Lösung zu Übungsblatt 4

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Lösung zu Übungsblatt 4
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sphericity:
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1-203.40
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:t 1-. 1-. 1- Erkl irungsgehal t de r vari abl en
mat MEig=g(11 /* speichern der rigenvektoren (gamma)*/
mat MEV=€(rv)
/*
spe'i che
rn
de
r
Ei genwe
rte (l ambda) */
sum 1 ength
Variable
length
I
0bs
I
200
Mean std.
21,4
,896
Dev.
.37 65542
seite
2
Min
213.8
0. 0000
0.0000
MAX
2l_6.
3
Bank. I og
. scalar lvar=r(var) /* speichern der Varianz der Ldnge*/
.
scal
ar
rho=MEi g [1, 1] * (rqrv [1, 1]
. display rho
=(25
-.201_36085
/lvar) n(L/2) /*eerechnung
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.
,
.
.
.
.
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saved)
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= 0,*4r,n
L.L.2 Sc reepl ot
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(file scree.gph
von rho(xl, ,yL)"
^,^^
u^i{
4rttc\'^:'--\
&
!,,l.an r,i,ntc,, vutll
* 1-. l-. 3 Test-Stati sti cs
mat UMat=l(5,1-,0)
mat pMat=J (5,1,0)
local lnbar=MEV[1,6]
local sumln=ln(urv[1,6])
forvalues k=2/6 {
local
2.
O*r-t,-r,^^ o\
k' ) ) o (exp ('1 npar ')+(L/ (' (.' :1) ) *MEV [1, 6- k'+1-] ) )
I nbar=l n ( ( (' k ' - t) / ('-suml
Ug .^^^]
n=' suinl n '+l n (tttEv [1, 6-' k'+1] )
I ochl
3.
mat UMat[-k'-1,1]=(200-(2't6+11)l6)o('k'*'lnbar'-'sumln' ;}
4.
5.
ptun
X)
-1) t' (' k'+2)
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t-1,
=
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1]
mat
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=chi
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7.
r
Jw.1
)
. matlist uMat
I
r1
r2
r3
r4
r5
cl-
910. 8444
973.9544
996. 9272
1067. 311430.L77
luR.
. matlist pMat
I
c1
r1
1-.6e-1-98
r4
r5
5.6e-21-9
3 .8e-291-
r2
r3
2 .6e-208
7 .9e-209
I
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scheme(s2mono)
ci
mean
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of a correlation matrix)
iigenvalues
. oraoh save scree2, rePlace
(file' scree2.gph saved)
- o t.Z.l vorhersaqe der lauptkomponenten
. toiainqpiot, sih6me(s2mono) factors(6) combined
. qraph save loadinq, rePlace
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, scoreplot, scheme(s2mono) factors(6) combined
. qraph save score, rePlace
(file score.gph saved)
.
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- cl ose
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: text
cloied'bn: 18 lun 20i-l- ' 14:06:28
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seite
5
theory-based confidence
interval of

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