Final from spring 2004

```Psy 524
Ainsworth
S’ 04
Final Exam
Psy 524
Ainsworth
You have exactly 2 hours.
You can use a calculator (I suggest it) and nothing else.
Good Luck and have a great summer!!!!
Name_______________________________________
Psy 524
Ainsworth
S’ 04
Final Part I
Discriminant Function Analysis (12 pts)
1. Purpose of the method (give at least 2 research questions).
2. How is discriminant function analysis different from MANOVA?
3. Describe the assumption of homogeneity of variance/covariance matrices.
4. Describe the use of classification scores in discriminant function analysis
(basically, how do you get them and what are they used for, no equations).
Logistic Regression (12 pts)
5. Purpose of the method (give at least 2 research questions).
Psy 524
Ainsworth
S’ 04
6. How is logistic regression different from discriminant function analysis?
When is it better to use discriminant function analysis over logistic regression?
7. What is a logit? How can you transform a logit into a probability?
8. What is the only real limitation with logistic regression? How does probit
analysis differ from logistic regression?
Factor Analysis (12 pts)
9. Purpose of the method (give at least 2 research questions).
10. What is the theoretical difference between principal components analysis and
factor analysis?
Psy 524
Ainsworth
S’ 04
11. What is the main purpose of factor rotation? Explain this. What is the
difference between orthogonal and oblique rotation?
12. Describe Kaiser’s criterion for picking the number of factors (hint: the default
method in SPSS). What is this criterion based on (why do we do it?)?
Structural Equation Modeling (14 pts)
13. How is confirmatory factor analysis different than exploratory factor analysis
(give me at least 2 ways)?
14. What are the differences between the structural and measurement parts of an
SEM model?
Psy 524
Ainsworth
S’ 04
15. What do we use as data in an SEM model? What formula do we use to
calculate the number of data points? How do we find the degrees of freedom
for a model?
16. Describe these matrices:
a)  (Beta)
b)  (Gamma)
c)  (Phi)
17. Define models that are:
a) overidentified
b) just identified
c) under identified
d) recursive.
Psy 524
Ainsworth
S’ 04
Final Part II
Output interpretation
Section 1 (12.5 pts) – Discriminant Function Analysis Predicting Political
Conservatism (Very Liberal, Liberal, Moderate, Conservative, Very Conservative)
Analysis Cas e Process ing S umm ary
Un weigh ted Cases
Va lid
N
Exc lude d Missing or ou t-of-ra nge g roup
cod es
At least one m issin g
discrimin ating vari able
Bo th mi ssing or ou t-of-ra nge
gro up co des a nd a t least one
missing discri mina ting v ariab le
To tal
To tal
119 4
Pe rcent
91. 1
25
1.9
86
6.6
5
.4
116
8.9
131 0
100 .0
Tes ts of Equa lity of Group M eans
Wil ks' La mbd a
F
df1
df2
Sig .
