cong thuc kt luong
DESCRIPTION
Cong Thuc Kinh Te LuongTRANSCRIPT
Trn Th Tuyt L - 0061
Trn Th Tuyt L - 00612010
BI TONHAI BINA BIN
1. Tnh
n = s mu
(Khuyn nn tnh ngay u bi dng dn, lc ny u c cn sng sut tnh ton ^_^ )
2. Xc nh PRF
3. Xc nh SRF
( SRF:
4. ngha ca cc h s hi quy
(ni ngha ca bin no th c nh cc bin cn li)V d ni ngha ca th c nh cc bin X2, X3,
X2 khng i, nu
X2Tng t cho cc bin cn li
5. Tng cc bnh phngTSS = 3 gi trESS = ny > 0RSS = TSS ESS TSS = ESS = RSS = TSS ESS
6. Tnh h s xc nh
7. H s xc nh hiu chnh c th m, trong trng hp ny, quy c Vi k l s tham s ca m hnhVd: (SRF) ( m hnh 3 bin
( k = 3, vi cc tham s Y, X1, X2
8. c lng ca Ci ny s tra bng kt qu ra
( dng S.E. of regression ( ct Std. Error, dng th 1 ( ct Std. Error, dng th 2
( ct Std. Error, dng th 3 .
9. Kim nh s ph hp m hnh SRF, mc ngha Phng php gi tr ti hn:
B1: Lp gi thit Ho: =0 ; H1: 0
B2: tra bng F, gi tr ti hn
B3: so snh F0 v F(1,n-2)
+ F0 > F(1,n-2): bc b H0 ( hm SRF ph hp vi mu + F0 < F(1,n-2): chp nhn H0F(1,n-2)
F(1,n-2)
Bc b
Chp nhn
F0 Phng php gi tr ti hn:
B1: Lp gi thit Ho: R2=0 ; H1: R2>0
B2: tra bng F, gi tr ti hn
B3: so snh F0 v F(k-1,n-k)
+ F0 > F(k-1,n-k): bc b H0 ( hm SRF ph hp vi mu + F0 < F(k-1,n-k): chp nhn H0F(k-1,n-k)
F(k-1,n-k)
Bc b
Chp nhn
F0
Phng php gi tr p-value:
(cch ny s lm khi cho sn bng kt qu)
Ly gi tr p-value ng vi F0 ( cui cng gc phi ch Prod(F-statistic))
Tin hnh so snh p-value v :
+ p-value < : bc b H0 ( hm SRF ph hp vi mu+ p-value > : chp nhn H0p-value
p-value
Bc b
Chp nhn
Phng php gi tr p-value:
(cch ny s lm khi cho sn bng kt qu)
Ly gi tr p-value ng vi F0 ( cui cng gc phi ch Prod(F-statistic))
Tin hnh so snh p-value v :
+ p-value < : bc b H0 ( hm SRF ph hp vi mu+ p-value > : chp nhn H0p-value
p-value
Bc b
Chp nhn
10. Kim nh gi thit bin c lp c nh hng ln bin ph thuc khng?Gi thit: H0: = 0 H1: 0
Phng php gi tr ti hn:
B1: Tnh: B2: Tra bng t-student gi tr B3: So snh v + > : bc b H0 ( bin c lp (X) nh hng ln bin ph thuc (Y)
+ < : chp nhn H0
Bc b
Chp nhn
Gi thit: H0: = 0 H1: 0
Phng php gi tr ti hn:
B1: Tnh: B2: Tra bng t-student gi tr B3: So snh v + > : bc b H0 ( bin c lp (X) nh hng ln bin ph thuc (Y) + < : chp nhn H0
Bc b
Chp nhn
Phng php p-value:Ly gi tr p-value tng ng vi bin c lp mnh ang xt
Tin hnh so snh p-value v :
+ p-value < : bc b H0 ( bin c lp (X) nh hng ln bin ph thuc (Y)+ p-value > : chp nhn H0p-value
p-value
Bc b
Chp nhn
Phng php p-value:Ly gi tr p-value tng ng vi bin c lp mnh ang xt
Tin hnh so snh p-value v :
+ p-value < : bc b H0 ( bin c lp (X) nh hng ln bin ph thuc (Y)
+ p-value > : chp nhn H0p-value
