#9.19{a) First, we standardize themethod to obtain the solutionloadings is the following:data,ofuse maximumanalysis.andfactorlikelihoodThe factor.,0"", ,m=3Factor1 Factor2 Factor30.572 0.598 0.5260.542 0.391 0.7200.700 0.545 0.3541.000 ° °0.591 0.303 0.3340.147 0.989 °0.413 0.512 0.728xlx2x3x4xSx6x7xlx2x3x4xSx6x7m=2Factor1 Factor20.777 0.5720.798 0.5420.621 0.700° -1.0000.420 0.5910.606 '0.1470.895 0.4132.6520.3790.803cXo~,'~x\ss loadings 2.970Proportion Var 0.424Cumulative Var 0.424,-. , \" --;c , ! ;, : \ ,(b) After rotating thefollowing factor loadings:2.6522.1401.5620.3790.3060.2230.3790.6840.908..: \:; 0"' '-' ...,.i. \: .~"- : ;;, ; !- i'loadings, we can obtainfactor~m=2 m=3Factor1 Factor2"; .Factor1 Factor2cx1 0.'852 '""0.::452 ,\cdc" xY\".0.783 0.408x2 0.868 ':"'0.419 f('i:,)c.;: x2 ;',0.904 0.343x3 0.717 ",0.602 !,~1:1, x3 0.~38 0.575 0x4 0.146 ,0.989 :-.. x4 0.234 0.972 0 ". cx5 0.502 0.523 ;QO. x5 0.530 0.484 0.193 {',tC);;,,~~~:t"",,!!!,1i!\[x6 0.621 0 x6 0.292 0 0.951 ""C~I",'c,x7 0.945 0.277 x7 0.913 0.210 0.293ss loadings 3.548 2.074 3.091 1.847 1.416Proportion Var 0.507 0.296 0.442 0.264 0.202Cumulative Var 0.507 0.803 0.442 0.705 0.908Comparing the two loading matrix, we can find that the loadings ofthe first two factors when m=3 is similar to the loadings of thei " , ;: .-; ,1~~~~for if the former islarge,the latter istwo factors when m=2,also relatively large.When m=2, we can interpret the first factor as the sale ability ofthe sales staff, for the factor loadings of the variables toevaluate their performance is relatively large. It is noticed thatthe loading for mathematics test is also large. We can regard itas the potential ability of the sales staff. We can interpret thesecond factor as creativity ability of a sales staff, for theloading for creativity test is large. When m=3, the first twofactors have the same interpretation as the two factors when m=2.For the third factor, we interpret it as the ability of abstractreasoning, for the factor loading for abstract reasoning test islarge, compared with others.We list our result in the following:m=3communality specific variance0.9615067 0.038467760.9649114 0.035090350.912362 0.087619211 4e-0100.5520339 0.44794540.9999489 0.000051069530.9630143 0.03700579~r4~i=\'1~i ,10-J;~~m=2communality specific variancex10.9310435 0.06898365x20.9295869 0.07037659x30.8765966 0.1233964x41 4e-010x50.525576 0.474428x60.3884295 0.6116298x70.9706386 0.0293693, ' -, "; -c " ' 'c ,~4",-" ,~+i:!~Q","~~: ~~"' .~QI.~.J.~;f!!Q4AAi Ac ,"7~ "",,!,,'", c "c ' ,LL + \T$ ci"T ,,,: \,m=2 ;"j.x1 x2 x3 ;x4 x5 x6 x7,x1 1.0000 "0.92960.88350:57200.6645 ro.5547 0.9312x2 0.9296 1.0000 0.8749 0.5415 0.6551 0,.5627 0.9371x3 0.88350.8749 1".0000(0.70040.67490.47920.8449x4 0.57200.5415 0.70041.00000.5907 0.14690.4126x5 0.66450.65510.67490:5907 1.0000 0.34130.6197x6 0.55470.56270.47920.1469 0.3413 1.0001 0.6025x7 0.93120.93710.8449 0.41260.61970.6025 1.0000-.