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Chng 7: M hnh hai bin: Kim nh gi thit

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1

CHNG 7: M HNH HAI BIN: KIM NH GI THUYT

CHNG 7

M HNH HAI BIN:

KIM NH GI THUYT

@ Phng Thanh Bnh, MB

Ta bit, l thuyt c in v suy lun thng k gm hai nhnh c mi quan h cht ch vi nhau; l, c lng v kim nh gi thuyt. Cho n y, v c bn ta hon thnh vic c lng cc tham s ca m hnh hi qui tuyn tnh theo phng php bnh phng b nht thng thng. Di cc gi nh m hnh hi qui tuyn tnh c in ta thy rng cc c lng ca cc tham s trong m hnh hi qui tuyn tnh: b1, b2, v

2

s

tha mn cc tnh cht thng k nh mong mun nh tuyn tnh, khng chch, phng sai b nht, ... Tuy nhin, nu dng li vic c lng th ta cha tr li c cu hi quan trng na l cc c lng ny gn vi cc gi tr thc B1, B2, v (2 nh th no. y l ni dung ch yu s c cp trong Chng 7. Chng 6 cng ch ra rng, cc c lng ny c gi tr thay i t mu ny qua mu khc, nn chng l cc bin ngu nhin. Nh vy, theo C th, chng ny s tp trung vo cc ni dung sau y:

ngha phn phi xc sut ca hng nhiu ui

Ti sao gi nh hng nhiu ui c phn phi chun

Tnh cht ca cc c lng OLS di gi nh phn phi chun

c lng khong ca cc c lng OLS

Khong tin cy cho cc h s hi qui B1 v B2

Khong tin cy cho phng sai nhiu (2

Kim nh gi thit v h s hi qui

Kim nh v phng sai nhiu

Gi tr xc sut p

nh gi kt qu hi qui

ng dng d bo

Nghin cu ng dng

TI SAO PHN PHI XC SUT CA HNG NHIU LI QUAN TRNG TRONG KIM NH GI THUYT?

Chng 6 ch ra rng cc c lng OLS l cc hm theo Yi, v Yi li l mt hm theo ui. Chnh v th, bit phn phi xc sut ca cc c lng OLS ta cn bit phn phi xc sut ca ui. Cc phng trnh (6.29) v (6.30) cho thy:

-

=

i

i

1

1

u

k

X

B

b

(6.30)

+

=

i

i

2

2

u

k

B

b

(6.29)

Ta bit rng ki, B1, v B2 u phi ngu nhin hay c nh, nn cc c lng b1 v b2 l mt hm tuyn tnh theo bin ngu nhin ui. V th, phn phi xc sut ca b1 v b2 s ph thuc vo gi nh v phn phi xc sut ca ui. Vic bit phn phi xc sut ca cc c lng OLS rt cn thit c th rt ra cc suy lun thng k v cc gi tr thc ca tng th, nn bn cht ca phn phi xc sut ca hng nhiu ui c cho l c mt vai tr ht sc quan trng trong kim nh gi thuyt.

HNG NHIU C GI NH THEO PHN PHI CHUN

T cc gi nh ca m hnh hi qui tuyn tnh c in cho thy hng nhiu ui c gi nh c phn phi chun nh sau:

Trung bnh: E(ui) = 0

(6.35)

Phng sai: E[(ui E(ui)]2 = E(ui)2 = (2

(6.36)

cov(ui,uj) = E{[(ui E(ui)][uj E(uj)]} = E(uiuj) = 0

(6.38)

Ni cch khc:

ui ~ N(0, (2)

(7.1)

To sao gi nh ui c phn phi chun?

1. Ta bit rng, ui c xem nh mt bin i din nh hng kt hp ca rt nhiu bin gii thch khng c a vo m hnh. Nh bit, nh hng ca mi bin khng c a vo vo m hnh l rt nh v ngu nhin. Theo nh l gii hn trung tm, nu c rt nhiu bin ngu nhin c phn phi c lp v ging nhau, th tng ca chng c phn phi chun.

2. Theo Gujarati (2003), mt bin dng ca nh l gii hn trung tm cho rng thm ch khi s bin khng phi l rt ln hoc cc bin ny khng hon ton c lp, th tng ca chng vn theo phn phi chun.

3. Vi gi nh phn phi chun, cc phn phi xc sut ca cc c lng OLS c th c suy ra mt cch d dng v cc c lng OLS l cc hm tuyn tn ca ui. Nu ui c phn phi chun th cc c lng OLS cng phn phi chun, nn vic kim nh gi thuyt tr nn d dng hn. Nh vy, nu mc ch s dng kt qu hi qui cho suy lun thng k th gi nh hng nhiu c phn phi chun c ngha ht sc quan trng.

4. Phn phi chun l mt phn phi tng i n gin ch c hai tham s l gi tr trung bnh v phng sai; v phn phi chun l mt phn phi rt ph bin c nghin cu rt nhiu trong thng k ton. Ngoi ra, rt nhiu hin tng kinh t x hi theo phn phi chun.

5. Khi ta lm vic vi cc mu nh th gi nh phn phi chun c vai tr ht sc quan trng v n khng ch gip ta suy ra phn phi xc sut ca cc c lng OLS m cn gip ta c th d dng s dng cc kim nh thng k t, F v (2. Ngoi ra, ta cng bit rng cc phn phi t, F, v (2 c mi quan h rt gn vi phn phi chun. c bit, cc phn phi ny c s dng rt nhiu trong cc kim nh gi thuyt trong kinh t lng.

TNH CHT CA CC C LNG OLS DI GI NH PHN PHI CHUN

Vi gi nh hng nhiu ui theo phn phi chun, ta d dng suy ra rng cc c lng OLS cng theo phn phi chun nh vo tnh cht ca phn phi chun cho rng mt tp hp tuyn tnh ca mt hay nhiu bin c phn phi chun s theo phn phi chun (Chng 3). Nh vy, ta c th khng nh li rng cc c lng OLS c cc tnh cht nh sau:

1. L cc c lng khng chch

2. L cc c lng c phng sai b nht

3. L cc c lng nht qun

4. b1 l mt hm tuyn tnh ca ui v c phn phi chun nh sau:

Trung bnh: E(b1) = B1

(6.40)

Phng sai: Var(b1) =

2

2

i

2

i

2

1

b

x

n

X

s

=

s

(6.47)

Hay c th vit li nh sau:

)

,

B

(

N

~

b

2

1

b

1

1

s

(7.2)

Theo tnh cht ca phn phi chun ta c th d dng chuyn b1 thnh mt bin chun ha Z nh sau:

1

b

1

1

B

b

Z

s

-

=

~ N(0,1)

(7.3)

5. b2 l mt hm tuyn tnh ca ui v c phn phi chun nh sau:

Trung bnh: E(b2) = B2

(6.41)

Phng sai: Var(b2) =

s

=

s

2

i

2

2

2

b

x

(6.44)

Hay c th vit li nh sau:

)

,

B

(

N

~

b

2

2

b

2

2

s

(7.4)

Theo tnh cht ca phn phi chun ta c th d dng chuyn b2 thnh mt bin chun ha Z nh sau:

2

b

2

2

B

b

Z

s

-

=

~ N(0,1)

(7.5)

*

2

B

Hnh 7.1: Phn phi xc sut ca b

1

v b

2

)

b

(

se

t

b

2

2

/

2

a

-

Hnh 7.2: Phn phi xc sut bin chun ha Z ca b

1

v b

2

6. T phng trnh (6.66):

(

)

-

=

2

i

2

i

i

2

i

2

i

x

y

x

y

e

v (6.67):

2

n

e

2

i

-

=

s

, ta thy

2

s

l mt hm tuyn tnh theo

2

i

e

, m

2

i

e

lm mt hm tuyn tnh theo

2

i

y

, v

2

i

y


2

i

u

. Cho nn,

)

/

)(

2

n

(

2

2

s

s

-


2

i

u

. Do ui c phn phi chun, nn theo tnh cht ca phn phi (2 c trnh by chng 3, th

)

/

)(

2

n

(

2

2

s

s

-

s c phn phi theo (2 vi s bc t do d.f. bng (n-2). Bit c phn phi ca

2

s

s gip ta rt ra cc suy lun v gi tr thc ca (2 t

2

s

.

7. (b1, b2) c phn phi c lp vi

2

s

. Tm quan trng ca tnh cht ny s c gii thch chng sau.

8. b1 v b2 c phng sai b nht trong s tt c cc c lng khng chch ca B1 v B2. Chnh v vy cc c lng OLS c gi l cc c lng khng chch tt nht.

Nh vy, vi gi nh hng nhiu ui c phn phi chun ui ~ N(0, (2), ta c th suy ra rng Yi, mt hm tuyn tnh ca ui, s theo phn phi chun vi:

E(Yi) = B1 + B2Xi

(7.6)

Var(Yi) = (2

(7.7)

Hay ta c th vit li nh sau:

Yi ~ N(B1+B2Xi, (2)

(7.8)

C LNG KHONG TIN CY

Nh cp chng 6, c lng khong c th c xem l cu ni gia c lng cc tham s tng th v kim nh gi thuyt. Quay li v d minh ha v hm cu nc ng chai Aquafina ta nhn thy h s b2 = -2.91 ch l mt gi tr c lng im ca gi tr thc nhng khng bit v tham s tng th B2. R rng rng nu ch da vo mt gi tr c lng nh vy th c th d b nghi vn v chng ta khng bit chc gi tr c lng ny ng tin cy nh th no. Nh Chng 4 cp, bi v c s doan ng mu; tc gi tr c lng s thay i t mu ny qua mu khc v him khi chng ging nhau, nn ch mt gi tr duy nht th c th s khc bit so vi gi tr thc, mc d bit rng nu ly mu lp i lp li th gi tr trung bnh ca cc gi tr c lng c k vng s bng gi tr thc tng th [E(b2) = B2]. Gujarati (2003) cho rng trong thng k s tinh cy ca mt c lng im c o bng sai s chun ca n. V th, thay v ch da v gi tr c lng im, ta c th xy dng mt khong xoay quanh c lng im. y l tng c bn ca c lng khong.

c th hn, gi s rng ta mun xem b2 gn vi B2 nh th no. Nh vy, c ngha l ta s tm hai s dng ( v (, vi 0 < ( < 1, sao cho xc sut khong ngu nhin (b2 - (, b2 + () cha gi tr thc B2 l 1- (.

Pr(b2 - ( ( B2 ( b2 + () = 1-(

(7.8)

Khong ngu nhin ny c gi l khong tin cy; 1-( c gi l h s tin cy; v ( c gi l mc ngha. Cc gi tr gii hn ca khong ngu nhin ny c gi l cc gii hn tin cy hay gi tr ph phn; trong , b2 - ( l gii hn chn di v b2 + ( l gii hn chn trn.