ITE M26 REA L MA N SHOUL D
AV OUD SHO WING WEA KNE SS
.93 5
20. 718
4
118 9
.00 0
ITE M29 FEM INIS TS T OO
AG GRE SSIV E
.87 3
43. 244
4
118 9
.00 0
ITE M31 TRA DITIO NAL SEX
RO LES
.83 7
57. 813
4
118 9
.00 0
ITE M33 WO MEN NOT GOO D
ENGINE ERS
.91 0
29. 448
4
118 9
.00 0
ITE M35 FEM ALE
AUTO-M ECHANICS
.94 6
17. 068
4
118 9
.00 0
ITE M37 ONE MEN SHO ULD
PL AY FO OTB ALL
.95 4
14. 267
4
118 9
.00 0
ITE M41 MAL E NURSE S
.89 8
33. 732
4
118 9
.00 0
ITE M42 MEN WHO CRY IN
PUBLIC
.93 8
19. 518
4
118 9
.00 0
ITE M44 HOM OSE XUA LITY
.75 4
96. 801
4
118 9
.00 0
ITE M45 WO MEN IN CO MBA T
.96 9
9.6 05
4
118 9
.00 0
ITE M47 FEM ALE FIGHTER
PIL OT
.95 4
14. 288
4
118 9
.00 0
ITE M49 WO MEN
SUPERV ISORS
.92 7
23. 552
4
118 9
.00 0
ITE M51 MAL E SE CRE TARI ES
.88 6
38. 362
4
118 9
.00 0
ITE M52 MAN WHO ST AYS
HO ME
.89 9
33. 457
4
118 9
.00 0
1. What does the box above (tests of equality of group means) tell you?
Psy 524
Ainsworth
S’ 04
Analysis 1
Box's Test of Equality of Covariance Matrices
Log Dete rminants
PS D ST UDE NTS P OLIT ICAL
SE RY
VE
LF-DE
LI BERA
SCRIPTIO
L
N
Ra nk
14
Log Dete rmin ant
-3.1 85
LIB ERAL
14
-4.3 76
MO DERATE
14
-4.7 10
CO NSERVAT IVE
14
-3.2 21
VE RY CONSE RVA TIVE
14
-1.9 27
Po oled within -grou ps
14
-3.2 99
Th e ranks and natu ral lo garith ms o f dete rmin ants p rinted are
tho se of the g roup covariance matrices.
Tes t Res ults
Bo x's M
F
806 .601
Ap prox.
1.8 06
df1
420
df2
997 11.0 25
Sig .
.00 2
Te sts nu ll hyp othe sis of equa l pop ulatio n covarian ce m atrice s.
1. What does Box’s M statistic test? Is it a problem here? Why or why not?
Summary of Canonical Discriminant Functions
Eigenv a lues
Fu nction
1
Eig enva lue % of Va riance Cu mula tive %
.48 0 a
88 .1
88 .1
Ca nonical
Co rrelat ion
.57 0
2
.03 6 a
6.6
94 .7
.18 6
3
.02 0 a
3.7
98 .4
.14 0
4
.00 9 a
1.6
10 0.0
.09 2
a. First 4 canon ical d iscrim inan t fun ction s were use d in t he an alysi s.
Wi lks' Lambda
Te st of Funct ion(s) Wi lks' La mbd a
1 t hroug h 4
.63 4
Ch i-squa re
53 9.639
df
Sig .
56
.00 0
2 t hroug h 4
.93 8
75 .581
39
.00 0
3 t hroug h 4
.97 2
33 .693
24
.09 0
4
.99 1
10 .126
11
.51 9
2. What do the two boxes above (Eigenvalues and Wilk’s Lambda) tell you
(annotate and interpret them)?
Psy 524
Ainsworth
S’ 04
Str uctur e Ma trix
Fun ction
1
2
3
4
ITE M44 HOM OSE XUA LITY
.82 2*
-.16 2
-.02 8
-.10 5
ITE M31 TRA DITIO NAL SEX
RO LES
.63 2*
.04 1
.26 8
.41 0
ITE M29 FEM INIS TS T OO
AG GRE SSIV E
.54 3*
-.17 5
-.36 6
.03 4
ITE M51 MAL E SE CRE TARI ES
.50 9*
.31 3
-.24 2
.07 3
ITE M52 MAN WHO ST AYS
HO ME
.47 4*
.20 3
.37 1
-.15 8
ITE M49 WO MEN
SUPERV ISORS
.39 3*
.31 3
.26 7
.15 9
ITE M26 REA L MA N SHOUL D
AV OUD SHO WING WEA KNE SS
.37 2*
.25 4
.22 1
.07 8
ITE M35 FEM ALE
AUTO-M ECHANICS
.34 5*
.03 2
-.09 1
.02 6
ITE M37 ONE MEN SHO ULD
PL AY FO OTB ALL
.31 4*
-.11 9
.00 1
.09 5
ITE M45 WO MEN IN CO MBA T
.25 1*
-.20 7
.05 6
.23 1
ITE M41 MAL E NURSE S
.45 6
.58 7*
-.18 8
-.28 9
ITE M33 WO MEN NOT GOO D
ENGINE ERS
.43 6
.45 6*
.12 3
-.14 7
ITE M42 MEN WHO CRY IN
PUBLIC
.34 8
.39 7*
-.28 4
.15 0
ITE M47 FEM ALE FIGHTER
PIL OT
.29 4
.35 2
.03 7
.50 5*
Po oled within -grou ps co rrela tions betwe en d iscrim inati ng va riable s and stan dard ized
can onical discrimin ant f uncti ons
Va riable s ord ered by ab solut e size of co rrela tion within function.
