eviews 的ADF检验得到了下表 帮忙解释下啊 我的数据平整吗？

Question

Group unit root test: Summary
Date: 05/09/09 Time: 14:51
Sample: 2001 2008
Series: CJBZ, GDP, JUMIN, ZHFU, SYLV
Exogenous variables: Individual effects
Automatic selection of maximum lags
Automatic selection of lags based on SIC: 0 to 1
Newey-West bandwidth selection using Bartlett kernel

Cross-
Method Statistic Prob.** sections Obs
Null: Unit root (assumes common unit root process)
Levin, Lin & Chu t* -3.08657 0.0010 5 31
Breitung t-stat 1.10723 0.8659 5 26

Null: Unit root (assumes individual unit root process)
Im, Pesaran and Shin W-stat -0.76657 0.2217 5 31
ADF - Fisher Chi-square 13.6978 0.1872 5 31
PP - Fisher Chi-square 11.3955 0.3275 5 35

Null: No unit root (assumes common unit root process)
Hadri Z-stat 0.71335 0.2378 5 40

** Probabilities for Fisher tests are computed using an asympotic Chi
-square distribution. All other tests assume asymptotic normality.

Answer 1

ADF - Fisher Chi-square 13.6978 0.1872 5 31
PP - Fisher Chi-square 11.3955 0.3275 5 35
看这一段，ADF值是13.69，P值是0.1872，表明无法拒绝原假设，即存在单位根，序列不平稳。PP检验的结果也是如此。