Psy 524 Ainsworth S’ 04 Final Exam Psy 524 Ainsworth Please write on the exam. 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 Short answer questions 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 of the discriminant functions (refer to your answer on number 3)? 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 & S nell R Na gelke rke R Sq uare Sq uare .21 2 .28 2 3. What is the difference between the Cox & 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 IN YOUR NE IGHB ORHOOD .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 sums of squared loadings? 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 your answer. 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 IN YOUR NE IGHB ORHOOD .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 IN YOUR NE IGHB ORHOOD .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 IN YOUR NE IGHB ORHOOD -.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? Explain your answer. 3. If there are 901 cases in the sample, how was the model chi-square calculated? Show your work. 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)?