bai giang quy hoach thuc nghiem

8/12/2019 Bai Giang Quy Hoach Thuc Nghiem

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NI DUNG MN HCQUY HOCH THC NGHIM

CHNG1. CC KHI NIM C BN1.1. M u1.2. Nhng khi nim c bn ca QHTN1.3. Cc nguyn tc c bn ca QHTN1.3.1. Nguyn tc khng ly ton b trng thi u vo1.3.2. Nguyn tc phc tp dn m hnh ton hc1.3.3. Nguyn tc i chng vi nhiu1.3.4. Nguyn tc ngu nhin ha (s dng ti u khng gian cc yu t)1.3.5. Nguyn tc ti u ca quy hoch thc nghim1.4. Thut ton (cc bc) ca QHTN1.4.1. Chn thng s nghin cu1.4.2. Lp k hoch thc nghim

1.4.3. Tin hnh th nghim nhn thng tin1.4.4. Xy dng v kim tra m hnh thc nghim1.4.5. Ti u ha hm mc tiu

- Xc nh ta im cc tr- Chuyn phng trnh b mt v dng chnh tc- Xc nh (kim tra) im cc tr thuc loi n o (cc i, cc tiu hay khng)?- Kim chng bng thc nghim

1.5. ng dng ca QHTN trong cc ng nh cng ngh1.5.1. Thit lp cc m t thng k

- Xc nh cc yu t nh hng- Xc nh cu trc h thc hin qu trnh ha l- Xc nh cc hm ton m t h- Xc nh cc tham s m t thng k- Kim tra s tng hp ca m t

1.5.2. Cc phng php k hoch ha thc nghim ch yu- K hoch bc mt hai mc ti u- K hoch bc hai

1.5.3. Xc nh cc gi tr ti u ca hm mc tiu1.5.4. Kt lun

1.6. M hnh ha1.6.1. M hnh1.6.2. M hnh ton1.6.3. Cc dng m hnh ton ca i tng cng ngh ha hc1.7. Ti u haCHNG2. PHNG PHP LA CHN CC YU T NH HNG2.1. La chn cc yu t vo (yu t c lp)2.1.1. Thng tin tin nghim


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2.1.2. Kt qu nghin cu l thuyt2.1.3. kin chuyn gia2.1.4. Cc thc nghim thm d, sng lc- Thc nghim thm d n yu t- Thc nghim thm d a yu t2.2. Phng php chuyn gia

2.2.1. Ni dung phng php2.2.2. V d2.3. Cc thc nghim sng lc theo phng n bo ha2.4. Cc thc nghim sng lc theo phng n cn i ngu nhin (phng nsiu bo ha)2.4.1. Xy dng k hoch thc nghim2.4.2. Xy dng biu sng lc2.4.3. Phn chia tun t cc yu t2.4.4. V d 2.2

2.5. Nhm cc yu t vo v chn mc tiu nh gi2.5.1. C s nhm cc yu t vo trong tng tp hp- Nhm cc yu t theo c tnh nh hng ca chng- Nhm cc yu t theo kh nng iu chnh trong thc nghim- Nhm cc yu t theo kh nng cho php v s lng yu t vo ca mi k hoch- Phn nhm cc yu t kt hp vi chn min quy hoch2.5.2. Chn mc tiu nh gi (cc yu t ra)2.6. nh hng ca cc tin ca phn tch hi quy n s la chn cc yu tc lp2.6.1. Tin v tnh n nh ca trng nhiu

2.6.2. Tin v tnh bt tng quan ca nhiu2.6.3. Tin v sai s iu chnh yu t vo2.6.1. Tin v tnh c lp tuyn tnh ca cc yu t nh h ngCHNG 3. TM TT MT S KT QU V KHI NIM CA XCSUT THNG K3.1. Bin ngu nhin3.2. Bng phn phi xc sut v hm xc sut3.3. Hm phn phi xc sut3.4. Hm mt xc sut

3.5. Cc c trng ca bin ngu nhin3.5.1. Mt3.5.2. K vng EX3.5.3. Phng sai DX3.5.4. lch chun X3.6. Mt s phn phi thng gp (trong quy hoch thc nghim)3.6.1. Phn phi Poa-xng3.6.2. Phn phi chun


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3.6.3. Phn phi Khi bnh phng3.6.4. Phn phi Student3.6.5. Phn phi Fisher-Snedecor3.7. Cc c trng mu3.7.1. K vng mu3.7.2. Phng sai mu thc nghim

3.7.3. Phng sai mu hiu chnh (iu chnh) mu thc nghim3.7.4. Phng sai ti hin (ti sinh)3.7.5. Phng sai d3.7.6. c lng tham s3.7.7. c lng khong3.8. Kim nh gi thuyt3.9. Kim nh thng k3.10. Sai s oCHNG4. CC PHNG PHP PHN TCH HI QUY TNG QUAN

4.1. Hi quy tuyn tnh mt thng s4.2. Hi quy Parabol4.3. Hi quy hm s m4.4. nh gi tnh mt thit ca lin h phi tuyn4.5. Phng php hi quy nhiu bin4.6. Phn tch hi quy di dng ma trn4.7. Lp phng trnh hi quy bng phng php Bradon4.8. Cc v d4.9. Phng php bnh phng cc tiu (nh nht)4.9.1. t bi ton

4.9.2. Ni dung ca phng php4.9.3. p dng cho hm mt bin4.9.4. p dng cho hm nhiu bin4.9.5. Cc v d c th trong thc tCHNG 5. QUY HOCH TRC GIAO5.1. Quy hoch trc giao v tnh cht5.1.1. M u5.1.2. nh ngha quy hoch trc giao (QHTG)5.1.3. Tnh cht ca QHTG

5.2. Quy hoch trc giao cp mt5.2.1. nh ngha5.2.2. Tnh cht5.2.3. Ma trn ca QHTG cp 15.2.4. M ha cc bin (i bin)5.2.5. Cc v d5.3. K hoch thc nghim bc 1 hai mc ti u5.3.1. K hoch bc 1 hai mc ti u ton phn (K hoch 2k)


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5.3.1. K hoch bc 1 hai mc ti u ring phn (K hoch 2k-p)5.3.3. Cc u im ca k hoch bc mt hai mc ti u5.3.4. Ti u ha bng phng php leo dc theo mt p tr (B mt biu din)5.3.5. Du hiu vng hu nh n nh (Vng dng)5.3.6. Cc v d5.4. Quy hoch trc giao cp 2

5.4.1. Khi nim v QHTG cp 25.4.2. Xy dng ma trn X (Ma trn thc nghim)5.4.3. Cc cng thc tnh ton5.4.4. Kim nh gi thuyt5.4.5. Cc v d5.5. K hoch thc nghim bc 25.5.1. M t vng phi tuyn (vng hu nh n nh)5.5.2. Cc k hoch bc hai trc giao5.5.3. Cc k hoch bc hai tm xoay

5.5.4. Cc v dCHNG 6. CC K HOCH THC NGHIM C BIT6.1. Phng php n hnh k hoch ha thc nghim v ti u ha6.2. K hoch ha tin ha cc thc nghim6.3. K hoch thc nghim khi nghin cu biu thnh phn-tnh cht6.3.1. Phng php mng n hnh6.3.2. K hoch mnh n hnh Scheffe6.3.3. K hoch trung tm n hnh6.3.4. K hoch thc nghim khi nghi n cu mt phn biu 6.3.5. K hoch ti u D

6.3.6. K hoch vi s cc tiu ha sai s c tnh h thng6.3.7. K hoch ha thc nghim khi nghi n cu quan h ph thuc ca tnh cht v ot l cc cu tCHNG 7. TI U HA THC NGHIM7.1. Quy hoch thc nghim tm cc tr7.1.1. t bi ton7.1.2. Phng php leo dc Box-Wilson7.1.3. Phng php n hnh u tm cc tr7.1.4. Quy hoch thc nghim gii bi ton nhiu mc tiu

7.2. Cc phng php ti u ha7.3. Ti u ha nh hm nguyn vng7.4. Phng php nghin cu b mt p tr7.4.1. Ti u ha bng phng php leo dc theo mt p tr (B mt biu din)7.4.2 Ti u ha bng phng php nghin cu b mt biu din (mt p tr)


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CHNG1. CC KHI NIM C BN1.1. M u(9)- Trong thc t ta thng gp cc bi ton nghin cu tnh cht hoc cht lng camt h thng, hoc hiu qu hot ng ca mt i tng ph thuc vo mt s ccyu t lin quan, chng hn nh:

+ Cht lng thp thng ph thuc vo nguyn liu, cc nguyn t ha hc, thi

gian nu luyn, nhit ,+ Sn lng cc sn phm v cht lng cc sn phm ph thuc vo cc yu tsn xut, cc ngun lc,thit b , my mc, vt liu,+ S ph thuc ca bin dng v o ng sut ca cc kim loi+ Mn ca chi tit my ph thuc v o ch lm vic ca vt liu, cht lng bmt.

- Trong cc bi ton k trn, nu c y cc thng tin t h ta c th xy dng ccm hnh gii tch cho h thng v vic kho st dng iu ca h thng hoc tm cctr c tin hnh theo cc phng php bit.

- Trong phn ny ta s xt phng php m hnh ha trong iu kin thiu thng tin,dng thc nghim xy dng m hnh v sau tm cch ti u n.Quy hoch thc nghim v phng php x l thc nghim l mt phng php tonhc c s dng rng ri trong hc tp v nghin cu, bao gm cc lnh vc:

+ L thuyt quy hoch v phng php thc nghim+ L thuyt h thng+ L thuyt thng k+ L thuyt ti u ha v ng dng.

Nhm chn mt chin lc ti u, trong iu kin cha hiu bit mt cch tondin mt qu trnh no tc ng vo qu trnh tin hnh thc nghim, ng thi

phi thu c:+ Cc s liu cn thit+ S lng th nghim t nht+ tin cy t ra trc+ Cng thc ton hc n gin nht+ t kt qu vi hiu qu kinh t, k thut cao nht.

- Cng vic ca quy hoch thc nghim l i xy dng m hnh hi quy dng:y=f(x1, x2,,xn, a1, a2,am)

Trong : a1, a2, ,am l cc tham s

x1, x2,,xn l cc bin s (thng s nh hng n qu trnh)Da vo cc kt qu thc nghim ta xc nh c cc tham s a1, a2,,am ta gi l i nhn dng m hnh thng k . Phng trnh nhn c gi l phngtrnh hi quy thc nghim ca h thng tng ng vi b th nghim cho.Phng trnh hi quy thc nghim ph thuc v o b n th nghim, phng phpnhn dng m hnh thng k. Nh vy cn phi c chin lc tc ng vo ccyu t vo, xy dng b n th nghim sao cho m hnh thu c:- t tin cy t ra


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- thng tin cn thit- Thun tin x l thng tin v tm cc tr- D dng s dng cc cng c tnh ton hin c.