*. Largest absol ute co rrela tion b etwe en ea ch variab le and any discrimina nt fu nction
3. What is the structure matrix? What does this tell you about the meaning
Classification Statistics
Cla ssifi cation Process ing Summ ary
Pro cesse d
Exclude d
Use d in Outp ut
13 10
Mi ssing or ou t-of-range
gro up co des
0
At least one missi ng
discrimi natin g variable
91
12 19
Psy 524
Ainsworth
S’ 04
Pri or Pr obabilities for G roups
PS D ST UDE NTS POLI TICA L
SE LF-DE SCRIPTIO N
VE RY L IBERAL
Ca ses Used in Ana lysis
Pri or
Sp ecifie d Pri or Eff ective Prio r
.20 0
Un weigh ted
80
We ighte d
80. 000
LIB ERA L
.20 0
295
295 .000
MO DERATE
.20 0
413
413 .000
CO NSERVAT IVE
.20 0
366
366 .000
VE RY CONSE RVA TIVE
.20 0
40
40. 000
1.0 00
119 4
119 4.00 0
To tal
4. Where do the prior probabilities come from (“Prior” column from the
table above)?
Function 2
Functions at Group Centroids
1.00
0.80
0.60
0.40
Very Liberal
0.20
0.00
-0.20
-0.40
-0.60
-0.80
-1.00
-2.00
Very Conservative
Liberal
Conservative
Moderate
-1.00
0.00
1.00
Function 1
5. What does function 2 tell you about separating the groups?
2.00
Psy 524
Ainsworth
S’ 04
Cla ssifi cation Re sults a
Pre dicte d Gro up M emb ershi p
Ori ginal
Co unt
PS D ST UDE NTS POL ITICA L
SE LF-DESCRIPTI ON
VE RY L IBERAL
VE RY L IBERAL
54
LIB ERA L
%
LIB ERA L
11
VE RY
MO DERATE CO NSE RVAT IVE CO NSE RVAT IVE
6
3
6
10 0
97
45
MO DERATE
50
80
12 9
CO NSE RVAT IVE
16
42
69
VE RY CONS ERVA TIVE
0
1
3
Un group ed cases
2
4
VE RY L IBERAL
67 .5
LIB ERA L
MO DERATE
CO NSE RVAT IVE
VE RY CONS ERVA TIVE
Un group ed cases
31
29 5
96
58
41 3
13 7
10 2
36 6
11
25
40
6
5
8
25
13 .8
7.5
3.8
7.5
10 0.0
33 .9
32 .9
15 .3
10 .5
7.5
10 0.0
12 .1
19 .4
31 .2
23 .2
14 .0
10 0.0
4.4
11 .5
18 .9
37 .4
27 .9
10 0.0
.0
2.5
7.5
27 .5
62 .5
10 0.0
8.0
16 .0
24 .0
20 .0
32 .0
10 0.0
6. Is this a good classification model? Explain you answer.
Section 2 (12.5 pts) - Logistic Regression of sexism (1 = sexist and 0 = not sexist) as
predicted by PSD (political self description; continuous with higher values meaning
more conservative), Gender (1 = male, 2 = female), Protestant (1 = yes and 0 = no)
and White (1 = yes and 0 = no).