~",~':!~:.:n=3x1 x2 x3 x4 x5 x6 x7x1 1.00000.9226 0.91240.57200.69460.6744 0.9256x2 0.92261.0000 0.84710.54150.67860.4654 0.9483x3 0.91240.8471 1.00000.70040.6968 0.64110.8257x4 0.57200.5415 0.7004 1.00000.5907 0.14690.4126x5 0.69460.67860.69680.5907 1.00000.3859 0.64192x6 0.67440.46540.6411 0.14690.3859 1.00000.5664x7 0.92560.94830.82570.4126 0.64190.5664 1.0000From the result, I prefer to choose m=3, because the residuals isless, and three factors can explain more variance.(d) We test whether our factor model is sufficient to explain thevariance. When m=2, from the SPLUS output we can find that thetest statistic is 117.09, associated with the degree of freedom of8, and the p-value is 0. So we should reject the null hypothesisat level 0.01.When m=3, we can find the test statistic from the SPLUS output. Itis 61.52, associated with the degree of freedom of 3, and the p-value is 2.78e-013. So we should reject the null hypothesis atlevelO.01.From the test, we can find that neither m=2 nor m=3 is enough todescribe the data. However, we cannot choose a larger m, for itwill give a negative degree of freedom. So we choose m=3.(e) Firstly, standardize x, and we get the standardized x is:(1.5215 -0.85160.46470.95691.1286 0.6730.4973)Then the predicted factor score isFactor30.70050.7024Factorl Factor2Weighted least squares method -0.3399 1.0795Regression method -0.3252 1.086ct':e~~~!$ ~~'11!::,' ~. :"l'~' ,~:;", ' 1i~ i,.--,.:- (}"'~~:;"'jlfu~l: ,", :~b~:,,;~", "-;::.:,:f-.:.:;~c e : '" ~-, c,.We use the maximum likelihood method to perform theanalysis. .Choose m~2, afterrotation; the loading matrix isFactorl Factor2loom 0.437 0.851200m 0.372 0.928400m 0.571 0.694800m 0.784 0.4661500m 0.915 0.3863000 0 887 0 409 .'" .., cm ..J,-'..j, ;-,., ..:.", .; :,\.-M th 0 786 0 423 "c" .--, ., "ara on. .~ :" c, ..i ,," -' c: .;:,~; .",.t~'-[t\4:5~; .~:"C;':;.~.:,,'~~~~~~~3~ss loadingsProportion VarCumulative Var3.5110.5020.5022.7810.3970.899; '0" ,. cJ'OC\-,"';'*!(')t')-dl=0" ;I(?ic'-(\I<\1000O"0~"0c8Q)(J)o0.'"~~"'0°cx>°°qp0""oooo011000'I;:;ii-""~.~000.0 e 0t61iOj;.)1i ,~ Cc- :;; ,-",,) '1 -!,J;.;);;8 -, ..r" U r- 1--'} t-C --, --,-0,~~g£0 O 0 'C';",:'.!,000<X>o0 000;c " 0 ; ; I I!-1 O 1 2 3First Factor, ~,;;' : -"We can calculate the factor scores, and plot the scores" -'.factor against the scores of the second factor. Obviously,are three outliers, namely, Cook Islands, Western Samoa,Democratic People's Republic of Korea.....~~and#9.29We use the maximum likelihood method to ,perform theanalysis. Choose m=2, after rotation, the loading matrix isfactor~~~~~4Factor1 Factor20.445 0.8490.388 0.9220.587 0.6870.810 0.4500.909 0.3990.879 0.4110.751 0.458lOOm2OOm4OOm8OOmlSOOm3OOOmMarathonFactor1 Factor2SS loadings 3.5111722.7824991Proportion Var 0.5015960.3974999Cumulative Var 0.5015960.8990959~~l::r ,:;;.;1;1'From the loadings matrix, we