1.2. Nhng khi nim c bn ca QHTN (13)1.3. Cc nguyn tc c bn ca QHTN(18)1.3.1. Nguyn tc khng ly ton b trng thi u vo

1.3.2. Nguyn tc phc tp dn m hnh ton hc1.3.3. Nguyn tc i chng vi nhiu1.3.4. Nguyn tc ngu nhin ha (s dng ti u khng gian cc yu t)1.3.5. Nguyn tc ti u ca quy hoch thc nghim1.4. Thut ton (cc bc) ca QHTN(21)1.4.1. Chn thng s nghin cu1.4.2. Lp k hoch thc nghim1.4.3. Tin hnh th nghim nhn thng tin1.4.4. Xy dng v kim tra m hnh thc nghim

1.4.5. Ti u ha hm mc tiu- Xc nh ta im cc tr- Chuyn phng trnh b mt v dng chnh tc- Xc nh (kim tra) im cc tr thuc loi n o (cc i, cc tiu haykhng)?- Kim chng bng thc nghim

1.5. ng dng ca QHTN trong cc ngnh cng ngh(25)1.5.1. Thit lp ccm t thng k

- Xc nh cc yu t nh hng- Xc nh cu trc h thc h in qu trnh ha l- Xc nh cc hm ton m t h

- Xc nh cc thams m t thng k- Kim tra s tng hp ca m t

1.5.2. Cc phng php k hoch ha thc nghim ch yu- K hoch bc mt hai mc ti u- K hoch bc hai

1.5.3. Xc nh cc gi tr ti u ca hm mc tiu1.5.4. Ktlun1.6. M hnh ha1.6.1. M hnh

- L mt i tng c mt ch th no trn c s ca s ng dng v cutrc v chc nng dng thay th cho mt nguyn bn tng ng c thgii quyt mt nhim v nht nh.

- Mt nguyn bn c th c nhiu m hnh ty thuc vo ch th cn gii quyt.1.6.2. M hnh ton

- Mt m hnh ton l biu din ton hc nhng mt ch yu ca 1 nguyn bntheo mt nhim v no , trong phm vi gii hn vi 1 chnh xc va v trong1 dng thch hp cho s vn dng.


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- Mt m hnh ton ca mt nguyn bn phi c 4 iu kin:+ Ch m t nhng mt chnh m ch th quan tm.+ M t trong phm vi gii hn.+ chnh xc va .+ Kh nng vn dng m hnh c lp trong iu kin c th.

1.6.3. Cc dng m hnh ton ca i t ng cng ngh ha hc

- Xt m hmh thng k thc nghim trong ho hc, cng ngh ha hcngita xy dng quan h gia cc i lng trn csthit lp cc quan h trn vic xl thng k nhng gi tr thc nghim.

- xc lp m t thng k ca i tng cng ngh ha hccn thc hinnhng bc sau:

+ Xc nh s cc yu t c lp nh hng ln h, tc l s yu t nh hng(k) ln 1 hay nhiu hm mc tiu.

+ Xc nh cu trc ca h s c m hnh ho.+ Xc nh cc hm ton m t cc qu trnh xy ra trong h, v thng l

hm nhiu bin v c biu din : y = f( x1, x2,,xk).+ Xc nh cc thng s m hnh theo s liu thc nghim.+ Kim tra s tng thch ca m hnh.

1.7. Ti u ha1.7.1. Khi nim

- L qu trnh tm kim iu kin tt nht (iu kin ti u) ca hm s cnghin cu.

- L qu trnh xc nh cc tr ca hm hay tm iu kin ti u tng ng thc hin 1 qu trnh cho trc.- nh gi im ti u cn chn chun ti u (l cc tiu chun cng ngh).

1.7.2. Cch biu din bi ton ti u - Gis mt h thng cng ngh c biu din di dng sau:

Y = F(x1,x2,...xk)Trong : (x1,x2,,xk) l vectcc thng s u vo.

Hm mc tiu : I = I (x1,x2,xk)Bi ton c biu din I opt = opt I (x1,x2,xk) =I (x1opt, x2opt,xkopt )

hoc I opt = max I ( x1,x2,xk) : i vi bi ton max.I opt = min I (x1,x2,xk) : i vi bi ton min.Iopt : hiu qu ti u.

(x1opt,x2opt,xk) nghim ti u hoc phng n ti u.1.7.3. Thnh phn c bn ca bi ton ti u 1.7.3.1. Hm mc tiu

- L hm ph thuc.- c lp ra trn cstiu chun ti u c la chn.

Hm mc tiu l hm th hin kt qu m ngi thc hin phi t c l tiuchun ti u dng hm, ph thuc vo yu t u vo, gi tr ca n cho php nhgi cht lng ca 1 nghin cu.


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1.7.3.2. Quan h gia cc i lng- Cc biu thc ton hc m t cc mi quan h gia tiu chun ti u ho

(hm mc tiu) v cc thng s nh hng (thng s cn ti u) n gi tr tiu chunti u ho ny.

- Cc quan h ny thng c biu din bng phng trnh cbn hoc mhnh thng k thc nghim (phng trnh hi qui).

- Quan h gia cc yu t nh hng vi nhau c biu din bng ng thchoc bt ng thc.1.7.3.3. Cc iu kin rng buc

- bi ton cng ngh c ngha thc t , cc biu thc m t iu kin rngbuc bao gm:

- iu kin bin.- iu kin ban u

1.7.4. Cc bc gii bi ton ti u:1.t vn cng ngh:xem xt cng ngh cn c gii quyt l cng ngh g v

chn ra nhng yu t nh hng chnh. Ch ra c hm mc tiu Y : YMAX,hoc YMIN2. Xy dng mi quan hgia cc yu t nh hng v hm mc tiu theo qui lutbit trc hoc m hnh thng k thc nghim.3. Tm thut gii:l phng php tm nghim ti u ca cc bi ton cng nghtrn cscc m t ton hc tng thch c thit lp. a s dn n tm cc trca cc hm mc tiu.4. Phn tch v nh gi kt quthu c

- Nu ph hp kim chng bng thc nghim- Nu khng ph hp xem li tng bc hoc lm li t vic t vn


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CHNG 2. PHNG PHP LA CHN CC YU T NH HNG2.1. La chn cc yu t vo (yu t c lp)(53)2.1.1. Thng tin tin nghim2.1.2. Kt qu nghin cu l thuyt2.1.3. kin chuyn gia2.1.4. Cc thc nghim thm d, sng lc

- Thc nghim thmd n yu t- Thc nghim thmd a yu t2.2. Phng phpchuyn gia (56)2.2.1. Ni dung phng php2.2.2. V d2.3. Cc thc nghim sng lc theo phng n bo ha (60)2.4. Cc thc nghim sng lc theo phng n cn i ngu nhin (phng nsiu bo ha) (64)2.4.1. Xy dng k hoch thc nghim

2.4.2. Xy dng biu sng lc2.4.3. Phn chia tun t cc yu t2.4.4. V d 2.22.5. Nhm cc yu t vo v chn mc tiu nh gi (75)2.5.1. C s nhm cc yu t vo trong tng tp hp- Nhm cc yu t theo c tnh nh hng ca chng- Nhm cc yu t theo kh nng iu chnh trong thc nghim- Nhm cc yu t theo kh nng cho php v s l ng yu t vo ca mi k hoch- Phn nhm cc yu t kt hp vi chn min quy hoch2.5.2. Chn mc tiu nh gi (cc yu t ra)

2.6. nh hng ca cc tin ca phn tch hi quy n s la chn cc yu tc lp(80)2.6.1. Tin v tnh n nh ca trng nhiu2.6.2. Tin v tnh bt tng quan ca nhiu2.6.3. Tin v sai s iu chnh yu t vo2.6.4. Tin v tnh c lp tuyn tnh ca cc yu t nh h ng


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CHNG3. TM TT MT S KT QU V KHI NIM CAXCSUT V THNG K NG DNG

3.1. Bin ngu nhin (BNN)+ nh ngha:L i lng ph thuc vo kt cc (kt qu) ca mt php th ngunhin no (gi tr ca n mang li mt cch ngu nhi n, s xut hin khng cbit trc).

K hiu: X, Y, x, y,l gi tr ca BNN VD: Nhit ca mt phn ng ha hc trong mt khong thi gian no l mtbin ngu nhin. Gi tr ca n nhn gi tr trong khong [t,T] no .+ Phn loi:BNN ri rc v BNN lin tc.

- BNN ri rc: tp gi tr ca n l mt tp hu hn hoc v hn m c ccphn t (s cuc in thoi trong mt n v thi gian, s tai nn giao thng, s laong ca mt cng ty,)

- BNN lin tc: tp gi tr ca n lp kn mt khong trn trc s(huyt p ca

bnh nhn, di ca chi tit my, nhit , p sut,)3.2. Bng phn phi xc sut v hm xc sut-i vi BNN ri rc, mi gi tr ca n c gn vi mt xc sut c trng

cho kh nng BNN nhn gi tr , Pi=P(X=xi)Pn=P(xn)=P(X=xn)

X=x x1 x2 xnP(x) P1 P2 Pn

Hm s p(x)=P(X=x), xX c gi l hm xc sut ca Xp(x)0 vi mi x v

( ) 1x

p x

(1.1)

Bng phn phi xc sut cha tng qut c trng cho 1 BNN ty , nht lBNN lin tc.3.3. Hm phn phi xc sut: F(x)F(x) c xc nh nh sau:

F(x)=P(X


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3.5.1. Mt(Mod)- Nu X l BNN ri rc th Mod(X) l im x0sao cho P(X=x0)= max P(X=xi),

tc l ti cxc sut xi xut hin ln nht.- Nu X l BNN lin tc vi hm mt f(x), Mod(X) l im x0 sao cho

f(x0)=max f(x) vi x[a,b]Mod(X)=x0

3.5.2. K vng:EX-Nu X l ri rc:

i

EX xiPi

vi P(xi)=Pi

-Nu X lin tc: ( )EX xf x dx

; f(x) l hm mt

K vng l gi tr trung bnh (theo xc sut) ca i lng ngu nhin3.5.3. Phng sai:DX (VX)

