay |
| 82 |
< |
%more attention to the statistical errors.} |
| 83 |
< |
|
| 84 |
< |
%The number of data events in region $D'$, which is defined in |
| 85 |
< |
%Section~\ref{sec:othBG} to be the same as region $D$ but with the |
| 86 |
< |
%$\met/\sqrt{\rm SumJetPt}$ requirement |
| 87 |
< |
%replaced by a $P_T(\ell\ell)/\sqrt{\rm SumJetPt}$ requirement |
| 88 |
< |
%is $N_{D'}=1$. Thus the BG prediction is |
| 89 |
< |
%$N_D = K \cdot K_{\rm fudge} \cdot N_{D'} = 1.5$ |
| 90 |
< |
%where we used $K=1.5 \pm xx$ and $K_{\rm fudge}=1.0 \pm 0.0$. |
| 91 |
< |
%Note that if we were to subtract off from region $D'$ |
| 92 |
< |
%the {\color{red} 0.4 $\pm$ 0.4} DY events estimated from |
| 93 |
< |
%Section~\ref{sec:othBG}, the background |
| 94 |
< |
%prediction would change to $N_D=0.9 \pm xx$ events. |
| 95 |
< |
%{\color{red} When we do this with a real |
| 96 |
< |
%$K_{\rm fudge}$, the fudge factor will be different |
| 97 |
< |
%after the DY subtraction.} |
| 98 |
< |
|
| 99 |
< |
As a cross-check, we use the $P_T(\ell \ell)$ |
| 100 |
< |
method to also predict the number of events in the |
| 101 |
< |
control region $125<{\rm SumJetPt}<300$ GeV and |
| 102 |
< |
\met/$\sqrt{\rm SumJetPt} > 8.5$. We predict |
| 103 |
< |
$5.6^{+x}_{-y}$ events and we observe 4. |
| 104 |
< |
The results of the $P_T(\ell\ell)$ method are |
| 105 |
< |
summarized in Figure~\ref{fig:victory}. |
| 88 |
> |
$N_D = K \cdot K_C \cdot N_{D'} = 3.07 \pm 2.17$ where $K=1.54 \pm xx$ |
| 89 |
> |
as derived in Sec.~\ref{sec:victory} and $K_C = 1$. |
| 90 |
> |
We next subtract off the expected DY contribution of |
| 91 |
> |
{\color{red} \bf 0.8 $\pm$ 0.8 (update DY estimate)} events, as calculated |
| 92 |
> |
in Sec.~\ref{sec:othBG}. This gives a predicted yield of |
| 93 |
> |
$N_D=1.8^{+2.5}_{-1.8}$ events, which is consistent with the observed yield of |
| 94 |
> |
1 event. |
| 95 |
> |
|
| 96 |
> |
|
| 97 |
|
|
| 98 |
|
\begin{figure}[hbt] |
| 99 |
|
\begin{center} |
| 137 |
|
Gaussian errors.} |
| 138 |
|
\begin{tabular}{l|c|c|c} |
| 139 |
|
\hline |
| 140 |
< |
& Predicted & Observed & Obs/Pred \\ |
| 140 |
> |
& Predicted & Observed & Obs/Pred \\ |
| 141 |
|
\hline |
| 142 |
< |
total SM MC & 0.96 & 1.41 & 1.46 \\ |
| 143 |
< |
data & 3.07 $\pm$ 2.17 & 1 & 0.33 \\ |
| 142 |
> |
total SM MC & 0.96 & 1.41 & 1.46 \\ |
| 143 |
> |
data & $N_D=1.8^{+2.5}_{-1.8}$ & 1 & 0.33 \\ |
| 144 |
|
\hline |
| 145 |
|
\end{tabular} |
| 146 |
|
\end{center} |
