\begin{table}%t2
\caption{\label{table2}The fractional crater detectability estimates calculated from data of Table~\ref{table1}: Col.~1 ($k$) is the cycle number. Columns~2 ($a_1$) and~3 ($a_2$) give the number of craters within cycle~$k$ divided by the total number of craters in samples $C_2(n=33)$ and $C_3(n=40)$. Columns~4 ($a_3$) and~5 ($a_4$) give the averages for pairs of two consecutive cycles in samples~$C_2$ and~$C_3$. Column~6 ($A_{{k}}=(a_3+a_4)/2$) is the mean of both samples.}
%\centering
\small
\begin{tabular}{c|rr|rr|c}
\hline \hline
$k$     & $C_2:a_1$ & $C_3:a_2$  & $C_2:a_3$ & $C_3:a_4$   &  $A_{{k}}=(a_3+a_4)/2$ \\
\hline
 1	& 4/33	& 3/40  & 6/33   & 7/40    & $A_1   = 0.1784$	\\
 2	& 8/33  & 11/40 & 6/33   & 7/40    & $A_2   = 0.1784$	\\
 \hline
 3	& 7/33  & 8/40  & 5/33   & 5.5/40  & $A_3   = 0.1445$	\\
 4	& 3/33  & 3/40  & 5/33   & 5.5/40  & $A_4   = 0.1445$	\\
 \hline
 5	& 3/33  & 6/40  & 3.5/33 & 5/40    & $A_5   = 0.1155$	\\
 6	& 4/33  & 4/40  & 3.5/33 & 5/40    & $A_6   = 0.1155$	\\
 \hline 
 7	& 2/33  & 1/40  & 1/33   & 1/40    & $A_7   = 0.0277$	\\
 8	& 0/33  & 1/40  & 1/33   & 1/40    & $A_8   = 0.0277$	\\
 \hline
 9	& 1/33  & 1/40  & 1/33   & 1.5/40  & $A_9   = 0.0339$	\\
10	& 1/33  & 2/40  & 1/33   & 1.5/40  & $A_{10}= 0.0339$	\\
\hline 
\end{tabular}
\end{table}