我做的是双因素模型,两个时间点,LCA后分了三个类别
第一个问题是五步法中应该第三步应该是算两个时间点的转换概率,但是没有跑出来,就直接做了第四步加入协变量的潜在转变分析。虽然跑出来了,但是不知道怎么看结果,优势比怎么看,协变量影响下转变概率的发生比。下面是我的命令
DATA:
FILE IS D:\Desktop\ws\ws.dat;
VARIABLE:
NAMES ARE gender ew sw pw qt qp rjgxmg yy jl dd kb pz jsbx other
EWB SWB PWB qt1 qp1 rj1 yy1 jl1 dd1 kb1 pz1 jsb1 other1;
CLASSES=C1(3) C2(3);
ANALYSIS: TYPE=MIXTURE;
MODEL: %OVERALL%
C2 on C1;
C1 on gender;
MODEL C1:%C1#1%
[ew] (1);
[sw] (2);
[pw] (3);
[qt] (4);
[qp] (5);
[rjgxmg] (6);
[yy] (7);
[jl] (8);
[dd] (9);
[kb] (10);
[pz] (11);
[jsbx] (12);
[other] (13);
%C1#2%
[ew] (14);
[sw] (15);
[pw] (16);
[qt] (17);
[qp] (18);
[rjgxmg] (19);
[yy] (20);
[jl] (21);
[dd] (22);
[kb] (23);
[pz] (24);
[jsbx] (25);
[other] (26);
%C1#3%
[ew] (27);
[sw] (28);
[pw] (29);
[qt] (30);
[qp] (31);
[rjgxmg] (32);
[yy] (33);
[jl] (34);
[dd] (35);
[kb] (36);
[pz] (37);
[jsbx] (38);
[other] (39);
MODEL C2:%C2#1%
[EWB] (1);
[SWB] (2);
[PWB] (3);
[qt1] (4);
[qp1] (5);
[rj1] (6);
[yy1] (7);
[jl1] (8);
[dd1] (9);
[kb1] (10);
[pz1] (11);
[jsb1] (12);
[other1] (13);
%C2#2%
[EWB] (14);
[SWB] (15);
[PWB] (16);
[qt1] (17);
[qp1] (18);
[rj1] (19);
[yy1] (20);
[jl1] (21);
[dd1] (22);
[kb1] (23);
[pz1] (24);
[jsb1] (25);
[other1] (26);
%C2#3%
[EWB] (27);
[SWB] (28);
[PWB] (29);
[qt1] (30);
[qp1] (31);
[rj1] (32);
[yy1] (33);
[jl1] (34);
[dd1] (35);
[kb1] (36);
[pz1] (37);
[jsb1] (38);
[other1] (39);
OUTPUT:TECH1 TECH8 TECH15;
部分结果
Categorical Latent Variables
C2#1 ON
C1#1 4.416 0.143 30.842 0.000
C1#2 2.013 0.131 15.322 0.000
C2#2 ON
C1#1 1.948 0.151 12.896 0.000
C1#2 1.761 0.139 12.698 0.000
C1#1 ON
GENDER -0.071 0.101 -0.704 0.481
C1#2 ON
GENDER 0.202 0.110 1.842 0.065
Intercepts
C1#1 2.490 0.177 14.039 0.000
C1#2 0.982 0.193 5.077 0.000
C2#1 -0.615 0.112 -5.510 0.000
C2#2 -0.420 0.104 -4.049 0.000
LOGISTIC REGRESSION ODDS RATIO RESULTS
(Est. - 1) Two-Tailed
Estimate S.E. / S.E. P-Value
Categorical Latent Variables
C1#1 ON
GENDER 0.931 0.094 -0.730 0.465
C1#2 ON
GENDER 1.224 0.134 1.668 0.095
TECHNICAL 15 OUTPUT
ESTIMATED CONDITIONAL PROBABILITIES FOR THE CLASS VARIABLES
EVALUATED AT THE SAMPLE MEAN FOR ALL COVARIATES
P(C1=1)=0.691
P(C1=2)=0.244
P(C1=3)=0.065
P(C2=1|C1=1)=0.889
P(C2=2|C1=1)=0.092
P(C2=3|C1=1)=0.020
P(C2=1|C1=2)=0.456
P(C2=2|C1=2)=0.431
P(C2=3|C1=2)=0.113
P(C2=1|C1=3)=0.246
P(C2=2|C1=3)=0.299
P(C2=3|C1=3)=0.455
这是我认为比较重要的一些结果,但是不知道怎么看。希望有好心人帮帮孩子吧。