kinh tế lương căn bản

Chng trnh Ging dy Kinh t Fulbright Nin kha 2010-2012

Cc phng php nh lng Bi c

Nhp mn Kinh t lng vi cc ng dng 5th ed. Ch. 3: M hnh hi qui tuyn tnh n

Ramu Ramanathan 1 Ngi dch: Thc oan Hiu nh: Cao Ho Thi

Chng 3

M Hnh Hi Quy Tuyn Tnh n

chng 1 pht biu rng bc u tin trong phn tch kinh t lng l vic thit lp m hnh m t c hnh vi ca cc i lng kinh t. Tip theo nh phn tch kinh t/ kinh doanh s thu thp nhng d liu thch hp v c lc m hnh nhm h tr cho vic ra quyt nh. Trong chng ny s gii thiu m hnh n gin nht v pht trin cc phng php c lng, phng php kim nh gi thuyt v phng php d bo. M hnh ny cp n bin c lp (Y) v mt bin ph thuc (X). chnh l m hnh hi quy tuyn tnh n. Mc d y l mt m hnh n gin, v v th phi thc t, nhng vic hiu bit nhng vn c bn trong m hnh ny l nn tng cho vic tm hiu nhng m hnh phc tp hn. Thc t, m hnh hi quy n tuyn tnh c th gii thch cho nhiu phng php kinh t lng. Trong chng ny ch a ra nhng kt lun cn bn v m hnh hi quy tuyn tnh n bin. Cn nhng phn khc v phn tnh ton s c gii thiu phn ph lc. V vy, i vi ngi c c nhng kin thc cn bn v ton hc, nu thch, c th c phn ph lc hiu r hn v nhng kt qu l thuyt.

3.1 M Hnh C Bn

Chng 1 trnh by v d v m hnh hi quy n cp n mi lin h gia gi ca mt ngi nh v din tch s dng (xem Hnh 1.2). Chn trc mt s loi din tch, v sau lit k s lng nh c trong tng th tng ng vi tng din tch chn. Sau tnh gi bn trung bnh ca mi loi nh v v th (quy c cc im c biu th l X). Gi thuyt c bn trong m hnh hi quy tuyn tnh n l cc tr trung bnh ny s nm trn mt ng thng (biu th bng + SQFT), y l hm hi quy ca tng th v l trung bnh c iu kin (k vng) ca GI theo SQFT cho trc. Cng thc tng qut ca m hnh hi quy tuyn tnh n da trn Gi thit 3.1 s l

GI THIT 3.1 (Tnh Tuyn Tnh ca M Hnh)

Yt = + Xt + ut (3.1)

trong , Xt v Yt l tr quan st th t (t = 1 n n) ca bin c lp v bin ph thuc, tip theo v l cc tham s cha bit v s c c lng; v ut l s hng sai s khng quan st c v c gi nh l bin ngu nhin vi mt s c tnh nht nh m s c cp k phn sau. v c gi l h s hi quy. (t th hin thi im trong chui thi gian hoc l tr quan st trong mt chui d liu cho.)

Thut ng n trong m hnh hi quy tuyn tnh n c s dng ch rng ch c duy nht mt bin gii thch (X) c s dng trong m hnh. Trong chng tip theo khi ni v m hi quy a bin s b sung thm nhiu bin gii thch khc. Thut ng hi quy xut pht t Fraccis Galton (1886), ngi t ra mi lin h gia chiu cao ca nam vi

Chng trnh Ging dy Kinh t Fulbright




chiu cao ca ngi cha v quan st thc nghim cho thy c mt xu hng gia chiu cao trung bnh ca nam vi chiu cao ca nhng ngi cha ca h hi quy (hoc di chuyn) cho chiu cao trung bnh ca ton b tng th. + Xb gi l phn xc nh ca m hnh v l trung bnh c iu kin ca Y theo X, l E(YtXt) = + Xt. Thut ng tuyn tnh dng ch rng bn cht ca cc thng s ca tng th v l tuyn tnh (bc nht) ch khng phi l Xt tuyn tnh. Do , m hnh ttt uXY ++= 2 vn c gi l hi quy quyn tnh n mc du c X bnh phng. Sau y l v d v phng trnh hi quy phi tuyn tnh Yt = + X + ut. Trong cun sch ny s khng cp n m hnh hi quy phi tuyn tnh m ch tp trung vo nhng m hnh c tham s c tnh tuyn tnh m thi. Nhng m hnh tuyn tnh ny c th bao gm cc s hng phi tuyn tnh i vi bin gii thch (Chng 6). nghin cu su hn v m hnh hi quy phi tuyn tnh, c th tham kho cc ti liu: Greene (1997), Davidson v MacKinnon (1993), v Griffths, Hill, v Judg (1993).

S hng sai s ut (hay cn gi l s hng ngu nhin) l thnh phn ngu nhin khng quan st c v l sai bit gia Yt v phn xc nh + Xt. Sau y mt t hp ca bn nguyn nhn nh hng khc nhau:

1. Bin b st. Gi s m hnh thc s l Yt = + Xt + Zt +vt trong , Zt l mt bin gii thch khc v vt l s hng sai s thc s, nhng nu ta s dng m hnh l Y = + Xt +ut th ut = Zt +vt. V th, ut bao hm c nh hng ca bin Z b b st. Trong v d v a c phn trc, nu m hnh thc s bao gm c nh hng ca phng ng v phng tm v chng ta b qua hai nh hng ny m ch xt n din tch s dng th s hng u s bao hm c nh hng ca phng ng v phng tm ln gi bn nh.

2. Phi tuyn tnh. ut c th bao gm nh hng phi tuyn tnh trong mi quan h gia Y v X. V th, nu m hnh thc s l tttt uXXY +++=

2 , nhng li c gi nh

bng phng trnh Y = + Xt +ut , th nh hng ca 2tX s c bao hm trong ut. 3. Sai s o lng. Sai s trong vic o lng X v Y c th c th hin qua u. V d,

gi s Yt gi tr ca vic xy dng mi v ta mun c lng hm Yt = + rt +vt trong rt l li sut n vay v vt l sai s tht s ( n gin, nh hng ca thu nhp v cc bin khc ln u t u c loi b). Tuy nhin khi thc hin c lng, chng ta li s dng m hnh Yt = + Xt +ut trong Xt = rt +Zt l li sut cn bn. Nh vy th li sut c o lng trong sai s Zt thay rt = Xt Zt vo phng trnh ban u, ta s c

Yt = +(Xt Zt) +vt = + Xt Zt + vt = + Xt + ut Cn lun lu rng tnh ngu nhin ca s hng ut bao gm sai s khi o lng li sut n vay mt cch chnh xc.

4. Nhng nh hng khng th d bo. D l mt m hnh kinh t lng tt cng c th chu nhng nh hng ngu nhin khng th d bo c. Nhng nh hng ny s lun c th hin qua s hng sai s ut.

Nh cp ban u, vic thc hin iu tra ton b tng th xc nh hm hi quy ca tng th l khng thc t. V vy, trong thc t, ngi phn tch thng chn mt mu bao gm cc cn nh mt cch ngu nhin v o lng cc c tnh ca mu ny thit lp hm hi quy cho mu. Bng 3.1 trnh by d liu ca mt mu gm 14





nh bn trong khu vc San Diego. S liu ny c sn trong a mm vi tn tp tin l DATA3-1. Trong Hnh 3.1, cc cp gi tr (Xt, Yt) c v trn th. th ny c gi l th phn tn ca mu cho cc d liu. Hnh 3.1 tng t nh Hnh 1.2, nhng trong Hnh 1.2 lit k ton b cc gi tr (Xt, Yt) ca tng th, cn trong Hnh 3.1 ch lit k d liu ca mu m thi. Gi s, ti mt thi im, ta bit c gi tr ca v . Ta c th v c ng thng + X trn biu . y chnh l ng hi quy ca tng th. Khong cch chiu thng xung t gi thc (Yt) n ng hi quy + X l sai s ngu nhin ut. dc ca ng thng () cng l Y/X, l lng thay i ca Y trn mt n v thay i ca X. V vy c din dch l nh hng cn bin ca X ln Y. Do , nu l l 0.14, iu c ngha l mt mt vung din tch tng thm s lm tng gi bn nh ln, mc trung bnh, 0.14 ngn la (lu n v tnh) hay 140 la. Mt cch thc t hn, khi din tch s dng nh tng thm 100 mt vung th hy vng rng gi bn trung bnh ca ngi nh s tng thm $14.000 la. Mc du l tung gc v l gi tr ca tr trung bnh Y khi X bng 0, s hng ny vn khng th c hiu nh l gi trung bnh ca mt l t trng. Nguyn nhn l v cng n cha bin b st v do khng c cch gii thch cho (iu ny c cp k hn trong Phn 4.5).

BNG 3.1 Gi tr trung bnh c lng v trung bnh thc t ca gi nh v din tch s dng (mt vung)

t SQFT Gi bn1 Gi trung bnh c lng2

1 1.065 199,9 200,386 2 1.254 288 226,657 3 1.300 235 233,051 4 1.577 285 271,554 5 1.600 239 274,751 6 1.750 293 295,601 7 1.800 285 302,551 8 1.870 365 312,281 9 1.935 295 321,316

10 1.948 290 323,123 11 2.254 385 365,657 12 2.600 505 413,751 13 2.800 425 441,551 14 3.000 415 469,351

HNH 3.1 Biu Phn Tn Ca Mu Trnh By Mi Lin H Gia Gi v SQFT

1 n v tnh: 1.000 la

2 Phng php tnh gi trung bnh c lng s c trnh by Phn 3.2





Y

X0

+ X

tX +

( )tt

YX ,

tu

tX

100

200

300

400

500

600

1000 1400 1800 2200 2600 3000

HNH 3.2 Phng Trnh Hi Quy ca Tng Th v ca Mu

Y

X

X +D

C

B

0 A

(Hoi qui tong the) + X

(Hoi qui mau)

tt XY += ( )ttt XYEX |=+

( )tt YX ,tu

tu

Mc tiu u tin ca mt nh kinh t lng l lm sao s dng d liu thu thp c c lng hm hi quy ca tng th, l, c lng tham s ca tng th v . K hiu l c lng mu ca v l c lng mu ca . Khi mi quan h trung bnh c lng l Y^ = ^ + ^X. y c gi l hm hi quy ca mu. ng vi mt gi tr quan st cho trc t, ta s c Y^ t = ^ + ^Xt. y l gi tr d bo ca Y vi mt gi tr cho trc l Xt. Ly gi tr quan st c Yt tr cho gi tr ny, ta s c c lng ca ut c gi l phn d c lng, hoc n gin l phn d, v k hiu l tu 1v c th hin trong phng trnh sau:

u^t = Yt Y^ t = Yt ^ ^Xt

Sp xp li cc s hng trn, ta c

1 Mt s tc gi v ging vin thch s dng a thay cho ^, b thay cho ^ v et thay cho u^t. Chng ta s dng du hiu

^ theo qui nh trong l thuyt thng k v n gip phn bit r rng gia gi tr tht v gi tr c lng v cng xc nh c thng s ang c c lng.





ttt uXY ++= (3.3)

Vic phn bit gia hm hi quy ca tng th Y = + X v hm hi quy ca mu XYt += l rt quan trng. Hnh 3.2 trnh by c hai ng v sai s v phn d (cn

nghin cu k vn ny). Lu rng ut l k hiu ch sai s, v tu l k hiu ch phn d.

BI TP 3.1 Xem xt cc phng trnh sau y:

a. tt uXY ++= b. tt uXY ++= c. tt uXY ++= d. XYt += e. tt uXY ++= f. tt uXY ++=

Gii thch k ti sao phng trnh (a) v (b) ng, nhng (c), (d), (e) v (f) sai. Hnh 3.2 rt c ch trong vic tr li cu hi ny.

3.2 c lng m hnh c bn bng phng php bnh phng ti thiu thng thng

Trong phn trc, nu r m hnh hi quy tuyn tnh c bn v phn bit gia hi quy ca tng th v hi quy ca mu. Mc tiu tip theo s l s dng cc d liu X v Y v tm kim c lng tt nht ca hai tham s ca tng th l v . Trong kinh t lng, th tc c lng c dng ph bin nht l phng php bnh phng ti thiu. Phng php ny thng c gi l bnh phng ti thiu thng thng, phn bit vi nhng phng php bnh phng ti thiu khc s c tho lun trong cc chng sau. K hiu c lng ca v l v , phn d c lng th bng

ttt XYu = . Tiu chun ti u c s dng bi phng php bnh phng ti thiu l cc tiu ha hm mc tiu

2

11

2 )(),( tnt

tt

nt

tt XYuESS ==

=

=

=

=

vi cc tham s cha bit l v . ESS l tng cc phn d bnh phng v phng php OLS cc tiu tng cc phn d bnh phng2. Cn nn lu rng ESS l khong

2 Rt d nhm khi gi ESS l tng ca cc phn d bnh phng, nhng k hiu ny c s

dng ph bin trong nhiu chng trnh my tnh ni ting v c t ti liu v Phn tch phng sai





cch bnh phng c o lng t ng hi quy. S dng khong cch o lng ny, c th ni rng phng php OLS l tm ng thng gn nht vi d liu trn th.

