资源描述
Degrees of Freedom = 38Minimum Fit Function Chi-Square = 204.78 (P = 0.0)Normal Theory Weighted Least Squares Chi-Square = 192.51 (P = 0.0)Estimated Non-centrality Parameter (NCP) = 154.5190 Percent Confidence Interval for NCP = (114.88 ; 201.67)Minimum Fit Function Value = 1.47Population Discrepancy Function Value (F0) = 1.1190 Percent Confidence Interval for F0 = (0.83 ; 1.45)Root Mean Square Error of Approximation (RMSEA) = 0.1790 Percent Confidence Interval for RMSEA = (0.15 ; 0.20)P-Value for Test of Close Fit (RMSEA 0.05) = 0.00Expected Cross-Validation Index (ECVI) = 1.7990 Percent Confidence Interval for ECVI = (1.50 ; 2.13)ECVI for Saturated Model = 0.95ECVI for Independence Model = 31.50Chi-Square for Independence Model with 55 Degrees of Freedom = 4356.65Independence AIC = 4378.65Model AIC = 248.51Saturated AIC = 132.00Independence CAIC = 4422.01Model CAIC = 358.87Saturated CAIC = 392.15Normed Fit Index (NFI) = 0.95Non-Normed Fit Index (NNFI) = 0.94Parsimony Normed Fit Index (PNFI) = 0.66Comparative Fit Index (CFI) = 0.96Incremental Fit Index (IFI) = 0.96Relative Fit Index (RFI) = 0.93Critical N (CN) = 42.52Root Mean Square Residual (RMR) = 0.020Standardized RMR = 0.0060Goodness of Fit Index (GFI) = 0.80Adjusted Goodness of Fit Index (AGFI) = 0.65Parsimony Goodness of Fit Index (PGFI) = 0.46首先生成测量模型后,点击output中的fit indices得到结果文件。如上,首先观察卡方为192.51 (p小于0.05)显著。其次非规准适配指标NNFI为0.94大于0.90,合适。再者平均残差协方差标准化的总和SRMR为0.0060,小于0.05,合适。第四,比较适配指数CFI为0.96,比较接近1,相当合适。第五、渐进残差均方和平方根RMSEA为0.020小于0.05非常好。二.统合模型分析由于模型过于复杂,导致lisel软件识别为矩阵不够正定。故将模型中的顾客满意度与物流服务成本之间的关系去掉,形成下图模型:顾客满意度物流组织网络化程度物流服务成本顾客满意度生成图为:1、估计(estimate)路径图IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIiiiiiiiiiiiiiiiiiiiiiiiiiiiiiIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII1_丄IIIl_IIIl_IIIIIIIl_IIIl_i i- I 厂i 111111亍11111i -TiiiiiiiiiiiiiiiiiiiiiiiiiiifiiiiiiiiiiiiiiiiiiiIiiiiiiiiifi,1IT111IT11IIT11ir T11iIT11ir - T2、标准路径图(采用固定因子载荷法,只是在标准路径中体现不出)3、T检验的路径图I丄IIT1但是在T检验中发现WF作为WC与GZ的中介变量并不显者。但是为了作为练习就假设它是显著的。点击“output中的indices”得到下面文字:Degrees of Freedom = 40 Minimum Fit Function Chi-Square = 228.80 (P = 0.0)Normal Theory Weighted Least Squares Chi-Square = 220.04 (P = 0.0)Estimated Non-centrality Parameter (NCP) = 180.0490 Percent Confidence Interval for NCP = (137.21 ; 230.38)Minimum Fit Function Value = 1.65Population Discrepancy Function Value (F0) = 1.3090 Percent Confidence Interval for F0 = (0.99 ; 1.66)Root Mean Square Error of Approximation (RMSEA) = 0.1890 Percent Confidence Interval for RMSEA = (0.16 ; 0.20)P-Value for Test of Close Fit (RMSEA 0.05) = 0.00Expected Cross-Validation Index (ECVI) = 1.9690 Percent Confidence Interval for ECVI = (1.65 ; 2.32)ECVI for Saturated Model = 0.95ECVI for Independence Model = 31.50Chi-Square for Independence Model with 55 Degrees of Freedom = 4356.65Independence AIC = 4378.65Model AIC = 272.04Saturated AIC = 132.00Independence CAIC = 4422.01Model CAIC = 374.52Saturated CAIC = 392.15Normed Fit Index (NFI) = 0.95Non-Normed Fit Index (NNFI) = 0.94Parsimony Normed Fit Index (PNFI) = 0.69Comparative Fit Index (CFI) = 0.96Incremental Fit Index (IFI) = 0.96Relative Fit Index (RFI) = 0.93Critical N (CN) = 39.69Root Mean Square Residual (RMR) = 0.020Standardized RMR = 0.0063Goodness of Fit Index (GFI) = 0.78Adjusted Goodness of Fit Index (AGFI) = 0.63Parsimony Goodness of Fit Index (PGFI) = 0.471. 计算卡方自由度比为=228.840=5.72,模型适配不佳,对真实数据反映不足,这主要是由于数据是随机生成并被操纵。2. 其次非规准适配指标NNFI为0.94大于0.90,合适。3. 再者平均残差协方差标准化的总和SRMR为0.0063,小于0.05,合适。4. 第四,比较适配指数 CFI 为 0.96,比较接近1,相当合适。5. 第五、渐进残差均方和平方根RMSEA为0.18大于1,不是很适配。第三、中介效应分析:Total and Indirect EffectsIndirect Effects ofKSI on ETAWCwr -GZ _ 0.13137GM -Total Effects of ETA on ETAWF GZ- GMWT -GZ_-ci_函-1.37GM -Standardized Total and Indirect EffectsStandardized Total Effects ofKSI on ETAWCVT -O.99|GZ 0.99GM二Standaidized Indirect Effects ofKSI on ETAWCWF -GZ 0.12GM -Standardized Total Effects of ETA on ETAWF GZ GMWF - - - GZ -0.13-GM-一Total Effects ojfKSI on ETAWF -1.29(0.06)-22.53GZ 1.07(0.03)42.35由上图可得a ,b估计系数的p值大于0.05,故中介效应不显著。GM 1.06但是为了进行sobel检验,假设a显著;b不显著(WC与WF之间为43.17a).zQb /sah其屮仇分別是2方的估计:心=加丘+咅二 町分别是空A的标准误。,Z的值越大越能说明显著。5 = 8:二=0.08, Z二ab/Sab=(-0.09)*(-0.13)/0.08=0.14625 比较小,故中介不明显。
展开阅读全文