DX=E(x-EX)2 ; (x-EX) l lch ca bin X so vi trung bnh ca n.Phng sai chnh l trung bnh ca bnh phng lch . Phng sai c trng cho

phn tn ca BNN quanh tr trung bnh ca n. Phng sai cng ln th btnh ca bin tng ng cng ln.2( )DX xi EX Pi (BNN ri rc)

2( ) ( )DX x EX f x dx

(BNN lin tc)

DX=E(X2)-(EX)2

3.5.4. lch chun:xx DX

3.6. Mt s phn phi thng gp (trong quy hoch thc nghim)

3.6.1. Phn phi Poa-xng:Bin ngu nhin X c gi l tun theo lut phn phi Poa-xng (X~P())nuhm xc sut ca n c dng:

( ) ( )!

xe

P X x P xx

; x=0,1,2, v l hng s

V d v phn phi Poa -xng: s cuc gi in thoi ca mt tng i trongmt ngy, s lng khch hng ca mt nh bng trong mt gi,l BNN c phnphi Poa-xng)3.6.2. Phn phi chun(Phn phi Gauss-c)X~N(a, 2) nu hm mt f(x) c dng:

2 2( ) /21( )2

x af x e

xR

EX=a v DX=2

3.6.3. Phn phi Khi bnh phngX2(n)Trong thng k ngi ta thng gp bi ton sau:

Cho n BNN c lp, X i (i=1,n), mi Xic phn phi chun dng X i N(0,1)


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Bin NN mi 2 21

n

i

X X i

l BNN lin tc, phn phi ca n mang t n phn phi khi

bnh phng vi n bc t do X2(n) c hm mt :1

2 2

2

( )

2 ( )

2

n x

n

x ef x

nr

(x>0, n>0) v f(x)=0 nu x0

r(x) l hm gama: 10

( ) x tr x t e dt

x>0

E(X2)=n, D(X2)=2n, ( n th PP X2(n) PP chun)

3.6.4. Phn phi StudentCho U, V l hai bin NN c lp, UN(0,1), V l BNN c phn phi khi bnh

phng X2(n bc t do), xt bin

( ) U

T n

Vn

Phn phi T(n) l PP Student vi n bc t do, c hm mt :1

2 2

1( )

2( ) 1. ( )

2

nnr

tf t

n nn r

r(x) l hm gamaE(T(n))=0, D(T(n))=n/(n-2)

P(t>tn)=/2 v P(t-t

n)=/2

3.6.5. Phn phi F isher-SnedecorCho Z1v Z2l hai bin NN c lp c phn phi khi bnh phng vi n1v n2bc t do, Z1X

2(n1), Z2X2(n2). Xt BNN:

F=(Z1/n1)/(Z2/n2) l PP Fisher-Snedecor vi n1, n2bc t do. K hiu F(n1, n2)

f(x)= 0 nu x0 v

1 2

2

1 2

2

( ) .

( 2 1 )

n n

n n

xf x C

n n x

nu x>0

y:

1 2

2 21 2( ) 1 22

1 2( ) ( )2 2

n nn n

r n n

Cn n

r r

E(F)= n1/(n2-2) ,2 2

2

2 2 ( 1 2 2)( )

1( 2 2) ( 2 4)

n n nD F

n n n

P(F< ( 1, 2)n nf )=

3.7. Cc c trng mu3.7.1. K vng mu(Trung bnh mu thc nghim)


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Mu dng 1: (x1, x2,,xn)

1

1 n

i

X xin

Mu dng 2:xi x1 x2 xk

ni n1 n2 nk

1

1 n

i

X xi nin

vi1

k

i

n ni

3.7.2. Phng sai mu th c nghim: S2

22

1

1( )

n

i

S xi X n

n l s ln o hay quan stxi l s o ca i lng x ln o th i

x

l trung bnh mu thc nghimHoc

22

1

1( )

k

i

S xi X nin

3.7.3. Phng sai mu hiu chnh ( iu chnh) mu th c nghi m

221

1

1( )

1

n

i

S xi xn

f=n-1 l bc t do c trng cho mu thc nghim

Hoc22

11

1( )

1

k

i

S xi x nin

(n-1): bc t do c trng cho mu thc nghim.3.7.4. Phngsai ti hin (ti sinh)

- Phng sai ti hin ca mt th nghim: gi s th nghim c lp i lp lim ln vi gi tr thu c l y1, y2,,ym. Khi :

22

1

1( )

1

m

th

i

S yi ym

vi m-1 bc t do, c trng cho kh nng bin i m khng

lm thay i h.m l s ln lp.-Phng sai ti hin ca N th nghim: thng s dng cho th nghim song

song (lp li)

2 2

1

1 Nth u

u

S SN

2

2

1

1( )

1

m

u ui u

i

S y ym

Suy ra:2

2

1 1

1( )

( 1)

N m

th ui u

u i

S y yN m


14/62

N l s th nghimm l s ln lp liPhng sai phn phi trung bnh cho tng th nghim c xc nh nh sau :

2 21( )th thS y Sm

V d: Tnh phng sai ti hin ca mt cuc th nghim tng ng vi nhng s liu

thc nghim thu c bng sau :

T bng s liu ta thy i = 1,2,3; u = 1,2,3,8; m = 3; N = 8. tnh phng sai ti hin ca mt cuc th nghim ta lp bng sau:

T kt qu bng 2, ta tnh phng sai ti hin ca cuc th nghim:2 2

1

1 1448

8

N

th u

u

S SN

Phng sai phn phi trung bnh cho mt th nghim:2 21 18( ) 6

3th thS y S

m

3.7.5. Phng sai d ( d)


15/62

- L hiu gia gi tr thc nghim thu c vi gi tr tnh theo phng trnh hiquy ca cc thng s ti u.

22

1

1( )

N

du u

u

S y yN L

N-L: t doL: s h s c ngha trong phng trnh hi quy

uy

l gi tr tnh theo phng trnh hi quy

uy

l gi tr trung bnh thc nghim ti th nghim th uyu l gi tr thc nghim trong iu kin khng l m th nghim lp.3.7.6. lch chun (SD)

- L tham s dng xc nh phn tn ca bin ngu nhin c cng n vvi n.

- Gi s S2v 21S l phng sai v phng sai iu chnh mu ngu nhin caX, khi S v S1 c gi l lch tiu chun iu chnh mu thc nghim ca X

v xc nh nh sau:2S S

21 1S S

3.7.7. Sai s chun (SE)- L t l gia lch chun trung bnh mu vi cn bc hai ca dung lng

mu:1S

SEN

- L thng s thng k quan trng nh gi mc phn tn ca mu chnh

n biu th sai s ca s trung bnh. Sai s y do s chnh lch chc c h thngca s liu m phng thc chn mu l mt trong nhng nguyn nhn chnh gynn.

- Mc ch chnh SE l xc nh mc phn tn ca gi tr trung bnh mu vgii hn tin cy ca mu thc nghim.3.7.8. ngha ca phng sai, lch chun, sai s chun

- Phng sai, lch chun, sai s chun gip cho ta nhn bit c mc ng u ca gi tr thc nghim.

- Nu phng sai, lch chun, sai s chun nh th cc gi tr thc nghim

tng i ng u v tp trung xung quanh gi tr trung bnh.3.7.9. c l ng tham sCho BNN X c lut phn phi xc sut bit nhng cha bit tham s no

. Ta phi xc nh da trn thng tin thu c t mt mu quan st (x1, x2,,xn)ca X.Qu trnh xc nh cha bit l qu trnh c lng tham s.

Gi tr tm c

gi l c lng ca .

l mt gi tr s c gi l clng im.


16/62

nh gi mt c lng l tt hay khng ngi ta phi so snh n vi thtnhng tht li cha bit nn phi a ra cc tiu chun nh gi cht lng ca

thng k

nh mt xp x tt nht ca .Cc tnh cht ca c lng im:

-c lng khng chch: Thng k

c gi l c lng khng chch ca

nu E(

)=. Suy ra E(

-)=0-c lng vng: Thng k

c gi l c lng vng ca nu

(x1,x2,,xn) khi n dn ti v cng (xc sut) bng .

-c lng hiu qu: : Thng k

c gi l c lng hiu qu ca nun l c lng khng chch vi phng sai b nht.3.7.10. c l ng khong

-Nu da vo thng k ngi ta xc nh c 1 v 2, sao cho vi mt xcsut cho trc tham s ri vo khong th (1, 2) gi l khong tin cyvi tin cy cho:

P(1< < 2)=1- l mc ngha(1- ) l tin cy(2-1): di khong tin cy-c lng im c mt nhc im l khng bit c chnh xc cng

nh xc sut c lng chnh xc. Nht l khi kch thc mu nh, s sai lchca c lng so vi gi tr tht kh ln.

- Gi s mt php o vi sai s tin cy nh sau:

|X- X

| = X= e tin cy P l xc sut kt qu cc ln o ri vo khong tin cy

(X- e


17/62

Chng ta cng mun xc sut xy ra hai sai lm trn cng b cng tt. Trongthc t khng th lm gim c c hai xc sut v tng th gim v ngc li.

Thng thng sai lm loi mt d kim sot v d tnh hn nn ngi ta haychn trc nh l ngng xc nh xc sut phm sai lm loi mt lun lunnh hn mt b .

= 0,1; 0,05; 0,01; 0,001;ph thuc yu cu thc t v nh nghin cu. l

mc ngha ca quy tc kim nh.3.9. Kim nh thng k

- BNN y ph thuc vo mu l mt thng k. Mi thng k cho bi mt hm camu (x1, x2,,xn); y=g(x1, x2,,xn) . Ty theo dng ca g m ta c cc thngk khc nhau. Cc thng k c th tun theo mt lut phn phi vi mt bc tdo no (da vo cc nh l v xc sut v thng k). kim tra ta trabng.

- VD: Xt hai mu ON1=(x1, x2, ,xn)ON2=(x1, x2,, xn)

2

2 211 22

2( )sF s s

s l mt thng k

- Bc t do: Cc i lng (N i lng) khng hon ton c t do m b rngbuc vi nhau bi m iu kin no . Do ch c N-m i lng t do. Dovy bc t do l (N-m)

* Cc bc kim nh gi thit:Bc 1:Chn mt thng k lin quan n Ho: g(x1, x2,,xn)Bc 2: Da vo cc nh l thng k bit thng k tun theo lut phn phinoBc 3: Chn mc ngha . Tra bng ta s c g.