p-value
Bc b
Chp nhn
11. Kim nh gi thit
Ho: = o ; H1: oVi mc ngha Phng php gi tr ti hn:
B1: Tnh: B2: Tra bng t-student gi tr B3: So snh v + > : bc b H0
+ < : chp nhn H0 ( c th xem = o
Bc b
Chp nhn
Phng php gi tr ti hn:
B1: Tnh: B2: Tra bng t-student gi tr B3: So snh v + > : bc b H0
+ < : chp nhn H0 ( c th xem = o
Bc b
Chp nhn
Phng php p-value:Ly gi tr p-value tng ng vi bin c lp mnh ang xt
Tin hnh so snh p-value v :
+ p-value < : bc b H0+ p-value > : chp nhn H0 ( c th xem = op-value
p-value
Bc b
Chp nhn
Phng php p-value:Ly gi tr p-value tng ng vi bin c lp mnh ang xt
Tin hnh so snh p-value v :
+ p-value < : bc b H0+ p-value > : chp nhn H0 ( c th xem = op-value
p-value
Bc b
Chp nhn
12. Xc nh khong tin cy ca
Vi mc ngha ( ko cho th ly =0,05)Tra bng t-student gi tr Tnh Khong tin cy ca :
Tra bng t-student gi tr Tnh tra bng kt qu
Khong tin cy ca :
13. Xc nh khong tin cy ca
Vi mc ngha ( ko cho th ly =0,05)Tra bng t-student gi tr Tnh Khong tin cy ca :
Tra bng t-student gi tr Tnh tra bng kt quKhong tin cy ca :
14. Xc nh khong tin cy ca phng sai var(Ui) = 2Vi tin cy (1 ) tin cy: 1 = a%
( = 100% - a%
Tra bng Chi-square cc gi tr:
Khong tin cy ca (2:
tin cy: 1 = a%
( = 100% - a%
Tra bng Chi-square cc gi tr:
Khong tin cy ca (2:
15. Kim nh gi thit
Ho: =o ; H1: oVi mc ngha Phng php gi tr ti hn
B1: Tnh
B2: So snh
+ < < chp nhn Ho, =o+ bc b Ho+ < bc b Ho
Bc b
Chp nhn
Bc b Phng php gi tr ti hn
B1: Tnh
B2: So snh
+ < < chp nhn Ho, =o+ bc b Ho+ < bc b Ho
Bc b
Chp nhn
Bc b
Phng php gi tr p-value
B1: Ly gi tr p-value trong bng kt qu
B2: So snh
+ < p-value < 1- ( chp nhn Ho, =o+ p-value < ( bc b Ho+ 1- < p-value ( bc b Ho
p-valuep-valuep-valueBc b
Chp nhn
Bc b
Phng php gi tr p-value
B1: Ly gi tr p-value trong bng kt qu
B2: So snh
+ < p-value < 1- ( chp nhn Ho, =o+ p-value < ( bc b Ho+ 1- < p-value ( bc b Ho
p-valuep-valuep-valueBc b
Chp nhn
Bc b
16. H s co gin, nghaEYX = Nu X(vd: thu nhp) tng 1% th Y (vd: chi tiu) tng EYX%
17. i n vTrong :
k1 : h s t l quy i gia n v c & mi ca Y
k2 : h s t l quy i gia n v c & mi ca X
= k1 = Trong :
ko : h s t l quy i gia n v c & mi ca Y
k1 : h s t l quy i gia n v c & mi ca X1
k2 : h s t l quy i gia n v c & mi ca X2= ko = =
18. D on (d bo) im
Dng???Khi cho Xo yu cu tnh YThay gi tr Xo vo phng trnh SRF:
D bo cho hi quy nhiu bin ch xt d bo im.
Thay gi tr , vo phng trnh SRF:
19. D on ( d bo) khongD on ( d bo) gi tr c bit
Dng???
Khi cho Xo v tin cy (1 ), yu cu c lng gi tr.
Thay gi tr Xo vo phng trnh SRF:
var() = var(Yo - = se() = Khong tin cy (1-)% ca Yo/Xo l:
D on (d bo) gi tr trung bnh
Dng???
