Cong Thuc KT Luong

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))


+ 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


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


+ 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



+ 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


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



+ p-value < : bc b H0+ p-value > : chp nhn H0 ( c th xem = op-value

p-value

Bc b

Chp nhn



+ 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

Documents

Cong Thuc KT Luong