Phng trnh (7.8) c gi l c lng khong v c lng khong ny a ra mt khong cc gi tr trong c th cha gi tr thc B2. c lng khong c cc tnh chc nh sau:

1. Phng trnh (7.8) khng ni rng xc sut B2 nm gia cc gi tr gii hn l 1- (. V B2, mc d khng bit, nhng c cho l c mt gi tr nh no , c th nm trong hoc ngoi khong . Phng trnh (7.8) ch cho rng, xc sut ca vic xy dng mt khong tin cy c cha B2 l 1-(.

2. Khong (7.8) l mt khong ngu nhin; ngha l, gi tr ca n s thay i t mu ny qua mu khc v n ph thuc vo b2, m b2 l mt bin ngu nhin.

3. V khong tin cy l ngu nhin, nn cc pht biu xc sut i km nn c hiu theo kiu di hi; ngha l, s mu c lp i lp li nhiu ln. Ni cch khc, nu s mu c lp i lp li nhiu ln, th, trung bnh, cc khong nh th s c 1-( trng hp c cha gi tr thc B2. V d, ta ly 100 mu nh th th c khong 95 khong tin cy c cha gi tr thc B2 nu ( bng 5%.

4. Lu rng, chng no b2 cha bit th khong tin cy (7.8) l ngu nhin. Nhng khi ta c mt mu c th v c mt gi tr b2 nht nh th khong (7.8) khng cn ngu nhin na, m l mt khong c nh. Trong trng hp ny, ta khng th pht biu rng xc sut mt khong tin cy c th cha B2 l 1-( v B2 nm trong hoc ngon khong c nh .

Vn t ra by gi l cc khong tin cy c xy dng nh th no? Nu ta bit phn phi xc sut hoc phn phi mu ca cc c lng OLS, th vic xy dng khong tin cy nh th s tr nn d dng. Nh phn tch trn, do ui c phn phi chun nn cc c lng OLS b1 v b2 theo phn phi chun v c lng OLS

2

s

theo phn phi (2. Trn c s ny, ta s xy dng khong tin cy cho B1, B2 v (2 nh sau.

KHONG TIN CY CHO CC H S HI QUI B1 V B2

Khong tin cy cho B2

Trn c s gi thuyt hng nhiu ui c phn phi chun, th bn thn cc c lng OLS b1 v b2 cng c phn phi chun vi trung bnh v phng sai nh phng trnh (7.2) v (7.4). Nh vy, theo tnh cht ca phn phi chun th bin

s

-

=

-

=

2

i

2

2

2

2

2

x

)

B

(b

)

b

(

se

B

b

Z

(7.9)

l mt bin chun ha v c phn phi chun vi trung bnh bng khng v phng sai bng mt. V th, chng ta c th s dng phn phi chun suy lun v B2 nu nh bit phng sai thc ca tng th

2

s

. Nu bit

2

s

, th vic suy lun thng k s tr nn r rng theo ng tnh cht ca phn phi chun. Nhng thc t him khi ta bit

2

s

, v ta ch c c lng khng chch ca n l

2

s

. Nu ta thay

2

s

bng

2

s

, th phng trnh (7.9) s c vit li nh sau:

s

-

=

-

=

x

)

B

(b

)

b

(

se

B

b

t

2

i

2

2

2

2

2

(7.10)

V ta cng bit rng, t s theo phn phi t vi s bc t do l n-k ( y l n-2). Thay v s dng phn phi chun, ta c th s dng phn phi t xy dng khong tin cy cho B2 nh sau:

Pr(-t(/2 ( t ( t(/2) = 1-(

(7.11)

Trong , gi tr t trong ngoc l gi tr t tnh ton bi cng thc (7.10) v t(/2 l bin t c c t phn phi t vi mc ngha l (/2 v n-2 bc t do; gi tr ny thng c gi l gi tr t ph phn mc ngha (/2. Th phng trnh (7.10) vo (7.11), ta c

a

-

=

-

-

a

a

1

t

)

b

(

se

B

b

t

Pr

2

/

2

2

2

2

/

(7.12)

Sp xp li phng trnh (7.12) ta c

a

-

=

+

-

a

a

1

)]

b

(

se

t

b

B

)

b

(

se

t

b

Pr[

2

2

/

2

2

2

2

/

2

(7.13)

Phng trnh (7.13) a ra mt khong tin cy 100(1-()% cho B2, v cng thc ny c th c vit li nh sau:

khong tin cy 100(1-()% cho B2 l

b2 ( t(/2se(b2)

(7.14)

Khong tin cy cho B1

Tng t nh B2, ta d dng xy dng khong tin cy cho B1 nh sau:

a

-

=

+

-

a

a

1

)]

b

(

se

t

b

B

)

b

(

se

t

b

Pr[

1

2

/

1

1

1

2

/

1

(7.14)

Phng trnh (7.14) a ra mt khong tin cy 100(1-()% cho B1, v cng thc ny c th c vit li nh sau:

khong tin cy 100(1-()% cho B2 l

b1 ( t(/2se(b1)

(7.15)

T hai phng trnh (7.13) v (7.15) ta nhn thy rng rng hay hp ca khong tin cy t l thun vi sai s chun ca c lng. Ngha l, sai s chun ca c lng cng ln th khong tin cy cng ln. Ni cch khc, sai s chun ca c lng cng ln th s khng chc chn ca vic c lng gi tr thc ca tham s tng th cng cao. Nh vy, sai s chun ca c lng thng c m t nh mt thc o s chnh xc ca c lng.

Tr li v d minh ha v ng cu nc ng chai Aquafina Chng 6, ta thy rng b1 = 54.8, se(b1) = 1.55 v b2 = -2.909, se(b2) = 0.25, v d.f. = 8. Gi s mc ngha ( = 5%; ngha l, khong tin cy 95%, t bng phn phi t (hoc tra bng hm =TINV(5%,8)) ta c gi tr t(/2 ph phn = t0.025 = 2.306. Thay cac gia tri nay vao cc cng thc (7.12) v (7.14), ta c cc khong tin cy 95% cho B2 v B1 nh sau:

(-3.486 ( B2 ( -2.333)

(7.16)

(51.226 ( B1 ( 58.374)

(7.17)

Ta gii thch khong tin cy cho B2 v B1 nh sau: Vi khong tin cy 95%, trong di hn, th 95 trong s 100 trng hp cc khong tin cy nh (7.16) v (7.17) s cha gi tr thc B2 v B1. Mt ln na, qu v cn lu rng ta khng th ni xc sut 95% mt khong tin cy nht nh, v d (-3.486, -2.333) cha gi tr thc B2 bi v by gi khong tin cy ny l c nh v khng cn ngu nhin na; v th, B2 c th nm trong hoc nm ngoi khong (-3.486, -2.333). iu ny c ngha l xc sut m mt khong tin cy nht nh c cha gi tr thc B2 c th l 1 hoc 0.

Thao tc xc nh khong tin cy trn Excel

Nu c lng trn Excel th trong bo co kt qu hi qui Excel c sn thng tin v khong tin cy mc nh 95% (hoc ta c th chn mt khong tin cy khc ngoi khong mc nh). Nhc li rng, sau khi m tp tin d liu Excel, ta thc hin nh sau:

1. Tools/Data Analysis

2. Chn Regression, v thy xut hin hp thoi nh sau:

3. Chn khi d liu ca bin Y (Input Y Range) v bin X (Input X Range), nu khi d liu c chn c c tiu ca bin Y v X th ta chn Labels, ri chn v tr cho kt qu c lng, chn OK, ta c kt qu nh sau:

( BNG 7.1: Kt qu hi qui Excel

Ngun: Tc gi, 2008

Thao tc vi Eviews

xy dng khong tin cy trn Eviews, trn ca s lnh ca Eviews qu v c th lm tng bc sau y:

Tnh gi tr t ph phn t(/2: scalar tc=@qtdist(1-(/2,d.f.)

Tnh sai s chun ca h s se(b2): scalar seb2=@stderrs(2)

Tnh gi tr chn di ca b2: scalar b2_lb=c(2)-tc*seb2

Tnh gi tr chn trn ca b2: scalar b2_ub=c(2)+tc*seb2

Khi quen cc lnh ny, qu v cng c th kt hp thc hin nhanh nh sau (v d mc ngha l 5%):

Tnh gi tr chn di: scalar b2_lb=c(2)-@qtdist(0.975,8)*@stderrs(2)

Tnh gi tr chn trn: scalar b2_ub=c(2)+@qtdist(0.975,8)*@stderrs(2)

KHONG TIN CY CHO (2

Nh cp tnh cht 6 ca cc c lng OLS di gi nh hng nhiu c phn phi chun, ta c bin:

2

2

2

)

2

n

(

s

s

-

=

c

(7.18)

theo phn phi (2 vi n-2 bc t do (m hnh hi qui n). V th, ta c th s dng phn phi (2 xy dng khong tin cy cho (2 nh sau:

a

-

=

c

c

c

a

a

-

1

)

Pr(

2

2

/

2

2

2

/

1

(7.19)

Trong (2 c xc nh theo cng thc (7.18) v

2

2

/

1

a

-

c

v

2

2

/

a

c

(cc gi tr (2 phn phn) l hai gi tr c ly t bng phn phi (2 vi n-2 bc t do. Thay th (2 cng thc (7.18) vo cng thc (7.19) v sp xp li, ta c

a

-

=

c

s

-

c

c

s

-

a

-

a

1

)

2

n

(

)

2

n

(

Pr

2

2

/

1

2

2

2

2

/

2

(7.20)

Cng thc ny cho mt khong tin cy 100(1-()% cho gi tr thc (2. minh ha, ta quay li v d minh ha v c lng phng trnh ng cu nc ng chai Aquafina Chng 6. Trong v d ny ta c

18

.

5

)

275

.

2

(

2

2

=

=

s

v d.f. = n-2 = 8. Nu chn mc ngha ( = 5%, t bng phn phi (2 ta thy

5346

.

17

2

025

.

0

=

c

(hoc =CHIINV(2.5%,8)) v

1797

.

2

2

975

.

0

=

c

(hoc = CHIINV(97.5%,8)). Cc gi tr ny cho thy rng xc sut ca mt gi tr (2 ln hn 17.5346 l 2.5% v ln hn 2.1797 l 97.5%. V th khong gia hai gi tr ny l khong tin cy 95% ca (2 nh trn HNH 7.2.

