Hello everyone,

I'm currently running a panel data consists of 124 observations. I want to do LM test and I'm having a problem where my Prob>chi2 = . Here is my regression:
xtreg srd genbod ownbod_w indboc polboc roa ta der age, re

Random-effects GLS regression Number of obs = 124
Group variable: id Number of groups = 59

R-sq: Obs per group:
within = 0.1532 min = 1
between = 0.0037 avg = 2.1
overall = 0.0273 max = 3

Wald chi2(7) = .
corr(u_i, X) = 0 (assumed) Prob > chi2 = .

------------------------------------------------------------------------------
srd | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
genbod | -.0470563 .0899987 -0.52 0.601 -.2234504 .1293379
ownbod_w | -.0403094 2.83212 -0.01 0.989 -5.591162 5.510543
indboc | .2781834 .1231357 2.26 0.024 .036842 .5195249
polboc | -.0151156 .0501099 -0.30 0.763 -.1133292 .083098
roa | -.2119119 .175164 -1.21 0.226 -.5552269 .1314032
ta | 3.85e-10 3.03e-10 1.27 0.203 -2.08e-10 9.79e-10
der | -.0028708 .0119586 -0.24 0.810 -.0263093 .0205677
age | .0007332 .0008396 0.87 0.383 -.0009124 .0023787
_cons | .3315796 .0636695 5.21 0.000 .2067897 .4563695
-------------+----------------------------------------------------------------
sigma_u | .10000396
sigma_e | .06281543
rho | .71707895 (fraction of variance due to u_i)
------------------------------------------------------------------------------

Variable TA (total assets)is in millions (IDR) and the rest of the variables are relatively small. My data looks something like this:
id indboc polboc roa ta der age ownbod_w
1. .25 0 .05 2.7e+07 .38 30 0
2. .25 0 .01 2.7e+07 .42 31 0
3. .33 .67 .02 3.0e+07 3.79 58 1.00e-05
4. .33 1 .02 3.7e+07 4.34. 59 1.00e-05



I have read several answers on Statalist regarding to this matter, but I can't find one that matches my exact problem. Is there any way I can still use TA without having to do log/ln? Because if I generate lnTA/logTA, I will have multicollinearity problem. I'm using Stata 14 for Mac. Thank you very much in advance.

Regards,
Bianca