罗教授: 您好!非常感谢您的回复,同时也为我没有把问题讲清楚,向您表达歉意!从回归方程本身看,如果一次项和二次项都显著特别是如果一次项的显著水平比二次项的高的话那么一次项的作用是起主导作用的,所以希望是二次项显著,而一次项不显著;但是如果从一元二次方程进行理解,一次项不显著,二次项显著,那么方程就变成了 Y=aX平方+c; 从图形看 变成了以Y轴为中心的倒U(a为负值),从Y的右边看(如果是问卷,就没有负数)这又不是一个典型倒U,而是一个减函数。我们最近探讨社会网络与创造力的关系时,其中发现 在加入控制变量之后,weak ties 对creativity 显著影响,然而将weak ties 和 weak ties square 放进回归方程之后,得到的是weak ties(p<0.01) 和weak ties square(p<0.05) 都对creativity 有显著影响。然而我们在 AMJ的Oh, Chung and Labianca(2004)的文章中发现其一次项是不显著的,但是将一次项与二次项同时加入之后,一次项又显著,一次项系数为正,而且二次项也显著,二次项系数为负数,为此就证明了倒U关系。因此,我们开始怀疑我们验证倒U关系的存在,特别希望罗教授授业解惑?如果存在倒U型关系,这个U型关系与线性型关系区别在何处?
Degrees of Freedom = 41
Minimum Fit Function Chi-Square = 152.21 (P = 0.00
Normal Theory Weighted Least Squares Chi-Square = 158.32 (P = 0.00)
Estimated Non-centrality Parameter (NCP) = 117.32
90 Percent Confidence Interval for NCP = (82.42 ; 159.80)
Minimum Fit Function Value = 0.51
Population Discrepancy Function Value (F0) = 0.39
90 Percent Confidence Interval for F0 = (0.28 ; 0.54)
Root Mean Square Error of Approximation (RMSEA) = 0.098
90 Percent Confidence Interval for RMSEA = (0.082 ; 0.11)
P-Value for Test of Close Fit (RMSEA < 0.05) = 0.00
Expected Cross-Validation Index (ECVI) = 0.70
90 Percent Confidence Interval for ECVI = (0.58 ; 0.84)
ECVI for Saturated Model = 0.44
ECVI for Independence Model = 7.98
Chi-Square for Independence Model with 55 Degrees of Freedom = 2355.75
Independence AIC = 2377.75
Model AIC = 208.32
Saturated AIC = 132.00
Independence CAIC = 2429.46
Model CAIC = 325.
Saturated CAIC = 442.23
Normed Fit Index (NFI) = 0.94
Non-Normed Fit Index (NNFI) = 0.94
Parsimony Normed Fit Index (PNFI) = 0.70
Comparative Fit Index (CFI) = 0.95
Incremental Fit Index (IFI) = 0.95
Relative Fit Index (RFI) = 0.91
Critical N (CN) = 128.16
Root Mean Square Residual (RMR) = 0.064
Standardized RMR = 0.064
Goodness of Fit Index (GFI) = 0.91
Adjusted Goodness of Fit Index (AGFI) = 0.86
Parsimony Goodness of Fit Index (PGFI) = 0.57