Bc 4: Da vo mu tnh g

:

- Nu g g

th bc b Ho

- Nu g g

th chp nhn Ho3.9.1. Kim tra s ng nht ca cc ph ng sai(kim tra hi t ca cc gi tr

th nghim). Phng php ny ch p dng trong phng n th nghim songsong (S dng phn phi Cochran).N l s th nghim trong mt cuc th nghim.f = m-1 l t do ng vi th nghim c phng sai ti hin ln nht.m l s ln lp ca th nghim c phng sai ti hin ln nht.Gb c tm thy bng vi mc ngha chn, l im gp nhau gia hng

biu th s th nghim N v ct biu th bc t do f.


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* Cc bc tin hnh kim tra- Xc nh i lng trung bnh t cc kt qu ca cc th nghim song song.- Xc nh cc phng sai thc nghim ( 2Su ) ti mi im th nghim

theo cng thc (2.9).

- Tnh tng cc phng sai 21

N

u

Su

- Tnh Gtn theo cng thc sau:

2

2

1

max; 1, 2,3,...,N

u

SuGtn u N

Su

2maxSu l gi tr cc i ca phng sai thc nghim th u.N l s th nghim trong mt cuc th nghim.- Tra bng Gb vi mc ngha chn, s th nghim N v t do f=m-1ca im thc nghim c phng sai ti hin ln nht.- So snh Gtn v Gb.

+ Nu Gtn < Gb : gi thit c chp nhn.+ Nu Gtn > Gb : gi thit khng c chp nhn.

- V d: Xc nh s bng nhau ca hai phng sai.Gi s ta c hai mu ly ra t hai tp gcON1: (x1, x2,,xn) ON1: (y1, y2,,yn)T y ta tnh c hai c trng mu:

212

111

1( )

1

N

i

i

s x xN

22

22

12

1( )

1

N

i

i

s y yN

a ra gi thuyt: 2 2 2 20 1 2 1 2: ( )H s s

i thuyt: 2 2 2 21 1 2 1 2: ( )H s s

Chn thng k:2

2 222 12

1

( )s

F s ss

Suy ra F c phn phi Fisher vi bc t do ca t l : N1-1 v ca mu N2-1- Chn mc ngha , tra bng ta tnh c F


19/62

- Da vo mu tnh c:2

221

sF

s

- Nu 0F F H

ng

- Nu 0F F H

sai (bc b H0)Minh ha: Gi s N

1=17, N

2=12. Gi s tm c

2 21 232,6; 63,68s s

Gi thuyt 2 2 2 20 1 2 1 2: ( )H s s

i thuyt:2

2 2 2 2 11 1 2 1 2 2

2

: ( ) : s

H s s Fs

Chn =0,01. Bc t 11, bc mu 16- Tra bng F= 3,61

- Tnh2

221

63,681,9 3,61

32,6

sF F

s

- Cng nhn H0nn2 2 2 21 2 1 2( )s s

Kim nh s bng nhau ca nhiu phng sai. Gi s cm tp gc X1, X2,,Xm.Ly ra Nmu (s ln th nghim) c cng dung lng m(s bin hoc s ln o lpli). Tnh c cc phng sai mu 2 2 21 2, ,..., ms s s

a ra gi thuyt: 2 2 20 1 2: ... mH s s s i thuyt: 2 2 21 1 2: ... mH s s s Chn thng k:

22ax

max2 2 21 2

, max...

mi

m

sG s s

s s s

Nu G ng th G c phn phi Cochran, bc t m-1, bc mu N. Chn , tra bngc G.

Nu G G

th cng nhn H0

Nu G G

th bc b H0Minh ha: c bn mu dung lng N=17. Gi s

2 2 2 21 2 3 40, 26; 0,36; 0, 4; 0, 42s s s s

Chn =0,05. Bc t 16, bc mu 4. Tra bng ta c G=0,4366.

Tnh

2ax

2 2 21 2

0,42

0,2917... 0, 26 0,36 0,4 0, 42

m

m

s

G s s s

V G G

nn cng nhn H03.9.2. Kim tra ngha ca cc h s trong ph ng trnh hi quy(Phn phi

Student, t

l gi tr tnh ton c, tra bng tm t). Nu t t

th loi cc h s

. Nu t t

th nhn cc h s .


20/62

- Mc ch ca kim tra ny l xem cc h s bj trong phng trnh hi qui ckhc 0 vi mt tin cy no hay khng.

- kim tra ngha ca cc h s trong phng trnh hi qui ta phi s dngchun Student (t).* Cc bc tin hnh kim tra:

- Tnh chun ttn theo cng thc: ttn = tj =

| |j

bj

b

S

bj l h s ng vi yu t th j trong PTHQ; j = 0,1,2,Sbj lch qun phng ca h s bj

- Tra bng tb (P,f) ng vi mc ngha P chn trc v f; f l bc t do ng viphng sai ti hin ca tng phng n m ngi nghin cu chn.

- So snh tj v tb+ Nu tj > tb h s bj c ngha v c gi li trong PTHQ.+ Nu tj < tb h s bj khng c ngha v loi khi PTHQ. Cc h s cn

li c tnh li theo phng phpbnh phng ti thiu cho ti khi tt c

chng u c ngha.3.9.3. Kim tra s t ng hp ca phng trnh hi quy thc nghim(PP Fisher, F

l gi tr tnh ton c, tra bng tm F). Nu F F

th m hnh tng hp

(tng thch). Nu F F

th m hnh khng tng hp ( khng tng thch).Khi ta phi: kim tra li cng vic tnh ton, kim tra li m hnh, chn lihm hi quy mc (bc) cao hn.- Dng PTHQ l do ngi nghin cu t chn v cc h s trong PTHQ c

xc nh da trn cc s kiu thc nghim. V vy cn phi xem xt m tton hc

c ph hp vi thc nghim hay khng, v ngi ta dng phn phi Fisher (F) vimt mc ngha no .* Cc bc tin hnh kim tra:- Vit PTHQ vi cc h s c ngha.- Tnh Ftn theo cng thc:

2

2tt

th

SFtn

S

Trong : Stt l phng sai tng thchSth l phng sai ti hin (ti sinh) vi phng n th nghim ti tm hoc vi phng

n th nghim song song (lp).- Fb tra bng fb (P, f1,f2) tc l ng vi mc ngha P chn v bc t do f1, f2- Tiu chun kim nh (so snh Ftn v Fb)

+ Nu Ftn < Fb th PTHQ va lp ph hp vi thc nghim.+ Nu Ftn > Fb th PTHQ va lp khng ph hp vi thc nghim v

lm tip cc cng vic sau:* Kim tra li cng vic tnh ton.* Xem li m hnh nghin cu ng cha.


21/62

* Chn m t ton hc (PTHQ) mc cao hn.3.10. Sai s o

- Trong thc nghim, nhng gi tr nhn c l gi tr gn ng ca mt gitr thc. x = xa gi l sai s o.Vi : a l gi tr thc ca mt vt

x l kt qu quan st c.

x l lch gia a v x.3.10.1. Sai s th

- L sai s phm phi do ph vnhng iu kin cn bn ca php o, hay ssut ca ngi thc hin dn n cc ln o c kt qu khc nhau nhiu. Khi phthin sai s th cn thc hin li cng vic (nu iu kin c ho php).

- Cch kh sai s th :+ Kim tra cc iu kin cbn c b vi phm hay khng.+ S dng mt phng php nh gi(phng php ton hc loi b sai

s th vi tin cy v hiu qu nht nh) , loi b hoc gi li nhng kt

qu khng bnh thng.3.10.2. Sai s h thng- L sai s khng lm thay i trong mt lot php o, m thay i theo mt

quy lut nht nh.- Nguyn nhn gy sai s: do khng iu chnh chnh xc dng c o, hoc mt

i lng lun thay i theo mt quy lut no , nh nhit Cc sai s ny cth pht hin, o c tm c nguyn nhn v hiu chnh c. Thng thng cckt qu thc nghim u xem nh pht hin sai s h thng v loi b.

- khc phc ngi ta t mt h s hiu chnh ng vi mi nguyn nhn.3.10.3. Sai s ngu nhin

- Sai s ngu nhin ca php o l i lng ngu nhin c trng bng lutphn phi th hin mi quan h gia cc gi tr c th c ca sai s v xc sut sais ngu nhin nhn cc gi tr y.

- L sai s cn li sau khi kh sai s th v sai s h thng.- Sai s ngu nhin do nhiu yu t gy ra, tc dng rt nh, khng th tch

ring ra, v th khng loi tr c. Nhng c th tm ra quy lut, tnh c nhhng ca chng n kt qu thc nghim. Vic xc nh nh hng ca chng davo cc hiu bit v quy lut phn phi cc i l ng ngu nhin.


22/62

CHNG 4. CC PHNG PHP PHN TCH H I QUY TNGQUAN4.1. Hi quy tuyn tnh mt thng s (33)4.2. Hi quy Parabol(37)4.3. Hi quy hm s m4.4. nhgi tnh mt thit ca lin h phi tuyn(38)4.5. Phng php hi quy nhiu bin (39)

4.6. Phn tch hi quy di dng ma trn(43)4.7. Lp phng trnh hi quy bng phng php Bradon(47)4.8. Cc v d(48)4.9. Phng php bnh phng cc tiu (nhnht)

Phng php bnh phng cc tiu l mt phng php rt c bn v hiu lc x l cc s liu thc nghim v xy dng m hnh thng k cho mt lp kh rngln cc i tng nghin cu thuc nhiu lnh vc khc nhau. Li gii ca ph ngphp BPCT l mt m hnh ton hc biu din mt cch gn ng i t ng thc.4.9.1. t bi ton

Gi s ta cn nghin cu mt i lng y trong mt h thng no . Thngthng, trong h y mt mt y ph thuc vo cc bin s c lp x1, x2,,xk c thiu khin c, mt khc y cn b nh hng ca cc tc ng ngu nhi n thngxuyn v khng iu khin c. (x1, x2,,xk) l cc bin vo hay cc nhn t,BNN gi l nhiu. y gi l hm mc tiu hay bin ra.Vn l phi tm mi quan h gia y v (x1, x2,,xk). Thng thng th t nhiu ctrc mt thng tin tin nghim v h thng ang xt. Bi vy ngi ta thng githit mi quan h gia y v (x1, x2,,xk) c dng:

1 2 1 2( , ,..., ; , ,..., )k my f x x x

Trong :dng ca hm f bit nhng cn m tham s 1, 2,, m cha bit.Nu ta gi thit thm rng 20,E D ngha l (0, )N th

1 2 1 2( , ,..., ; , ,..., )k my f x x x

2Dy

Hm y

gil hm phn hi ca y. Phng trnh trn gi l phng trnh hi quy lthuyt ca y theo (x1, x2,, xk) tm mi quan h tht gia y v (x1, x2,,xk) ngi ta tin hnh N th nghim vlp thnh mt bngNi x1 x2 xk y1 X11 X12 X1k y12 X21 X22 X2k Y2 N xN1 xN2 xNk yNBi ton t ra l trn c s cc s liu thu c hy tm mt hm s

( 1, 2,..., )y f x x xk

biu din gn ng tt nht hm y

v tm mt c lng tt nht cho2 .