Khi yu cu d on m khng cho tin cy (1 )
Khi cho Xo v tin cy (1 ), yu cu c lng gi tr trung bnh.Thay gi tr Xo vo phng trnh SRF:
var( = se() = Khong tin cy (1-)% ca E(Yo/Xo) l:
20. So snh R2Ch so snh c khi tha 3 iu kin sau:
1. Cng c mu n.2. Cng s bin c lp.
(nu ko cng s bin c lp th dng )
3. Cng dng hm bin ph thucCh so snh c khi tha 3 iu kin sau:
1. Cng c mu n.2. Cng s bin c lp.(nu ko cng s bin c lp th dng ) 3. Cng dng hm bin ph thuc
21. Thm bin vo m hnh, vi mc ngha B1: tnh R2 (3 bin) ; (3 bin) ; R2 (2 bin) ; (2 bin)
B2: So snh (3 bin) v (2 bin) Nu (3 bin) < (2 bin): khng thm bin vo m hnh
Nu (3 bin) > (2 bin): c th thm bin vo m hnh, cn lm thm cng vic sau: kim nh bin thm vo c ngha ko, sau mi chc chn c thm bin vo ko?CNG VIC KIM NH THC HIN GING CNG THC S 10
( NGHA H S HI QUY V H S CO GIN CA CC M HNH1. M hnh tuyn tinh:
Y = + *X
ngha h s hi quy: Nu X tng 1 n v th Y tng n v (Vi iu kin cc yu t khc khng i)
EYX =
, ta tnh lc u
ngha h s co gin: Nu X tng ln 1% th Y tng ln EYX%2. M hnh lin-log:
Y = + *logX
ngha h s hi quy: Nu X tng ln 1% th Y tng ln n v (Vi iu kin cc yu t khc khng i)
EYX =
ngha h s co gin: Nu X tng ln 1% th Y tng ln EYX%3. M hnh log-lin:logY = + *X ngha h s hi quy: Nu X tng ln 1 n v th Y tng ln % (Vi iu kin cc yu t khc khng i)
EYX = = ngha h s co gin: Nu X tng ln 1% th Y tng ln EYX% 4. M hnh tuyn tnh log:logY = + *logX
ngha h s hi quy: Nu X tng 1% th Y tng % (Vi iu kin cc yu t khc khng i)EYX = = ngha h s co gin: Nu X tng ln 1% th Y tng ln EYX% 5. M hnh nghch o:Y = + * ngha h s hi quy: X tng ln th Y cng tng ln theo, nhng Y i a l n v (Vi iu kin cc yu t khc khng i)
EYX =
ngha h s co gin: Nu X tng ln 1% th Y tng ln EYX%( TRNH BY KT HI QUY
= ;n = ???
se = ;R2 = ???
t = t(t(;Fo = ???
TSS = ??? ; ESS = ??? ; RSS = ??? ; = ???
( C BNG KT QU HI QUY Const t p-valueVariableCoefficientStd. Errort-StatisticProb.
C ( 14.321681.11628312.829790.0001
X1 ( -2.2587410.320460-7.0484380.0009
X2 ( 1.2377620.3425863.6129970.0153
R-squared ( R20.909573Mean dependent var ( 9.000000
Adjusted R-squared ( 0.873402S.D.dependent var ( SY2.878492
S.E. of regression ( 1.024183
Sum squared resid ( RSS5.244755
F-statistic ( Fo25.14667
Prob(F-statistic) ( p-value(Fo)0.002459
( THAY I S HNG DC V S HNG TUNG GC KHI NO??? (cu ny c th chim 1)1. Thay i s hng h s gc (s hng gc) khi thm D vo 2. Thay i s hng tung gc khi thm D vo
Ta c 3 trng hp nh sau:
phi gii ma trn, nhng iu ny ko phi lo
NHN XT:
Lm sao nh ht cng thc???? QUOTE Hc cng thc hm a bin thui, nh ci k ca cng thc ci ny chnh l s tham s ca phng trnh. ( Vy l hm 2 bin thay k=2, hm 3 bin thay k=3, . (tha l xong phn cng thc *_^)
Luyn tp nh th no???? ( n ti dng no th xem cng thc cho chc (tha l oki ri ^_^)
MO:
Cch ni ngha h s hi quy:
a.1 Tham s no c log th n v l %, cn li th dng n v bi cho
a.2 Tham s X c log, Y ko log th ni ngha ca Y nh h s l QUOTE
a.3 Tham s X ko log, Y c log th ni ngha ca Y nh h s l QUOTE
H s co gin EYX: t cng thc gc EYX = QUOTE , tham s no c log th gi tr trung bnh ca tham s = 1