Logistic Regression
De pende nt Va riabl e Enc oding
Ori ginal Valu e Inte rnal Value
.00
No
Ye s
Block 0: Beginning Block
Cla ssifi cation Table a,b
Pre dicte d
Se xism
Ste p 0
Ob serve d
Se xism
NO
59 1
0
Ye s
57 8
0
a. Co nstan t is in clude d in the m odel .
b. Th e cut value is .5 00
Ye s
NO
Overall Perce ntag e
80
22
a. 37 .0% o f orig inal group ed cases correctly cl assifi ed.
1.0 0
To tal
Pe rcent age Correct
10 0.0
.0
50 .6
Psy 524
Ainsworth
S’ 04
Va riable s in the E quati on
B
Ste p 0 Co nstan t
-.02 2
S.E .
.05 8
Wa ld
.14 5
df
Sig .
1
.70 4
Exp (B)
.97 8
Va riable s not in the Equation
Sco re
Ste p 0 Va riable s
df
Sig .
Po litical Self
De script ion
163 .451
1
.00 0
Ge nder
117 .512
1
.00 0
7.3 21
1
.00 7
.01 9
1
.88 9
253 .123
4
.00 0
Pro testa nt
Wh ite
Overall Statistics
1. What type of model is block 0 testing? Is the block 0 model doing a good
job? How do you know?
Block 1: Method = Enter
Om nibus Tes ts of Model Coeffic ients
Ste p 1
Ste p
Ch i-squa re
27 7.922
df
Sig .
4
.00 0
Blo ck
27 7.922
4
.00 0
Mo del
27 7.922
4
.00 0
2. What is the block 1 model? What does this box tell you about the model?
Model S umm ary
Ste p
1
-2 L og li keliho od
134 2.51 2
Co x &amp; S nell R Na gelke rke R
Sq uare
Sq uare
.21 2
.28 2
3. What is the difference between the Cox &amp; Snell and Nagelkerke R square
estimates?
Psy 524
Ainsworth
S’ 04
Hosmer and Lemeshow Te st
Ste p
1
Ch i-squa re
8.8 42
df
Sig .
8
.35 6
Contingency Table for Hosm er and Lem eshow Te st
SE XISM = no
SE XISM = Ye s
Ste p 1 1
Ob serve d
99
Exp ecte d
98. 087
Ob serve d
15
Exp ecte d
15. 913
To tal
2
95
94. 133
24
24. 867
119
3
72
82. 000
46
36. 000
118
4
100
92. 557
48
55. 443
148
5
57
57. 535
49
48. 465
106
6
31
32. 122
38
36. 878
69
7
44
45. 038
65
63. 962
109
8
41
34. 677
59
65. 323
100
9
34
32. 189
95
96. 811
129
10
18
22. 664
139
134 .336
157
114
4. What does the Hosmer and Lemeshow Test tell you about the model?
And how do you know (what test)?
Cla ssifi cation Table a
Pre dicte d
SE XISM
Ste p 1
Ob serve d
SE XISM
Overall Perce ntag e
No
No
42 3
Ye s
16 8
Pe rcent age Correct
71 .6
Ye s
18 2
39 6
68 .5
70 .1
a. Th e cut value is .5 00
5. Is this model doing a good job of classifying the respondents? Why or
why not?
Psy 524
Ainsworth
S’ 04
Va riable s in the E quation
B
.84 3
S.E .
.07 6
Wa ld
12 4.316
ST UDSE X
-1. 330
.13 6
PROTE ST
.27 8
.14 0
WHITE
-.2 09
Co nstan t
-.4 30
Ste p 1 a PS D
df
Sig .