23/62

Hm s y

c gi l m hnh thng k ca h thng thc ta ang nghin cu vc gi l phng trnh hi quy thc nghim. gii quyt bi ton ny ngi tadng phng php bnh phng cc tiu. u im ca phng php ny l khngcn bit ti lut phn phi xc sut ca bin ngu nhi n y. Theo phng php BPCT,bi ton dn v vic xc nh cc tham s 1, 2,, m sao cho tng bnh phng

cc sai s l nh nht. 2

1

( ) minN

i i

i

y y

Trong :yil cc kt qu th nghim (i=1,N)

1 2 1 2( , ,..., ; , ,..., ), ( 1, )i i i ik my f x x x i N

-cc gi tr l thuytV cc tham s 1, 2,, m cha bit nn tng bnh phng cc sai s l mt hm

ca cc tham s . K hiu l S( 1, 2,, m). Gi i iy y

Ta c:2

1 21

( , ,..., ) minN

m i

i

S

Ch : Trng hp ring hay gp ca cc hm hi quy thc nghim l:

1 1 1 2 1 2( , ,..., ) ... ( , ,..., )k m m k y g x x x g x x x

Trong : 1 1 2 1 2( , ,..., ),..., ( , ,..., )k m kg x x x g x x x l nhng hm s bit:

0 1 1 2 2 ... k ky x x x

2 20 1 1 2 2 12 1 2 11 1 22 2y x x x x x x

4.9.2. Ni dung ca phng php* Trng hp tuyn tnh:Bi ton: Gi s

0 1 1 2 2 ... k ky x x x

(0, )N

0 1 1 2 2 ... k ky x x x

Ni x0 x1 x2 xk y1 1 X11 X12 X1k y12 1 X21 X22 X2k Y2 N 1 xN1 xN2 xNk yNBi ton t ra:Xc nh j= bjsao cho:


24/62

2 2 20 1 ij

1 0 1

( , ,..., ) ( ) ( ) minN k N

k i i i j i

i j i

S y y y x

Bn cht: Tm mt ng thng d sao cho: Tng bnh phng cc lch gia tung ca ng th nghim vi ng thng l b nht.Da vo iu kin cn ca cc tr:

0, ( 0, )j

s

j k

Ta c: ij ij1 0

2 [(y - )x ] 0, ( 0, )N k

i j

i jj

sx j k

ij ij1 0

[(y - )x ] 0, ( 0, )N k

i j

i j

x j k

y l h k+1 phng trnh v k+1 n (0, 1, 2,, k) v c gii bng phngphp ma trn.

11 12 1

21 22 2

1 2

1 ...( )

(1) ...( )..............................

..............................

..............................

(1) ...( )

k

k

N N Nk

x x x

x x x

X

x x x

1

2.

.

.

y

y

Y

yN

1

2

.

.

.

N

y

y

Y

y

1

2

.

.

.

N

1

2

.

.

.

N

0

1

.

.

.

b

b

B

bk

Y=X +Y X

( ) ( )T TS Y Y Y Y

( ) 0TX Y X T TX X X Y

Bi ton tr thnh tm vect =B tha mn cc phng trnh trn. Khi S l b nht.

Gi s1 1

( ) 0 ( ) ( )T T T

X X B X X X Y

p dng cc nh l phn tch thng k li gii nhn c.Xt m hnh (hnh v): gi thit:

a. 0 1 1 2 2 ... k ky x x x b. (0, )N

0 1 1 2 2 ... k ky x x x


25/62

tm c 0 1 1 2 2 ... k ky b b x b x b x Cng bit c bjl c lng trng ca jCng bit c iy l c lng trng ca jy

Cn kim tra li cc gi thit:- Xem (0, )N

- Xem y c tuyn tnh hay khng? V nhng bin no nh hng trc tip n y(cc j no khc 0?)(1). Kim tra (0, )N * Thng thng, ta coi E =0 l chp nhn c. Nu E 0 th tnh tin li trc ta c bng khng.t 0 0 ( ) 0a E E a a a . Cch lm ny khng lm nh hng n ktqu nh gi.

Kim tra D=2nhng 2cha bit. Gi thit 2:oH D

i thit 2:oH D Ti mi im th nghim th I (i=1,N) ta lp li n (m) ln o c n(m) gi tr cayi1, yi2,,yin(m). Ta coi l mt mu v tnh c:

22

ij ij1 1

1 1; ( ) , 1,

1

n n

i i

j j

y y s y y i Nn n

Cc 2i

s u l c lng trng ca 2D . Vy nu H0ng th N s 2is phi bngnhau. kim nh H0ta s kim nh s bng nhau ca N ph ng sai.

2 2 21 2 ... Ns s s

Dng thng k:2

2axmax2 2 2

1 2

, max...

mi

N

sG s s

s s s

G tun theo phn phi Cochran, chn c ngha . Tra bc t = n-1, bc mu =N. Tac G. Tnh c G .

Nu G G th cng nhn H0ng. Dng phng sai ti sinh2 2

1

1 Nts i

i

S SN

c

lng 2.Nu G G th bc b H0.

(2). Kim nh s ph hp ca y va tm c.

a. Xem s bin c bng k khng?Gi s hm y l tuyn tnh, xem xt xem nhng bin n o nh hng trc tip ny, tc l ta xem xt xem nhng jno bng khng.Gi thit H0: j=0. i thit H1: j0Chn thng k


26/62

( 0)j j jbj jj j

b bt

sb sb

Theo nh l ca xc sut v thng k ng dng th tbjtun theo phn b Studentbc N-k-1Chn mc ngha tra bng Student ta c t.Tnh tb

jda vo mu,

2 2 1

j ts jj

s b s C

-Nu | |jtb t th cng nhn H0, suy ra j=0, chng t xjkhng nh hng n y

-Nu | |jtb t th bc bH0(cng nhn H1), suy ra j0, chng t xjkhng nh

hng n y. Nu c j=0 th cc h s cn li phi tnh ton li v cc h s tngquan ln nhau.b. S ph hp ca y

dng th nghim lp (ti sinh) tnh 2 21

1 Nts i

i

s SN

l c lng ca 2 khng

ph thuc dng ca y.

Mt khc: 2

2

1

1( )

1

N

du i i

i

s y yN k

cng l c lng ca 2nhng ph thuc

dng ca y.Nu y ph hp vi m hnh nghin cu th hai phng sai bng nhau.Gi thit: Ho: 2 2ts dus si thit: H1: 2 2ts dus s

Chn thng k2

2 22

( ) isdu du tsts

sF s s F her

s

Bc t: N-k-1, bc mu N(n-1)Chn tra bng tm F. Tnh Fda vo mu.Nu F F th cng nhn H0, tc l m hnh ph hp

Nu F F th bc b H0, tc l m hnh khng ph hp.Phi gi thit li m hnh.C th khng phi tuyn tnh m l bc hai, bc 3.Trong trng hp khng c iu kin tin hnh lp li th nghim th c th sosnh phng sai d 2dus vi phng sai trung bnh

2ys

2

2 1

( )

1

N

i

i

y

y y

s N

Theo tiu chun Fisher:2

2

y

du

SF

S bc t N-1, bc mu N-k-1, nu F cng nh hn

Fth phng sai ti sinh cng chnh xc .Th d:Tm m hnh biu din s ph thuc gia y v hai bin x1, x2trn c s bng quan stsau: (Th nghim lp li n=6)TT X1 X2 Y(1) Y(2) Y(3) Y(4) Y(5) Y(6) Y


27/62

1 2 1 10,5 9,5 9 11 10,6 9,4 102 2 2 13 11 11,5 12,5 12,2 11,8 123 8 10 18 16 16,6 17,4 17 17 174 2 4 13 12,5 13 13,5 14 12 135 6 8 15 15 14,5 15,5 14 16 156 3 4 10,5 9,5 10 11 10 9 107 5 7 14,2 13,8 13,6 14,4 15 13 148 3 3 12,5 11,5 11 13 12,2 11,8 129 9 10 17 15,6 15 16,4 16,5 15,5 1610 10 11 19 17 17,5 18,5 18,2 17,8 18

Nh vy:

10

12

.

..

16

18

Y

(1) 2 1

(1) 2 2

...............

..............................

(1) 9 10

(1) 10 11

X

1 2 1

1 1 . . . . . . . . 1 1 1 0 5 0 6 01 2 2

2 2 . . . . . . 9 1 0 . . . . . . . . . . . . . . . 5 0 3 3 6 3 9 8

1 9 1 0( 1 ) 2 . . . . . . 1 0 1 1 6 0 3 9 8 4 8 0

1 1 0 1 1

TX X C

10

1 1 ........ 1 1 13712

2 2 ...... 9 10 ... 756

16(1) 2 ...... 10 11 908

18

TX Y

1 1

(2876) 120 2601

( ) 120 1200 980

det( ) 7160 260 980 960

TC X X

C

19,3871

( ) ( ) 0,1286

(0,6174)

T TB X X X Y

Vy phng trnh hi quy c dng: 9,3871 0,1286 1 0,6174 2y x x Kim nh 2D trn bng ta c yi


28/62

Ta tnh cc 2is , y n=6

2 2 2 2 2 2 21

1 2,82(0,5) ( 0,5) 1 1 (0, 4) ( 0, 4) 0,564

5 5s

Tng t ta c:

2

2

2,58

0,5165s

2

3

2,32

0,4645s

242,50

0,55

s 252,50

0,55

s

262,50

0,55

s 272,40

0,4805

s

282,58

0,5165

s 292,82

0,5645

s

2102,82

0,5645

s 2ax 0,564ms

10

2

1

0, 564 0, 5640,10915,168

i

i

Gs

Chn mc ngha =0,05; bc t m=n-1=5, bc mu f=N=10. Tra bng ta cG=0,3029. Ta c G G . Vy Hoph hp.Ta dng phng sai ti sinh c lng 2

