作者mld7558 (amy)
看板Statistics
标题Re: [问题] lisrel配适度分析问题
时间Fri May 4 16:07:42 2012
※ 引述《mld7558 (amy)》之铭言:
※ 引述《mld7558 (amy)》之铭言:
目前正在进行配适分析,
但配适指标的卡方、RMSEA与NNF皆未达到标准,
因此要针对MI量表将不适的题项删除,
但MI量表只出现"No Non-Zero Modification Indices for LAMBDA-X"
所以想请问:
为何MI量表没有出现?
是哪里发生错误?
如何才能达到配饰指标呢?
谢谢!!!!!!!!!
报表内容:
title :AT
data NI=5 NO=384 MA=KM
KM SY FI=C:\Users\amy\Desktop\0417\LISREL2\AT\AT.COR
LA; 1 2 3 4 5
MODEL NX=5 NK=1 LX=FU,FI TD=DI,FR
LK; ATITUDE
FREE LX(1,1) LX(2,1) LX(3,1) LX(4,1) LX(5,1)
FIX PH(1,1)
VALUE 1 PH(1,1)
PD
OUTPUT SE TV RS MR FS EF SS SC MI AD>50
AT
Number of Input Variables 5
Number of Y - Variables 0
Number of X - Variables 5
Number of ETA - Variables 0
Number of KSI - Variables 1
Number of Observations 384
LISREL Estimates (Maximum Likelihood)
LAMBDA-X
ATITUDE
--------
1 0.74
(0.05)
16.21
2 0.65
(0.05)
13.60
3 0.78
(0.04)
17.30
4 0.80
(0.04)
18.04
5 0.84
(0.04)
19.24
THETA-DELTA
1 2 3 4 5
-------- -------- -------- -------- --------
0.45 0.58 0.40 0.36 0.30
(0.04) (0.05) (0.04) (0.03) (0.03)
11.54 12.48 10.96 10.46 9.43
Goodness of Fit Statistics
Degrees of Freedom = 5
Minimum Fit Function Chi-Square = 102.65 (P = 0.0)
Normal Theory Weighted Least Squares Chi-Square = 114.27 (P = 0.0)
Estimated Non-centrality Parameter (NCP) = 109.27
90 Percent Confidence Interval for NCP = (78.13 ; 147.84)
Minimum Fit Function Value = 0.27
Population Discrepancy Function Value (F0) = 0.29
90 Percent Confidence Interval for F0 = (0.20 ; 0.39)
Root Mean Square Error of Approximation (RMSEA) = 0.24
90 Percent Confidence Interval for RMSEA = (0.20 ; 0.28)
P-Value for Test of Close Fit (RMSEA < 0.05) = 0.00
Expected Cross-Validation Index (ECVI) = 0.35
90 Percent Confidence Interval for ECVI = (0.27 ; 0.45)
ECVI for Saturated Model = 0.078
ECVI for Independence Model = 2.65
Chi-Square for Independence Model with 10 Degrees of Freedom = 1005.54
Independence AIC = 1015.54
Model AIC = 134.27
Saturated AIC = 30.00
Independence CAIC = 1040.29
Model CAIC = 183.78
Saturated CAIC = 104.26
Normed Fit Index (NFI) = 0.90
Non-Normed Fit Index (NNFI) = 0.80
Parsimony Normed Fit Index (PNFI) = 0.45
Comparative Fit Index (CFI) = 0.90
Incremental Fit Index (IFI) = 0.90
Relative Fit Index (RFI) = 0.80
Critical N (CN) = 57.30
Root Mean Square Residual (RMR) = 0.062
Standardized RMR = 0.062
Goodness of Fit Index (GFI) = 0.8
Adjusted Goodness of Fit Index (AGFI) = 0.68
Parsimony Goodness of Fit Index (PGFI) = 0.30
Modification Indices and Expected Change
No Non-Zero Modification Indices for LAMBDA-X
No Non-Zero Modification Indices for PHI
Modification Indices for THETA-DELTA
1 2 3 4 5
-------- -------- -------- -------- --------
1 - -
2 64.82 - -
3 1.66 7.23 - -
4 9.64 25.70 4.21 - -
5 16.89 19.53 1.49 77.93 - -
Expected Change for THETA-DELTA
1 2 3 4 5
-------- -------- -------- -------- --------
1 - -
2 0.25 - -
3 0.04 0.08 - -
4 -0.09 -0.15 -0.06 - -
5 -0.12 -0.13 -0.04 0.27 - -
Completely Standardized Expected Change for THETA-DELTA
1 2 3 4 5
-------- -------- -------- -------- --------
1 - -
2 0.25 - -
3 0.04 0.08 - -
4 -0.09 -0.15 -0.06 - -
5 -0.12 -0.13 -0.04 0.27 - -
Standardized Solution
LAMBDA-X
ATITUDE
--------
1 0.74
2 0.65
3 0.78
4 0.80
5 0.84
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 140.128.118.210
1F:推 sk2allwin:你这个看来要拉TD45 05/04 17:02
2F:→ sk2allwin:没办法去删题,你的题目看来信度都不错 05/04 17:04
3F:→ mld7558:可是我的适配性太低了耶!!我怕模型到时候会被质疑。。 05/04 17:18
4F:推 sk2allwin:呃,我在下面有回文。 05/04 17:47
5F:→ mld7558:嗯!我试过你的方法 配适度逐渐达到标准了 05/04 18:20
6F:→ mld7558:想请问是看哪个量表可以得知要拉TD45呢?谢谢!!!! 05/04 18:21
7F:推 sk2allwin:下面有回文罗,请过目。 05/04 18:31