Trc quan hn, gi s ta chn mt tp hp nhng gi tr v , l mt ng thng X . C th tnh c lch ca Yt t ng thng c chn theo phn d c lng XYu tt = . Sau bnh phng gi tr ny v cng tt c cc gi tr bnh phng ca ton b mu quan st. Tng cc phn d bnh phng ca cc tr quan st [c xem nh tng bnh phng sai s (ESS)] do s bng 2tu . Tng ng vi mt im trn ng thng s c mt mt tr tng bnh phng sai s. Phng php bnh phng ti thiu chn nhng gi tr v sao cho ESS l nh nht.

Vic bnh phng sai s t c hai iu sau. Th nht, bnh phng gip loi b du ca sai s v do xem sai s dng v sai s m l nh nhau. Th hai, bnh phng to ra s bt li cho sai s ln mt cch ng k. V d, gi s phn d ca mu l 1, 2, 1 v 2 ca h s hi quy chn trc tr v chn trc. So snh cc gi tr ny vi mt mu khc c phn d l 1, 1, 1 v 3. Tng gi tr sai s tuyt i c hai trng hp l nh nhau. Mc d mu chn th hai c sai s tuyt i thp hn t 2 n 1, iu ny dn n sai s ln khng mong mun l 3. Nu ta tnh ESS cho c hai trng hp th ESS ca trng hp u l 10 (12 + 22+ 12+ 22), ESS cho trng hp sau l 12 (12 + 12+ 12+ 32). Phng php bnh phng ti thiu p t s bt li ln cho sai s ln v do ng thng trong trng hp u s c chn. Phn 3.3 s tip tc trnh by nhng c tnh cn thit khc ca phng php cc tiu ESS.

Phng Php Thch Hp Cc i

Phn ny ch cp s v phng php thch hp cc i. Phng php ny s c trnh by chi tit phn 2.A.4. Phn 3.A.5 s trnh by nguyn tc p dng m hnh hi quy tuyn tnh n. Mc d phng php thch hp cc i da trn mt tiu chun ti u khc, nhng cc thng s c lng vn ging nh cc thng s c lng phng php OLS. Ni n gin, phng php thch hp cc i chn c lng sao cho xc sut xy ra ca mu quan st l ln nht.

Phn tho lun trc cho thy nu thc hin hai phng php c lng v khc nhau mt cch chnh xc th u dn n cng mt kt qu. Nh vy th ti sao cn phi xem xt c hai phng php? Cu tr li l trong cc chng sau, ta s thy rng khi mt s gi thit ca m hnh c gim nh, th thc t, hai phng php c lng khc nhau s cho kt qu khc nhau. Mt phng php khc c th cho kt qu khc na, l phng php cc tiu tng sai s tuyt i tu . Nhng phng php ny khng c dng ph bin trong kinh t lng v kh tnh ton.

Phng Trnh Chun

Trong phn 3.A.3 ca ph lc, phng php OLS c chnh thc p dng. Phn ny cho thy rng iu kin cc tiu ESS vi v s theo hai phng trnh sau y, c gi l phng trnh chun (khng c lin h g n phn phi chun).





=== ttttt XnYXYu )()(0 (3.4)

)]([)( ttttt XYXuX = = 0 (3.5) Trong Phng trnh (3.4), cn lu rng = n bi v mi s hng s c mt v

c n s hng. Chuyn v cc s hng m trong Phng trnh (3.4) sang phi v chia mi s hng cho n, ta c

+= tt Xn

Yn

11 (3.6)

(1/n)Yt l trung bnh mu ca Y, k hiu l Y , v (1/n)Yt l trung bnh mu ca X, k hiu l X . S dng kt qu ny thay vo Phng trnh (3.6), ta c phng trnh sau

XY += (3.7)

ng thng ^ +^ X l ng c lng v l ng hi quy ca mu, hoc ng thng thch hp. C th thy rng t Phng trnh (3.7) ng hi quy ca mu i qua im trung bnh ( )YX , . Trong Bi tp 3.12c, ta s thy rng tnh cht ny khng m bo tr khi s hng hng s c trong m hnh.

T Phng trnh (3.5), cng tt c theo tng s hng, v a v ra lm tha s chung, ta c

0)( 2 = tttt XXYX hay

+= 2)( tttt XXYX (3.8)

Li Gii v Phng Trnh Chun

thun li cho vic p n v hai phng trnh chun, cc tnh cht sau y l rt cn thit. Nhng tnh cht ny c chng minh trong Ph lc Phn 3.A.2

TNH CHT 3.1 Sxx = (Xt X)2 = Xt2 nX)2 = Xt2 1n(Xt)

2

TNH CHT 3.2 Sxy = (Xt X)(Yt Y) = (XtYt) n XY

= XtYt [(Xt ) (Yt) / n]





T Phng trnh (3.7),

== tt Xn

Yn

XY 11 (3.9)

Thay vo (3.8)

+

=

2)(11 tttttt XXXn

Yn

YX

Nhm cc s hng c tha s :

( )( ) ( )

+

=

n

XX

n

YXYX tt

tttt

22

Tm ta c ( )( )

( )

=

n

XX

n

YXYX

tt

tttt

22

S dng k hiu n gin c gii thiu Tnh cht 3.1 v 3.2, c th c din t nh sau

xx

xy

SS

= (3.10) trong

( )n

XXS ttxx

22 = (3.11)

v ( )( )n

YXYXS ttttxy

= (3.12)

K hiu Sxx v Sxy c th c nh mt cch trc quan nh sau, nh ngha XXx tt = v YYy tt = , trong k hiu thanh ngang ch trung bnh ca mu. Do

xt v yt k hiu lch gia X v Y so vi gi tr X v Y trung bnh. Kt qu sau y s c chng minh phn Ph lc Phn 2.A.1 v 3.A.2.

xt = 0

( )2222 1)( === ttttxx Xn

XXXxS (3.13)





( )( )[ ] === ttttttttxy YXn

YXYYXXyxS 1))(( (3.14)

Sxy l tng cc gi tr ca xt nhn yt . Tng t, Sxx tng cc gi tr ca xt nhn xt , hay tng ca xt bnh phng

Phng trnh (3.9) v (3.10) l li gii cho phng trnh chun [(3.4) v (3.5)] v cho ta c lng v ca mu cho tham s v ca tng th.

Cn lu rng khng th xc nh c c lng ca trong Phng trnh (3.10) nu 0)( 22 === XXxS ttxx . Sxx bng khng khi v ch khi mi xt bng khng, c ngha l khi v ch khi mi Xt bng nhau. iu ny dn n gi thuyt sau y

GI THIT 3.2 (Cc Gi Tr Quan St X L Khc Nhau) Khng phi l tt c gi tr Xt l bng nhau. C t nht mt gi tr Xt khc so vi nhng gi tr cn li. Ni cch khc, phng sai ca mu 2)(

11)( XX

nXVar t

= khng

c bng khng.

y l mt gi thit rt quan trng v lun lun phi tun theo bi v nu khng m hnh khng th c lng c. Mt cch trc quan, nu Xt khng i, ta khng th gii thch c ti sao Yt thay i. Hnh 3.3 minh ha gi thuyt trn bng hnh nh. Trong v d v a c, gi s thng tin thu thp ch tp trung mt vo loi nh c din tch s dng l 1.500 mt vung. th phn tn ca mu s c th hin nh Hnh 3.3. T th c th thy r rng d liu ny khng y cho vic c lng ng hi quy tng th +X.

HNH 3.3 V D v Gi Tr X Khng i Y

X0 1,500

V d 3.1

Theo thut ng c dng ph bin trong kinh t lng, nu ta s dng d liu trong Bng 3.1 v thc hin hi quy Y (GI) theo s hng hng s v X (SQFT), ta c th xc nh c mi quan h c lng (hay hm hi quy ca mu) l

tt XY 13875351,0351,52 += . tY l gi c lng trung bnh (ngn la) tng ng





vi Xt. (xem Bng 3.1). H s hi quy ca Xt l nh hng cn bin c lng ca din tch s dng n gi nh, mc trung bnh. Do vy, nu din tch s dng tng ln mt n v, gi trung bnh c lng k vng s tng thm 0,13875 ngn la ($138.75). Mt cch thc t, c mi 100 mt vung tng thm din tch s dng, gi bn c lng c k vng tng thm, mc trung bnh, $ 13.875.

Hm hi quy ca mu c th c dng c lng gi nh trung bnh da trn din tch s dng cho trc (Bng 3.1 c trnh by gi trung bnh ct cui.) Do , mt cn nh c din tch 1.800 mt vung th gi bn k vng trung bnh l $302.551[ = 52,351 + (0,139 1.800)]. Nhng gi bn thc s ca cn nh l $285.000. M hnh c lng gi bn vt qu $17.551. Ngc li, i vi mt cn nh c din tch s dng l 2.600 mt vung, gi bn trung bnh c lng l $413.751, thp hn gi bn thc s $505.000 mt cch ng k. S khc bit ny c th xy ra bi v chng ta b qua cc yu t nh hng khc ln gi bn nh. V d, mt ngi nh c sn vn rng v/ hay h bi, s c gi cao hn gi trung bnh. iu ny nhn mnh tm quan trng trong vic nhn din c cc bin gii thch c th nh hng n gi tr ca bin ph thuc v a cc nh hng ny vo m hnh c thit lp. Ngoi ra, rt cn thit trong vic phn tch tin cy ca cc c lng ca tung v h s dc trong Phng trnh (3.1), v mc thch hp ca m hnh i vi d liu thc t.

BI TP 3.2 Sao chp hai ct s liu trong Bng 3.1 vo mt bng mi. Trong ct u tin ca bng tnh sao chp cc gi tr v Yt (GI) v Xt (SQFT) trong ct th hai. S dng my tnh v tnh thm gi tr cho hai ct khc. Bnh phng tng gi tr trong ct th hai v in gi tr vo ct th ba (x). Nhn ln lt tng gi tr ct th nht vi gi tr tng ng ct hai v in kt qua vo ct th t (XtYt). Tip theo, tnh tng ca tng ct v nh gi cc tng sau y:

753.26= tX 515.462.552

= tX

9,444.4= tY 5,985.095.92

= tY

trnh tnh trng qu nhiu v sai s lm trn, cn s dng cng nhiu s thp phn cng tt. Sau , tnh Sxy t Phng trnh (3.12) v Sxx t Phng trnh (3.11). Cui cng, tnh theo (3.10) v theo (3.9) v kim tra li nhng gi tr trnh by ban u.

3.3 Tnh cht ca cc c lng

Mc d phng php bnh phng cho ra kt qu c lng v mi quan h tuyn tnh c th ph hp vi d liu sn c, chng ta cn tr li mt s cu hi sau. V d, c tnh thng k ca v ? Thng s no c dng o tin cy ca v ? Bng cch no c th s dng v kim nh gi thuyt thng k v thc hin d bo? Sau y chng ta s i vo tho lun tng vn trn. S rt hu ch nu bn n li Phn 2.6, phn ny a ra tm tt v nhng tnh cht cn thit ca thng s c lng.





Tnh cht u tin cn xem xt l khng thin lch. Cn lu rng trong Phn 2.4 cc thng s c lng v ? t thn chng l bin ngu nhin v do tun theo phn phi thng k. Nguyn nhn l v nhng ln th khc nhau ca mt cuc nghin cu s cho cc kt qu c lng thng s khc nhau . Nu chng ta lp li nghin cu vi s ln th ln, ta c th t c nhiu gi tr c lng. Sau chng ta c th tnh t s s ln m nhng c lng ny ri vo mt khong gi tr xc nh. Kt qu s s cho ra phn phi ca cc c lng ca mu. Phn phi ny c gi tr trung bnh v phng sai. Nu trung bnh ca phn phi mu l thng s thc s (trong trng hp ny l hoc ), th y l c lng khng thin lch. khng thin lch r rng l iu lun c mong mun bi v, iu c ngha l, mc trung bnh, gi tr c lng s bng vi gi tr thc t, mc d trong mt s trng hp c bit th iu ny c th khng ng.

C th ni rng thng s c lng OLS ca v a ra trong Phn 3.2 c tnh cht khng thin lch. Tuy nhin, chng minh iu ny, chng ta cn t ra mt s gi thuyt b sung v Xt v ut. Cn nh rng, mc d Gi thit 3.1 c th v c gim nh phn sau, nhng Gi thuyt 3.2 v 3.3 l lun lun cn thit v phi tun theo. Sau y l cc gi thit b sung cn thit.

GI THIT 3.3 (Sai S Trung Bnh bng Zero) Mi l u mt bin ngu nhin vi E(u) = 0

Trong Hnh 3.1 cn lu rng mt s im quan st nm trn ng + X v mt s im nm di. iu ny c ngha l c mt gi tr sai s mang du dng v mt s sai s mang du m. Do + X l ng trung bnh, nn c th gi nh rng cc sai s ngu nhin trn s b loi tr nhau, mc trung bnh, trong tng th. V th, gi nh rng ut l bin ngu nhin vi gi tr k vng bng 0 l hon ton thc t.

GI THIT 3.4 (Cc Gi Tr X c Cho Trc v Khng Ngu Nhin) Mi gi tr Xt c cho trc v khng l bin ngu nhin. iu ny ngm ch rng ng phng sai ca tng th gia Xt v ut, Cov(Xt, ut) = E(Xt, ut) E(Xt)E(ut) = XtE(ut) XtE(ut) = 0. Do gia Xt v ut khng c mi tng quan (xem nh ngha 2.4 v 2.5).

