结构方程论坛SEM-Structural·Equation·Modeling » 临时文件区 » 请教：警告: TD 5,5 may not be identified.

因子分析 发表于 2011-7-2 17:06

请教：警告: TD 5,5 may not be identified.

<br /> 如题，该问题应该从何处下手去排查？<br /> W_A_R_N_I_N_G: TD 5,5 may not be identified.<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Standard Errors, T-Values, Modification Indices,<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; and Standardized Residuals cannot be computed.<br /> <br /> &nbsp; &nbsp;Goodness of Fit Statistics<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;Degrees of Freedom = 29<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Minimum Fit Function Chi-Square = 98.53 (P = 0.00)<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;Normal Theory Weighted Least Squares Chi-Square = 87.86 (P = 0.00)<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;Estimated Non-centrality Parameter (NCP) = 58.86<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; 90 Percent Confidence Interval for NCP = (34.41 ; 90.94)<br /> <br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Minimum Fit Function Value = http://bbs.pinggu.org/0.47<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Population Discrepancy Function Value (F0) = 0.28<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;90 Percent Confidence Interval for F0 = (0.16 ; 0.44)<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Root Mean Square Error of Approximation (RMSEA) = 0.099<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;90 Percent Confidence Interval for RMSEA = (0.075 ; 0.12)<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;P-Value for Test of Close Fit (RMSEA &lt; 0.05) = 0.00054<br /> <br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Expected Cross-Validation Index (ECVI) = 0.77<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; 90 Percent Confidence Interval for ECVI = (0.66 ; 0.93)<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; ECVI for Saturated Model = 0.63<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;ECVI for Independence Model = 2.20<br /> <br /> &nbsp; &nbsp;&nbsp; &nbsp;Chi-Square for Independence Model with 55 Degrees of Freedom = 437.31<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Independence AIC = 459.31<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;Model AIC = 161.86<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Saturated AIC = 132.00<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Independence CAIC = 507.12<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Model CAIC = 322.71<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;Saturated CAIC = 418.91<br /> <br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;Normed Fit Index (NFI) = 0.77<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Non-Normed Fit Index (NNFI) = 0.66<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Parsimony Normed Fit Index (PNFI) = 0.41<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Comparative Fit Index (CFI) = 0.82<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Incremental Fit Index (IFI) = 0.83<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Relative Fit Index (RFI) = 0.57<br /> <br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;Critical N (CN) = 106.19<br /> <br /> <br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Root Mean Square Residual (RMR) = 0.076<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;Standardized RMR = 0.076<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Goodness of Fit Index (GFI) = 0.93<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; Adjusted Goodness of Fit Index (AGFI) = 0.84<br /> &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;Parsimony Goodness of Fit Index (PGFI) = 0.41

angel 发表于 2011-7-2 17:11

这个通常与TD矩阵没有多大关系，与你的模型太复杂有关系。

结构方程 发表于 2011-7-2 17:16

我检验了数据的正态性，发现很多指标的非正态分布，于是我用mintab做了个体分布识别，发现这些指标几乎都不能显著的归属于任何一种分布，在这种情况下，我用spss做了正态性检验，根据偏度和峰度指标以及直方图的判断，选择了个别正态性稍好的指标进行COX-BOX转换，结果又多了一个符合正态性的指标。接着用eviews做了截面数据的回归，发现符合统计检验的最后就剩下四个变量了。同时请教了学校的统计老师，老师建议用smart-pls试试，但是smartpls好像取消了国内用户的访问权限，所以现在请求哪位大侠把smart-pls借我用用，非常感谢！

angel 发表于 2011-7-2 17:21

自己顶一下了

查看完整版本: 请教：警告: TD 5,5 may not be identified.