Upload
tai-sonsurk
View
221
Download
0
Embed Size (px)
Citation preview
8/7/2019 final muti
http://slidepdf.com/reader/full/final-muti 1/5
CFA and EFA
1)
Exploratory factor analysis (EFA) could be described as orderly simplification of interrelated
measures. EFA, traditionally, has been used to explore the possible underlying factor structure of a
set of observed variables without imposing a preconceived structure on the outcome (Child, 1990).
By performing EFA, the underlying factor structure is identified.
Exploratory Factor Analysis
C onfirmatory factor analysis ( C FA) is a statistical technique used to verify the factor structure of a set
of observed variables. CFA allows the researcher to test the hypothesis that a relationship between
observed variables and their underlying latent constructs exists. The researcher uses knowledge of
the theory, empirical research, or both, postulates the relationship pattern a priori and then tests
the hypothesis statistically. The process of data analysis with EFA and CFA will be explained.
Examples with FACTOR and CALIS procedures will illustrate EFA and CFA statistical techniques.
Confirmatory Factor Analysis
CFA and EFA are powerful statistical techniques. An example of CFA and EFA could occur with the
development of measurement instruments, e.g. a satisfaction scale, attitudes toward health,
customer service questionnaire. A blueprint is developed, questions written, a scale determined, the
8/7/2019 final muti
http://slidepdf.com/reader/full/final-muti 2/5
ins
¡ ¢ £ ent pilot teste ¤
¥ data collected, and CF¦
completed. The blueprint identifies the factor
structure or what we think it is. However, some questions may not measure what we thought they
should. If the factor structure is not confirmed, EF¦
is the ne§
t step. EF¦
helps us determine what
the factor structure looks like according to how participant responses. Exploratory factor analysis is
essential to determine underlying constructs for a set of measured variables.
2) Basic step for C ̈ A
1. Developing a theoretically based model.
3 In confirmatory factor analysis can be illustrated by a synthesis of the
principal components common factor analysis. 3 For example, brand awareness, brand loyalty and brand image are
the sub-factor of brand equity.
2. Checking the assumption.
3 Outlier
âëå outlier ë spss úâëå
8/7/2019 final muti
http://slidepdf.com/reader/full/final-muti 3/5
3 Multivariate Normality.
- Select data file
- Analyze properties test normal and outlier
- calculate estimate
- View text assessment of normality
cr.
Assessment of normality (Group number 1)
Variable min max skew c.r. kurtosis c.r.
CL3 3.000 5.000 -.008 -.041 -.095 -.262
CL2 3.000 5.000 .086 .472 -.390 -1.067CL1 3.000 5.000 -.031 -.169 .148 .404
CS3 2.000 5.000 .132 .720 -.316 -.866
CS2 2.000 5.000 -.255 -1.394 -.182 -.499
CS1 2.000 5.000 .204 1.120 -.738 -2.020
PM1 3.000 5.000 -.101 -.554 -.884 -2.422
PM2 3.000 5.000 -.209 -1.142 -.807 -2.211
PM3 2.000 5.000 -.379 -2.077 -.362 -.990
SQ1 3.000 5.000 .122 .666 -.344 -.942
SQ2 2.000 5.000 -.164 -.898 .845 2.314
SQ3 3.000 5.000 .268 1.466 .237 .648SQ4 3.000 5.000 .166 .909 -.004 -.011
SQ5 3.000 5.000 .180 .984 .846 2.317
Multivariate 16.003 5.072
èâõèèõèúèåúë è -1.96 ÷è 1.96 õè normal
øå: ûü åâ Observations farthest from the centroid õå Mahalanobis d-
squared
ø
âõ üûãö
úúø
spss
âõsave
õ
úspss
8/7/2019 final muti
http://slidepdf.com/reader/full/final-muti 4/5
3. Evaluating goodness-of-fit criteria
3 Absolute fit (GFI, RMSEA)
GFI and AGFI must be greater than 0.9.
RMSEA should be less than 0.05 while 3 Incremental fit (TLI, NFI)
NFI and TLI (NNFI) should be greater than 0.9. 3 Parimonious fit (CMIN/DF)
n > 100
It is the ratio of the chi-square divided by the degrees of freedom. Accepted at 1-3 or 1-5
Ho: Model is fit to data.
Ha: Model is not fit to data.
P must be greater than 0.05 in order to accept Ho. CMIN/DF must be less than 2.00.
4. Interpreting and modifying the model3 Unstandardized and standardized 3 Model respecification
2) Basic step for SEM
1. üõ estimate õúø ââ .05 ø p value â
éâ hypothesis
Ho: XX has no eff ect on YY
8/7/2019 final muti
http://slidepdf.com/reader/full/final-muti 5/5
Ha: XX has eff ect in YY
2 ø run model fit ü