Exp(B)
2.3 23
1
.00 0
95 .531
1
.00 0
.26 4
3.9 64
1
.04 6
1.3 21
.16 4
1.6 21
1
.20 3
.81 1
.31 4
1.8 76
1
.17 1
.65 1
a. Va riable (s) en tered on step 1 : PS D, ST UDS EX, P ROT EST, WHI TE.
6. Write the logistic regression equation for this model (think about it first).
Would you remove any of the predictors? Why?
Section 3 (12.5 pts) - Factor Analysis of some social psychological data related to
prejudice and discrimination
Factor Analysis
Communalities
Ini tial
.17 5
Extracti on
.56 1
ITE M27 RA CIAL EQUALIT Y
.42 9
.45 9
ITE M28 BLA CK P RES IDENT
OF USA
.56 4
.62 0
ITE M30 BLA CK NEIG HBO ORS
.52 2
.67 1
ITE M32 FO REIG HNERS
.37 0
.34 9
ITE M34 INT ERRACIA L DA TING
SHOUL D BE AVO IDED
.61 3
.65 8
ITE M36 EA CH E THNI C GROUP
ST AY IN ITS OWN PLA CE
.53 9
.60 8
ITE M38 TO O MA NY B LACK
ST UDE NTS ON CAMP US
.48 9
.55 0
ITE M39 INCREA SED
EQ UAL ITY
.38 0
.42 5
ITE M40 WHITE S UPE RIORITY
.53 1
.64 1
ITE M43 BUSING
.20 6
.27 4
ITE M46 A B LACK
SUPERVISO R
.47 2
.53 0
ITE M48 ME XICA N
IM MIGRANT S
.37 4
.33 8
ITE M50 INT ERRACIA L
MA RRIA GE
.61 8
.85 2
ITE M25 AFFIRM ATIV E ACTION
Extracti on M ethod : Pri ncipa l Axi s Factorin g.
1. What is the difference between the initial and extraction columns in the
table above? Where do the values come from?
Psy 524
Ainsworth
S’ 04
Total Va rianc e Ex plained
Ini tial E igenvalue s
Fa ctor
1
Extractio n Su ms o f Squ ared Load ings
To tal
% of Va riance Cu mula tive %
6.1 16
43 .688
43 .688
To tal
% of Va riance Cu mula tive %
5.6 85
40 .605
40 .605
Ro tation Sum s of Squa red L oadin gs
To tal
% of Va riance Cu mula tive %
2.7 66
19 .760
19 .760
2
1.2 49
8.9 23
52 .610
.72 8
5.1 98
45 .803
1.9 36
13 .829
33 .588
3
1.0 46
7.4 74
60 .084
.71 4
5.0 98
50 .901
1.8 88
13 .486
47 .074
4
.83 2
5.9 46
66 .031
.40 9
2.9 21
53 .822
.94 5
6.7 48
53 .822
5
.78 7
5.6 22
71 .653
6
.63 7
4.5 48
76 .200
7
.57 0
4.0 72
80 .272
8
.53 3
3.8 07
84 .080
9
.50 3
3.5 96
87 .675
10
.40 5
2.8 92
90 .567
11
.39 6
2.8 25
93 .393
12
.37 2
2.6 58
96 .050
13
.32 8
2.3 41
98 .391
14
.22 5
1.6 09
10 0.000
Extractio n Me thod : Prin cipal Axis Fact oring .