Theo trc gic, nu X v u c mi tng quan, th khi X thay i, u cng s thay i. Trong trng hp ny, gi tr k vng ca Y s khng bng + X. Nu gi tr X l khng ngu nhin th gi tr k vng c iu kin ca Y theo gi tr X s bng + X. Kt qu ca vic vi phm Gi thit 3.4 s c trnh by trong phn sau, c bit l khi nghin cu m hnh h phng trnh (Chng 13). Tnh cht 3.3 pht biu rng khi hai gi thit c b sung, thng s c lng OLS l khng thin lch.

TNH CHT 3.3 ( Khng Thin Lch)

Trong hai gi thit b sung 3.3 v 3.4, [E(ut) = 0, Cov(Xt, ut) = 0], thng s c lng, thng s c lng bnh phng ti thiu v l khng thin lch; ngha l

( ) =E , v ( ) =E .





CHNG MINH (Nu c gi khng quan tm n chng minh, c th b qua phn).

T Phng trnh (3.10), ( ) ( )xxxy SSEE = . Nhng theo Gi thuyt 3.4, Xt l khng ngu nhin v do Sxx cng khng ngu nhin. iu ny c ngha l khi tnh gi tr k vng, cc s hng lin quan n Xt c th c a ra ngoi gi tr k vng. V vy, ta c

( ) ( )xy

xx

SES

E 1 = . Trong Phng trnh (3.12), thay Yt t Phng trnh (3.1) v thay bng n .

( ) ( )( )

++++=

n

uXnXuXXS ttttttxy

(3.15)

( ) ( )( )

++=

n

uXn

XXuXXX tttttttt

22

( ) ( )( )

+

=

n

uXuX

n

XX tttt

tt

2

xuxx SS += trong Sxx c cho bi Phng trnh (3.13) v

( )( )n

uXuXS ttttxu

= (3.16)

( ) ttttt uXXuXuX ==

X l trung bnh mu ca X, Xt l khng ngu nhin, X xut hin mi s hng, v k vng ca tng cc s hng th bng tng cc gi tr k vng. Do vy,

( ) ( ) ( ) ( ) ( ) 0=== ttttttxu uEXuEXuEXuXESE

theo Gi thit 3.3. Do , E(Sxy) = Sxx, ngha l ( ) == xxxy SSEE )( . Nh vy l c lng khng thin lch ca . Chng minh tng t cho ^. Cn nhn thy rng vic chng minh khng thin lch ph thuc ch yu vo Gi thit 3.4. Nu E(Xtut) 0, c th b thin lch.

BI TP 3.3 S dng Phng trnh (3.9) chng minh rng l khng thin lch. Nu r cc gi thuyt cn thit khi chng minh.





Mc du khng thin lch lun l mt tnh cht lun c mong mun, nhng t bn thn khng thin lch khng lm cho thng s c lng tt, v mt c lng khng thin lch khng ch l trng hp c bit. Hy xem xt v d sau v mt thng s c lng khc l ~ = (Y2 Y1)/(X2 X1). Lu rng ~ n gin l dc ca ng thng ni hai im (X1, Y1) v (X2, Y2). Rt d nhn thy rng ~ l khng thin lch

( ) ( )12

12

12

1122

12

12~

XXuu

XXuXuX

XXYY

+=

++++=

=

Nh ni trc y, cc gi tr X l khng ngu nhin v E(u2) = E(u1) = 0. Do , ~ l khng thin lch. Thc ra, ta c th xy dng mt chui v hn ca cc thng s c lng khng thin lch nh trn. Bi v ~ loi b cc gi tr quan st t 3 n n, mt cch trc quan y khng th l mt thng s c lng tt. Trong Bi tp 3.6, tt c cc gi tr quan st c s dng th thit lp cc thng s c lng khng thin lch khc, nhng tng t nh trn y khng phi l l thng s c lng khng thin lch tt nht. Do , rt cn c nhng tiu chun b sung nh gi tt ca mt thng s c lng.

Tiu chun th hai cn xem xt l tnh nht qun, y l mt tnh cht ca mu ln c nh ngha trong Phn 2.6 (nh ngha 2.10). Gi s ta chn ngu nhin mt mu c n phn t v i tm v . Sau chn mt mu ln hn v c lng li cc thng s ny. Lp li qu trnh ny nhiu ln c c mt chui nhng thng s c lng. Tnh nht qun l tnh cht i hi cc thng s c lng vn ph hp khi c mu tng ln v hn. c lng ~ c trnh by trn r rng l khng t c tnh nht qun bi v khi c mu tng ln khng nh hng g n thng s ny. Tnh cht 3.4 pht biu cc iu kin mt c lng c tnh nht qun.

TNH CHT 3.4 (Tnh Nht Qun)

Theo Gi thit (3.2), (3.3) v (3.4), c lng bnh phng ti thiu c tnh cht nht qun. Do , iu kin t c tnh nht qun l E(ut) = 0, Cov(Xt, ut) = 0 v Var(Xt) 0.

CHNG MINH (Nu c gi khng quan tm, c th b qua phn ny.)

T Phng trnh (3.15) v (3.10)

nSnS

xx

xu

//

+= (3.17)

Theo quy lut s ln (Tnh cht 2.7a), Sxu/n ng quy vi k vng ca chnh n, l Cov(X, u). Tng t, Sxx/n ng quy vi Var(X). Do vy dn ti iu, nu n hi t n v





cng, s ng quy vi + [Cov(X,u)/Var(X), v s bng nu Cov(X,u) = 0 ngha l nu X v u khng tng quan. Nh vy, l c lng nht qun ca .

Mc d l khng thin lch v nht qun, vn c nhng tiu chun cn b sung bi c th xy dng c lng nht qun v khng thin lch khc. Bi tp 3.6 l mt v d v loi c lng . Tiu chun s dng tip theo l tnh hiu qu (nh ngha trong Phn 2.6). Ni mt cch n gin, c lng khng thin lch c tnh hiu qu hn nu c lng ny c phng sai nh hn. thit lp tnh hiu qu, cn c cc gi thit sau v ut.

GI THIT 3.5 (Phng sai ca sai s khng i) Tt c gi tr u c phn phi ging nhau vi cng phng sai 2, sao cho ( ) 22)( == tt uEuVar . iu ny c gi l phng sai ca sai s khng i (phn tn u).

GI THIT 3.6 (c Lp Theo Chui) Gi tr u c phn phi c lp sao cho Cov(ut, us) = E(utus) = 0 i vi mi t s. y c gi l chui c lp.

Cc gi thit trn ngm ch rng cc phn d phn c phn phi ging nhau v phn phi c lp (iid). T Hnh 1.2 ta thy rng ng vi mt gi tr X s c mt gi tr phn phi Y xc nh phn phi c iu kin. Sai s ut l lch t trung bnh c iu kin + Xt. Gi thit 3.5 ngm nh rng phn phi ca ut c cng phng sai (2) vi phn phi ca us cho mt quan st khc s. Hnh 3.4a l mt v d v phng sai ca sai s thay i (hoc khng phn tn u) khi phng sai thay i tng theo gi tr quan st X. Gi thuyt 3.5 c gim nh trong Chng 8. Phn 3.6 Ph chng c trnh by m t ba chiu ca gi thuyt ny.

Gi thit 3.6 (s c gim nh trong Chng 9) ngm nh rng l ut v us c lp v do vy khng c mi tng quan. C th l, cc sai s lin tip nhau khng tng quan nhau v khng tp trung. Hnh 3.4b l mt v d v t tng quan khi gi thuyt trn b vi phm. Ch rng khi cc gi tr quan st k tip nhau tp trung li, th c kh nng cc sai s s c tng quan.

HNH 3.4 V D v Phng Sai Ca Sai S Thay i v T Hi Quy Y

X

a. Phng sai ca sai s thay i





Y

X

b. T hi quy

TNH CHT 3.5 (Hiu qu, BLUE v nh l Gauss-Markov)

Theo Gi thit 3.2 n 3.6, c lng bnh phng ti thiu thng thng (OLS) l c lng tuyn tnh khng thin lch c hiu qu nht trong cc c lng. V th phng php OLS a ra c Lng Khng Thin lch Tuyn Tnh Tt Nht (BLUE).

Kt qu ny (c chng minh trong Phn 3.A.4) c gi l nh l GaussMarkov, theo l thuyt ny c lng OLS l BLUE; ngha l trong tt c cc t hp tuyn tnh khng thin lch ca Y, c lng OLS ca v c phng sai b nht.

Tm li, p dng phng php bnh phng ti thiu (OLS) c lng h s hi quy ca mt m hnh mang li mt s tnh cht mong mun sau: c lng l (1) khng thin lch, (2) c tnh nht qun v (3) c hiu qu nht. khng thin lch v tnh nht qun i hi phi km theo Gi thuyt E(ut) = 0 v Cov(Xt, ut) = 0. Yu cu v tnh hiu qu v BLUE, th cn c thm gi thuyt, Var(ut) = 2 v Cov(ut, us) = 0, vi mi t s.

3.4 Chnh Xc ca c Lng v Mc Thch Hp ca M Hnh

S dng cc d liu trong v d v a c ta c lng c thng s nh sau 351.52 =v 13875,0 = . Cu hi c bn l cc c lng ny tt nh th no v mc thch hp ca hm hi quy mu XYt 13875351,0351,52 += vi d liu ra sao. Phn ny s tho lun phng php xc nh thng s o lng chnh xc ca cc c lng cng nh ph hp.

Chnh Xc ca Cc c Lng

T l thuyt xc sut ta bit rng phng sai ca mt bin ngu nhin o lng s phn tn xung quanh gi tr trung bnh. Phng sai cng b, mc trung bnh, tng gi tr ring bit cng gn vi gi tr trung bnh. Tng t, khi cp n khong tin cy, ta bit rng phng sai ca bin ngu nhin cng nh, khong tin cy ca cc tham s cng b. Nh vy, phng sai ca mt c lng l thng s ch chnh xc ca mt c lng. Do vic tnh ton phng sai ca v l lun cn thit.





Do v thuc vo cc gi tr Y, m Y li ph thuc vo cc bin ngu nhin u1, u2, , un, nn chng cng l bin ngu nhin vi phn phi tng ng. Sau y cc phng trnh c rt ra trong Phn 3.A.6 phn ph lc ca chng ny.

( )xxS

EVar222

)( =

== && (3.18)

( )[ ] 2222

)( xx

t

nSX

EVar === (3.19)

( )( )[ ] 2

),(

xxSXECov === (3.20)

trong Sxx c nh ngha theo Phng trnh (3.11) v 2 l phng sai ca sai s. Cn lu rng nu Sxx tng, gi tr phng sai v ng phng sai (tr tuyt i) s gim. iu ny cho thy s bin thin X cng cao v c mu cng ln th cng tt bi v iu cho chng t chnh ca cc thng s c c lng.

Cc biu thc trn l phng sai ca tng th v l n s bi v 2 l n s. Tuy nhin, cc thng s ny c th c c lng bi v 2 c th c c lng da trn mu. Lu rng tt XY += l ng thng c lng. Do , ttt XYu = l mt c lng ca ut, v l phn d c lng. Mt c lng d thy ca 2 l nut /

2 nhng c lng ny ngu nhin b thin lch. Mt c lng khc ca 2 c cho sau y (xem chng minh Phn 3.A.7)

2

222

==n

us

t (3.21)

L do chia t s cho n 2 th tng t nh trng hp chia chi-square cho n 1, c tho lun trong Phn 2.7. n 1 c p dng do ( ) xxi c iu kin l bng 0. p dng chia cho n 2, cn c hai iu kin bi Phng trnh (3.4) v (3.5). Cn bc hai ca phng sai c lng c gi l sai s chun ca phn d hay sai s chun ca hi quy. S dng c lng ny, ta tnh c cc c lng ca phng sai v ng phng sai ca v . Cn bc hai ca phng sai c gi l sai s chun ca h s hi quy v k hiu s v s . Phng sai c lng v ng phng sai ca h s hi quy c lng bng

xxSs

22

= (3.22)

22

2

xx

t

nSX

s

= (3.23)

2 xxS

Xs =

(3.24)





Tm li: Trc tin, cn tnh h s hi quy c lng v bng cch p dng Phng trnh (3.9) v (3.10). Kt qu cho cho mi quan h c lng gia Y v X. sau tnh gi tr d bo ca Yt theo tt XY += . T , ta c th tnh c phn d tu theo

tt YY . Sau tnh ton c lng ca phng sai ca ut da theo Phng trnh (3.21). Thay kt qu vo Phng trnh (3.18), (3.19) v (3.20), ta c gi tr phng sai v ng phng sai ca v .

Cn lu rng cng thc tnh phng sai ca phn d s2 c cho trong Phng trnh 3.21 c ngha, cn c iu kin n > 2. Khng c gi thuyt ny, phng sai c c lng c th khng xc nh c hoc m. iu kin tng qut hn c pht biu trong Gi thuyt 3.7, v bt buc phi tun theo.

GI THIT 3.7 (n > 2) S lng quan st (n) phi ln hn s lng cc h s hi quy c c lng (k). Trong trng hp hi quy tuyn tnh n bin, th iu kin n > 2 khng c.

V d 3.2 Sau y l sai s chun trong v d v gi nh, Sai s chun ca phn d = s = = 39,023 Sai s chun ca 285,37

== s

Sai s chun ca 01873,0

== s ng phng sai gia v 671,0

== s Thc hnh my tnh Phn 3.1 ca Ph chng D s cho kt qu tng t.

