结构方程论坛SEM-Structural·Equation·Modeling » lisrel » 问题解答 » Lisrel结果解读（来自网友刘华）

结构方程爱好者 发表于 2011-7-5 07:28

Lisrel结果解读（来自网友刘华）

<p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Degrees of Freedom = 1168（自由度）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Minimum Fit Function Chi-Square = 3066.43 (P = 0.0)　（大于0.1可接受，但对样本数非常敏感，样本过大极容易被拒绝）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Normal Theory Weighted Least Squares Chi-Square = 4258.93 (P = 0.0)（大于0.1可接受，但对样本数非常敏感，样本过大极容易被拒绝）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Estimated Non-centrality Parameter (NCP) = 3090.93（没有统计检验的准则作为依据）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 90 Percent Confidence Interval for NCP = (2895.16 ; 3294.15)</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Minimum Fit Function Value = http://bbs.pinggu.org/11.84</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Population Discrepancy Function Value (F0) = 11.93</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 90 Percent Confidence Interval for F0 = (11.18 ; 12.72)</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Root Mean Square Error of Approximation (RMSEA) = 0.10（小于0.05非常好，小于0.08比较好，小于0.1尚可接受，大于0.1不可接受）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 90 Percent Confidence Interval for RMSEA = (0.098 ; 0.10)</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P-Value for Test of Close Fit (RMSEA &lt; 0.05) = 0.00</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Expected Cross-Validation Index (ECVI) = 17.27（常用于评鉴模式复核效度的问题，即在同一个母群体下，类似的样本之间，模式复核效度的可能性。值越小越好，并没有一个可以决定模式是否可以接受的范围值）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 90 Percent Confidence Interval for ECVI = (16.51 ; 18.05)</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ECVI for Saturated Model = 9.85</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ECVI for Independence Model = 61.96（Expected Cross-Validation Index (ECVI) 小于 ECVI for Independence Model ，且小于 ECVI for Saturated Model ，表示假设模式可以接受）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Chi-Square for Independence Model with 1225 Degrees of Freedom = 15947.65<br/>Independence AIC = 16047.65</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Model AIC = 4472.93</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Saturated AIC = 2550.00（Model AIC 小于Independence AIC，也小于Saturated AIC ，模式可接受）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Independence CAIC = 16275.69</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Model CAIC = 4960.92</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Saturated CAIC = 8364.87</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Normed Fit Index (NFI) = 0.81（大于0.9时表示良好的适配）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Non-Normed Fit Index (NNFI) = 0.86（大于0.9时表示良好的适配）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Parsimony Normed Fit Index (PNFI) = 0.77（当比较不同的模式时，0.06至0.09的差别，被视为是模式间具有实质的差异存在。不做模式比较时，可采用大于0.5为模式通过与否的标准）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Comparative Fit Index (CFI) = 0.87（大于0.9时表示良好的适配）</p><p>&Oslash;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Incremental Fit Index (IFI) = 0.87（大于0.9时表示良好的适配）</p>

数据分析 发表于 2011-7-5 07:33

it is so great!

统计分析 发表于 2011-7-5 07:38

太好了，很有帮助。谢谢。

mischina 发表于 2011-7-5 07:43

<strong> 2# <i>lifi1241</i> </strong> <br /> <br /> it is so helpful!

51jijin 发表于 2011-7-5 07:48

感谢楼主！～

freeshuju 发表于 2011-7-5 07:53

很有帮助,谢谢楼主

lisrel 发表于 2011-7-5 07:58

谢谢楼主的分享

偏最小二乘 发表于 2011-7-5 08:03

太好了，很有帮助。谢谢。

查看完整版本: Lisrel结果解读（来自网友刘华）