2 5,168 0,516810ts

S

Kim nhs ph hp ca y

Kim nh 2 2 10;

j bj ts jjs s C

20

0

28760,5168 0, 2075

71600,4555

b

b

s

s

21

1

12000,5168 0,08661

71600,2942

b

b

s

s

22

2

9600,5168 0,06206

71600,2492

b

b

s

s

0

1

2

9,3871; 20,6083

0,4555

0,12850,4367

0,2942

0,61742,4775

0,2492

j

bj b

bj

b

b

bt t

s

t

t

Chn =0,05 bc N-k-1=10-3=7t=1,9Ta c tb1


29/62

4.9.3. p dng cho hm mt binGi s

a. 0 1y x b. (0, )N Lm N th nghim ta c bng sau:

N x0 x y1 1 x1 y12 1 x2 y2. N 1 xN yNTnh ton bjTa c th tnh B bng vic i gii h phng trnh:

0 10

0 11

[y ( )] 0

[y ( )] 0

N

i i i

i

N

i i

i

x x

x

Theo phng php bnh phng cc tiu:

1

2

1 1

1 1( )

1 1 ............. 1 (1) 2

1 2 .... ............ ( )( )(1)( )

N

i

iT

N N

i i

i i

xN x

xC X X

x x xNx x

xN

2

1 11 1

2 2

1 1 1

( )( )1

( ) ( ) ( )( )

N N

i i

i iT

N N N

i i ii i i

x x

C X XN x x x N

1

1

1

11 1 ..........(1)

.( 1)( 2)..( )

. ( )

N

iT

N

i i

i

y

y yi

X Yx x xN

x y

N

2

1 1 1 1 102 2

1 1

1 1 11

2 2

1 1

( ) ( )( )

( )

N N N N

i

T T i i i iN N

i i

i i

N N N

i i i

N N

i i

i i

yi x xi xiyi

B X X X Y bN x x

N xiyi xi yi

b

N x x

d tnh ton ta lp bng sau:


30/62

TT xi yi xi2 xiyi1 X1 Y1 X12 X1y12 X2 Y2 X22 X2y2. . . . .. . . . .N xN yN xN2 xNyNCng

1

N

i

xi

1

N

i

yi

2

1

N

i

xi

1

N

i

xiyi

Kim nh:Ging nh trng hp k bin, lu :

2

2 2 10

2 2

1 1

( )

N

ib ts N N

i i

xi

s s

N xi xi

2 21

2 2

1 1

( )b ts N N

i i

Ns s

N xi xi

2

2

1

1( )

2

N

du i i

i

s y yN

Th d: Tm cng thc lin h bn ca si v m ca khng kh: m khng kh

(x)

bn ca si (y) bn

TBTT (1) (2) (3) (4) (5) (6)1 36 2 2,1 2,1 1,9 2 1,9 22 38 2,4 2,6 2,5 2,5 2,4 2,6 2,53 40 2,2 2,4 2,2 2,4 2,3 2,3 2,34 58 2,8 2,9 3,0 2,6 2,7 2,8 2,85 70 3,0 3,1 3,1 3,2 3,1 3,1 3,16 80 3,1 3,3 3,1 3,1 3,1 2,9 3,17 82 3,2 3,2 3,2 3,2 3,2 3,2 3,2

8 93 2,9 2,9 3,0 3,1 3,0 3,1 3,0

p dng cng thc ta c:b0=1,667, b1=0,0174Vy: y=1,667+0,0174xKim nh:* Kim tra D=2. Theo bng ta c n=6 th nghim ti mi im i (i=1,8). Ta tnh cc

2i

s (i=1,N)


31/62

62 21 1 1

1

1( ) 0,008

6 1 tts y y

Tng t ta c: 22 0,008s ;2

3 0,008s ;2

4 0,02s ;2

5 0,004s ;2

6 0,016s ;2

7 0s ;2

8 0,008s

Smax2=0,02

82

1

0,02 0,020,27780,072

i

i

Gs

Chn =0,05, bc t n-1=6-1=5, bc mu N=8. G=0,3595Ta c: G G . Vy D=

2

Dng phng sai ti sinh c lng 22 2

1

1 Nts

i

S siN

2 0,072 0,0098ts

S

* Kim tra 0; jj bjbj

bt

s

2

2 2 10

2 2

1 1

20

343770,009 0,011

28007( )

0,105

N

ib ts N N

i i

bo b

xi

s s x

N xi xi

s s

2 21

2 2

1 1

1

10

80,009 0,0000026

28007( )

0,0158

1,667 0,017415,87; 11,012

0,1058 0,00158

b ts N N

i i

b

bb

Ns s x

N xi xi

s

t t

Chn =0,05; bc t do ca phng sai ti sinh:m=N(n-1)=8(6-1)=40t=1,68Ta c:

0 0

0 1

00

b

b

t t

t t

Vy phng trnh hi quy c c b0v b1.* Kim nh s ph hp ca y:

2

2

1

1( ) 0,048

2

N

du i i

i

s y yN


32/62

Ta tnh 2 0,009tss Chn thng k:

2

2isdu

ts

sF F her

s

Bc t N-2=8-2=6; Bc mu N(n-1)=8(5)=40 0,048

5,33340,009F

Chn =0,05, F=2,3 suy ra F F . Vy m hnh tuyn tnh khng ph hp.4.9.4. Tuyn tnh ha mt s hm smt binphi tuyna. Hm s m: y=abx+

xy ab . Hy xc nh a, blg lg lgy a x b

t lga=0; lgb= 1Ta c: Y= 0 + 1X

p dng cng thc ca PP bnh phng cc tiu ta tm c b0v b1. T suy ra av b.b. Hm s 20 1 2y x x Phng php 1: Tuyn tnh hat Y= y ; X1=x; X2=x

2; Y= 0+ 1X1+ 2X2p dng cng thc ca PP bnh phng cc tiu vi k=2 bin, ta tnh c b0; b1; b2Phng php 2: p dng trc tip phng php bnh phng cc tiu ta tm c b0,b1v b2.C th m rng n a thc bc n.

c. Hm s ax ( 0; 0)b

y a x Logarit hai v:lg lg lgy a b x

Y= 0+ 1X (Trong 0 1lg ; lg ; ; lgy Y a b x X )p dng cng thc ca PP bnh phng cc tiu ta tm c b0v b1

d. Hm s ax

xy

b

Ta c:

1 1a b

xy

t 0 11 1; ; ;Y X a bxy

Khi : Y= 0+ 1Xp dng cng thc ca PP bnh phng cc tiu ta tm c b0v b14.9.5. Tuyn tnh ha mt s hm s nhi u bin phi tuyn

1 2 30

1 2 3 ...a a a any a x x x xn

Logarit hai v:lny=lna0+ a1lnx1+a2lnx2++anlnxn


33/62

i bin s v dng: Y=A 0+a1X1+a2X2++anXnp dng cng thc ca PP bnh phng cc tiu ta tm c cc h s b i

Bi tp:1. Xc nh phng trnh hi quy thc nghim vi cc s liu cho bng sau:

xi 1 2 3 4 5 6 7 8yi 2,35 2,41 2,6 2,73 2,90 3,11 3,25 3,45

p s: y=2,2285+0,1381x2. Xc nh phng trnh hi quy thc nghim vi cc s liu cho bng sau:

xi 2 3 4 5 6 7yi 11,52 15,12 18,47 22,05 25,61 28,05

p s: y=4,5759+3,4913x


34/62

CHNG 5. QUY HOCH TRC GIAO5.1. Quy hoch trc giao v tnh cht5.1.1. M uGi s ta lm N th nghim o y, ta c bng s liu sau:TT X1 X2 .. xk y1 X11 X12 X1k Y1

2 X21 X22 . X2k Y2. . . . . N xN1 xN2 xNk yNGi thit:

0 1 1 2 2 ... k ky x x x

(0, )N

0 1 1 2 2 ... k ky x x x

(Phng trnh hi quy l thuyt)Bng phng php BPCT ta tnh c:

1

( ) ( )T T

B X X X Y

Trong :

11 12 1

21 22 2

1 2

1 ...( )

(1) ...( )

..............................

..............................

..............................

(1) ...( )

k

k

N N Nk

x x x

x x x

X

x x x

1

2

.

.

.

y

y

Y

yN

0

1

.

.

.

b

b

B

bk

Thay B vo phng trnh ta c:0 1 1 2 2 ... k ky b b x b x b x l phng trnh hi quy thc nghim. X l ma trn dng

tnh ton.By gi ta xt vn liu c th b tr cc th nghim sao cho:- S th nghim t nht- Tnh ton gn- Bo m mc chnh xc5.1.2. nh ngha quy hoch tr c giao (QHTG)QHTG l quy hoch b tr cc th nghim sao cho ma trn:

11 12 1

21 22 2

1 2

1 ...( )(1) ...( )

..............................

..............................

..............................

(1) ...( )

k

k

N N Nk

x x xx x x

X

x x x

C tnh cht:


35/62

ij1

0N

im

i

x x

i l ch s th nghim; m, j l ch s bin (m, j=0,k): Tch hai vect ct bt

k bng khng. l tnh cht trc giao.

Khi m=0 th x i0=1 vi mi i nn suy ra: ij0

0; 0N

i

x j

: Tng cc phn t trong mt

ct bt k (tr ct 1) u bng khng .5.1.3. Tnh cht ca QHTG

Da trnphng php BPCT -ch khc l ch ng b tr cc th nghim. Do cc kt qu ca phng php BPCT u p dng c cho QHTG. C th, da v occ nh l ca phng php BPCT ta suy ra tnh cht ca QHTG.* Cng thc tnh cc bjn gin (Da theo PP BPCT) -(Khng chng minh)

01

ij ;1

1

11,

N

i

i

N

j ii

b yN

b x y j k N

5.2. Quy hoch trc giao cp mt5.2.1. nh nghaMt cch b tr th nghim sao cho quy hoch trc giao v c thm tnh cht:

2 2ij

1

( 0, )N

j

i

C x N j k

Tng bnh phng cc phn t ca mt ct ng bng s th nghim (N).5.2.2. Tnh chtKhi :

01

ij ;1

1

11,

N

i

i

N

j i

i

b yN

b x y j k N

5.2.3. Ma trn ca QHTG cp 1Khi k=2 (s bin) ta c m hnh:

y=b0+b1x1+b2x2(1)( 1)( 1)

(1)( 1)( 1)

(1)( 1)( 1)

(1)( 1)( 1)

X

S th nghim: N=22=4Khi k=3 ta c:

y=b0+b1x1+b2x2+b3x3


36/62

(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)

