\begin{table}%t5 \caption{\label{tabfits}Fits to the data ($b = 0.25$).} \small%\centerline { \begin{tabular}{lrrrr} \hline\hline %&&&& \\[-8pt] Relationship & Slope & Y intercept & $\chi^2$ & Corr.? \\ \hline $T$ vs. 1 + $z$ &$ -2.00 \pm 0.2 $&$ 6.1 \pm 0.5 $& 90.6 &Y \\ Log $T$ vs. Log (1 + $z$) &$ -1.80 \pm 0.2 $&$ 0.9 \pm 0.06 $& 100.8 &Y \\ $T$ vs. $r$ &$ -4.00 \pm 1.4 \times 10^{-3}$ &$ 2.5 \pm 0.3 $& 181.6 &N \\ Log $T$ vs. Log $r$ &$ -0.25 \pm 0.11 $&$ 1 \pm 0.2 $& 183 &N \\ $E_{\rm T}$ vs. 1 + $z$ &$ 1.70 \pm 0.3 $&$ -1.1 \pm 0.5 $& 92.8 &Y \\ Log $E_{\rm T}$ vs. Log (1 + $z$) &$ 1.80 \pm 0.2 $&$ -0.1 \pm 0.06 $& 100.8 &Y \\ $E_{\rm T}$ vs. $r$ &$ 1.40 \pm 1.4 \times 10^{-3}$ &$ 1.5 \pm 0.2 $& 138 &N\\ Log $E_{\rm T}$ vs. Log $r$ &$ 0.25 \pm 0.11 $&$ -0.2 \pm 0.2 $& 183 &N \\ $n_{\rm a}$ vs. $r$ &$ -1.90 \pm 0.9 \times 10^{-4}$ &$ 0.09 \pm 0.04 $& 347.3 &N\\ Log $n_{\rm a}$ vs. Log $r$ &$ -1.90 \pm 0.2 $&$ 3.4 \pm 0.5 $& 561.9 &Y \\ Log $n_{\rm a}$ vs. Log (1 + $z$) &$ -1.90 \pm 1.1 $&$ -0.3 \pm 0.3 $& 1213.6 &N \\ $v/c$ vs. $r$ &$ 1.30 \pm 0.3 \times 10^{-4}$ &$ 0.006 \pm 0.003 $& 741.1 &N \\ Log $v/c$ vs. Log $r$ &$ 0.70 \pm 0.1 $&$ -3 \pm 0.2 $& 740.9 &N \\ $v/c$ vs. 1 + $z$ &$ 0.02 \pm 0.004 $&$ -0.01 \pm 0.006 $& 747.5 &N\\ Log $v/c$ vs. Log (1 + $z$) &$ 1.50 \pm 0.4 $&$ -1.88 \pm 0.1 $& 966.9 &N \\ $L_j$ vs. $r$ &$ 0.01 \pm 0.03 $&$ 16 \pm 5 $& 364.8 &N \\ Log $L_j$ vs. Log $r$ &$ 0.60 \pm 0.2 $&$ 0.3 \pm 0.5 $& 710.6 &N \\ $L_j$ vs. 1 + $z$ &$ 36.50 \pm 9 $&$ -38 \pm 13 $& 280.1 &N \\ Log $L_j$ vs. Log (1 + $z$) &$ 3.80 \pm 0.5 $&$ 0.72 \pm 0.11 $& 361.4 &Y \\ \hline \end{tabular}} \medskip The best fits to the radio source relationships for $b = 0.25$. Column~1: the relationship. $v/c$~is the lobe expansion velocity, $r$~is the core-hotspot separation, $L_j$ ($\times$$10^{44}$~erg~s\mone) is the beam power, $n_{\rm a}$ ($\times 10^{-3}$~cm \mthree) is the ambient density, $T_{\rm T}$ is the source lifetime, and $E_{\rm T}$ is the total energy. Column~2: the best fit slope. Column~3: the best fit Y intercept. Column~4: the $\chi^2$ for 60~points and 58~degrees of freedom. Column~5: is there a correlation? The uncertainty of the best fit parameter is multiplied by $\sqrt{(\chi^2/58)}$ to obtain the normalized uncertainty in the best fit parameter, and this is compared with the value of the best fit parameter to determine whether the correlation is significant. \end{table}