Eviews结果如下怎么判断有没有单位根啊?有两个P值小于0.05,而有两个大于0.05,拒绝还是接受原假设?Pool unit root test:Summary \x05Series:YBJ,YTJ,YHEBEI,YSHX,YNMG,YLN,YJL,YHLJ,YSHH,YJS,YZHJ,YAH,YFJ,YJX,YSHD,YHENAN,YHUBEI,Y

Eviews结果如下怎么判断有没有单位根啊?有两个P值小于0.05,而有两个大于0.05,拒绝还是接受原假设?Pool unit root test:Summary \x05Series:YBJ,YTJ,YHEBEI,YSHX,YNMG,YLN,YJL,YHLJ,YSHH,YJS,YZHJ,YAH,YFJ,YJX,YSHD,YHENAN,YHUBEI,Y
Eviews结果如下怎么判断有没有单位根啊?有两个P值小于0.05,而有两个大于0.05,拒绝还是接受原假设?
Pool unit root test:Summary \x05
Series:YBJ,YTJ,YHEBEI,YSHX,YNMG,YLN,YJL,YHLJ,YSHH,YJS,YZHJ,
YAH,YFJ,YJX,YSHD,YHENAN,YHUBEI,YHUNAN,YGD,YGX,
YHAINAN,YCHQ,YSCH,YGZH,YYN,YSHXI,YGS,YQH,YNX,YXJ
Date:02/25/12 Time:09:44\x05
Sample:1997 2009\x05\x05
Exogenous variables:Individual effects
Automatic selection of maximum lags\x05
Automatic lag length selection based on SIC:0 to 1
and Bartlett kernel\x05\x05
\x05\x05\x05\x05
\x05\x05\x05\x05
\x05\x05\x05Cross-\x05
Method\x05Statistic\x05Prob.**\x05sections\x05Obs
Null:Unit root (assumes common unit root process)
Levin,Lin & Chu t*\x05-6.78531\x05 0.0000\x05 30\x05 356
\x05\x05\x05\x05
Null:Unit root (assumes individual unit root process)
Im,Pesaran and Shin W-stat \x05-1.36644\x05 0.0859\x05 30\x05 356
ADF - Fisher Chi-square\x05 70.7329\x05 0.1619\x05 30\x05 356
PP - Fisher Chi-square\x05 91.2017\x05 0.0058\x05 30\x05 360
\x05\x05\x05\x05
\x05\x05\x05\x05
** Probabilities for Fisher tests are computed using an asymptotic Chi
-square distribution.All other tests assume asymptotic normality.

Eviews结果如下怎么判断有没有单位根啊?有两个P值小于0.05,而有两个大于0.05,拒绝还是接受原假设?Pool unit root test:Summary \x05Series:YBJ,YTJ,YHEBEI,YSHX,YNMG,YLN,YJL,YHLJ,YSHH,YJS,YZHJ,YAH,YFJ,YJX,YSHD,YHENAN,YHUBEI,Y
p值过大,则不能拒绝原假设（与你选的显著水平有关）,原假设是序列为单位根过程（非平稳）.