)

b

(

se

t

b

2

2

/

2

a

+

Hnh 7.2: Phn phi chi bnh phng vi 8 bc t do

Th cc thng tin trn vo phng trnh (7.20), ta c khong tin cy 95% cho (2 nh sau:

1.038 ( (2 ( 8.350

(7.21)

Gii thch khong tin cy ny nh sau: Nu ta thit lp cc khong tin cy 95% cho (2, trong di hn, th 95 trong 100 trng hp cc khong tin cy nh vy s c cha gi tr thc (2.

Thao tc vi Eviews

tnh gi tr (2 ph phn, th trn ca s lnh ca Eviews ta nhp:

scalar chisq0025=@qchisq(0.025,8) = 2.1797 => (20.025,8

scalar chisq0975=@qchisq(0.975,8) = 17.5346 => (20.975,8

KIM NH GI THUYT

Sau khi tho lun vn c lng im v c lng khong, by gi chng ta s xem xt vn cn li ca suy lun thng k: kim nh gi thuyt. Theo Gujarati (2003), vn kim nh gi thuyt thng k c th c pht biu mt cch n gin nh sau: Mt kt qu nghin cu nht nh c ph hp vi mt gi thuyt no hay khng? Khi nim ph hp y c ngha l gi tr c lng c gn vi gi tr c gi thuyt hay khng trn c s ta c khng bc b hay bc b gi thuyt . V vy, nu l thuyt hay kinh nghim trc y cho ta c s tin rng h s dc thc B2 trong v d minh ha ng cu nc ng chai Aquafina l -2, th h s c lng t mu quan st b2 = -2.909 c ph hp vi gi thuyt trn hay khng? Nu ng, ta khng bc b gi thuyt cho rng B2 = -2; ngc li ta bc b gi thuyt B2 = -2. Theo ngn ng thng k, gi thuyt c pht biu c gi l gi thuyt khng v c k hiu thng nht l H0. Gi thuyt khng thng c kim nh i li vi mt gi thuyt khc v gi thuyt khc ny c k hiu thng nht l H1. V d, gi thuyt khc cho rng gi tr thc B2 khc -2.

L thuyt kim nh gi thuyt quan tm n vic pht trin cc nguyn tc hay th tc quyt nh xem c bc b hay khng bc b gi thuyt khng. Nh cp Chng 3, c hai cch tip cn c tnh cht b sung cho nhau l khong tin cy v kim nh ngha. C hai cch tip cn ny cho rng mt bin (thng k hay c lng) ang xem xt theo mt phn phi xc sut nht nh no v kim nh gi thuyt lin quan n vic a ra cc pht biu hay li khng nh v (cc) gi tr ca (cc) tham s phn phi . V d, ta bit rng vi gi nh b2 c phn phi chun vi trung bnh l B2 v phng sai l

2

2

b

s

nh phng trnh (6.44). Nu ta gi thit rng B2 = -2, ta ang a ra mt khng nh v mt trong cc tham s ca phn phi chun, tc gi tr trung bnh.

PHNG PHP KHONG TIN CY

Kim nh hai pha

minh ha cch tip cn kim nh gi thuyt theo khong tin cy, ta li xem xt v d v ng cu nc ng chai Aquafina. Kt qu c lng cho thy b2 = -2.909, v nu c gi thuyt nh sau:

H0: B2 = -2

H1: B2 ( -2

Ngha l, dc thc ca ng cu l -2 di gi thuyt khng nhng nh hn hoc ln hn -2 di gi thuyt khc. Nh vy, gi thuyt khng l mt gi thuyt n, trong khi gi thuyt khc l mt gi thuyt hp; v c bit nh mt gi thuyt hai ui/hai pha. Thng th mt gi thuyt hai ui nh vy phn nh s tht rng ta khng c sn thng tin k vng t c s l thuyt hoc cc kt qu nghin cu trc y v chiu hng ca gi thuyt khc so vi gi thuyt khng.

Nh vy, gi tr b2 quan st c t mu c ph hp vi gi thuyt H0 hay khng? tr li cu hi ny, ta cn n khong tin cy nh c thit lp phng trnh (7.16). Ta bit rng trong di hn th cc khong tin cy nh khong (-3.486, -2.333) s cha gi tr thc B2 vi xc sut 95%. Trong di hn (khi mu c ly lp i lp li nhiu ln) th cc khong tin cy nh th cung cp ta mt khong gii hn trong c th cha gi tr thc B2 vi mt h s tin cy, v d, 95%. Nu B2 di gi thuyt H0 nm trong khong tin cy 100(1-()%, th ta khng bc b gi thuyt H0. Ngc li, nu B2 di gi thuyt H0 nm ngoi khong tin cy, th ta bc b gi thuyt H0.

Trong v d minh ha v ng cu nc ng chai, ta nhn thy rng gi thuyt H0: B2 = -2 r rng nm ngoi khong tin cy 95% nh phng trnh (7.16). V th, ta c th bc b gi thuyt cho rng h s dc thc ca phng trnh ng cu nc ng chai l -2, vi khong tin cy l 95%. Lu rng, nu gi thuyt H0 l ng, th xc sut ta chp nhn gi tr -2.909 khong 5% v y chnh l xc sut chp nhn sai lm loi I.

Trong thng k, khi ta bc b gi thuyt khng, ngha l ta ni rng kt qu nghin cu ca ta l c ngha thng k. Ngc li, khi ta khng bc b gi thuyt khng, ngha l ta ni rng kt qu nghin cu ca ta l khng c ngha thng k. Thng thng, ta hay s dng ba mc ngha l 1%, 5%, v 10%. Tuy nhin, sau ny ta thy rng gi tr xc sut p s rt hu ch v ch cn nhn vo gi tr xc sut p, ta c th kt lun mt h s c lng c ngha thng k mc ngha l bao nhiu.

( HNH 7.3: Khong tin cy 100(1-()% cho B2

000

.

0

)

613

.

11

t

(

=

>

Kim nh mt pha

i khi ta c mt c s k vng rt vng chc t thc t cc nghin cu trc y hay t l thuyt rng gi thuyt khc l mt pha ch khng phi hai pha nh va c tho lun trn. Khi ny th gi thuyt ca ta c th s l nh sau:

H0: B2 ( -2

H1: B2 > -2

Mc d th tc kim nh loi gi thuyt ny vi cch tip cn khong tin cy c th c thc hin mt cch d dng theo cng thc phng trnh (7.13), nhng Gujarati (2003) cho rng s dng cch tip cn kim nh mc ngha p dng cho loi kim nh ny s d dng hn rt nhiu. Tm li, qui trnh thc hin kim nh mt h s hi qui theo cch tip cn khong tin cy c th c thc hin theo cc bc sau y:

1. Xc nh gi thuyt H0 v H1

2. Chn mc ngha ( v xy dng khong tin cy 100(1-()%

3. So snh gi tr gi thuyt H0 vi khong tin cy trn ri ra quyt nh nh sau

Nu gi tr ca gi thuyt H0 nm trong khong tin cy trn, ta khng bc b gi thuyt H0.

Nu gi tr ca gi thuyt H0 nm ngoi khong tin cy trn, ta bc b gi thuyt H0.

PHNG PHP KIM NH NGHA

Kim nh ngha ca cc h s hi qui: Kim nh t

Mt cch tip cn khc v b sung cho cch tip cn da vo khong tin cy l kim nh ngha do Fisher, Neyman v Pearson cng tm ra. Ni cung, mt kim nh ngha l mt th tc trong cc kt qu c lng t mu c s dng kim chng ng hay sai ca mt gi thuyt khng. tng ch o ca kim nh theo cch ny l kim nh ngha ca mt thng k kim nh v phn phi mu ca thng k di gi thuyt khng. Quyt nh chp nhn hay bc b gi thuyt khng c thc hin trn c s gi tr ca thng k kim nh c c t d liu mu. y l cch kim nh ph bin c p dng cho nhiu c lng khc nhau ty vo loi thng k kim nh v phn phi xc sut ca c lng l g. Lu , cc thng k kim nh y c ngha l cc c lng OLS nh b2 hay

2

s

. Trng hp ang xt th thng k kim nh l cc h s hi qui, vn c bit c phn phi mu l phn phi chun.

Bit rng, b2 c phn phi chun vi trung bnh l B2 v phng sai l

s

=

s

2

i

2

2

2

b

x

nh phng trnh (7.4). Cho nn ta c th d dng suy ra bin phn phi chun ha Z c phn phi chun vi trung bnh l 0 v phng sai l 1, nh phng trnh (7.5). Tuy nhin, thc t ta khng th bit thng tin v phng sai nhiu ((2), thay vo ta ch c th c c lng ca n l

2

s

. Chnh v th, theo l thuyt thng k, ta s c bin t c tnh theo cng thc (7.10) nh sau:

(

)

s

-

=

s

-

=

-

=

x

)

B

(b

x

)

B

b

(

)

b

(

se

B

b

t

2

i

2

2

2

i

2

2

2

2

2

(7.10)

s theo phn phi t vi n-2 bc t do. Nu gi tr thc B2 c xc nh di gi thuyt H0, th gi tr t theo cng thc (7.22) c th c tnh ton d dng t d liu mu, v v th n c th c vai tr nh mt thng k kim nh. V do thng k kim nh ny theo phn phi t, nn ta c th xc nh khong tin cy nh sau:

a

-

=

-

-

a

a

1

t

)

b

(

se

B

b

t

Pr

2

/

2

*

2

2

2

/

(7.22)

Trong ,

*

2

B

l gi tr ca B2 di gi thuyt H0 v -t(/2 v t(/2 l cc gi tr t ph phn c t bng phn phi t vi mc ngha (/2 v n-2 bc t do (c th tnh theo cng thc =TINV((,n-2) trn Excel. Nh vy, phng trnh (7.23) c th c vit li nh sau:

a

-

=

+

-

a

a

1

)]

b

(

se

t

B

b

)

b

(

se

t

B

Pr[

2

2

/

*

2

2

2

2

/

*

2

(7.23)

y l khong tin cy trong xc sut n c cha b2 l 1-(, khi bit B2 =

*

2

B

. Theo ngn ng kim nh gi thuyt thng k, khong tin cy 100(1-()% nh phng trnh (7.23) c gi l vng chp nhn gi thuyt H0 v (cc) vng bn ngoi khong tin cy ny c gi l (cc) vng bc b hay vng ph phn gi thuyt H0. Lu , nh c trnh by trc y, cc im gii hn ca khong tin cy ny c gi l cc gi tr ph phn.

Mi quan h gia cc cch tip cn khong tin cy v kim nh ngha c th c phn tch da trn s so snh gia hai cng thc (7.13) v (7.23). Trong cch tip cn khong tin cy, ta c gng thit lp mt khong vi mt xc sut nht nh no c cha gi tr thc (nhng khng bit) B2, trong khi , vi cch tip cn mc ngha, ta gi thuyt mt gi tr nht nh no ca B2 v c gng xem xt gi tr b2 tnh ton t mu c nm trong khong tin cy theo gi tr c gi thuyt hay khng.