Mc d c cc i lng o lng s hc v chnh xc ca cc c lng, t thn cc o lng ny khng s dng c bi v cc o lng ny c th ln hoc nh mt cch ty tin bng cch n gin l thay i n v o lng (xem thm Phn 3.6). Cc o lng ny c s dng ch yu trong vic kim nh gi thuyt, ti ny s c tho lun chi tit Phn 3.5.

Thch Hp Tng Qut

Hnh 3.1 cho thy r rng khng c ng thng no hon ton thch hp vi cc d liu bi v c nhiu gi tr d bo bi ng thng cch xa vi gi tr thc t. c th nh gi mt mi quan h tuyn tnh m t nhng gi tr quan st c tt hn mt mi quan h tuyn tnh khc hay khng, cn phi c mt o lng ton hc thch hp. Phn ny s pht trin cc thng s o lng .

Khi thc hin d bo v mt bin ph thuc Y, nu ta ch c nhng thng tin v cc gi tr quan st ca Y c c t mt s phn phi xc sut, th c l cch tt nht c th l l c lng gi tr trung bnh Y v phng sai s dng ( )[ ] ( )1 22 = nYYtY . Nu cn d bo, mt cch n gin, ta c th s dng gi tr trung bnh bi v khng cn thng tin no khc. Sai s khi d bo quan st th t bng YYt . Bnh phng gi tr ny





v tnh tng bnh phng cho tt c mu, ta tnh c tng phng sai ca Yt so vi Y l ( )2 YY . y l tng bnh phng ton phn (TSS). lch chun ca mu ca Y o lng phn tn ca Yt xung quanh gi tr trung bnh ca Y, ni cch khc l phn tn ca sai s khi s dng Y lm bin d bo, v c cho nh sau ( )1 = nTSSY

Gi s ta cho rng Y c lin quan n mt bin X khc theo Phng trnh (3.1). Ta c th hy vng rng bit trc gi tr X s gip d bo Y tt hn l ch dng Y . C th hn l, nu ta c cc c lng v v bit c gi tr ca X l Xt, nh vy c lng ca Yt s l tt XY += . Sai s ca c lng ny l ttt YYu = . Bnh phng gi tr sai s ny v tnh tng cc sai s cho ton b mu, ta c c tng bnh phng sai s (ESS), hay tng cc bnh phng phn d, l ESS = 2tu . Sai s chun ca cc phn d l )2( = nESS . Gi tr ny o lng phn tn ca sai s khi s dng tY lm bin d bo v thng c so snh vi Y c cho trn xem xt mc gim xung l bao nhiu. Bi v ESS cng nh cng tt, v mc gim xung cng nhiu. Trong v d a ra, 498,88 =Y v 023,39 = , gim hn phn na so vi gi tr ban u.

Phng php ny khng hon ton tt lm, tuy nhin bi v cc sai s chun rt nhy cm i vi n v o lng Y nn rt cn c mt thng s o lng khc khng nhy cm vi n v o lng. Vn ny s c cp sau y.

HNH 3.5 Cc Thnh Phn ca Y

Y

X0

( )tt YX ,

tu

XY +=

YYt

Y

X

tY

tX

YYt

Thng s o lng tng bin thin ca tY so vi Y (l gi tr trung bnh ca tY ) cho ton mu l ( )2 YYt . c gi l tng bnh phng hi quy (RSS). Phn 3.A.8 cho thy

( ) ( ) += 222 ttt uYYYY (3.25)





Do vy, TSS = RSS + ESS. Lu rng ttt uYYYY )()( += . Hnh 3.5 minh ha cc thnh phn trn. Phng trnh (3.25) pht biu rng cc thnh phn cng c bnh phng. Nu mi quan h gia X v Y l cht ch, cc im phn tn (Xt, Yt) s nm gn ng thng X + . ni cch khc ESS s cng nh v RSS cng ln. T s

TSSESS

TSSRSS

= 1

c gi l h s xc nh a bin v k hiu l R2. Thut ng a bin khng p dng trong hi quy n bin bi v ch c duy nht mt bin ph c lp X. Tuy nhin, do biu thc R2 trong hi quy n bin cng ging nh trong hi quy a bin nn y chng ta dng cng thut ng

( ) TSSRSS

TSSESS

YYu

Rt

t==

=

1

12

2 10 2 R (3.26)

R rng rng, R2 nm gia khong t 0 n 1. R2 khng c th nguyn v c t s v mu s u c cng n v. im quan st cng gn ng thng c lng, thch hp cng cao, ngha l ESS cng nh v R2 cng ln. Do vy, R2 l thng s o lng thch hp, R2 cng cao cng tt. ESS cn c gi l bin thin khng gii thch c bi v tu l nh hng ca nhng bin khc ngoi Xt v khng c trong m hnh. RSS l bin thin gii thch c. Nh vy, TSS, l tng bin thin ca Y, c th phn thnh hai thnh phn: (1) RSS, l phn gii thch c theo X; v (2) ESS, l phn khng gii thch c. Gi tr R2 nh ngha l c nhiu s bin thin Y khng th gii thch c bng X. Ta cn phi thm vo nhng bin khc c nh hng n Y.

Ngoi ngha l mt t l ca tng bin thin ca Y c gii thch qua m hnh, R2 cn c mt ngha khc. l thng s o lng mi tng quan gia gi tr quan st Yt v gi tr d bo )(

ttYYtrY . Cn xem li phn trnh by v h s tng quan ca mu v ca

tng th Phn 2.3 v 3.5. Phn 3.A.9 trnh by

22

2 )()(

)(R

TSSRSS

YVarYVarYYCov

r

tt

ttYY === (3.26a)

Nh vy, bnh phng h s tng quan n bin gia gi tr quan st Yt v gi tr d bo tY bng phng trnh hi quy th s cho ra kt qu bng vi gi tr R

2 c nh ngha

trong Phng trnh (3.26a). Kt qu ny vn ng trong trng hp c nhiu bin gii thch, min l trong hi quy c mt s hng hng s.

C mt thc mc ph bin v thch hp tng th, l bng cch no xc nh rng R2 l cao hay thp?. Khng c mt quy nh chun hay nhanh chng kt lun v R2 nh th no l cao hay thp. Vi chui d liu theo thi gian, kt qu R2 thng ln bi v c nhiu bin theo thi gian chu nh hng xu hng v tng quan vi nhau rt nhiu. Do , gi tr quan st R2 thng ln hn 0.9. R2 b hn 0.6 v 0.7 c xem l thp. Tuy nhin, i vi d liu cho, i din cho dng ca mt yu t thay i vo mt





thi im no , th R2 thng thp. Trong nhiu trng hp, R2 bng 0.6 hoc 0.7 th cha hn l xu. y n gin ch l thng s o lng v tnh y ca m hnh. iu quan trng hn l nn nh gi m hnh xem du ca h s hi quy c ph hp vi cc l thuyt kinh t, trc gic v kinh nghim ca ngi nghin cu hay khng.

V d 3.3 Trong bi tp v gi nh, TSS, ESS v R2 c cc gi tr sau (xem li kt qu Phn thc hnh my tnh 3.1):

TSS = 101.815 ESS = 18.274 R2 = 0,82052

Nh vy, 82,1% bin thin ca gi nh trong mu c gii thch bi din tch s dng tng ng. Trong chng 4, s thy rng thm vo cc bin gii thch khc, nh s lng phng ng v phng tm s ci thin thch hp ca m hnh.

3.5 Kim nh Gi Thuyt Thng K

Nh lc u, kim nh gi thuyt thng k l mt trong nhng nhim v chnh ca nh kinh t lng. Trong m hnh hi quy (3.1), nu bng 0, gi tr d bo ca Y s c lp vi X, ngha l X khng c nh hng i vi Y. Do , cn c gi thuyt = 0, v ta k vng rng gi thuyt ny s b bc b. H s tng quan () gia hai bin X v Y o lng tng ng gia hai bin. c lng mu ca c cho trong Phng trnh (2.11). Nu = 0, cc bin khng c tng quan nhau. Do cng cn kim nh gi thuyt = 0. Phn ny ch tho lun phng php kim nh gi thuyt i vi v . Kim nh gi thuyt i vi p s c trnh by phn sau. Cn lu rng, trc khi tip tc phn tip theo, bn nn xem li Phn 2.8 v kim nh gi thuyt v Phn 2.7 v cc loi phn phi.

Kim nh gi thuyt bao gm ba bc c bn sau: (1) thit lp hai gi thuyt tri ngc nhau (Gi thuyt khng v Gi thuyt ngc li), (2) a ra kim nh thng k v phn phi xc sut cho gi thuyt khng, v (3) a ra quy lut ra quyt nh bc b hay chp nhn gi thuyt khng. Trong v d v gi nh, Gi thuyt khng l Ho : = 0. Bi v chng ta k vng rng s dng, Gi thuyt ngc li l H1: 0. thc hin kim nh ny, v sai s chun c lng s c s dng a ra thng k kim nh. a ra phn phi mu cho v , m iu ny nh hng gin tip n cc s hng sai s ngu nhin u1, u2, un (xem Phng trnh 3.15), cn b sung mt gi thuyt v phn phi ca ut.

GI THIT 3.8 (Tnh Chun Tc ca Sai S) Mi gi tr sai s ut tun theo phn phi chun N(0, 2) , ngha l mt c iu kin ca Y theo X tun theo phn phi N( + X, 2).

Nh vy, cc s hng sai s u1, u2, un c gi nh l c lp v c phn phi chun ging nhau vi gi tr trung bnh bng khng v phng sai bng 2. Gi thit 3.8 l gi thit cn bn trong kim nh gi thuyt thng k. Bng 3.2 s trnh by tm tt tt c cc





gi thit c a ra. Nhng s hng sai s tha cc Gi thit t 3.2 n 3.8 th c xem l sai s ngu nhin hay sai s do nhiu trng.

BNG 3.2 Cc Gi Thit ca M Hnh Hi Quy Tuyn Tnh n Bin 3.1 M hnh hi quy l ng thng vi n s l cc h s v ; l

Yt = + Xt + ut, vi t = 1, 2, 3, n. 3.2 Tt c cc gi tr quan st X khng c ging nhau; phi c t nht mt gi tr khc

bit. 3.3 Sai s ut l bin ngu nhin vi trung bnh bng khng; ngha l, E(ut) = 0. 3.4 Xt c cho v khng ngu nhin, iu ny ngm nh rng khng tng quan vi ut;

ngha l Cov (Xt, ut) = E(Xtut) E(Xt)E(ut)= 0. 3.5 ut c phng sai khng i vi mi t; ngha l Var(ut) = ( ) 22 =tuE 3.6 ut v us c phn phi c lp i vi mi t s, sao cho Cov(ut, us) = E(ut us). 3.7 S lng quan st (n) phi ln hn s lng h s hi quy c c lng ( y n

> 2). 3.8 ut tun theo phn phi chun ut ~ N(0, 2), ngha l ng vi gi tr Xt cho trc, Yt ~

N( + Xt, 2). Xc nh Tr Thng K Kim nh

Phn ny chng minh rng kim nh thng k ( ) 0 stc = tun theo phn phi Student t, theo gi thuyt khng, vi bc t do l n 2 (bi v ta ang c lng hai tham s v ). Lu rng Gi thuyt 3.7 rt cn chc chn rng bc t do l dng.

CHNG MINH (c gi khng quan tm n ngun gc vn , c th b qua phn ny).

Trc ht cn xem xt cc tnh cht sau

TNH CHT 3.6

a. v c phn phi chun. b. ( ) [ ] 2222 )2( = nu t c phn phi chi-bnh phng vi bc t do n2. c. v c phn phi c lp vi 2 .

Tnh cht 3.6a xut pht t thc t l v l nhng t hp tuyt tnh ca ut v ut c phn phi chun. chng minh tnh cht b v c, nn tham kho ti liu Hogg v Graig (1978, trang 296-298). Tn dng cc kt qua ta c

),,(~ 2 N ),,(~ 2 N 2 22

~

2

n

t Xu





trong 2 v 2 l phng sai ca v theo Phng trnh (3.18) v (3.19). Bng cch chun ha phn phi ca thng s c lng ngha l tr cho trung bnh v chia cho lch chun) ta c

( ) ),1,0(~

N

( ) ),1,0(~

N

( ) 22

2

~

22

nXn

Trong phn 2.7, phn phi t c nh ngha l t s ca s chun chun ha trn cn bc hai ca mt chi-square c lp vi n. Thay vo cho v p dng phng trnh (3.18), (3.19) v (3.22), ta c

( ) ( ) ( )2

21

2

2

~

=

=

= nts

t

trong

===

xxxx SSs

s l sai s chun c lng ca theo Phng trnh (3.22). t c trnh by trn l tr thng k kim nh da trn quy lut ra quyt nh c thit lp sau ny. Kim nh ny c gi l kim nh t. Cc bc kim nh thng k phn ra trong hai trng hp kim nh mt pha v kim nh hai pha c trnh by sau y.

Quy Tc Ra Quyt nh

Kim nh t-test mt pha

BC 1 H0: = 0 H1: 0 BC 2 Kim nh thng k l ( ) 0 stc = , c tnh da trn mu. Theo gi

thuyt khng, kim nh thng k c phn phi t vi bc t do l n 2. Nu tc tnh c l ln, ta c th nghi ng rng s khng bng 0. iu ny dn n bc tip theo.

BC 3 Trong bng tra phn phi t trang ba trc ca sch, tra bc t do l n 2. V chn mc ngha () v xc nh im t*n2() sao cho P(t > t*) = .

BC 4 Bc b H0 nu tc > t*. Nu gi thuyt ngc li < 0 , tiu chun kim nh bc b H0 l nu tc < t*.

