ISQS 6348 Final Open notes, no books. Points out of 100 in parentheses. Y 1 ε 2

Transcription

1 ISQS 6348 Fnal Open notes, no books. Ponts out of 100 n parentheses. 1. The followng path dagram s gven: ε 1 Y 1 ε F Y 1.A. (10) Wrte down the usual model and assumptons that are mpled by ths dagram. Soluton: Y 1 = βf + ε 1 Y = ε where Var(F) = 1, F s uncorrelated wth the ε s, and Cov(ε 1,ε )=φ. 1.B. (10) Fnd the covarance matrx of that s mpled by the model and assumptons. Y F β 0 β 1 0 β 1 0 Soluton: = ε1 Y 0 0 1, so Cov = 0 ψ1 φ 1 ε Y φ ψ 0 1

2 β + ψ1 φ =. φ ψ C. (10) Suppose the covarance matrx s Cov =. Usng 1.B., dentfy the Y 1..3 parameters of your model n 1.A. that are dentfable, and state whch parameters are not dentfable. β + ψ1 φ Soluton: Equatng to, we can dentfy φ = 1. and φ ψ 1..3 ψ =.3. However, there are nfntely many choces for β and ψ 1 that gve β + ψ 1 =4., so β and ψ 1 are not dentfable. d.... (0) Suppose Y N0(µ,Σ), =1,,n. How does ths statement relate to the data matrx, and to numercal and graphcal summares of the data n the data matrx? By data matrx, I mean my usual depcton of the data set, where I show OBS n the frst column. State the varous connectons. Soluton: The data matrx looks lke ths: OBS Y 1 Y Y 0 1 y 11 y 1 y 1,0 y 1 y y,0 3 y 31 y 3 y 3,0 n y n1 y n y n,0 d... The statement Y N0(µ,Σ) states that the twenty-dmensonal row vectors of the data matrx are randomly sampled from a multvarate normal dstrbuton wth some 0-dmensonal mean µ and 0x0 covarance matrx Σ. For example, (y 11 y 1 y 1,0 ) s the result of one random selecton from N 0 (µ,σ), (y 1 y y,0 ) s the result of another, ndependent random selecton from N 0 (µ,σ), etc. If you look at the 0 hstograms for each of the varables, they wll show bell shapes. If you look at the 0*19/ scatterplots of pars of the varables, such as (Y 1, Y ), (Y 1, Y 3 ) etc., they wll show ellptcal shapes. If you construct the χ q-q plot of the n Mahalanobs dstances aganst ther expected Ch-Square quantles (df=0), t wll show the appearance of a 45 degree lne.

3 If you calculate the 0-dmensonal sample mean vector Y you calculate the 0x0 sample covarance matrx to Σ, both by the large of large numbers. n = 1 n = Y / n t wll be close to µ, and f = 1 S = ( Y Y)( Y Y)'/( n 1), t wll be close 3. Consder the followng SAS code: data measurement; do = 1 to 10000; F = rannor(0); e1 = rannor(0); e = rannor(0); e3 = rannor(0); x1 =.5 * F + sqrt(1-.5**) * e1 ; x =.5 * F + sqrt(1-.5**) * e ; x3 =.5 * F + sqrt(1-.5**) * e3; AverageX = (X1 + X + X3)/3; output; end; run; 3.A. (10) Fnd the true relablty of AverageX usng theory. Soluton: The relablty s the squared correlaton between AverageX and what t s measurng, namely, F. Frst we fnd the covarance matrx of AverageX and F, Note that AverageX =.5F + (1/3) sqrt(1-.5**)e1 + (1/3) sqrt(1-.5**)e + (1/3) sqrt(1-.5**)e3, so F AverageX.5.75 / 3.75 / 3.75 / 3 ε1 = F ε ε3 and hence AverageX.5.75 / 3.75 / 3.75 / / 3 0 Cov = F / / 3 0

4 / 3.75 / 3.75 / 3.75 / = = / / 3 0 So.5 Corr(AverageX, F) = =.5, and hence the relablty s.5 1 { Corr(AverageX, F) } = (.5) =.5 3.B. (5) How could you use the data n the fle WORK.MEASUREMENT to fnd the usual estmated lower bound on the true relablty of AverageX? Soluton: proc corr data=work.measurement alpha; var x1 x x3; run; In other words, calculate Cronbach s Alpha usng the X data. Note that ths gves a lower bound on the relablty of the summate, but the relablty of the summate s dentcal to the relablty of the average, so t also gves a lower bound on the relablty of the average. 3.C. (5) How could you use the data n the fle WORK.MEASUREMENT to fnd an estmate of the true relablty of AverageX? Soluton: True relablty s the squared correlaton between the measure and the object beng measured. You can estmate ths by calculatng the sample correlaton coeffcent f you have both quanttes. In ths case you do, snce you smulated the data yourself. So fnd the correlaton of AverageX wth F; these varables already exst n the fle work.measurement: proc corr data=work.measurement; var AverageX F; run; The output would gve you the estmated correlaton between the AverageX and F; square t to get the estmated relablty. 4. The followng output was obtaned from PROC CALIS. I show ths just to refresh you r memory. Do not use the output drectly n your answers.

5 Answer the followng questons genercally, rather than wth specfc reference to the numerc values shown n the output. Gve formulas and explanatons. 4.A. (5) What s the null hypothess tested by the ch-square statstc? Soluton: The null hypothess s H 0 : Σ=Σ(θ), e that the true covarance matrx obeys the restrctons mpled by the model. 4.B. (5) How are the degrees of freedom for the ch-square test obtaned? Soluton: Typcally df = p(p+1)/ {number of parameters n θ}, e, the number of free covarance parameters n the unrestrcted model mnus the number of model parameters used to determne the covarance matrx n the restrcted model. 4.C. (10) How are RMSEA and the ch-square statstc smlar? How are they dfferent? Soluton: Smlar: Both are related to the Ch Square statstc. RMSEA s a normalzaton of the ch-square statstc. Dfferent: The RMSEA s used to remove the dependency of the Chsquare statstc on the sample sze. We know that E(χ ) = df + N, where N s the sample sze and denotes dscrepancy between Σ and Σ(θ) (whch does not depend on N). RMSEA provdes an estmate of /df. 4.D. (10) How are GFI and RMR smlar? How are they dfferent? Soluton: Smlar: They both measure dscrepancy between the observed (unrestrcted) covarance matrx ( ˆΣ ) and the ftted (restrcted) covarance matrx ( Σ ( ˆ θ )). Dfferent: Smaller values of RMR ndcate less dscrepancy, whle larger GFI (closer to 1.0) ndcate less dscrepancy. Also, RMR s unweghted by the sze of the covarances (hence s an absolute measures) whle GFI s weghted by the sze of the covarances (hence s a relatve measure).