LZ补上数据和代码。
LZ首先对vol_m_11根据median进行分组，组别的dummy为volume_c。volume_c = 1说明低于median。
然后LZ做了univariate分析：
proc univariate data=merge_all;
class volume_c;
var bharMM0;
run;
发现volume_c = 1的组比=0的组bharMM0高。

然后LZ做了如下回归：
    proc surveyreg data=merge_all;
      cluster permno;
      model bharMM0 = price facshr facshr2 vol_m_11 log_size year1990 year1991 year1992 year1993
      year1994 year1995 year1996 year1997 year1998 year1999 year2000 year2001 year2002 year2003
      year2004 year2005 year2006 year2007 year2008 year2009 year2010 year2011 year2012;
    run;
这个回归的结果和上面univariate的相反，于是LZ又把vol_m_11替换成了volume_c，回归结果和univariate
的又相同了...求指导
数据在此https://bbs.pinggu.org/a-1689165.html