Tr li v d minh ha v ng cu nc ng chai, ta thy b2 = -2.909, se(b2) = 0.25, v d.f. = 8. Nu gi nh mc ngha ( = 5%, th t(/2 = 2.306. Nu ta c gi thuyt nh sau:

H0: B2 =

*

2

B

= -2

H1: B2 ( -2

Th vo phng trnh (7.23) ta c:

95

.

0

]

25

.

0

*

306

.

2

2

b

25

.

0

*

306

.

2

2

Pr[

2

=

+

-

-

-

95

.

0

]

424

.

1

b

577

.

2

Pr[

2

=

-

-

(7.24)

X

Hnh 7.4: Khong tin cy 95% ca b

2

di gi thit H

0

: B

2

=-2

2.5%

2.5%

Nh vy, gi tr b2 tnh ton t d liu mu bm vng ph phn, nn ta bc b gi thuyt khng rng B2 = -2.

Trong thc t ta khng cn c lng khong tin cy (7.23), m ta ch cn tnh gi tr t tnh ton

)

b

(

se

B

b

2

*

2

2

-

=

trong cng thc (7.22) v xem gi tr ny nm trong hay ngoi hai gi tr t ph phn vi mt mc ngha xc nh. Cch kim nh ny rt nhanh v tin li v hu ht cc phn mm kinh t lng u c bo co cc thng tin v b2, se(b2), bc t do d.f., v thm ch gi tr t tnh ton cho gi thuyt H0: B2 = 0. Nu gi thuyt H0 cho B2 l mt gi tr khc, v d B2 = -2, th ta cn phi tnh li gi tr t tnh ton nh sau:

636

.

3

25

.

0

)

2

(

909

.

2

t

-

=

-

-

-

=

(7.25)

Vi mc ngha ( = 5%, ta c hai gi tr t ph phn vi d.f. = 8 ln lc l -2.306 v 2.306 (=TINV(5%,8)), nn gi tr t tnh ton (-3.636) nm vng bc b gi thuyt H0, xem Hnh 7.5.

i

Y

Hnh 7.5: Khong tin cy 95% ca t di gi thit H

0

: B

2

=-2

2.5%

2.5%

Lu rng, nu gi tr c lng ca B2 (= b2) bng vi gi tr gi thuyt ca B2, th gi tr t tnh ton s bng khng. Tuy nhin, khi gi tr c lng ca b2 cng khc gi tr gi thuyt ca B2, th gi tr tuyt i ca t tnh ton, (t(, s cng ln. Nh th, khi gi tr tuyt i ca t tnh ton cng ln s l bng chng bc b gi thuyt H0. D nhin, ta c th lun s dng bng phn phi t (hay n gin nht l dng =TINV((, d.f.)) xc nh xem mt gi tr t tnh ton l ln hay nh; v cu tr li s ty thuc vo s bt t do v mc ngha l bao nhiu. ngi c hnh dung c ti sao quyt nh chp nhn hay bc b mt gi thuyt H0 li ty thuc vo s bc t do v mc ngha, ti xin trnh by mt phn ca Bng phn phi t (c tnh bng hm =TINV((, d.f.)) nh Bng 7.1.

Do ta s dng phn phi t, nn qui trnh kim nh va trnh by trn c gi l kim nh t. Theo ngn ng ca cc kim nh mc ngha, mt kim nh c cho l c ngha v mt thng k nu gi tr ca thng k kim nh (gi tr t tnh ton) nm trong vng ph phn/bc b gi thuyt H0. Trong trng hp ny, gi thuyt H0 b bc b. Ngc li, mt kim nh c cho l khng c ngha v mt thng k nu gi tr ca thng k kim nh nm trong vng chp nhn gi thuyt H0. Trong trng hp ny, gi thuyt H0 khng b bc b. Trong v d ca ta, kim nh t c ngha thng k v v th ta bc b gi thuyt H0 cho rng B2 = -2.

( BNG 7.1: Bng phn phi t

t ph phn

Mc ngha (()

Loi kim nh

5%

2.5%

1%

0.5%

0.1%

Mt ui

10%

5%

2%

1%

0.2%

Hai ui

S bc t do

1

6.314

12.706

31.821

63.657

318.31

2

2.920

4.303

6.965

9.925

22.327

3

2.353

3.182

4.541

5.841

10.215

4

2.132

2.776

3.747

4.604

7.173

5

2.015

2.571

3.365

4.032

5.893

6

1.943

2.447

3.143

3.707

5.208

7

1.895

2.365

2.998

3.499

4.785

8

1.860

2.306

2.896

3.355

4.501

9

1.833

2.262

2.821

3.250

4.297

10

1.812

2.228

2.764

3.169

4.144

11

1.796

2.201

2.718

3.106

4.025

12

1.782

2.179

2.681

3.055

3.930

13

1.771

2.160

2.650

3.012

3.852

14

1.761

2.145

2.624

2.977

3.787

15

1.753

2.131

2.602

2.947

3.733

16

1.746

2.120

2.583

2.921

3.686

17

1.740

2.110

2.567

2.898

3.646

18

1.734

2.101

2.552

2.878

3.610

19

1.729

2.093

2.539

2.861

3.579

20

1.725

2.086

2.528

2.845

3.552

21

1.721

2.080

2.518

2.831

3.527

22

1.717

2.074

2.508

2.819

3.505

23

1.714

2.069

2.500

2.807

3.485

24

1.711

2.064

2.492

2.797

3.467

25

1.708

2.060

2.485

2.787

3.450

26

1.706

2.056

2.479

2.779

3.435

27

1.703

2.052

2.473

2.771

3.421

28

1.701

2.048

2.467

2.763

3.408

29

1.699

2.045

2.462

2.756

3.396

30

1.697

2.042

2.457

2.750

3.385

35

1.690

2.030

2.438

2.724

3.340

40

1.684

2.021

2.423

2.704

3.307

45

1.679

2.014

2.412

2.690

3.281

50

1.676

2.009

2.403

2.678

3.261

55

1.673

2.004

2.396

2.668

3.245

60

1.671

2.000

2.390

2.660

3.232

65

1.669

1.997

2.385

2.654

3.220

70

1.667

1.994

2.381

2.648

3.211

80

1.664

1.990

2.374

2.639

3.195

90

1.662

1.987

2.368

2.632

3.183

100

1.660

1.984

2.364

2.626

3.174

120

1.658

1.980

2.358

2.617

3.160

Lu rng qui trnh kim nh va c nu trn l qui trnh kim nh ngha 2 pha/2 ui, trong ta xem hai ui hai u ca phn phi xc sut tng ng l cc vng bc b gi thuyt H0, ngha l ta s bc b gi thuyt H0 nu gi tr thng k kim nh (gi tr t tnh ton) nm mt trong hai ui ny. Tuy nhin, nu cc kinh nghim trc y cho rng h s co gin ca cu nc ng chai theo gi l ln hn -2 (ngha l t co gin hn). Trong trng hp ny ta c:

H0: B2 ( -2

H1: B2 ( -2

Nh vy, H1 by gi l mt gi thuyt mt pha. Qui trnh thc hin loi kim nh mt pha ny cng ging vi qui trnh thc hin kim nh hai pha, nhng ngoi tr gii hn tin cy hay gi tr ph phn cn trn by gi s l t( = t0.05. Nh vy, lc ny vi s bc t do l 8, th gi tr t trang bng s mc ngha 5% cho mt pha hoc 10% cho hai pha (1.86). Nu s dng hm trn Excel, ta s thc hin nh sao: =TINV(10%,8) = 1.86. Vi gi tr t ph phn l 1.86, th gi tr gi tr tin cy cn di ca B2 s l -2 - 1.86 * 0.25 = -2.465. Nh vy, c gi tr t tnh ton l -3.636 hoc b2 = -2.909 u nm vng ph phn, nn ta bc b gi thuyt H0. Hnh 7.6 v Bng 7.2 minh ha qui tc quyt nh khi s dng phng php kim nh mc ngha.

( BNG 7.2: Qui tc quyt nh vi kim nh ngha t

Loi kim nh

H0: Gi thuyt khng

H1: Gi thuyt khc

Qui tc quyt nh: Bc b gi thuyt H0 nu

Hai ui

*

2

2

B

B

=

*

2

2

B

B

.

f

.

d

,

2

/

t

t

a

>

ui phi

*

2

2

B

B

*

2

2

B

B

>

.

f

.

d

,

t

t

a

>

ui tri

*

2

2

B

B

*

2

2

B

B

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l cc gi tr t ph phn cc mc ngha ( v (/2 vi d.f. l s bc t do, d.f. = s quan st - s h s c lng trong m hnh hi qui./ppNgun: Gujarati, 2003, trang 133./ppp2.5%/p

/pppHnh 7.6: Kim nh mc ngha trng hp mt ui/p

p5%/p

/ppThao tc vi Eviews/ppKim nh ngha ca cc h s hi qui trn Eviews c th c thc hin nh sau: /pp ui phi/pp Xc nh gi thuyt H0 v H1 (gi s H0: B2 ( 0, H1: B2 0)/pp Chn mc ngha (, v d 5%/pp Tnh gi tr t ph phn: scalar tc95=@qtdist(0.95,8)/pp Tnh gi tr t tnh ton: scalar tstat=(c(2)-0)/@stderrs(2)/pp So snh tstat vi tc95 /pp ui tri/pp Xc nh gi thuyt H0 v H1 (gi s H0: B2 ( 0, H1: B2 < 0)/pp Chn mc ngha (, v d 5%/pp Tnh gi tr t ph phn: scalar tc05=@qtdist(0.05,8)/pp Tnh gi tr t tnh ton: scalar tstat=(c(2)-0)/@stderrs(2)/pp So snh tstat vi tc05/pp Hai ui/pp Xc nh gi thuyt H0 v H1 (gi s H0: B2 = 0, H1: B2 ( 0)/pp Chn mc ngha (, v d 5%/pp Tnh gi tr t ph phn ui phi: scalar tc975=@qtdist(0.975,8)/pp Tnh gi tr t ph phn ui tri: scalar tc025=@qtdist(0.025,8)/pp Tnh gi tr t tnh ton: scalar tstat=(c(2)-0)/@stderrs(2)/pp So snh tstat vi tc975 hoc tc025/p

pKim nh ngha ca (2: Kim nh (2 /ppTrn c s phn tch phn tnh cht ca cc c lng OLS ta bit rng, do hng nhiu c phn phi chun nn bin sau y: /pp div class="embedded" id="_1287012382"/p2/p