X

S th nghim: N=23=85.2.4. M ha cc bin ( i bin)

trn, cc phn t ca ma trn X l +1 v -1. Nhng khong bin thin cacc bin m ta nghin cu ni chung khc vi [ -1,+1]. Vy ta phi tm cch b tr thnghim sao cho X c tnh cht trc giao.Gi bin thc t l Zj; j=1,k; ZjminZj Zjmax

Vn t ra l i bin xj=1.Gi :min ax ax min 0

0 ; ;2 2

m m

j j j j j j

j j j

j

Z Z Z Z Z ZZ Z x

Z

Khi :min

0

max

1

0

1

j j j

j j j

j j j

Z Z x

Z Z x

Z Z x

Kim nh cc kt qu:Vic kim nh cc gi thit thng k hon ton ging nh trnh by trong phn PPBPCT, bao gm:* Kim tra D=2.V 2 khng bit nn ta dng phng sai ti sinh c lng 2. Nu khng c iukin lm nhiu th nghim ti mi im th ta c th tnh 2tss nh sau:Lm n0th nghim ti tm, tc l x1=x2==xk=0

o c1 2 3 00 0 0 0; ; ;......;

ny y y y

02

0 010

0

0010

1

( )11

nt

tst

nt

t

s y yn

y yn

Bc t do l n0-1* Kim ta j=0 ta lm nh c.Ch : Khi c jno bng khng, ngha l xjthc s khng nh hng trc tipn y nn s bin thay i. Trong trng hp chung th phi lm li vi m hnh mi


37/62

(khng c xjtham gia). Nhng trong quy hoch trc giao th cc bjkhng tng quannn khng phi lm li th nghim v khng phi tnh li cc bj0 c.* Kim nh s ph hp ca m hnh: y c th dng phng sai ti sinh bng cch lm n th nghim ti cc im

nhng cng c th lm n0th nghim ti tm.5.2.5. Th d: Tm mi quan h gia y v z1, z2, z3100z1200; 20z260; 10z330TT X0 X1 X2 X3 Z1 Z2 Z3 y1 + - - - 100 20 10 22 + + - - 200 20 10 63 + - + - 100 60 10 44 + + + - 200 60 10 85 + - - + 100 20 30 106 + + - + 200 20 30 187 + - + + 100 60 30 8

8 + + + + 200 60 30 12

Gi s m hnh tuyn tnh:

0 1 1 2 2 3 3y x x x

Tnh cc h s ca:

0 1 1 2 2 3 3y b b x b x b x

Theo phng php QHTG:

0

1

1 11

2 21

3 31

18,5

12,5

10,5

13,5

N

i

iN

i i

i

N

i i

i

N

i i

i

b y

N

b x yN

b x yN

b x yN

1 2 38,5 2,5 0,5 3,5y x x x

* Kim nh j=0Lm ba th nghim tm (n0=3). Khi quy hoch th nghim ta lm sn lun tu khng i n y mi lm. Ta c:


38/62

1 2 30 0 0

0

2 2 2 2

0

1

2

3

8; 9; 8,8

8,6

1[(8-8,6) (9 8,6) (8,8 8,6) ] 0, 28

3 1

0,28 0,55

0,55 0, 28

8,542,5

0, 2

2,512,5

0,2

8

| 0,5 |2,5

0,2

3,517,5

0,2

j

ts

ts

tsbj

bj b

bj

b

b

b

y y y

y

S

S

ssN

bt t

s

t

t

t

Chn =0,05, bc t do n0-1=2. Tra bng ta c t=2,92. 2=0 v |tb2|< t

Phng trnh hi quy dng: 1 38,5 2,5 3,5y x x

* Kim nh s ph hp ca m hnh:Tnh phng sai d:

22

1

1( )

( 1)

N

du i i

i

S y yN k

ngha: Phng sai d cng b cng t t. Dng phng sai d tnh c lng 2

.

282

1

1 2 3

4 5 6

7 8

22 2

2

1( )

8 3

2,5; 7,5; 2,5

7,5; 9,5; 14,5

9,5; 14,5

26 5, 25, 2; 0, 28; 18,57

5 0, 28

du i i

i

dudu ts

ts

S y y

y y y

y y y

y y

sS S F

s

Yi iy i iy y 2( )i iy y2 2,5 -0,5 0,256 7,5 -1,5 2,254 2,5 1,5 2,258 7,5 0,5 0,2510 9,5 0,5 0,2518 14,5 3,5 12,258 9,5 -1,5 2,25


39/62

12 14,5 -2,5 6,25Chn =0,01Bc t (v1, f1)=8-3=5 (Bc t do ca phng sai d=N-l (s h s c ngha)Bc mu (v2, f2)=n0-1=2(Bc t do ca phng sai lp)Tra bng ta c F=99,3V F F Ph hpCui cng phi i v bin tht:

0

31

1 3

201508,5 2,5 3,5

50 10

0,05 0,35 6

j j

j

j

z z zzx y

z

y z z

5.2.6. Vi nhn xt v kt lun v kh nng ca QHTG cp 1(1). QHTG cp 1 c th xc nh cc h s ca phng trnh hi quy dng:

ij

0

k

j

j

y x

Nu kim nh khng ph hp ta c th gi thit m hnh c dng bc hai khng honchnh:

0 1 1 2 2 12 1 2y x x x x

ij

0 , 0

k k

j j j

j i ji j

y x b x

(Tng qut)

Khi ta t x3=x1x2 ta c:

0 1 1 2 2 3 3y x x x a v tuyn tnh ba bin. Khi ma trn quy hoch l :N X0 X1 X2 X31 + - - +2 + + - -3 + - + -4 + + + +Ta thy X tha mn tnh cht ca QHTG cp 1. Tnh ton nh trn ta xc nh ccc bj(j=0,3). Kim nh nh c.(2). Gp m hnh dng:

20 1 1 2 2 ...

n

n ny a a x a x a x th t bin mi:2

1 2; ;....; n

nx x x x x x ta c:

0 1 1 2 2 ... n ny a a x a x a x

m hnh tuyn tnh n bin.Dng phng php QHTG cp 1 xc nh cc h s. Nu khng c th dng PPBPCT.(3). Mt s trng hp m hnh dng y=f(x), cng c th bin i tr v tuyn tnh(bng phng php i bin, t n ph v do dng c QHTG cp 1).Vic lm khng phi ch i vi h m mt bin, m ngay c i vi hm nhiubin s min l iu kin cho php . Tc l sau khi bin i th tnh cht trc giao cama trn X vn c bo m.


40/62

(4). Nu trng hp bc hai khng ho n chnh v kim nh khng ph hp th taphi gi thit m hnh l bc hai hon chnh.

2 20 1 1 2 2 12 1 2 11 1 22 2

2ij

0 1, 1

k k k

j j i j jj j

j i j ji j

y x x x x x x

y x x x x

Trng hp ny nu ta cng bin i nh trn th X khng cn tnh cht trc giaona. Ta phi quy hoch th nghim theo kiu khc. l QHTG cp 2.(5). trn ta xt quy hoch vi X=2kgi l quy hoch ton phn.i khi ngi ta xt quy hoch ring phn. Khi N=2k-1, N=2k-2, ., N=2k-p.Xt trng hp: N=2k-1;* Khi k=3, khi

0 1 1 2 2 3 3y x x x

N=22=4

13

(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)

X

hoc 23

(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)(1)( 1)( 1)( 1)

(1)( 1)( 1)( 1)

X

Ct x3c th xc nh bng cch:x3=x1x2 ( 13X ) hoc x3=-x1x2(

23X )

* Khi k=4, khi

0 1 1 2 2 3 3 4 4y x x x x

Ct x4c th xc nh bng cch:x4=x1x2x4=-x1x2x4=x1x3x4=-x1x3x4=x2x3x4=-x2x3x4=x1x2x3x4=x1x2x35.2.7. Th d v QHTN ring phn:Nghin cu qu trnh bin tnh nhm bng Molipden (Mo). Tham s ra l y (s ht

nhm/1cm2

).Cc tham s vo:Z1: khi lng Mo a vo (%)Z2: Nhit qu nung (

oC)Z3: Thi gian qu nung (pht)Z4: c tnh cht nh tnh v ch nhn hai gi tr:- Lm ngui nhanh- Lm ngui chm


41/62

Gi tr gc ca cc tham s, cn trn v cn di ca cc bin v Zj cho trong bngsau:Yu t Hm lng Mo

(%)Nhit qunung (oC)

Thi gian qunung (pht)

Tc ngui

t bin Z1 Z2 Z3 Z4Gi tr gc 0jZ

(Mc c s)

0,40 840 60 -

jZ 0,15 100 60 -

Cn trn jZ 0,55 940 120 Lm nguinhanh

Cn di jZ 0,25 740 0 Lm nguichm

a. M ha v lp ma trn thc ngh im:0

; 1, 2,3j

j j

j

z zx j

z

y c 4 yu t nh hng. Thng thng phi tin hnh N=24=16 th nghim.

(1.3)0

1

1

0

jj j

jj j

j j j

x z z

x z z

x z z

Nhng giai on u, khi cha tm vng ti u m ch xy dng m hnh. nbin z4ch c tnh cht nh tnh nn ta lm th nghim ring phn.N=2

k-1=24-1=8Gi s m hnh l tuyn tnh:

0 1 1 2 2 3 3 4 4y x x x x

xy dng ma trn thc nghim ri ng phn ta t:x4=x1x2x3 hay 1=x1x2x3x4Ta lm lun ba th nghi m tm: n0=3.TT X0 X1 X2 X3 X4 y1 + - - - - 642 + + - - + 903 + - + - + 694 + + + - - 1305 + - - + + 36

6 + + - + - 957 + - + + - 818 + + + + + 1009 0 0 0 0 0 8010 0 0 0 0 0 8211 0 0 0 0 0 78


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Ch : c ma trn z ta vn dng quy tc:

0

1

1

0

jj j

jj j

j j j

x z z

x z z

x z z

b. Tnh bj8

01

8

ij1

1 2 3 4

1 81,38

1

8

20,0; 11,9; 5,1; 9, 4

i

i

j i

i

b y

b x y

b b b b

Vy y=83,1+20x1+11,9x2-5,1x3-9,4x4c. Kim nh D=2

Tnh phng sai ti sinh theo th nghim lp tm:3

2 2 2 20 0

10

1 1( ) [(80-80) +(82-80) +(78-80) ] 41 2ts tt

s y yn

Bc t do ca phng sai ti sinh: n0-1=3-1=2Dng phng sai ti sinh c lng 2

d. Kim nh j=0;2

2 4; 0,5;8

0,5 0,71

j tsbj bj

j

bj

b st s

sb N

s

Chn mc ngha =0,05. Tra bng Student vi bc t do m=n0-1=2 ta c t=2,92.