2. For factor 1 the percent of variance is 43.688, how is that calculated?
Why do the eigenvalues change from the initial values to the extraction
Scree Plot
7
6
5
4
3
Eigenvalue
2
1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Factor Number
3. Based on the scree plot how many factors would you extract? Explain
Psy 524
Ainsworth
S’ 04
Fa ctor M atri x a
Fa ctor
1
2
3
4
ITE M25 AFFIRM ATIV E ACTION
.26 5
.57 8
.31 4
.24 0
ITE M27 RACIAL EQUALIT Y
.65 4
-.1 25
.12 0
.02 0
ITE M28 BLA CK P RES IDENT
OF USA
.75 4
.04 2
.10 4
-.1 96
ITE M30 BLA CK NEIG HBOO RS
.71 5
-.0 14
.20 1
-.3 44
ITE M32 FOREIG HNERS
.58 8
.05 1
-.0 04
-.0 15
ITE M34 INT ERRACIA L DA TING
SHOULD BE AVO IDED
.69 1
.03 9
-.4 09
.10 7
ITE M36 EACH E THNI C GROUP
ST AY IN ITS OWN PLA CE
.73 6
-.1 84
-.0 37
.17 6
ITE M38 TO O MA NY B LACK
ST UDENTS ON CAMP US
.68 6
-.1 48
.16 8
.17 4
ITE M39 INCREA SED
EQ UALI TY
.61 0
-.0 79
.17 2
.13 2
ITE M40 WHITE S UPE RIORITY
.71 4
-.2 97
.04 2
.20 2
ITE M43 BUS ING
.32 2
.41 0
.02 2
.04 4
ITE M46 A B LACK
SUPERV ISO R
.67 1
.04 2
.16 2
-.2 28
ITE M48 ME XICA N
IM MIGRANT S
.56 7
.04 7
-.0 96
-.0 72
ITE M50 INT ERRACIA L
MA RRIA GE
.71 6
.22 3
-.5 34
-.0 57
Extractio n M ethod : Prin cipa l Axis Factoring .
a. Att empt ed to extra ct 4 facto rs. Mo re th an 2 5 iteration s req uired .
(Co nvergence=.00 6). E xtraction was te rmin ated.
4. Identify this matrix (factor matrix). What are the values in the matrix?
Psy 524
Ainsworth
S’ 04
Rotated Factor M atrixa
Fa ctor
1
2
3
4
ITE M25 AFFIRM ATIV E ACTION
.10 4
.06 8
-.0 12
.73 8
ITE M27 RACIAL EQUALIT Y
.52 8
.36 8
.18 9
.08 9
ITE M28 BLA CK P RES IDENT
OF USA
.39 5
.59 0
.29 0
.17 9
ITE M30 BLA CK NEIG HBOO RS
.34 1
.71 5
.17 8
.10 8
ITE M32 FOREIG HNERS
.35 2
.32 5
.30 1
.17 0
ITE M34 INT ERRACIA L DA TING
SHOULD BE AVO IDED
.39 1
.15 4
.68 9
.08 1
ITE M36 EACH E THNI C GROUP
ST AY IN ITS OWN PLA CE
.65 5
.23 8
.34 6
.05 3
ITE M38 TO O MA NY B LACK
ST UDENTS ON CAMP US
.65 3
.27 9
.15 7
.14 3
ITE M39 INCREA SED
EQ UALI TY
.55 0
.27 5
.13 2
.17 2
ITE M40 WHITE S UPE RIORITY
.72 8
.22 9
.24 4
-.0 12
ITE M43 BUS ING
.04 5
.15 3
.22 8
.44 3
ITE M46 A B LACK
SUPERV ISO R
.33 8
.58 9
.20 1
.17 0
ITE M48 ME XICA N
IM MIGRANT S
.28 7
.32 8
.36 8
.11 1
ITE M50 INT ERRACIA L
MA RRIA GE
.20 2
.25 9
.85 0
.14 7
Extractio n M ethod : Prin cipa l Axis Factoring .
Ro tation Met hod: Varim ax with K aiser Norm alization .
a. Ro tation con verge d in 6 iteration s.
5. What type of rotation is this? Orthogonal or Oblique?
6. Calculate the communality for the item 25 (affirmative action) using this
matrix (show your work). Calculate the eigenvalue for the second factor
based on this matrix (show your work).