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p2/p

pn/p

p(/p

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p=/p

pc/p

(7.26)/ppc phn phi (2 vi s bc t do l n 2. Trong v d minh ha v ng cu nc ng chai, ta c div class="embedded" id="_1286523586"/p2/p

p/p

ps/p

= (2.2754)2 = 5.18 v s bc t do l 8. Nu ta c gi thuyt sau y:/ppH0: (2 = 10/ppH1: (2 ( 10/pp kim nh gi thuyt H0 ny ta s dng thng k kim nh l thng k (2 vi gi tr tnh ton nh cng thc (7.26). Thay cc gi tr bit vo cng thc (7.26) ta c gi tr (2 tnh ton s l:/pp div class="embedded" id="_1286524742"/p142/p

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(7.27)/ppNu vi mc ngha ( = 5%, th cc gi tr (2 ph phn s ln lt l 2.1797 v 17.5346 (nh Hnh 7.2). Do gi tr (2 tnh ton nm gia hai gi tr (2 ph phn ny, nn ta khng bc b gi thuyt H0. Qui trnh kim nh nh vy c gi l kim nh ngha ca (2. /pppHnh 7.2: Phn phi chi bnh phng vi 8 bc t do/p

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pMT S VN THC TIN CA KIM NH GI THUYT/pp ngha ca vic chp nhn hay bc b mt gi thuyt/ppNu trn c s ca mt kim nh ngha, v d kim nh t, ta quyt nh chp nhn gi thuyt khng, th c ngha ta ang ni rng trn c s d liu mu ta khng c l do g bc b gi thuyt c, ch ta khng ni rng gi thuyt khng l ng m khng c bt c hoi nghi no. Ti sao? tr li cu hi ny, ta quay li v d v ng cu nc ng chai v gi s rng H0: B2 = -2.5. Vi h s c lng t d liu mu b2 = -2.909 v se(b2) = 0.25, th gi tr t tnh ton s l (-2.909 (-2.5))/0.25 = -1.636, ta kt lun h s c lng khng c ngha thng k mc ngha ( = 5%. V th, ta chp nhn H0. Nhng by gi gi s ta gi nh H0: B2 = -3, v tnh c gi tr t tnh ton l (-2.909 (-3))/0.25 = 0.364. Vi gi tr t tnh ton ny th h s c lng vn khng c ngha thng k. V by gi ta cng chp nhn H0. Nh vy, trong hai gi thuyt ny th gi thuyt no thc s ng? Ta thc s khng bit. V th, khi chp nhn mt gi thuyt khng ta lun lun nn hiu rng c mt gi thuyt khc c th s cng tng thch vi d liu mu. Cho nn, tt nht l ta nn ni c th chp nhn gi thuyt khng, hn l chp nhn gi thuyt khng. /ppGi thuyt khng Bi = 0 v nguyn tc t = 2 /ppMt gi thuyt khng c s dng ph bin nht trong cc nghin cu thc nghim l H0: B2 = 0; ngha l, h s dc bng khng. Mc ch ca loi gi thuyt ny l nhm xem c mi quan h no gia bin ph thuc (Y) v mt bin gii thch (X) no hay khng. Nu kt qu cho thy khng c mi quan h no gia Y v X, th vic kim nh mt gi thuyt, v d H0: B2 = -2, l v ngha./pp/ppGi thuyt H0 ny c th c kim nh mt cch d dng bng phng php khong tin cy hay kim nh mc ngha nh trnh by trn. Nhng thng thng ngi ta c th kim nh nhanh bng cch p dng nguyn tc t=2 nh sau:/pp( BNG 7.4: Nguyn tc t = 2/ppNguyn tc t=2. Nu s bc t do l 20 hoc cao hn v nu mc ngha c chn l ( = 5%, th gi thuyt H0: B2 = 0 c th b bc b nu gi tr tuyt i ca gi tr t tnh ton (b2/se(b2)) ln hn 2./p

pNgun: Gujarati, 2003, trang 134/ppTt c cc phn mm kinh t lng u c bo co gi tr t tnh ton cho loi gi thuyt ny. Cho nn, ta ch cn so snh gi tr t tnh ton vi gi tr t ph phn mt mc ngha xc nh, hoc n gin vi t = 2./ppHnh thnh gi thuyt khng v gi thuyt khc/ppKhi xc nh gi thuyt khng v gi thuyt khc, th vic kim nh ngha thng k khng cn l mt iu huyn b na. Nhng ta xc nh cc gi thuyt ny nh th no? Thng thng hin tng ang c nghin cu s gi ln bn cht ca gi thuyt khng v gi thuyt khc. /ppXem xt ng th trng vn (CML) ca l thuyt v danh mc nh trnh by phn tnh hung ng dng hi qui n c lng h s beta (Chng 6) ta c Ei = B1 + B2(i, trong , E = sut sinh li k vng ca danh mc v ( = lch chun ca sut sinh li, thc o v mc ri ro. Theo l thuyt ti chnh, sut sinh li v ri ro c k vng c mi quan h ng bin - ri ro cng cao, sut sinh li k vng s cng cao. Nn gi thuyt c th l H0: B2 = 0 v H1: B2 0. y l loi kim nh mt ui./ppMt v d khc v nhu cu tin trong nn kinh t. Nh ta s c bit cc chng sau, mt trong nhng nhn t quyt nh nhu cu tin l thu nhp. Cc nghin cu trc y v hm cu tin cho rng h s co gin ca cu tin theo thu nhp trong khong 1 1.3 (Gujarati, 2003). V th, trong mt nghin cu mi v cu tin, nu ta gi nh H0: B2 = 1 v H1: B2 ( 1. y l loi kim nh hai ui. /ppTrong kinh t hc vi m ta ni rt nhiu v h s co gin ca cu theo chnh gi l m; ngha l gi tng th lng cu gim. i vi go, th cu tng i t co gin theo gi vi h s co gin 1 < B2 < 0. V theo kt qu nghin cu giai on 2006 2008 ca IMF th h s co gin ny l khong -0.38. Gi s ta ang thc hin mt nghin cu v co gin ca go Vit Nam th gi thuyt c th s l H0: B2 = -0.38 v H1: B2 < -0.38. y cng l loi kim nh mt ui./ppTm li, cc k vng l thuyt hoc mt nghin cu thc hin trc y hoc c hai c th c s dng lm c s xy dng cc gi thuyt. Nhng cho d cc gi nh ny c hnh thnh nh tho no i na th mt iu cc k quan trng m ngi lm nghin cu cn lu l phi thit lp cc gi thuyt trc khi thc hin cng vic nghin cu thc nghim ca mnh. /ppChn mc ngha (/ppNh ta va c cp rng vic bc b hay khng bc b mt gi thuyt khng (H0) ph thuc rt nhiu vo (. Lu , mc ngha ( l mc ngha hay xc sut chp nhn sai lm loi I ngha l xc sut bc b gi thuyt ng. Nh cp Chng 4, vi mt mu d liu nht nh, hai loi sai lm hay s dng trong thng k l sai lm loi I v sai lm loi II (xc sut chp nhn gi thuyt sai) lun c s nh i. Trn thc t, ngi ta thng chn cc mc ngha 1%, 5%, v 10%. Tuy nhin, vic s dng my tnh ngy cng tr nn ph bin th chng ta c th chn bt k mc ngha no, v d 2%, 3%, hay 6%. Vn t ra y l nn chn mc ngha no l thch hp nht cho tng trng hp nht nh. trnh kh khn ny, ngi ta c xu hng s dng phng php gi tr xc sut (p value) ca thng k kim nh. /ppGi tr xc sut p: Mc ngha chnh xc/pp/ppMt khi gi tr thng k kim nh bit t mt kt qu c lng (v d, -11.613), th ti sao ta khng tm gi tr xc sut sao cho gi tr thng k kim nh ln hn hoc bng gi tr thng k kim nh c tnh ton t kt qu c lng? Ni cch khc, sao ta khng tm xc sut sao cho gi tr tuyt i ca t (kim nh hai ui) ln hn hoc bng (hoc ch ln hn l ) 11.613? Khi bit xc sut l bao nhiu th ta d dng so snh n vi mc ngha mong mun. Gi tr xc sut ny c tnh rt n gin trn Excel nh sau: =TDIST(X,d.f.,tails) =TDIST(11.613,8,2) = 0.000. y, X l gi tr thng k mun tnh xc sut ( y l thng k t, vi gi tr t tnh ton l 11.613); d.f. l s bc t do; v tails ngha l tnh xc sut mt ui hay hai ui, nu mt ui ta chn s 1, nu hai ui, ta chn s 2. Tht qu d dng phi khng! Mt lu nho nh cn nhc qu v y nh sau. Bt k mt phn phi xc sut no cng c dng mt hm y = f(x), vi X l gi tr thng k kim nh, v d t, trc honh v y l gi tr xc sut trc tung (din tch ca vng ph phn). Khi ta c X th ta th X vo hm y tnh xc sut, v ta s dng hm XDIST, v d TDIST, FDIST, Ngc li, khi ta bit y (tc bit gi tr xc sut), th ta cng d dng tm c gi tr X bng cch s dng hm ngc, tc XINV, v d TINV, FINV, /ppQuay li cng vic hin ang c cp, th gi tr xc sut c tnh nh vy c gi l gi tr xc sut p. Gi tr xc sut ny cng c gi l mc ngha chnh xc hay mc ngha quan st c; hay cng c th gi l xc sut chnh xc chp nhn sai lm loi I. Ni theo ngn ng thng k, gi tr xc sut p c nh ngha l mc ngha thp nht ti mt gi thuyt H0 c th b bc b. Nu d liu mu khng ng h gi thuyt khng, th gi tr tuyt i ca t tnh ton di gi thuyt khng s ln v v th gi tr xc sut p c c gi tr s nh, v ngc li. Ni cch khc, vi mt c mu nht nh, nu gi tr tuyt i ca t tng, gi tr xc sut p s gim./pp C mi quan h g gia gi tr xc sut p v mc ngha ( hay khng? Gujarati (2003) cho rng nu chng ta c gi thi quen c nh mt mc ngha ( bng vi gi tr xc sut p ca mt thng k kim nh (v d, thng k t), th s khng c g mu thun gia hai gi tr ny. Ni cch khc, tt nht l nn t b thi quen c nh mc ngha ( bng mt mc no v hy lm quen vi vic chn gi tr xc sut p ca thng k kim nh. Tt nht l hy cho qu v t quyt nh c bc b gi thuyt H0 hay khng trn c s gi tr xc sut p c tnh. iu ny rt t nhin v ty thuc vo mc khc khe hay d gii ca chnh qu v i vi vn m minh ang nghin cu. Gi s, sau khi tnh c gi tr xc sut p l 12%, nu qu v mun bc b H0 ti mc ngha chnh xc bng 12% cng c. Tng t, nu qu v chn gi tr xc sut p l 0.05% (ngha l khng chp nhn sai lm ln hn nm ln trong 10.000 ln. iu ny khng c vn g c. Ni chung, mi ngi chng ta c cch suy ngh ring, ty thuc qu v l ngi thch ri ro hay s ri ro./ppRt nhiu ngi l nghin cu hay sinh vin c thi quen tr nn ph bin l so snh gi tr xc sut p vi mt mc ngha ( no , v d 5%, quyt nh xem nn bc b hay khng bc b gi thuyt H0 mc ngha 5%. Cch lm nh vy khc no vic bit mn trn tay l 9 lng nhng li c mun t ln cn 1kg xem n c nh hn 1kg hay khng. iu ny tht ra cng khng thnh vn , nhng qu v hy t hi xem cch lm nh th c phi l qu my mc khng nh!/ppLu , vi bt k gi thuyt H0 no ta cng c mt gi tr t tnh ton nht nh v hin nhin ta u c th bit gi tr xc sut p./ppThao tc vi Eviews/pp tnh gi tr xc sut p, th trn ca s lnh ca Eviews ta thc hin nh sau:/pp ui phi: scalar pval=1-@ctdist(tstat,d.f.)/pp ui tri: scalar pval=@ctdist(tstat,d.f.)/pp Hai ui:/pp Scalar leftpval=@ctdist(-abs(tstat),d.f.)/pp Scalar rightpval=@1-@ctdist(abs(tstat),d.f.)/pp Scalar pval2=leftpval+rightpval/p