Kim nh trn c minh ha bng hnh nh qua Hnh 3.6 (k hiu c s dng ch mc ngha trnh nhm ln vi ch tung ). Nu tc ri vo din tch in m trong hnh v (c gi l vng ti hn) ngha l tc >t*. Trong trng hp , gi thuyt khng s b bc b v kt lun c rng ln hn 0 rt nhiu.





HNH 3.6 Kim nh Mt Pha vi H0: = 0 H1: 0

Chap nhan Ho Bac bo Ho

Dien tch a

0

f(tn-2)

tn-2 t*n-2(a)

V d 3.4 Trong v d v gi nh, ta c 0 = 0. Do , stc = , l kim nh thng k n gin v l t s gia h s hi quy c lng trn sai s chun. T s c gi l tr thng k t. Cc c lng l 13875,0 = , v theo v d 3.2 ta bit 01873,0

=s . Do , tr thng

k t c tnh s l tc = 0,13875/0,01873 = 7.41. Bc t do bng n 2 = 14 2 = 12. Cho mc ngha l 1%, ngha l = 1%. Tra bng phn phi t, ta c t*n2=2,681 . Do tc > t*, gi thuyt H0 b bc b v kt lun c rng ln hn zero mt cch ng k vi mc ngha 1%. Lu rng h s ny vn c ngha trong trng hp mc ngha ch l 0,05% bi v t*12(0,0005) = 4,318.

Tr thng k t i vi c cho bi tc = 52,351/37,285 = 1.404 nh hn t*12(0,0005) = 1.782. Do khng th bc b H0 nhng thay vo c th c th kt lun rng khng ln hn zero xt v mt thng k vi mc ngha 5%. Cc im khng ngha hai im sau. Th nht, X = 0 th hon ton nm ngoi khong mu v do c lng Y khi X = 0 khng ng tin cy (xem thm Phn 3.9). Th nh, t Hnh 3.1 c th thy rng c im hai bin l khng y gii thch bin thin gi ca cc gi tr quan st. Trong chng 4 s cho thy bao hm c nh hng trung bnh ca bin b b st v tnh phi tuyn, khi X bng 0. Cc nh hng trn s lm cho khng c ngha.

Mt S Lu khi S Dng Kim nh t-Test

Mc d kim nh t-test rt hu ch trong vic xc nh ngha thng k ca cc h s, tuy nhin rt d nhm ln gia cc ngha ca kim nh. V d, V d 3.4 kim nh t-test i vi khng th bc b gi thuyt khng l = 0. Nh vy c phi kim nh ny chng minh rng = 0 hay khng? Cu tr li l khng. C th chc chn rng, theo tp d liu v m hnh c m t, khng c bng chng no cho thy > 0. Trong chng 4, s cp kim nh t-test cho nhiu h s hi quy. Nu mt trong nhng h s ny khng c ngha (ngha l, khng th bc b gi thuyt rng h s bng 0), iu khng c ngha l bin tng ng khng c nh hng g n bin ph thuc hoc bin khng quan trng. Vn ny s c tho lun y trong chng sau. Trong chng 5 s thy rng khi m hnh thay i, mc ngha ca h s cng thay i. Do , cn thc hin k cc kim nh gi thuyt a ra v khng nn vi v kt lun m khng





xt n m hnh v nhng phn tch thm v cc kim nh chun on cn thit a ra mt kt lun ngha (n nh theo c im m hnh).

Phng Php p-value trong Kim nh Gi thuyt

Kim nh t-test c th c thc hin theo mt phng php khc tng ng. Trc tin tnh xc sut bin ngu nhin t ln hn tr quan st tc, ngha l

p-value = P(t>tc ) = P (sai lm loi I)

Xc sut ny (c gi l p-value) l phn din tch bn phi tc trong phn phi t (xem Hnh 3.7) v l xc sut sai lm loi I ngha l xc sut loi b gi thuyt H0. Xc sut ny cng cao cho thy hu qu ca vic loi b sai lm gi thuyt ng H0 cng nghim trng. p-value b ngha l hu qu ca vic loi b gi thuyt ng H0 l khng nghim trng (ngha l, xc sut xy ra sai lm loi I l thp) v do c th yn tm khi bc b H0. Nh vy, quy lut ra quyt nh l khng bc b H0 nu p -value qu ln, v d: ln hn 0,1, 0,2, 0,3. Ni cch khc, nu p-value ln hn mc ngha , c th kt lun rng h s hi quy khng ln hn 0 mc ngha . Nu p-value nh hn , gi thuyt H0 b bc b v kt lun c rng ln hn 0 mt cch ng k.

thy c s tng ng ca hai phng php, lu rng trn Hnh 3.7 nu xc sut P(t>tc ) b hn mc ngha , th im tng ng l tc phi nm bn phi im t*n-2(). Ngha l tc ri vo min bc b. Tng t, nu xc sut P(t>tc ) ln hn mc ngha , th im tng ng l tc phi nm bn tri im t*n-2() v do ri vo min chp nhn. Sau y l cc bc b sung trong phng php p-value nh sau:

HNH 3.7 Kim nh Gi thuyt theo Phng Php p-value

Bac bo Ho neu p- value< a

0

f(tn-2)

tn-2 t* tc

BC 3a Tnh xc sut (k hiu p-value) t ln hn tc , ngha l tnh phn din tch bn phi gi tr tc.

BC 4a Bc b H0 v kt lun rng h s c ngha nu p-value b hn mc ngha c chn.





Tm li, c xem l ln hn 0 mt cch ng k nu tr thng k t ln hay p-value l b, mc nh th no l ln v b s c quyt nh bi ngi nghin cu. Phng php ph bin trong kim nh gi thuyt l xc nh gi tr mc t*. Tuy nhin theo hng php tnh p-value, li cn tnh ton phn din tch mt u ng vi gi tr tc cho trc. Ngy cng c nhiu phn mm my tnh tnh ton sn p-value (chng trnh SHAZAM v ESL c gii thiu trong sch ny) v do phng php ny d ng dng d dng. Tuy nhin, cn cn thn kim tra li gi tr p-value l dng cho kim mt pha hay kim nh hai pha.

V d 3.4a p dng phng php p-value cho v d v gi nh, ta tnh xc sut t ln hn gi tr quan st = 7.41. S dng ESL tnh ton ta c p < 0,0001 (tham kho phn kt qu trong phn Thc hnh my tnh 3.1). iu c ngha l, nu ta bc b gi thuyt khng, th c hi xy ra sai lm loi I b hn 0,01%, v do hon ton yn tm khi bc b Ho v kt lun c rng ln hn 0. i vi tham s , p-value bng 0,093, ngha l P(t>1,404) = 0,093. Nu H0: = 0 b bc b, xc sut xy ra sai lm loi I l 9,3%, ln hn 5%. Do , khng th bc b H0 mc ngha 5%, ngha l ta c cng kt lun nh trong phng php u, l mc ngha 5%, khng ln hn zero xt v mt thng k. Nh vy phng php p-value c mt u im l, ta bit c chnh xc mc m h s c ngha v c th nh gi xem mc ngha ny thp hay khng xem xt bc b H0. Cui cng, khng cn lo lng i vi cc gi tr 0,01, 0,05 v 0,1.

Kim nh t-test Hai Pha

Bao gm cc bc sau:

BC 1 H0: = 0 H1: 0 BC 2 Kim nh thng k l ( ) 0 stc = , c tnh da trn mu. Theo gi

thuyt khng, kim nh thng k c phn phi t l tn-2. BC 3 Trong bng tra phn phi t trang ba trc ca sch, tra bc t do l n 2

v chn mc ngha () v xc nh im t*n2() sao cho P(t>t*) = /2 (phn na mc ngha).

BC 3a p dng phng php - value, tnh gi tr p - value = P(t > tc hoc t < tc ) = 2P(t > |tc |)

do phn phi t i xng. BC 4 Bc b H0 nu |tc |> t* v kt lun khc vi 0 mt cch ng k mc

ngha . BC 4a Bc b H0 nu p-value < , mc ngha ny.

Kim nh trn c minh ha bng hnh nh qua Hnh 3.8. Bc t do trong trng hp ny bng n2. Nu tr thng k t (tc ) ri vo vng din tch en, gi thuyt khng b bc b v kt lun c rng khc vi 0. gi tr t* = 2 c s dng l quy lut nh gi mc ngha ca tr thng k t mc 5% (kim nh hai pha). Bi v t* gn bng 2 vi bc t do l 25.





HNH 3.8 Kim nh Hai Pha vi H0: = 0 H1: 0

Dien tch a/2

0

f(tn-2)

tn-2 t*n-2(a/2)

Dien tch a/2

Chapnhan Ho Bac bo HoBac bo Ho

-t*n-2(a/2)

V d 3.5 Theo cch tnh ny tc trong v d gi nh c gi tr nh cch tnh theo t-test, 41.7 = v

404.1 = . Tra bng gi tr t, ta c 055.3)005.0(*12 =t , iu ny c ngha l din tch ca c 2 pha tng ng vi gi tr 3.055 l 0.01. Bi i vi th tc>t* do ta c th loi gi thuyt H0 v kt lun c rng khc vi mc ngha 1%. i vi th

179.2)025.0(t *12 = ln hn gi tr tc. Do ta khng th bc b gi thuyt H0 (lu rng ta ang dng kim nh gi tr mc ngha 5%). T bc 3a ta c th suy ra c gi tr p-value i vi )404.1(2 >= tP = 0.186 (lu gi tr p-value tng ng vi tc trong trng hp kim nh 2 pha s gp 2 ln gi tr ca n trong trng hp kim nh 1 pha). Do sai lm loi I c gi tr 18.6% l khng th chp nhn c nn ta khng th bc b gi thuyt H0: = 0. iu ny c ngha l khng c ngha v thng k trong khi li c.

BI TP 3.4 Trong v d gi nh, hy kim nh gi thuyt H0: = 0.1 v gi thuyt H1: 0.1 ln lt mc ngha 0.05 v 0.01.

BI TP 3.5 Chng minh rng nu mt h s c ngha mc 1% th h s ny cng s c ngha mc cao hn.

BI TP 3.6 Hy chng minh rng nu mt h s khng c ngha mc 10% th h s ny cng s khng c ngha bt k mc ngha no thp hn 10%.

Kim nh 2

Mc d thng k kim nh mc ngha phng sai sai s 2 khng ph bin nhng vn c trnh by y trong phn ny. Kim nh 2 gm cc bc sau:





BC 1 H0: 2 = 02 H1: 2 02

BC 2 Tr kim nh l 0

2

2

)2(

= nQc . Sau tra bng phn phi Chi-square

vi bc t do n-2. Nu Q c gi tr ln ta c th nghi ng rng 2 khng bng 02

BC 3 Trong bng tra phn phi Chi-square trang ba trc ca sch, tra gi tr ca Q*n-2() sao cho din tch bn phi bng .

BC 4 Bc b H0 mc ngha nu Qc> Q*n-2().

Nguyn nhn tng qut lm cho kim nh ny khng ph bin l do ngi kim nh khng c thng tin s cp ban u v gi tr ca 2 s dng trong gi thuyt H0.

Kim nh Thch Hp

Ta c th thc hin kim nh thch hp. Gi p l h s tng quan tng th gia X v Y c nh ngha Phng trnh (2.7). Theo phng trnh (2.11), ta thy gi tr c lng p2 c xc nh bi )/(22 yyxxxyxy SSSr = trong Sxx v Sxy c nh ngha theo Phng trnh (3.8) v (3.9), v

( )TSSYY

n

YYS t

ttyy ==

=

22

2 )( (3.27)

Phn 3.A.10 ngi ta chng minh rng r2xy bng vi R2 (iu ny ch ng trong trng hp hi qui n bin m thi). Phn kim nh gi thuyt 2.8 trnh by phng php kim nh gi thuyt cho rng X v Y khng c mi tng quan. Kim nh ny gi l kim nh F (F-test). Kim nh F-test gm cc bc sau:

BC 1 H0: xy = 0 H1: xy 0 BC 2 Tr thng k kim nh l Fc = R2(n 2)/(1 R2). Fc cng c th c tnh

theo cng thc sau Fc = RSS(n 2)/ESS. Theo gi thuyt H0, tr thng k ny tun theo phn phi F vi 1 bc t do t s v n 2 bc t do mu s.

BC 3 Tra bng F theo 1 bc t t s v n 2 bc t do mu s tm gi tr F*1, n 2 () sao cho phn din tch v pha phi ca F* l , mc ngha.

BC 4 Bc b gi thuyt H0 (ti mc ngha ) nu Fc > F*.

Nn lu rng gi thuyt H0 trn s khng hp l khi c nhiu gi tr X. Nh s c trnh by chng 4, kim nh F vn c s dng nhng H0 s khc.

V d 3.6 Trong v d gi nh, R2 = 0,82052. Fc = 0,82052(14 2)/(1 0,82052) = 54,86. Theo v d 3.5, ESS = 18.274, v RSS = TSS ESS = 83.541. V vy Fc cn c th c tnh theo cng thc khc nh bc 2: Fc = 83.541 (14 2)/18.274 = 54,86. Bc t do ca t





s l 1, ca mu s l 12. Vi mc ngha = 5%, tra bng A.4b ta c F*1, 12(0.05) = 4,75. V Fc > F* chng ta bc b (ti mc ngha 5%) gi thuyt H0 cho rng X v Y khng tng quan. Thc ra, v Fc > F*1, 12(0.01) (tra bng A.4a), gi thuyt H0 cng b bc b ti mc ngha 1%. Nh vy, mc d gi tr R2 kh nh hn 1, n cng khc 0 mt ng k.