Fa ctor Trans form ation Matr ix
Fa ctor
1
1
2
3
4
.65 6
.53 2
.48 6
.22 5
2
-.4 75
.02 3
.22 1
.85 1
3
.24 6
.32 2
-.8 46
.34 8
4
.53 3
-.7 82
-.0 11
.32 2
Extractio n Me thod : Prin cipa l Axis Fact oring .
Ro tation Met hod: Varim ax with K aiser Norm aliza tion.
7. What is this matrix (factor transformation matrix) used for?
Psy 524
Ainsworth
S’ 04
Fa ctor S core Coe fficie nt M atrix
Fa ctor
1
2
3
4
ITE M25 AFFIRM ATIV E ACTION
.01 0
-.1 01
-.0 90
.65 0
ITE M27 RACIAL EQUALIT Y
.12 1
.04 7
-.0 34
-.0 07
ITE M28 BLA CK P RES IDENT
OF USA
-.0 44
.30 1
-.0 12
.04 0
ITE M30 BLA CK NEIG HBOO RS
-.0 91
.53 5
-.1 30
-.0 71
ITE M32 FOREIG HNERS
.02 8
.04 8
.03 7
.03 6
ITE M34 INT ERRACIA L DA TING
SHOULD BE AVO IDED
.11 7
-.1 71
.22 2
-.0 41
ITE M36 EACH E THNI C GROUP
ST AY IN ITS OWN PLA CE
.27 6
-.1 12
.03 9
-.0 46
ITE M38 TO O MA NY B LACK
ST UDENTS ON CAMP US
.25 5
-.0 53
-.0 80
.04 5
ITE M39 INCREA SED
EQ UALI TY
.17 0
-.0 17
-.0 81
.05 1
ITE M40 WHITE S UPE RIORITY
.40 3
-.1 31
-.0 54
-.1 25
ITE M43 BUS ING
-.0 50
.00 4
.01 8
.21 3
ITE M46 A B LACK
SUPERV ISO R
-.0 28
.26 5
-.0 81
.02 5
ITE M48 ME XICA N
IM MIGRANT S
.00 2
.06 2
.03 4
.01 1
ITE M50 INT ERRACIA L
MA RRIA GE
-.3 05
-.0 27
.83 2
.04 5
Extractio n M ethod : Prin cipa l Axis Factoring .
Ro tation Met hod: Varim ax with K aiser Norm alization .
8. What is this matrix (factor score coefficient matrix) used for?
Psy 524
Ainsworth
S’ 04
Section 4 (12.5 pts) – Structural Equation Model using EQS (selected output).
/LABELS
V1=CICI102; V2=CICI108; V3=CICI112; V4=QDIA104R; V5=QDIA111R;
V6=QDIA129; V7=MEIMO104; V8=MEIMM106; V9=MEIMO109; V10=MEIMM111;
V11=MEIMM114; V12=MEIMO117; V13=MEIMM118; V14=MEIMO119;
/EQUATIONS
/VARIANCES
V1 =
1F1 + E1;
F1 = *;
V2 =
*F1 + E2;
E1 = *;
V3 =
*F1 + E3;
E2 = *;
V4 =
1F2 + E4;
E3 = *;
V5 =
*F2 + E5;
E4 = *;
V6 =
*F2 + E6;
E5 = *;
V7 =
1F3 + E7;
E6 = *;
V8 =
1F4 + E8;
E7 = *;
V9 =
*F3 + E9;
E8 = *;
V10 =
*F4 + E10;
E9 = *;
V11 =
*F4 + E11;
E10 = *;
V12 =
*F3 + E12;
E11 = *;
V13 =
*F4 + E13;
E12 = *;
V14 =
*F3 + E14;
E13 = *;
F2 =
*F1 + *F3 + *F4 + D2;
E14 = *;
F3 =
*F1 + D3;
D2 = *;
F4 =
*F1 + D4;
D3 = *;
D4 = *;
/COVARIANCES
e13,e10 = *;
1. Based on the syntax above draw a structural diagram (typical SEM
diagram).