pPHN TCH PHNG SAI/ppTrong phn ny, chng ta s nghin cu phn tch hi qui trn gc phn tch phng sai v gii thiu n qu v mt cch khc c tnh h tr v b sung rt hu ch khi xem xt vn suy lun thng k./ppNh bit Chng 6, tng bin thin ca Yi quanh gi tr trung bnh c th c biu din bng ng thc sau:/ppdiv class="embedded" id="_1283378978"/p/p

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pLu , trong kt qu Eviews ch bo co cc thng tin v gi tr F tnh ton, gi tr xc sut p, RSS, v R2. Cho nn, mun lp bng ANOVA nh trn, ta cn thc hin mt vi tnh ton nh. Nhng iu khng thc s cn thit. Ngc li, i vi hi qui trn Excel (Tools/Data Analysis/Regression) th kt qu hi qui c bo co ring v ANOVA. /ppVi gi tr F tnh ton l 134.855, v gi tr F ph phn mc ngha 5% l =FINV(5%,1,8) = 5.317, hoc =FINV(1%,1,8) = 11.259, hoc gi tr xc sut p = FDIST(134.855,1,8) = 0.0000028, th ta d dng bc b gi thuyt H0: B2 = 0. Nh vy, ngoi thng k t, thng k F cung cp mt phng php kim nh khc cho gi thuyt H0: B2 = 0 vi cng mt kt qu quyt nh nh kim nh t. Ti y, c l qu v s t hi rng ti sao chng ta khng da vo kim nh t v khng nht thit phi quan tm n kim nh F? Gujarati (2003) cho rng i vi m hnh hi qui n th ta khng cn thit phi quan tm n kim nh F. Nhng trong cc m hnh hi qui bi th kim nh F s cung cp nhiu ng dng quan trng cho vic kim nh cc gi thuyt thng k. /pp/ppThao tc vi Eviews/ppQu v c th tnh gi tr F tnh ton, gi tr F ph phn, v gi tr xc sut p bng cc hm trn Eviews nh sau (v d mc ngha ( = 5%):/pp Tnh gi tr F tnh ton:/pp Tnh TSS: scalar TSS=(@sumsq(Y)-10*(@mean(Y))^2)/pp Tnh RSS: scalar RSS=@ssr/pp Tnh ESS: scalar ESS=TSS-RSS/pp Tnh F tnh ton: scalar fval=(ESS/1)/(RSS/8)/pp Tnh gi tr F ph phn: scalar fc=@qfdist(95%,1,8)/pp Gi tr xc sut p: scalar pp=1-@cfdist(fval,1,8)/pp So snh gi tr F tnh ton vi gi tr F ph phn, hoc s dng gi tr xc sut p ra quyt nh bc b hay chp nhn gi thuyt H0./p

pTRNH BY KT QU HI QUI/ppC nhiu cch trnh by kt qu phn tch hi qui, nhng thng thng ngi s trnh by theo mu sau:/ppdiv class="embedded" id="_1286843980"/pi/p

pY/p

p/p

= 54.80 2.157Xi

r2 = 0.9757

(7.31)/pp se = (1.554) (0.251)

d.f. = 8/pp t = (35.255) (-11.613)

F1,8 = 134.855/pp p = (0.000) (0.000)

DW = 1.095/ppIn phng trnh (7.31), cc con s trong ngoc n dng th nht l cc sai s chun ca cc h s hi qui; cc con s dng th hai l cc gi tr t tnh ton theo cng thc (7.10) di gi thuyt khng cho rng gi tr thc ca tng th ca mi h s hi qui bng khng [(-2.157 0)( 0.251 = -11.613], v cc con s dng th ba l cc gi tr xc sut p tnh ton t kt qu c lng./ppBng cch trnh by cc gi tr xc sut p ca cc gi tr t tnh ton t kt qu c lng, chng ta c th thy ngay mc ngha chnh xc ca mi gi tr t c lng. V th, di gi thuyt khng cho rng h s tung gc thc ca tng th bng khng, th xc sut chnh xc c mt gi tr t ln hn hay bng 35.255 l 0.000 (0%). y l mt mc xc sut chp nhn sai lm loi I cc k nh. Ch nhn vo cc gi tr xc sut p nh vy, ta c th d dng cho rng cc h s thc ca tng th ca m hnh trn l khc khng. Trong kt qu bo co c trnh by s bc t do d.f. nhm gip ta bit s bc t do ca RSS l bao nhiu, t ta c th d dng thc hin cc kim nh gi thuyt (t, F, DW). Nhn vo gi tr F chng ta d dng nh gi c cc bin gii thch trong m hnh (thng l hi qui bi) c nh hng n bin ph thuc Y hay khng ( l kim nh gi thuyt ng thi s c trnh by Chng 8). Ngoi ra, gi tr thng k DW (s c trnh by Chng 13) s cung cp mt kim nh rt quan trng trong phn tch hi qui l kim nh xem m hnh hi qui c hin tng tng quan chui hay khng. Lu , chng ta cng c th trnh by nguyn bng kt qu Eviews km theo mt s li gii thch tng ng. /ppNH GI KT QU PHN TCH HI QUI/ppTr li Hnh 1.1 v xem v d minh ha c lng m hnh hi qui ng cu nc ng chai c phi l mt m hnh tt cha? Gujarati (2003) cho rng xem mt m hnh c tt hay khng, ta cn xem xt cc tiu ch sau y. Th nht, du ca cc h s c lng c ph hp vi l thuyt kinh t hay cha? Theo l thuyt, dc ca ng cu, B2, phi c du m. V v d ca ta l ng. Th hai, nu l thuyt cho rng mi quan h gia gi v lng cu khng ch m m cn c ngha thng k. Vy kt qu c lng ca ta vn tha mn. Th ba, m hnh hi qui c gii thch tt cho bin thin ca lng cu hay khng? Kt qu cho thy r2 rt cao, chng t bin gi di dng tuyn tnh c v ph hp. Nh vy, m hnh ta chn gii thch hnh vi tiu dng c v rt tt. Tuy nhin, trc khi c kt lun cui cng, chng ta cn phi xem liu m hnh ny c tha mn cc gi nh ca m hnh hi qui tuyn tnh c in hay khng. Chng ta s cha xem xt tt c cc gi nh c trnh by Chng 6, v y ch l mt m hnh hi qui n. Nhng c mt gi nh quan trng m chng ta cn xem xt trc. l, tnh phn phi chun ca cc hng nhiu ui. Lu rng, cc kim nh t, (2, v F u da vo gi nh cho rng cc hng nhiu ui c phn phi chun. Cho nn, nu iu ny khng xy ra th cc qui trnh kim nh trn tr nn khng c gi tr i vi mt mu tng i nh nh v d ca chng ta. C nhiu cch kim nh tnh chun, nhng y ta ch xt hai cch in hnh./pp th tn sut Histogram ca phn d/pp th tn sut l mt cng c hnh v n gin c s dng xem hnh dng ca hm PDF ca mt bin ngu nhin. Nu nhn vo th ta thy phn phi xc sut ca phn d c dng mt phn phi hnh chung, ta c th tin rng phn d c th c phn phi chun. Thc vy, t kt qu hi qui ng cu nc ng chai (by gi ta xem mt mu c 55 quan st), ta thy phn d ca m hnh ny c v c phn phi chun./pp( HNH 7.3: th tn sut ca phn d/ppp0/p

p2/p

p4/p

p6/p

p8/p

p10/p

p12/p

p-6-4-2024/p

pSeries: RESID/p

pSample 1 55/p

pObservations 55/p

pMean 4.76e-15/p

pMedian 7.11e-15/p

pMaximum 5.000000/p

pMinimum -7.000000/p

pStd. Dev. 2.775555/p

pSkewness -0.256974/p

pKurtosis 2.614992/p

pJarque-Bera 0.945022/p

pProbability 0.623435/p

/pp c th trn, ngay sau kt qu hi qui ta vo Quick/Graph/ vo nhp vo hp thoi tn bin RESID, chn OK, v thc hin mt s chnh sa nh v khung, font ch, ta c c th nh trn. /ppKim nh JB (Jarque-Bera)/ppTheo Gujarati (2003), kim nh JB v tnh chun l mt kim nh p dng cho c mu ln. Vi gi thuyt H0 cho rng phn d ca m hnh hi qui c phn phi chun, trc ht ta phi tnh cc thng k v nghing (S) v nhn (K) ca phn d, sau tnh gi tr thng k JB theo cng thc sau y: /ppdiv class="embedded" id="_1287172614"/p/p