Trnh By Cc Kt Qu Hi Quy

Cc kt qu ca phn tch hi quy c trnh by theo nhiu cch. Theo cch thng thng, ngi ta s vit phng trnh c lng km vi cc tr thng k t di mi h s hi quy nh sau:

SQFT13875,0351,52GIA +=

(1,404) (7,41)

821.02 =R 12.f.d = 023.39 =

Mt cch khc l in cc sai s chun di cc h s hi quy:

SQFT13875,0351,52GIA +=

(37.29) (0.019)

Nu nhiu m hnh hi quy c c lng, vic trnh by kt qu dng bng nh Bng 4.2 s thun tin hn.

Vic tch tng cc bnh phng ton phn ra thnh cc thnh phn thng c tm tt dng bng Phn Tch Phng Sai (ANOVA) Bng 3.3.

3.6 Thang o v n V o

Gi s chng ta tnh GI theo n v ng la thay v theo ngn ng la. Ct GI bng 3.1 s cha cc gi tr nh 199.900, 228.000, v.v. Nhng c lng ca h s hi quy, cc sai s chun ca chng, R2, v.v. s b nh hng nh th no bi s thay i n v ny? Cu hi ny s c kho st y v GI v SQFT c tnh cc n v khc nhau. u tin chng ta chy li m hnh.

GI = + SQFT + u

Gi GI* l gi tnh theo la thng. Nh vy GI* = 1.000 GI. Nhn mi s hng trong phng trnh vi 1.000 v thay GI* vo v tri. Chng ta c

GI* = 1.000 + 1.000SQFT + 1.000u = GI* = * + *SQFT + u*

Nu chng ta p dng phng php OLS cho phng trnh ny v cc tiu ha (u*t)2, chng ta s tm c cc gi tr c lng ca * v *. D dng nhn thy rng cc h s hi quy mi s bng cc h s c nhn vi 1,000. Nh vy, thay i thang o ca ch bin ph thuc trong m hnh hi quy lm cho thang o ca mi h s hi quy thay i theo tng ng. V u* = 1,000u, cc phn d v sai s chun cng s c nhn





ln 1.000. Tng cc bnh phng s c nhn thm 1 triu (1.000 bnh phng). Cn lu rng cc tr thng k t, F, v R2 s khng b nh hng v chng l cc t s trong yu t thang o s trit tiu.

BNG 3.3 Phn Tch Phng Sai Ngun Tng bnh phng

(SS) Bc t do

(d.f.) Bnh

phng trung bnh (SSd.f.)

F

Hi quy (RSS) 2)( YYt = 83.541 1 83.541 8654ESS

2nRSS,

)(=

Sai s (ESS) 2tu = 18.274 N 2 = 12 1.523 Tng (TSS) 2)( YYt = 101.815 N 1 = 13 7.832

Tc ng ca vic thay i thang o ca mt bin c lp s ra sao? Gi s SQFT c tnh theo n v trm mt vung thay v theo mt vung thng thng, nhng GI c tnh theo n v ngn la nh trc. Gi SQFT l bin tnh theo trm mt vung. Vy SQFT= 100SQFT. Thay vo phng trnh ban u ta c:

GI = + 100SQFT + u

R rng theo phng trnh ny, nu chng ta hi quy GI theo mt hng s v SQFT, h s duy nht s b nh hng l h s ca SQFT. Nu l h s ca SQFT, th

100' = . Sai s chun ca n cng s nhn vi 100. Tuy nhin, tt c cc s o khc ESS, gi tr thng k t, F, R2 chng hn s khng b nh hng. Tm li, trong mt m hnh hi quy tuyn tnh, nu thang o ca mt bin c lp thay i cc h s hi quy ca n v cc sai s chun tng ng s thay i tng ng nhng cc tr thng k khc s khng thay i.

C l do chnh ng thay i thang o ca cc gi tr sao cho cc s sau khi thay i s khng ln cng khng qu nh v tng t vi cc gi tr ca cc bin khc. iu ny l v cc s c gi tr ln s ln t cc sai s v cc s nh s gy ra sai s lm trn, c bit l khi tnh gi tr tng bnh phng, vic ny s lm nh hng xu n chnh xc ca kt qu.

hiu mt cch thc t hu qu ca vic thay i n v, hy Thc Hnh My Tnh phn 3.2 ph lc D.

BI TP 3.7 Gi s chng ta t mt bin mi X* = SQFT 1.000 (ngha l, X* l phn din tch vung trn 1.000) v c lng m hnh GI = a + bX* + v. Gii thch bng cch no bn c th tm c a v b t v m khng phi c lng li m hnh mi.

3.7 ng dng: c Lng ng Engel Biu Din Quan H Gia Chi Tiu cho Chm Sc Sc Khe v Thu Nhp.





Trong phn ny, chng ta s trnh by mt ng dng tp dt vi m hnh hi quy hai bin. D liu c s dng l chui d liu cho cho 50 bang v qun Columbia (n = 51), d liu c thu thp t cun Tm Lc Thng K M nm 1995 (Statistical Abstract of the US). Cc gi tr ca d liu thc c tp tin DATA3-2. Cc bin l:

EXPHLTH = Chi tiu tng hp (n v t la) cho chm sc sc khe ca bang vo nm 1993, Bng 153, trang 111, khong t 0,998-9,029.

INCOME = Thu nhp c nhn (n v t la) ca bang vo nm 1993, Bng 712, trang 460, khong t 9,3-64,1.

M hnh l ng Engel tm c v d 1.4 v c p dng vi tng chi tiu cho chm sc sc khe ca M l hm s theo tng thu nhp c nhn. Phn ng Dng My Tnh 3.3 (xem ph lc bng D.1) trnh by hng dn tm ra kt qu. Bn ch thch ca bo co in t my tnh, s dng chng trnh ESL v tp tin PS3-3.ESL, c trnh by bng 3.4. Phn c in m l nhp lng ca chng trnh v cc phn in nghing l cc nhn xt v kt qu. Bn nn tm hiu cc ch thch ny cn thn v s dng chng trnh hi quy bn c chy li cc kt qu ny (tp tin PS3-3.SHZ cha cc dng lnh s dng phn mm SHAZAM). Di y l m hnh c lng cng vi tr thng k mu t trong ngoc n, v p-value (gi tr xc sut p) trong ngoc vung:

INCOME141652,0176496,0EXPHLTH += (0.378) (49.272)

[0.707] [t*) = 0,025. iu ny tng ng vi P(- t* t t*) = 0,95. Nh vy,

BNG 3.4 Bo Co t My Tnh Km Theo Ch Gii cho Phn 3.7 Cc lnh ESL c in m v cc nhn xt c in nghing. Danh sch bin





(0) Hng s (1) exphlth (2) income

( th ca mc chi tiu theo thu nhp cho thy c s quan h cht ch gia hai bin)

EXPHLTH 94,178 | o | | 78,648 + | | | o | 52,764 + | o | o | o | o o 26,881 + o | o o | o | o ooo | oooo o 0,998 + ooo | +++++++ 9,3 income 683,5

c lng OLS vi 51 quan st 1-51 Bin ph thuc EXPHLTH

Bin H s Sai s chun T-stat 2 Prob(t >T) (0) hng 0,176496 0,467509 0,377525 0,707414 (1) income 0,141652 0,002875 49,271792 T) l vng din tch hai u phn phi t chn bi gi tr kim nh t v l gi tr p-value hoc xc sut sai lm loi I (i vi kim nh 2 pha). Nu p-value nh (trong trng hp ny, nh hn 0,0001), chng ta an ton khi bc b gi thuyt Ho rng = 0, v kt lun rng h s ca bin thu nhp l khc 0 ng k. Gi tr p-value ca s hng hng s bng 0,707414 gi rng nu chng ta bc b gi thuyt Ho cho rng = 0, chng ta c th phm phi sai lm loi I trong 70,7 % s ln. V mc sai lm ny qu cao, chng ta khng th bc b gi thuyt Ho. Nh vy chng ta kt lun rng s hng hng s khng khc 0 ng k. Lu rng trong v d 1.4, vic suy din l thuyt ra ng Engel m ch rng khng c s hng hng s. S hng hng s khng c ngha l ph hp vi kt qu theo l thuyt. Xu hng chi tiu cn bin cho vic chm sc sc khe ly t thu nhp l





0,141652; ngha l, vi mi khon tng thu nhp 100 la, chng ta c th k vng cc c nhn s chi trung bnh 14,17 la cho chm sc sc khe.

Gi tr R2 (R-square) ch ra rng 98% s bin i ca chi tiu c gii thch bi bin thu nhp. S khc nhau gia gi tr R2 Hiu chnh v Khng hiu chnh s c gii thch chng 4 cng vi cc gi tr thng k mu chn m hnh.

Gi tr thng k mu Durbin-Watson v h s tng quan chui bc nht s c gii thch chng 9, nhm gii quyt s vi phm gi thit 3.6 cho rng cc s hng sai s ca hai quan st l khng tng quan. Gi tr trung bnh ca bin ph thuc l Y v S.D. l lch chun ca Sy

Gi tr trung bnh ca bin ph thuc

15,068863 S.D. ca bin ph thuc 17,926636

Tng bnh phng sai s (ESS)

317,898611 Sai s chun ca phn d 2,547102

R- bnh phngkhng hiu chnh

0,980 R- hiu chnh 0,980

Tr thng k F 2427,709468 p-value = Prob(F>2427.709)





12 3,485 4,057766 -0,57276526 13 3,433 3,476992 -0,0439923 14 3,747 4,652705 -0,90570543 15 4,4 4,666871 -0,26687065 16 3,878 3,916114 -0,03811407 17 5,197 4,341071 0,85592937 18 4,118 4,426062 -0,30806194 19 6,111 5,672601 0,43839884 20 6,903 7,301601 -0,39850129 21 6,187 5,686766 0,50023362 22 7,341 7,485749 -0,14474913 23 7,999 8,533975 -0,53497529 24 8,041 7,967367 0,07353344 25 12,216 13,250993 -1,034993 26 10,066 11,027054 -0,96105374 27 9,029 8,84561 0,1833899 28 10,384 9,256401 1,127599 29 10,635 10,276297 0,35870284 30 12,06 10,318793 1,741207 31 13,014 10,276297 2,737703 32 14,194 13,619289 0,57471128 33 15,154 16,96228 -1,80828 34 14,502 14,32755 0,17445035 35 16,203 13,477637 2,725363 36 15,949 14,68168 1,26732 37 15,129 16,395672 -1,256672 38 16,401 15,701576 0,69942416 39 23,421 20,985202 2,435798 40 6,682 20,036133 -13,354133 41 20,104 19,002072 1,101928 42 18,241 18,56295 -0,32194997 43 25,741 30,093438 -4,352438 44 27,136 27,756177 -0,62017675 45 33,456 31,042507 2,413493 46 34,747 37,516012 -2,769012 47 41,521 36,439456 5,081544 48 44,811 40,320726 4,490274 49 49,816 49,0465 0,7694999 50 67,033 64,004971 3,028029 51 94,178 96,995765 -2,817765

ststPts

tP **** (95.0)( +==

T y c th rt ra rng khong tin cy 95% ca v ln lt l st * v st *





V d 3.7

Trong v d v gi nh, sai s chun ca v l s = 37,285 v s = 0,18373. ng thi, t bng t, ta c t*12(0,025) = 2,179. Do , khong tin cy 95% l i vi : 52,351 (2,179x37,285) = (-28,893; 133,595) i vi : 0,13875 (2,179x0,018373) = (0,099; 0,179)

Lu rng cc khong tin cy ny l tng i rng. y l du hiu cho thy m hnh hi quy tuyn tnh thch hp rt km vi tp d liu. Mt m hnh hi quy thch hp s cho khong tin cy hp hn.

BI TP 3.8 Xc nh khong tin cy ca v trong Phn ng Dng 3.7

3.9 D Bo

Nh cp trc y, mt trong cc ng dng ph bin ca m hnh hi quy l d bo (ch ny s c tho lun chi tit hn chng 11). Trong v d gi nh, chng ta c th t cu hi gi bn d bo ca mt ngi nh c din tch 2,000 mt vung s l bao nhiu. M hnh hi quy c lng l XY 13875.0351.52 += . Nh vy, khi X = 2,000, gi tr d bo caY l 52,351 + (2,000x0,13875) = 329,851. V gi c tnh theo n v ngn la, gi tr d bo ny cng c n v ngn la. V vy, theo m hnh, gi trung bnh c lng ca mt cn h din tch 2,000 mt vung l 329.851 la. Mt cch tng qut, d dng nhn thy nu X c gi tr X0 th gi tr d bo ca Y0 s l

00

XY += . Gi tr trung bnh c iu kin ca bin d on Y cho trc X = X0 l

)()()()( 0000 XXYEXEXEXXYE ==+=+==

Nh vy 0Y l gi tr d bo c iu kin khng thin lch ca gi bn trung bnh ti X0.