Psy 524
Ainsworth
S’ 04
GOODNESS OF FIT SUMMARY FOR METHOD = ML
INDEPENDENCE MODEL CHI-SQUARE
INDEPENDENCE AIC =
MODEL AIC =
4176.91881
89.52370
= 4358.919 ON 91 DEGREES OF FREEDOM
INDEPENDENCE CAIC =
MODEL CAIC =
3648.79983
-322.52518
CHI-SQUARE =
231.524 BASED ON
71 DEGREES OF FREEDOM
PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS
.00000
THE NORMAL THEORY RLS CHI-SQUARE FOR THIS ML SOLUTION IS 225.349.
FIT INDICES
----------BENTLER-BONETT
NORMED FIT INDEX =
.947
BENTLER-BONETT NON-NORMED FIT INDEX =
.952
COMPARATIVE FIT INDEX (CFI)
=
.962
BOLLEN
(IFI) FIT INDEX
=
.963
MCDONALD (MFI) FIT INDEX
=
.915
LISREL
GFI FIT INDEX
=
.965
LISREL
AGFI FIT INDEX
=
.949
ROOT MEAN-SQUARE RESIDUAL (RMR)
=
.161
STANDARDIZED RMR
=
.058
ROOT MEAN-SQUARE ERROR OF APPROXIMATION (RMSEA)
90% CONFIDENCE INTERVAL OF RMSEA (
.043,
=
.050
.057)
ITERATIVE SUMMARY
ITERATION
1
2
3
4
5
6
PARAMETER
ABS CHANGE
.917223
.349358
.167457
.026958
.003896
.000779
ALPHA
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
FUNCTION
2.02774
.99908
.28508
.25754
.25726
.25725
2. Assuming the model had no problems, did the model fit the data?
3. If there are 901 cases in the sample, how was the model chi-square
Psy 524
Ainsworth
S’ 04
PARAMETER INDIRECT EFFECTS
-------------------------QDI
=
-.467*CIC
- .579 D3
.098
.084
-4.745@
-6.882@
+
.305 D4
.052
5.920@
4. How can this be interpreted (parameter indirect effects)?
MULTIVARIATE LAGRANGE MULTIPLIER TEST BY SIMULTANEOUS PROCESS IN STAGE
1
PARAMETER SETS (SUBMATRICES) ACTIVE AT THIS STAGE ARE:
PVV PFF PEE PDD GVV GVF GFV GFF BVV BVF BFV BFF
CUMULATIVE MULTIVARIATE STATISTICS
---------------------------------STEP
---1
2
3
4
5
6
7
8
9
10
11
PARAMETER
----------V8,V7
V6,V13
F4,V10
V13,V9
V5,F3
V6,V2
V2,V9
V13,V11
F3,V7
F2,V10
V4,V3
CHI-SQUARE
---------51.210
74.588
95.152
116.668
132.846
144.681
156.491
165.047
170.908
176.152
180.031
D.F.
---1
2
3
4
5
6
7
8
9
10
11
PROB.
----.000
.000
.000
.000
.000
.000
.000
.000
.000
.000
.000
UNIVARIATE INCREMENT
-----------------------------HANCOCK'S
SEQUENTIAL
CHI-SQUARE
PROB. D.F.
PROB.
-------------- -------51.210
.000
71
.963
23.377
.000
70
1.000
20.564
.000
69
1.000
21.516
.000
68
1.000
16.178
.000
67
1.000
11.835
.001
66
1.000
11.810
.001
65
1.000
8.556
.003
64
1.000
5.861
.015
63
1.000
5.244
.022
62
1.000
3.879
.049
61
1.000
5. What is the lagrange multiplier test used for? If the path in step 1
above was added to the model how would the chi-square change and
by how much (roughly)?
```