p/p

p/p

p/p

p/p

p/p

p-/p

p+/p

p=/p

p24/p

p)/p

p3/p

pK/p

p(/p

p6/p

pS/p

pn/p

pJB/p

p2/p

p2/p

(7.32)/ppTrong , n = c mu, S = h s nghing, v K = h s nhn. Theo l thuyt thng k, mt bin c phn phi chun khi S = 0 v K = 3. Nh vy, kim nh tnh chun JB l mt kim nh ng thi vi H0: S = 0 v K = 3. Nu iu ny xy ra, JB s c k vng s bng khng./ppVi gi thuyt H0 cho rng phn d c phn phi chun, Jarque v Bera cho rng thng k JB nh cng thc (7.32) s c phn phi (2 vi s bc t do l 2 (ti sao?). Gujarati (2003) cho rng nu gi tr xc sut p ca thng k JB tng i thp, ngha l gi tr thng k JB rt khc khng, ta c th bc b gi thuyt H0 cho rng phn d c phn phi chun. Ngc li, nu gi tr xc sut p tng i cao, ngha l gi tr thng k JB rt gn vi 0, ta s khng bc b gi thuyt H0./ppVi kt qu nh trong bng thng k trn Hnh 7.3 ta c th kt lun rng phn d trong m hnh hi qui ca chng ta c th c phn phi chun v cc kt lun kim nh cc phn trc, trong chng mc no , c th tin cy./ppThao tc vi Eviews/ppQu v c th thc hin kim nh JB bng cc lnh trn Eviews theo cc hng dn sau y:/pp Tnh h s nghing: scalar s=@skew(RESID)/pp Tnh h s nhn: scalar k=@kurt(RESID)/pp Tnh thng k JB: scalar jb=55*((1/6)*s^2+(1/24)*(k-3)^2)/pp Tnh gi tr (2 ph phn: scalar chic=@qchisq(95%,2)/pp Tnh gi tr xc sut p: scalar pjb=@1-@cchisq(jb,2)/p

pNG DNG D BO/ppKt qu phng trnh hi qui (7.31) nh sau:/ppdiv class="embedded" id="_1286843980"/pi/p

pY/p

p/p

= 54.80 2.157Xi

r2 = 0.9757

(7.31)/pp se = (1.554) (0.251)

d.f. = 8/pp t = (35.255) (-11.613)

F1,8 = 134.855/pp p = (0.000) (0.000)

DW = 1.095/ppTrong ,div class="embedded" id="_1287175719"/pi/p

pY/p

p/p

l c lng ca gi tr k vng thc ca Yi (E(Yi)) tng ng vi gi tr X cho trc. Kt qu hi qui ny c th c dng lm g? Nh cp Chng 1, mt trong nhng mc tiu quan trong ca phn tch hi qui l ng dng cho d bo. Theo Gujarati (2003) c hai loi d bo:/pp D bo gi tr trung bnh c iu kin ca Y theo mt gi tr X cho trc, v d X0; ngha l mt im trn ng hi qui tng th. Loi d bo ny c gi l d bo trung bnh./pp D bo mt gi tr Y c bit no theo X0; ngha l xung quanh gi tr E(Y) c th c rt nhiu gi tr Y. Loi d bo ny c gi l d bo c bit./ppD BO TRUNG BNH/ppGi s gi bn X0 = 12 v ta mun d on gi tr trung bnh ca tng th ti mc gi ny s l bao nhiu, ngha l E(Y(X0=12). Kt qu c lng t phng trnh (7.31) cho thy gi tr c lng im ca d on trung bnh ny l div class="embedded" id="_1287175888"/p0/p

pY/p

p/p

nh sau:/ppdiv class="embedded" id="_1287175914"/p0/p

pY/p

p/p

= b1 + b2X0/pp

= 54.8 2.909 *(12) /pp

= 19.89/ppTrong :div class="embedded" id="_1287175914"/p0/p

pY/p

p/p

l c lng ca E(Y(X=X0). Nhiu nghin cu chng minh rng c lng im ny l mt c lng tuyn tnh khng chch tt nht (BLUE)./ppVdiv class="embedded" id="_1287175914"/p0/p

pY/p

p/p

l mt c lng, nndiv class="embedded" id="_1287175914"/p0/p

pY/p

p/p

c th khc gi tr thc ca n trn ng hi qui tng th. Chnh lch gia hai gi tr ny l sai s d bo. nh gi sai s d bo ny, chng ta cn tm phn phi mu cadiv class="embedded" id="_1287175914"/p0/p

pY/p

p/p

. Cho Xi = X0, gi tr d on trung bnh thc Ediv class="embedded" id="_1287176364"/p0/p

pY/p

p(/p

(div class="embedded" id="_1287176367"/p)/p

pX/p

p0/p

nh sau:/ppE(Y0(X0) = B1 + B2X0

(7.33)/ppTa c lng (7.33) t:/pp div class="embedded" id="_1287176466"/p0/p

pY/p

p/p

= b1 + b2X0

(7.34)/ppVy gi tr k vng cadiv class="embedded" id="_1287176516"/p0/p

pY/p

p/p

khi Xi = X0 s l:/ppdiv class="embedded" id="_1287176566"/p)/p

pY/p

p/p

p(/p

pE/p

p0/p

= E(b1) + X0E(b2) /pp

= B1 + B2X0

(7.35)/ppBi v b1 v b2 l cc c lng khng chch nn/ppdiv class="embedded" id="_1287176627"/p)/p

pY/p

p/p

p(/p

pE/p

p0/p

= E(Y0(X0) = B1 + B2X0

(7.36)/ppNh vy,div class="embedded" id="_1287176466"/p0/p

pY/p

p/p

l mt c lng khng chch ca E(Y0(X0)/ppKhong tin cy cho gi tr d bo trung bnh/ppTa c:/ppdiv class="embedded" id="_1287176748"/p)/p

pY/p

p/p

p(/p

pVar/p

p0/p

= div class="embedded" id="_1287176887"/p(/p

p)/p

p2/p

p0/p

p0/p

p)/p

pY/p

p/p

p(/p

pE/p

pY/p

p/p

pE/p

p-/p

/pp

= E[b1 + b2X0 B1 B2X0]2/pp = E[(b1-B1) + X0(b2-B2)]2/pp = E[(b1-B1)2 +div class="embedded" id="_1287176930"/p2/p

p0/p

pX/p

(b2-B2)2 + 2X0(b1-B1)(b2-B2)]2/pp = Var(b1) + div class="embedded" id="_1287176937"/p2/p

p0/p

pX/p

Var(b2) + 2X0Cov(b1,b2)

(7.37)/ppTrong :/ppdiv class="embedded" id="_1287176949"/ps/p

p/p

p/p

p=/p

p2/p

pi/p

p2/p

pi/p

p1/p

px/p

pn/p

pX/p

p)/p

pb/p

pvar(/p

(6.47)/ppdiv class="embedded" id="_1287176953"/p/p

ps/p

p=/p

p2/p

pi/p

p2/p

p2/p

px/p

p)/p

pb/p

pvar(/p

(6.44)/ppCov(b1,b2) = E{[b1-E(b1)][b2-E(b2)]}/pp

= E(b1-B1)(b2-B2)/pp(Do div class="embedded" id="_1287177158"/pX/p

pb/p

pY/p

pb/p

p2/p

p1/p

p-/p

p=/p

v div class="embedded" id="_1287177162"/pX/p

pB/p

pY/p

p)/p

pb/p

p(/p

pE/p

p2/p

p1/p

p-/p

p=/p

nn/ppdiv class="embedded" id="_1287177166"/p)/p

pB/p

pb/p

p(/p

pX/p

p)/p

pb/p

p(/p

pE/p

pb/p

p2/p

p2/p

p1/p

p1/p

p-/p

p-/p

p=/p

p-/p

)/pp

= div class="embedded" id="_1287177170"/p2/p

p2/p

p2/p

p)/p

pB/p

pb/p

p(/p

pX/p

p-/p

p-/p

/pp

= div class="embedded" id="_1287177173"/p)/p

pb/p

pvar(/p

pX/p

p2/p

p-/p

(6.68)/ppT (7.37), (6.44), (6.47), v (6.68) ta c:/ppdiv class="embedded" id="_1287181940"/p2/p

p0/p

pY/p

p/p

p0/p

p)/p

pY/p

p/p

p(/p

pVar/p

ps/p

p=/p

= div class="embedded" id="_1287177358"/p/p

p/p

p/p

p/p

p/p

p/p

p/p

p-/p

p-/p

p+/p

ps/p

p2/p

p2/p

p0/p

p2/p

p)/p

pX/p

pX/p

p(/p

p)/p

pX/p

pX/p

p(/p

pn/p

p1/p

(7.38)/ppBng cch thaydiv class="embedded" id="_1287177420"/p2/p

ps/p

bngdiv class="embedded" id="_1287178809"/p2/p

p/p

ps/p

ta c/pp div class="embedded" id="_1287181895"/p2/p

p0/p

pY/p

p/p

p/p

ps/p

= div class="embedded" id="_1287181974"/p/p

p/p

p/p

p/p

p/p

p/p

p/p

p-/p

p-/p

p+/p

ps/p

p2/p

p2/p

p0/p

p2/p

p)/p

pX/p

pX/p

p(/p

p)/p

pX/p

pX/p

p(/p

pn/p

p1/p

p/p

(7.39)/pp

div class="embedded" id="_1287178856"/p)/p

pY/p

p/p

p(/p

pse/p

p)/p

pY/p

p/p

p(/p

pE/p

pY/p

p/p

pt/p

p0/p

p0/p

p0/p

p-/p

p=/p


pY/p

p/p

p(/p

pse/p

p)/p

pX/p

pB/p

pB/p

p(/p

pX/p

pb/p

pb/p

p0/p

p0/p

p2/p

p1/p

p0/p

p2/p

p1/p

p+/p

p-/p

p+/p

(7.40)/pptheo phn phi t vi n- 2 bc t do. V th phn phi t c th c s dng suy ra cc khong tin cy cho gi tr k vng thc E(Y0(X0)./pp Pr[div class="embedded" id="_1287178970"/p0/p

pY/p

p/p

- t(/2se(div class="embedded" id="_1287178987"/p0/p

pY/p

p/p

) ( Y0 ( div class="embedded" id="_1287179161"/p0/p

pY/p

p/p

+ t(/2se(div class="embedded" id="_1287178987"/p0/p

pY/p

p/p

)] = 0.95

(7.41)/ppKhi X0 cng xa gi tr trung bnh th sai s d bo cng ln v khong tin cy cng rng. iu ny c ngha nu d bo c thc hin qu xa phm vi ca mu, tin cy ca d bo s gim. Nu X0 = div class="embedded" id="_1287179028"/pX/p

, khong tin cy s hp nht./ppVi X0 = 12, th div class="embedded" id="_1287176748"/p)/p

pY/p

p/p

p(/p

pVar/p

p0/p

= div class="embedded" id="_1287180600"/p17/p

p./p

p3/p

p5/p

p./p

p82/p

p)/p

p5/p

p./p

p5/p

p12/p

p(/p

p10/p

p1/p

p*/p

p18/p

p./p

p5/p

p2/p

p0/p

p=/p

p/p

p/p

p/p

p/p

p/p

p/p

p-/p

p+/p

, v div class="embedded" id="_1287179984"/p)/p

pY/p

p/p

p(/p

pse/p

p0/p

= 1.78. Vy khong tin cy 95% ca gi tr E(Y/X0) = B1 + B2X0 c tnh nh sau:/pp19.89 2.306*1.78 ( E(Y0/X=12) ( 19.89 + 2.306*1.78/pp