Khong Tin Cy cho Gi Tr D Bo Trung Bnh

V v c c lng c sai s, gi tr d bo 0Y cng chu sai s. xt n yu t ny, chng ta tnh sai s chun v khong tin cy cho gi tr d bo trung bnh. Di y l c lng ca phng sai ca gi tr d bo (xem chng minh Phn 3.A.11 )

+=xx

Y SXX

ns

2022 )(1

0

(3.28)

Khong tin cy ca gi tr d bo trung bnh l





],[00

*

0*

0 YY stYstY +

trong t* l gi tr ngng ca phn phi t. Lu rng khi X0 cng lch xa gi tr trung bnh X , th

0Ys cng ln v khong tin cy tng ng cng rng. iu ny c ngha rng

nu d bo c thc hin qu xa khi phm vi ca mu, tin cy ca d bo s gim i. Nu X0 = X , khong tin cy s hp nht. Hnh 3.9 cho nim v di tin cy vi cc gi tr X0.

HNH 3.9 Di Khong Tin Cy ca Cc Gi Tr D Bo

Khong Tin Cy cho D Bo im

Phng sai mu trnh by phn trc dng d bo gi tr trung bnh. Bn cnh chng ta cng mun tm phng sai ca sai s d bo cho cc gi tr thc Y0 tng ng vi X0. Cng thc di y c ly t Ph lc 3.A.12:

2

202

02

00

)(11)( Yxx

u sSXX

nuVars >

++== (3.29)

trong 000 YYu = l sai s ca d bo im. Khong tin cy c tnh theo 0us thay v

0Ys . Khi c mu ln, s hng th hai v th ba trn s khng ng k so vi 0us

mt gi tr gn bng . Ngoi ra, t* cng gn bng 2 trong trng hp mc ngha 95%. Nh vy, khong tin cy ca mu c kch thc ln l 20 Y

V d 3.8





Trong v d gi nh, chng ta c 2

0Ys = 111,555 v 2

0us

=1634,353 v khong tin cy tng ng khi X0 = 2.000 s l (307, 353) v (242, 418). Khong tin cy vi c mu ln l (252,408). (Xem phn Thc Hnh My Tnh 3.4 chy li kt qu ny).

Chng ta nn chn loi khong tin cy no trong s hai loi trn? V quan tm chnh l sai s d bo i vi gi tr thc Y0, phng trnh (3.29) thng c s dng. Lu rng khong tin cy theo phng trnh ny rng hn nhiu khong tin cy da trn phng trnh (3.28)

So Snh Cc Gi Tr D Bo

Cc nh phn tch kinh t v kinh doanh thng s dng nhiu hn mt m hnh d bo. Mt s o thng dng so snh nng lc d bo ca cc m hnh khc nhau l sai s bnh phng trung bnh (hoc i khi ngi ta s dng cn bc hai ca n, v c gi l cn bc hai sai s bnh phng trung bnh).

Gi Yft l gi tr d bo ca bin ph thuc cho quan st t, v Yt l gi tr thc. Sai s bnh phng trung bnh c tnh nh sau:

2nYY

MSE2

tf

t

=

)( MSERMSE =

Nu hai m hnh c s dng d bo Y, m hnh no c MSE nh hn s c nh gi l m hnh tt hn cho mc ch d bo.

Mt s o hu ch khc l sai s phn trm tuyt i trung bnh (MAPE)

=

t

ftt

Y

YY100

n1MAPE

S o ny ch c ngha nu tt c cc gi tr Y u dng (xem Phn ng Dng 3.11). Mt cch khc, chng ta c th tnh sai s phn trm bnh phng trung bnh (MSPE) hoc cn ca n

=

2

t

ftt

YYY

100n1MSPE

MSPERMSPE =

Mt phng php khc nh gi m hnh v nng lc d bo ca n l thc hin d bo hu mu. Theo phng php ny, ngi phn tch s khng s dng mt s quan st cui cng (chng hn, 10% s quan st cui cng) trong vic c lng m hnh, nhng s s dng cc c lng thng s t tp quan st u tin d bo Yt cho phn mu dnh. Sau chng ta c th tnh MSE v MAPE cho giai on hu mu. M hnh no c cc gi tr o lng ny thp hn s tt hn cho mc ch d bo.

3.10 Tnh Nhn Qu trong M Hnh Hi Quy

Khi nh m hnh dng Y = + X + u, chng ta ngm gi nh rng X gy ra Y. Mc d R2 o thch hp, n khng th c s dng xc nh tnh nhn qu. Ni cch





khc, vic X v Y tng quan cht vi nhau khng c ngha rng s thay i X dn n s thay i Y hay ngc li. V d, h s tng quan gia s lng knguru ca c v tng dn s nc ny c th l rt cao. Phi chng iu ny c ngha rng s thay i mt bin s lm cho bin kia thay i? R rng l khng, v y chng ta c mt trng hp tng quan gi to. Nu chng ta hi quy mt trong cc bin vi bin cn li, chng ta s c s hi qui gi to. Ly mt v d khc thc t hn, gi s chng ta hi quy s lng v trm trong mt thnh ph vi s hng hng s v s nhn vin cnh st (X) v sau quan st thy h s gc c lng c gi tr dng, c ngha rng c tng quan thun gia X v Y. Phi chng iu ny c ngha rng vic tng s lng cnh st s lm tng s v trm, do ngm ko theo phi c chnh sch gim lc lng cnh st? R rng kt lun ny l khng th chp nhn c. iu xy ra c th l mi quan h nhn qu l ngc li, c ngha l thnh ph nn thu thm cnh st v s v trm tng ln, v nh vy vic hi quy X theo Y l hp l hn. Tuy nhin, trong thc t, hai bin s c xc nh kt hp v do chng ta nn nh r hai phng trnh, mt vi Y theo X v cc bin khc v phng trnh cn li vi X theo Y v cc bin khc. Vic xc nh ng thi cc bin s c trnh by chi tit chng 13. Nh s thy chng ny cc c lng thu c bng cch b qua tnh ng thi s b sai lch v khng nht qun. Cng c th l s tng quan cao quan st c gia X v Y c th hon ton l do cc bin khc v khng bin no trong s chng c th trc tip gy ra cc bin cn li. Nhng v d ny nhn mnh tm quan trng ca vic cn nhc k lng bn cht c ch hnh vi tim n l g, tc l, qu trnh pht d liu l g (DGP), v lp m hnh mt cch ph hp. L thuyt kinh t, kin thc ca nh phn tch v cc hnh v tim n, kinh nghim qu kh, v.v. phi gi m hnh nn phi c xc nh nh th no. Tuy nhin, c th kim nh phng hng ca s nhn qu mt cch r rng (chi tit s trnh by chng 10). c gi quan tm n vn ny c th tham kho bi vit ca Granger (1969) v Sims (1972).

minh ha tm quan trng ca vic xc nh chnh xc s nhn qu, gi s chng ta o ngc v tr ca X v Y v c lng m hnh:

Xt = * + *Yt + vt (3.1) Liu chng ta c th tm c ng thng ging nh trc khng? Cu tr li, ni chung, l khng. V th tc bnh phng nh nht c p dng cho phng trnh (3.1) s cc tiu ha tng bnh phng ca cc lch ng t ng thng (xem hnh 3.10). Tri li, ng thng nghch cc tiu ha tng bnh phng ca cc lch ngang vt. Tm Yt theo Xt, Phng trnh (3.1) c th c vit li nh sau:

'''

***

* 1tt

t

tt vXvXY ++=

+

=

Vic cc tiu ha 2tu , lm tng t nh vi phng trnh (3.1), v cc tiu ha 2tvs thng cho ra cc kt qu khc nhau. C th hn, gi tr c lng ca s khc vi gi tr t phng trnh (3.1).

HNH 3.10 Cc Tiu Ha Tng Bnh Phng theo Trc Tung v Trc Honh





V d 3.9

Quan h c lng khi 2tu c cc tiu ha l (xem Phn Thc Hnh My Tnh 3.5)

SQFT13875035152GIA ,, +=

Khi quan h nhn qu c o ngc v 2tv c cc tiu ha, chng ta c

GIA913666538533SQFT ,, +=

Nghch o quan h c lng th hai v biu din GI l hm ca SQFT, ta c

SQFT169.0645.5SQFT913666.5

1913666.5

385.33GIA +=+=

Lu rng du ca s hng hng s b nghch o v dc l hon ton khc.

Nh vy di iu kin g th hai ng c lng s nh nhau? tr li cu hi ny, u tin p dng OLS cho phng trnh (3.1); ngha l cc tiu ha 2tv . Hon i X v Y trong phng trnh 3.10, ta c:

'

*

1

== yyxy

SS





V do xyyy SS /'

= . c lng bnh phng nh nht lm cc tiu 2tu l xxxy SS / = . ' bng , iu kin l

12

==

yyxx

xy

xy

yy

xx

xy

SSS

hoacSS

SS

Nhng v tri ca phng trnh th hai l rxy2, bnh phng ca h s hi quy n gia X v Y (nh ngha phng trnh 2.11). Nh vy, iu kin cn l X v Y phi tng quan hon ho. Tnh cht 2.4d ni rng nu tn ti s tng quan hon ho gia hai bin, th phi tn ti mt quan h tuyn tnh chnh xc gia chng. V vy, s thch hp gia X v Y phi hon ho th chng ta mi nhn c cng mt ng hi quy cho d chng ta p dng OLS cho phng trnh (3.1) hay (3.1). Nhn chung, s tng quan gia X v Y s khng hon ho, chnh v vy chng ta s khng nhn c cng mt ng thng hi quy. iu ny nhn mnh tm quan trng ca vic xc nh ng hng quan h nhn qu thay v vic chn thiu suy xt bin X v Y.

Nh c minh ha trc y trong v d v ti phm, quan h nhn qu c th theo c hai chiu, tnh hung ny c gi l phn hi. Quan h gia gi bn v lng bn cng l v d ca hin tng ny. V gi v lng bn c xc nh cng lc bi quan h tng tc gia cung v cu, cho nn ci ny c th nh hng ci kia. Tng t, hin tng phn hi cng c tm thy trong quan h gia thu nhp tng hp v tiu dng hay u t. Nhng tnh hung ny s c trnh by ch m hnh hi quy h phng trnh chng 13.

3.11 ng Dng: Quan H gia Bng Sng Ch v Chi Ph cho Hot ng Nghin Cu v Pht Trin (R&D)

Phn ny s trnh by mt v d din tp khc v phn tch hi quy. D liu dng trong v d ny tp tin DATA3.3, m s cp n cc bin sau:

PATENTS = S ng dng bng sng ch c ghi nhn, n v ngn, giao ng t 84,5 - 189,4

R&D = Chi ph cho nghin cu v pht trin, n v t la 1992, c xc nh bng t s gia chi ph theo la hin hnh v ch s gim pht tng sn phm quc ni gp (GDP), giao ng t 57,94 n 166,7.

D liu theo nm ly trong vng 34 nm t 1960 n 1993 cho ton b nc M. Ngun c trnh by ph lc D.

Nu mt quc gia chi nhiu hn cho hot ng nghin cu v pht trin, chng ta c th k vng rng quc gia ny s t c nhiu ci tin c bo v thng qua lut bng sng ch hn. Do , chng ta k vng tn ti mt quan h dng gia s lng bng sng ch c ban b v chi tiu cho R&D. Mc d hiu qu ca hot ng nghin cu v pht trin s tr vi nm sau khi d n c bt u, n gin ha chng ta b qua hin tng ny. nhng chng sau chng ta s kho st hiu ng tr ca cc bin c lp v s quay li v d ny.





M hnh hi quy tuyn tnh c lng c trnh by di y km vi cc tr thng k mu t trong ngoc n (Phn Thc Hnh My Tnh 3.6 hng dn cch lp li kt qu ca phn ny v Bng 3.5 trnh by kt qu.)

DR792057134SANGCHE &,, += (5,44) (13,97)

R2 = 0,859 d.f. = 32 Fc (1,32) =195,055 172,11 =

kim nh m hnh v s ngha tng th, chng ta s dng tr thng k F, c gi tr bng 195,055. Theo gi thuyt H0 th s bng sng ch v chi ph cho R&D l khng tng quan, Fc tun theo phn phi F vi bc t do t s l 1 v bc t do mu s l 32 (= 34 2). T bng A.4a (cng trong ba sau) chng ta c nhn xt rng gi tr ngng F(1,32) mc ngha 1% nm gia 7,31 v 7,56. V Fc cao hn nhiu so vi gi tr ny, chng ta kt lun rng s bng sng ch v chi ph cho R&D l tng quan ng k. Kt lun ny c cng c thm thng qua gi tr thng k mu t. Kim nh hai u vi mc ngha 1%, bng t trong ba trc ca quyn sch (hay Bng A.2) cho thy gi tr ngng vi 32 bc t do nm gia 2,704 v 2,75. V gi tr quan st tc cao hn nhng gi tr ny nhiu chng ta kt lun rng c s hng tung gc v dc c gi tr khc 0 mt cch ng k. S o thch hp R2 cho bit m hnh gii thch c 85,9% s bin i ca bin ph thuc. Mc d y dng nh l mt s thch hp tt, tuy nhin chng ta thy t hnh 3.11 rng m hnh khng hon ton th hin s bin i thc t v s bng sch ch. ng thng hi quy l ng lin v n khng i din y bn cht ng cong ca d liu quan st. Chnh v iu ny m hnh s d bo rt km s lng bng sng ch ti nhiu nm.

im ny c nu ra r hn Bng 3.5, bng ny c nhiu tr thng k hu ch khc. Ct th t l gi tr trung bnh c lng ( )tY , ct nm l gi tr phn d c tnh bng gi tr quan st tr i gi tr trung bnh c lng ( )ttt YYu = v ct cui cng l sai s phn trm tuyt i (APE), c tnh bng 100 tt Yu / . Gi tr d bo trnh by bng 3.5 c lm trn n 1 ch s thp phn. V d liu gc v s bng sng ch ch c mt s thp phn, nn vic c gng c c cc gi tr d bo c chnh xc n hn mt s thp phn l khng c ngha.

HNH 3.11 S Bng Sng Ch Theo Chi Ph cho R&D ca Nc M

S bng sng ch (ngn)

Ch ph R&D (t)





BNG 3.5 Bo Co My Tnh c Ch Thch cho Phn ng Dng ca Phn 3.11.

Cc lnh ELS c in m v cc li nhn xt c in nghing Danh sch cc bin (0) Hng s (1) Nm (2) R&D (3) PATENTS (SNG CH)

Thi on: 1, quan st ln nht: 34, phm vi quan st: sut 1960-1993, hin hnh 1960-1993 (c lng m hnh theo OLS)

c lng theo OLS s dng 34 quan st t 1960-1993 Bin ph thuc PATENTS

Bin H s Sai s chun T stat 2Prob(t > |T|)

(0) Hng (2) R&D

34,571064 0,791935

6,357873 0,056704

5,437521 13,966211

< 0,0001*** < 0,0001***

Gi tr trung bnh ca bin ph thuc

119,238235 S.D. ca bin ph thuc

29,305827

Tng bnh phng sai s (ESS)

3994,300257 Sai s chun ca phn d

11,172371

R-bnh phng khng hiu chnh

0,859 R- bnh phng hiu chnh

0,855

Tr thng k F 195,055061 p-value = Prob(F>2427.709)





1965 80,00 100,4 97,9 2,5 2,49 1966 84,82 93,5 101,7 -8,2 8,77 1967 86,84 93,0 103,3 -10,3 11,08 1968 88,81 98,7 104,9 -6,2 6,28 1969 88,28 104,4 104,5 -0,1 0,10 1970 85,29 109,4 102,1 7,3 6,67 1971 83,18 111,1 100,4 10,7 9,63 1972 85,07 105,3 101,9 3,4 3,23 1973 86,72 109,6 103,2 6,4 5,84 1974 85,45 107,4 102,2 5,2 4,84 1975 83,41 108,0 100,6 7,4 6,85 1976 87,44 110,0 103,8 6,2 5,64 1977 90,11 109,0 105,9 3,1 2,84 1978 94,50 109,3 109,4 -0,1 0,09 1979 99,28 108,9 113,2 -4,3 3,95 1980 103,64 113,0 116,6 -3,5 3,19 1981 108,77 114,5 120,7 -6,2 5,41 1982 113,96 118,4 124,8 -6,4 5,41 1983 121,72 112,4 131,0 -18,5 16,55 1984 133,33 120,6 140,2 -19,6 -16,25 1985 144,78 127,1 149,2 -22,1 17,39 1986 148,39 133,0 152,1 -19,1 14,36 1987 150,90 139,8 154,1 -14,3 10,23 1988 154,36 151,9 156,8 -4,9 3,23 1989 157,19 166,3 159,1 7,2 4,33 1990 161,86 176,7 162,8 13,9 7,87 1991 164,54 178,4 164,9 13,5 7,57 1992 166,70 187,2 166,6 20,6 11,00 1993 165,20 189,4 155,4 24,0 12,67

Nhiu gi tr APE ln hn 5%, v trong mt s nm chng vt qua 10%, y l t l kh ln. Chng ta cng quan st thy rng cc im phn tn co cm li vi nhau trong cc nm t 1966-1977, ch ra rng mt yu t no khc hn l chi ph R&D gy ra s thay i v s bng sng ch. Do , quan st k hn cc kt qu ch cho thy s xc nh sai m hnh. Trong chng 6, chng ta s dng tp d liu ny c lng m hnh ng cong v s xem xt xem liu vic xc nh ny c th hin tt hn cc bin i quan st c v s bng sng ch khng.

TM TT

Mc d m hnh hi quy tuyn tnh n hai bin c s dng trong chng ny, nhng hu ht cc kha cnh c bn ca vic tin hnh phn tch thc nghim c cp. Tht hu ch khi tm tt li cc kt qu c tho lun t u n gi.

Mt m hnh hi quy tuyn tnh n l Yt = + Xt + ut (t = 1, 2, , n). Xt v Yt l quan st th t ln lt ca bin c lp v bin ph thuc, v l cc thng s ca tng th khng bit s c c lng t d liu ca X v Y, ut s hng sai s khng quan st c, y l cc bin ngu nhin vi cc tnh cht c cp di y, n l tng s quan st. dc () c din dch l nh hng cn bin ca s tng mt n v gi tr Xt ln Yt , + Xt l tr trung bnh c iu kin ca Y cho trc X = Xt.

Th tc bnh phng nh nht thng thng (OLS) cc tiu ha tng bnh phng sai s 2tu v tnh ton cc c lng (k hiu v ) ca s hng tung gc v





dc . Yu cu duy nht thc hin vic c lng cc thng s theo OLS l n c gi tr nh nht bng 2 v t nht mt trong nhng gi tr ca X l khc nhau ngha l, khng phi tt c cc gi tr ca X l nh nhau.

Nu ut l bin ngu nhin c gi tr trung bnh bng 0, v Xt cho trc v khng ngu nhin, th E(ut) = 0 v E(Xtut) = 0. Cc phng trnh chun l = 0tu v = 0ttuX . Li gii ca cc phng trnh ny cho kt qu l cc c lng theo OLS ca v .

Di cc gi nh va nu ra, cc c lng theo OLS l khng thin lch v nht qun. S nht qun c gi nguyn ngay c nu Xt l ngu nhin, min l Cov(X, u) = 0 v 0 < Var(X) < - ngha l, min l X v u khng tng quan v X khng l hng s.

Nu cc gi tr u tun theo phn phi c lp v tng t nhau (iid) vi mt phng sai xc nh, v cng s l cc c lng khng thin lch tuyn tnh tt nht (BLUE); tc l, trong s tt c t hp tuyn tnh khng thin lch ca cc gi tr ca Y, v

c phng sai nh nht. Kt qu ny c gi l nh l Gauss-Markov v c ngha rng, ngoi tnh cht khng thin lch v nht qun, cc c lng theo OLS cng l cc c lng hiu qu nht. Nu cc gi tr ca u tun theo phn phi chun c lp v tng t nhau N(0, 2), cc c lng theo OLS cng l cc c lng thch hp nht (MLE).

T v , gi tr d bo ca Yt (k hiu l tY ) thu c bng tt XY += , v phn d c c lng bng ttt YYu = . Sai s chun ca cc phn d l mt c lng

ca lch chun v c tnh theo cng thc [ ] 2/12 )2( = nut . T cc kt qu ny, ta c th suy ra sai s chun ca v ( s v s ). Cc sai s chun cng nh, chnh xc ca cc c lng ca cc thng s cng ln. S bin i ca X cng ln cng tt v iu ny c khuynh hng ci thin chnh xc ca cc c lng ring l.

Cc bc tin hnh kim nh i thuyt mt u v c tin hnh nh sau:

BC 1 H0: = 0 H1: > 0 BC 2 Tr thng k kim nh l ( ) 0 / stc = , trong s l sai s chun

c lng ca . Theo gi thuyt H0, gi tr ny tun theo phn phi t vi n 2 bc t do.

BC 3 Tra bng t vi gi tr ng vi n 2 bc t do v mt mc ngha cho trc (chng hn ), v tm im t*n-2() sao cho P(t> t*) = .

BC 4 Bc b H0 ti mc ngha nu tc > t*. Nu gi thuyt ngc li H1 l < 0, H0 s b bc b nu tc < - t*.

Kim nh c th c thc hin theo mt cch tng ng. Cc bc 3 v 4 c iu chnh nh sau:

BC 3a Tnh xc sut (k hiu l p-value) sao cho t > |tc|. BC 4a Bc b H0 v kt lun l h s c ngha nu p-value nh hn mt mc

ngha no ().





Cc bc kim nh gi thuyt ngc li H1 c tnh hai pha c thc hin nh sau:

BC 1 H0: = 0 H1: 0 BC 2 Tr thng k kim nh l ( ) 0 / stc = . Theo gi thuyt H0, gi tr tun

theo phn phi t vi n 2 bc t do. BC 3 Tra bng t vi gi tr ng vi n 2 bc t do v mt mc ngha cho trc

(chng hn ), v tm im t*n-2(/2) sao cho P(t> t*) = /2 (mt na ca mc ngha).

BC 4 Bc b H0 ti mc ngha nu |tc| > t*.

Cc bc hiu chnh thc hin kim nh theo phng php p-value nh sau: BC 3a Tnh p-value = 2P(t > |tc|). BC 4a Bc b H0 nu p-value nh hn mt mc ngha no ().

Tr thng k o lng thch hp ca mt m hnh l R2 = 1- (ESS/TSS), trong

= 2tuESS v2

t YYTSS

= . R2 c gi tr t 0 n 1. Gi tr ny cng cao

thch hp cng tt. R2 mang hai ngha: (1) n l t l ca tng phng sai ca Y m m hnh gii thch, v (2) n l bnh phng ca h s tng quan gia gi tr quan st (Yt) ca bin ph thuc v gi tr d bo ( )tY .

Kim nh v thch hp tng th ca m hnh c th c thc hin bng cch s dng gi tr R2. Cc bc c tin hnh nh sau (xy l h s tng quan ca tng th ca hai bin X v Y):

BC 1 H0: xy = 0 H1: xy 0 BC 2 Tr thng k kim nh l Fc = R2(n 2)/(1 R2). Theo gi thuyt H0, tr

thng k ny tun theo phn phi F vi 1 bc t do t s v n 2 bc t do mu s.

BC 3 Tra bng F theo t s 1 bc t do v mu s n 2 bc t do v mt mc ngha cho trc (chng hn ) tm g tr F* sao cho: P(F>F*) = .

BC 4 Bc b gi thuyt H0 (ti mc ngha ) nu Fc > F*.

Khong tin cy 95% ca c xc nh nh sau: ( ) ** , stst +

D bo c iu kin ca Y, cho trc X bng X0, l 0 XY += . Phng sai ca n (php o tin cy ca d bo) t l thun vi khong cch ca X0 so vi gi tr trung bnh X . Nh vy, X0 cng xa khi gi tr trung bnh ca X, gi tr d bo cng km tin cy.





Thay i thang o ca bin ph thuc dn n thay i tng ng thang o ca mi h s hi quy. Tuy nhin, cc gi tr R2 v tr thng k t s khng i. Nu thang o ca mt bin c lp thay i, h s hi quy ca n v cc h sai s chun tng ng b thay i cng thang o, tuy nhin tt c cc tr thng k khc khng thay i.

Vic xc nh chnh xc quan h nhn qu l ht sc quan trng trong m hnh hi quy. Gi thit chun l X gy ra Y. Tuy nhin, nu X v Y c tro i, v m hnh c c lng bng Xt = * + *Yt + vt, ng thng hi quy ni chung s khc vi ng c xc nh t m hnh Yt = + Xt + ut.

THUT NG

Analysis of variance (ANOVA) Phn tch phng sai Best linear unbiased estimator (BLUE) c lng khng thin lch tuyn tnh tt

nht Coefficient of multiple determination H s xc nh bi Conditional mean of Y given X Gi tr trung bnh iu kin ca Y bit trc

X Critical region Vng ngng (vng ti hn) Data-generating process (DGP) Qu trnh pht d liu Engel curve ng cong Engel Error sum of square (ESS) Tng bnh phng sai s Estimated residual Phn d c lng Explained variation S bin i gii thch c Feedback Phn hi Fitted straight line ng thng thch hp F-test Kim nh F Gauss-Markov theorem nh l Gauss-Markov Goodness of fit khp Heteroscedasticity Phng sai ca sai s thay i Homoscedasticity ng phng sai sai s (tnh cht phng

sai ca sai s khng thay i) Joinly determined c xc nh cng lc Linear estimator c lng tuyn tnh Marginal effect of X on Y Hiu ng cn bin ca X ln Y Mean absolute percent error (MAPE) Sai s phn trm tuyt i trung bnh Mean squared error (MSE) Sai s bnh phng trung bnh Mean squared percentage error (MSPE) Sai s phn trm bnh phng trung bnh Method of least square Phng php bnh phng ti thiu Nonlinear regression model M hnh hi quy phi tuyn Normal equation Phng trnh chun Ordinary least squares (OLS) Bnh phng ti thiu thng Population parameters Tham s ca tng th Population regression function Hm hi quy ca tng th Population regression line ng hi quy ca tng th Population variance Phng sai ca tng th Postsample forecast D bo hu mu





p-value Gi tr p Regression coefficients Cc h s hi quy Regression sum of squares (RSS) Tng bnh phng hi quy Residual Phn d Root mean squared error Cn bc hai ca sai s bnh phng trung

bnh Sample estimate c lng ca mu Sample regression line ng hi quy ca mu Sample regression function Hm hi quy ca mu Sample scatter diagram Biu phn tn ca mu Serial correlation Tng quan chui Serial independence c lp chui Significanly different from zero Khc 0 mt cch ng k Significanly greater from zero Ln hn 0 mt cch ng k Simple linear regression model M hnh hi quy tuyn tnh n Spurious correlation Tng quan gi to Spurious regression Hi quy gi to Standard error of a regression coefficient

Sai s chun ca h s hi quy

Standard error of the regression Sai s chun ca hi quy Standard error of the

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