15.78 ( E(Y0/X=12) ( 23.99/ppD BO C BIT/ppNu ta mun d bo gi tr Y c bit, v d Y0, tng ng vi mt gi tr X cho trc, v d X0, th Y0 c xc nh nh sau:/pp

Y0 = B1 + B2X0 + u0

(7.42)/ppTa d on Y0 khi /ppdiv class="embedded" id="_1287179168"/p0/p

pY/p

p/p

= b1 + b2X0

(7.43)/ppSai s d bo, Y0 - div class="embedded" id="_1287179168"/p0/p

pY/p

p/p

c xc nh nh sau:/pp

Y0 -div class="embedded" id="_1287179168"/p0/p

pY/p

p/p

= B1 + B2X0 + u0 b1 b2X0/pp

= (B1 b1) + (B2 b2)X0 + u0

(7.44)/ppV th, vi b1 v b2 l cc c lng khng chch, ta c /pp

E(Y0 -div class="embedded" id="_1287179168"/p0/p

pY/p

p/p

) = E(B1 b1) + E(B2 b2)X0 + E(u0) = 0

(7.45)/ppKhong tin cy cho gi tr d bo c bit/ppLy bnh phng hai v ca phng trnh (7.45) ta c:/pp

E(Y0 -div class="embedded" id="_1287179168"/p0/p

pY/p

p/p

)2 = [E(B1 b1) + E(B2 b2)X0 + E(u0)]2

(7.46)/ppt f = Y0 -div class="embedded" id="_1287179168"/p0/p

pY/p

p/p

, ta /pp Var(f) = Var(b1) + div class="embedded" id="_1287182379"/p2/p

p0/p

pX/p

Var(b2) + 2X0Cov(b1,b2) + Var(u0) (7.47)/pp Var(f) = div class="embedded" id="_1287182232"/p2/p

pf/p

ps/p


pY/p

p/p

p(/p

pVar/p

p)/p

pX/p

pX/p

p(/p

p)/p

pX/p

pX/p

p(/p

pn/p

p1/p

p1/p

p0/p

p2/p

p2/p

p0/p

p2/p

p>

-

-

+

+

s

(7.48)

2

f

s

=

-

-

+

+

s

2

2

0

2

)

X

X

(

)

X

X

(

n

1

1

(7.49)

2

f

s

=

2

2

b

2

0

2

2

)

X

X

(

n

s

-

+

s

+

s

(7.50)

Ta c:

)

Y

Y

(

se

Y

Y

t

0

0

0

0

-

-

=

(7.51)

cng theo phn phi t vi bc t do d.f. = n-2. V th phn phi t c th c s dng rt ra cc suy lun thng k v gi tr thc Y0. Vi X0 = 12, ta c khong tin cy cho gi tr Y0 ti X0 = 12 nh sau:

Vi X0 = 12, th

)

Y

(

Var

0

=

35

.

8

5

.

82

)

5

.

5

12

(

10

1

1

*

18

.

5

2

0

=

-

+

+

, v

)

Y

(

se

0

= 2.89. Vy khong tin cy 95% ca gi tr Y0 c tnh nh sau:

19.89 2.306*2.89 ( Y0 ( 19.89 + 2.306*2.89

13.23 ( E(Y0/X=12) ( 26.55

Cc khong tin cy ca hai loi d bo ny c minh ha Hnh 7.4, trong , khong tin cy ca d bo c bit rng hn khong tin cy ca d bo trung bnh.

Thao tc thc hin d bo trn Eviews

Qu v c th d bo lng cu vi mt mc gi cho trc, v d X = 12, theo mt trong hai cch sau:

Cch tnh n gin bng cng thc

T kt qu hi qui, ta chn View/Representations/ v copy phng trnh, dn vo mn hnh lnh, ri th X bng 12 tnh gi tr c lng theo cng thc sau: scalar Yhat = 54.8 - 2.90909090909*12

Tnh gi tr t ph phn: scalar tc=@qtdist(0.975,8)

Tnh s quan st n: scalar n=@regobs

Tnh phng sai ca phn d: scalar vare=(@se)^2

Tnh phng sai ca b2: scalar varb2=(@stderrs(2))^2

Tnh X trung bnh: scalar Xbar=@mean(X)

Tnh phng sai ca d bo: scalar varf=vare+vare/n+((12-Xbar)^2)*varb2

Tnh sai s ca d bo: scalar sef=@sqrt(varf)

Tnh gi tr chn di ca d bo: scalar Yhat_lb=Yhat-tc*sef

Tnh gi tr chn trn ca d bo: scalar Yhat_lb=Yhat+tc*sef

Thc hin bng lnh forecast trn Eviews

T i tng tp tin Eviews, nhp p vo Range

Thay i s quan st t 10 ln 11

Nhp p vo bin X, chn Edit+/-

Nhp gi tr mi X = 12 vo quan st th 11

Chn Edit+/- kt thc bin tp

Quay li ca s kt qu c lng, nhp vo forecast

Ta c kt qu sau y:

10

20

30

40

50

60

1234567891011

YF 2 S.E.

Forecast: YF

Actual: Y

Forecast sample: 1 11

Included observations: 10

Root Mean Squared Error 2.035146

Mean Absolute Error 1.836364

Mean Abs. Percent Error 5.017189

Theil Inequality Coefficient 0.025621

Bias Proportion 0.000000

Variance Proportion 0.014406

Covariance Proportion 0.985594

( H NH 7.4: Khong tin cy ca d bo

Lu , nu qu v s dng Eviews 6.0 th kt qu hi qui c khc kt qu hi qui trn Eviews 5.0 hay 5.1 mt cht. y l kt qu c lng vi Eviews 6.0:

NGHIN CU NG DNG:

C LNG H S BETA

Chng 6 ta s dng phng php OLS c lng m hnh CAPM cho trng hp c phiu VinaMilk v ta c kt qu nh sau:

Do m hnh ch c 28 quan st, nn ta cn nh gi phn d ca m hnh ny c m bo gi nh phn phi chun hay khng. T kt qu hi qui trn Eviews, ta c th kim nh theo mt trong hai cch sau y:

S dng thng k JB

T bng kt qu hi qui, ta chn View/Residual Tests/Histogram Normality Test v ta c kt qu sau y:

0

1

2

3

4

5

6

-0.1-0.00.10.2

Series: Residuals

Sample 2006M02 2008M08

Observations 28

Mean -3.97e-18

Median -0.005601

Maximum 0.185969

Minimum -0.097673

Std. Dev. 0.067031

Skewness 0.818430

Kurtosis 3.534869

Jarque-Bera 3.459627

Probability 0.177318

Nh vy, vi xc sut p = 17.73% l tng i cao, nn ta kt lun rng phn d ca m hnh hi qui ca chng ta c th c phn phi chun v cc kim nh thng k nh t, F c th p dng c.

Ngoi ra, ta cng c th s dng gin t tng quan kim nh phn d c theo phn phi hay khng. T kt qu hi qui, ta chn View/Residual Tests/Correlogram Squared Residuals v c kt qu nh sau:

Vi h s t tng quan r1 = 0.196 khng c ngha thng k mc ngha ( = 5%, nn ta c th ni rng phn d ca m hnh l mt chui dng v s c phn phi chun.

Vigi tr t tnh ton ca h s beta l 10.287 ln hn gi tr t ph phn mc ngha 1% vi d.f. = 26 l 2.779 (=TINV(1%,26)). iu ny c ngha h s b2 = 0.846 khc khng mt cch c ngha thng k mc ngha 1%. H s xc nh r2 = 0.803 cho bit Rm gii thch c 80.3% bin thin ca Ri quanh gi tr trung bnh. Trn ca s lnh ca Eviews ta nhp cc lnh sau y v xc nh c khong tin cy 95% cho B2 nh sau:

Gi tr chn di: scalar b2_lb=c(2)-@qtdist(0.975,26)*@stderrs(2) = 0.676

Gi tr chn trn: scalar b2_ub=c(2)+@qtdist(0.975,26)*@stderrs(2) = 1.014

Vi h s t tng quan gia Ri v Rm l 0.896 (scalar rim=@cor(log(vnm/vnm(-1)),log(vni/vni(-1))) th cc h s beta chn trn v chn di s c tnh nh sau:

H s beta chn di: scalar b2a_lb= b2_lb/rim = 0.676/0.896 = 0.755

H s beta chn trn: scalar b2a_ub= b2_ub/rim = 1.014/0.896 = 1.132

Nh vy, chi ph s dng vn cho cng ty VinaMilk c th nm trong khong 14.95% (scalar coc_lb=10.8%+b2a_lb*5.5%) n 17.02% (scalar coc_ub=10.8%+b2a_ub*5.5%), vi mc trung bnh khong 15.99% (xem Chng 6).

B1

B2

b1

b2

0

0

Z

Z

2.1797

17.5346

95%

2.5%

2.5%

(2

f((2)

Cc gi tr B2 nm trong khong ny th ph hp vi gi thuyt H0 vi khong tin cy 100(1-()%. V th, khng bc b gi thuyt H0 nu B2 nm trong vng ny.

EMBED Equation.3

EMBED Equation.3

b2=-2.909 vng ph phn ny

Vng ph phn

-2.577

-1.424

b2

Vng ph phn

Vng chp nhn 95%

t =-3.636 vng ph phn ny

2.306

t

-2.306

-1.860

t

-2.465

b2

b2=-2.909 vng ph phn ny

t =-3.636 vng ph phn ny

Vng chp nhn 95%

Vng chp nhn 95%

[B2*-1.86*se(b2)]

2.1797

17.5346

95%

2.5%

2.5%

(2

f((2)

H0: B2 = 0

Con s ny c ngha xc sut gi tr tuyt i ca t ln hn 11.613. Ni cch khc Pr EMBED Equation.3

Pr(F>134.855)

Y

X

EMBED Equation.3

Khong tin cy ca gi tr Y trung bnh

Khong tin cy ca gi tr Y c bit

EMBED Equation.3

Din tch di ng phn phi chun v gia hai gi tr trung bnh tr/cng mt lch khong 68%; gia tr/cng hai lch khong 95%; v gia trung bnh tr/cng ba lch khong 99.7.

Lu , EMBED Equation.3 l mt gi tr nht nh bit di gi thuyt H0. Cho nn, cng thc (7.22) v (7.12) c im khc nhau. Hn na, khong tin cy cng thc (7.23) cng khc khong tin cy cng thc (7.13).

PAGE

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