0 2

1 3

4

81,3 11,9114,5; 16,7

0,71 0,71

20 5,128, 2; 7, 2

0,71 0,71

13, 2;| | ; 0,1, 2,3;

b b

b b

b bj

t t

t t

t t t j

Nn mi h s cng c ngha.e. Kim tra s ph hp ca m hnh yTnh phng sai d:

282 2

1 1

2

2

1 1( ) ( ) 8

( 1) 8 5

is

N

du i ii i

i i

du

ts

S y y y yN k

sF F her

s

Bc t m=N-(k+1)= 3Bc mu: n=n0-1=2Chn mc ngha =0,05. Tra bng ta c F=19,2


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8 24

F

Ta c: F F Vy m hnh ph hp.Trng hp chung khi dng quy hoch th nghim ring phn:Ty theo iu kin c th ngi ta s dng (chn) s nguyn p thch hp. Thngthng khi k bng 4 hoc 5 ta c th chn p=1, khi k bng 6 hoc bng 7 ta chn p=2;khi k=8 hoc 9 ta chn p=3. Vic chn n y cn bo m s th nghim ln hn stham s cn xc nh trong m hnh thng k.Cch tin hnh lp phng nth nghim nh sau:

- u tin ta chn trong k nhn t (bin vo) k-p nhn t chnh v k-p nhn tchnh nhn cc gi tr nh trong quy hoch th nghim ton phn.

- Sau , gi tr ca p nhn t cn li, ti mi th nghim s ph thuc v o gi trca ccnhn t chnh ti th nghim , theo p h thc gi l p h thc sinh.Cc h thc sinh ny cn t chn sao cho ma trn th nghim ca m hnhthng k vn gi c tnh cht trc giao.

Mun vy thng thng c th chn p h thc sinh dng:1 2 ...j j j jsx x x x

Trong : xjkhng phi l cc nhn t chnh v xj1, xj2,xjsl cc nhn t chnh.

5.3. K hoch thc nghim bc 1 hai mc ti u(84)5.3.1. K hoch bc 1 hai mc ti u ton phn (K hoch 2k) (84)5.3.1. K hoch bc 1 hai mc ti u ring phn (K hoch 2k-p) (90)5.3.3. Cc u im ca k hoch bc mt hai mc ti u(95)

5.4. Quy hoch trc giao cp 25.4.1. Khi nim v QHTG cp 2Khi kim nh m hnh tuyn tnh hoc m hnh cp hai khng y m thy khngph hp th vic sdng quy hoch trc giao cp 1 khng hiu qu. Ta phi xt nQHTG cp 2.Xt m hnh bc hai y :

20 ij

1 1, 1

(0, )

k k k

j j i j jj j

j i j ji j

y x x x x

N

y Ey

Ta c:

20 ij

1 1, 1

k k k

j j i j jj j

j i j ji j

y x x x x

2D


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Vn t ra l phi quy hoch th nghim th no c m hnh thng k y biudin gn ng tt nht y.Ngi ta ngh mt cch b tr th nghim nh sau gi l QHTG cp 2. Gm ba loith nghim:Loi 1:Gm n1=2

k-1hoc 2k-pth nghim ging nh QHTG cp 1.Loi 2:Gm n0th nghim tm (x1=0, x2=0, ,xk=0) ng vi 0 0 01 2( , ,..., )kz z zLoi 3:Gm nk=2k th nghim b tr trn cc trc ta cch gc ta mt on>0 sao cho ma trn X trc giao, tc l ly xj=.Vy tng s th nghim l: N=2k+n0+2kTa s xt cc vn sau y:

- Xy dng ma trn X- Tm cc h s- Kim nh kt qu

5.4.2. Xy dng ma trn X (Ma trn thc nghim) theo di phng php, ta xt k=2. Khi k>2 cch lm c ng tng t.

2 20 1 1 2 2 12 1 2 11 1 22 2y x x x x x x Lc ny: n1=2

2=4; nk=2.2=4. Ma trn X c dng:2 2

0 1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( )

(1 ) . . ( ) . . ( ) . . . ( ) . . . ( ) . . . ( )

(1 ) . . ( ) . . ( ) . . . ( ) . . . ( ) . . . ( )

(1 ) . . ( ) . . ( ) . . . ( ) . . . ( ) . . . ( )

(1 ) . . ( ) . . ( ) . . . ( ) . . . ( ) . . . ( )

(1 ) . . ( 0 ) . . ( 0 ) . . . ( 0 ) . . . . ( 0 ) . . . . ( 0 )

( 1 ) ( ) . ( 0 ) . .

x x x x x x x

X

2

2

2

2

. ( 0 ) . . . . ( ) . . ( 0 )(1 ) ( ) . ( 0 ) . . . ( 0 ) . . . . ( ) . . ( 0 )

(1 ) . . ( 0 ) . ( ) . . . ( 0 ) . . . . ( 0 ) . . ( )

(1 ) . . ( 0 ) . ( ) . . . ( 0 ) . . . . ( 0 ) . . ( )

Nu ta chn v hai ct 2 21 2;x x khng kho th ma trn khng c tnh cht trc giao v:

20 uj

1

2ui

1

0; 1, 2

0( )

N

u

u

N

uj

u

x x j

x x i j

gii quyt kh khn ngi ta lm nh sau:a. Chn :

2 1.2 2 ;k kN k

b. Lm php bin i:'

2' 2 ; 1,

j j

jj j

x x

x x x j k

Trong :


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2 2uj

1

22

1

2 2

N

j

u

k

j

x xN

xN

Vy:' 2 21

(2 2 );k

j jx x kN Vi k=2 ta c:

' 2 2 2

'1

'1

'1

'1

1

1 2(2 2.1)

9 32 1

1, 2,3, 4; 13 3

2 25; 0

3 3

2 16,7; 13 32 2

8,9; 03 3

j j j

u

u

u

u

x x x

u x

u x

u x

u x

Hon ton tng t ta tnh c '2xMa trn X khi :

' '

0 1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( )

1 1(1 ) . . ( ) . . ( ) . . . ( ) . . . ( ) . . . ( )

3 3

1 1(1 ) . . ( ) . . ( ) . . . ( ) . . . ( ) . . . ( )3 3

1 1(1 ) . . ( ) . . ( ) . . . ( ) . . . ( ) . . . ( )

3 3

1 1(1 ) . . ( ) . . ( ) . . . ( ) . . . ( ) . . . ( )

3 3

2 2(1 ) . . ( 0 ) . . ( 0 ) . . . ( 0 ) . ( ) ( )

3 3

(1 ) ( 1 ) .

x x x x x x x

X

1 2

( 0 ) . . . ( 0 ) . . . ( ) . . ( )3 3

1 2(1 ) ( 1 ) . ( 0 ) . . . ( 0 ) . . . ( ) . . ( )

3 3

2 1(1 ) . . ( 0 ) . ( 1 ) . . . ( 0 ) . ( ) . . ( )

3 3

2 1(1 ) . . ( 0 ) . ( 1 ) . . . ( 0 ) . ( ) . . ( )

3 3

Ma trn ny r rng c tnh cht trc giao.5.4.3. Cc cng thc tnh ton


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Ta vn s dng cc cng thc m ha v cc cng thc khc nh c.Lu :

01

uj1

2uj

1

'ui

1 1

2 ' 2ui uj uj

1 1

1

( 1, )

;( ) ( )

N

u

u

N

u

u

j N

u

N N

uj u uj u

u uij jjN N

u u

b yN

x x

b j kx

x x y x y

b b

x x x

5.4.4. Kim nh gi thuyt

- kim nh D=2ta vn dng phng sai ti sinh tnh theo cc th nghi m tm.

- kim nh xem cc h s c bng 0 hay khng ta phi tnh cc phng sai.2 2 2

2 2 20

2 2uj ui uj

1 1

22

' 2uj

1

; ;( )

( )

ts ts tsb bj bijN N

u u

tsbjj N

u

s s ss s s

Nx x x

ss

x

kim nh s ph hp ca m hnh ta vn so snh hai phng sai ti sinh ( bit)

v phng sai d c tnh nh sau:

22

1

1( )

( 1)

N

du i i

i

s y yN k

Trong k l cc s hng cha bin cn li trong m hnh ca y .5.4.5. Cc v dTh d:Cn xc nh cc iu kin t c phn hy cc i hp cht Boratbng hn hp cc axit sunfuric v axit phtphoric. Bc phn tch ca y ph thuc v occ yu t sau:Z1: Nhit phn ng

Z2: Thi gian phn ngZ3: T lca axit phtphoric (%)Z4: Nng ca axit phtphoric ( % P2O5)Gi tr chnh v khong bin i ca cc yu t cho trong bng sau:TT Z1 Z2 Z3 Z4

0j

z 55 37,5 80 32,8

jz 25 25,5 20 18,8


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30Z180 15Z260 60Z3100 14Z451,6

T cc th nghim s b thy rng, cc iu kin ti u tin hnh qu trnh nm trongmin bin i cc thng s ta d ng QHTG cp 2.

4 4

20 12 1 2 13 1 3 14 1 4 23 2 3 24 2 4 34 3 4

1 1j j jj j

j j

y x x x x x x x x x x x x x x

lp ma trn thc nghim ta chuyn t z sang x :0

j j

j

j

z zx

z

S th nghim vi k=4 l 24+1+2.4=252 3

' 2 4 2 2 4 2

25.2 2 2 1, 414

1 1 4(2 2 ) (2 2.2)

25 5j j j jx x x x

N

TT X0 X1 X2 X3 X4 '1x'2x

'3x

'4x y

1 +1 -1 -1 -1 -1 0,2 0,2 0,2 0,2 86,9

2 +1 +1 -1 -1 -1 0,2 0,2 0,2 0,2 403 +1 -1 +1 -1 -1 0,2 0,2 0,2 0,2 664 +1 +1 +1 -1 -1 0,2 0,2 0,2 0,2 54,45 +1 -1 -1 +1 -1 0,2 0,2 0,2 0,2 76,66 +1 +1 -1 +1 -1 0,2 0,2 0,2 0,2 55,77 +1 -1 +1 +1 -1 0,2 0,2 0,2 0,2 918 +1 +1 +1 +1 -1 0,2 0,2 0,2 0,2 47,69 +1 -1 -1 -1 +1 0,2 0,2 0,2 0,2 74,110 +1 +1 -1 -1 +1 0,2 0,2 0,2 0,2 52

11 +1 -1 +1 -1 +1 0,2 0,2 0,2 0,2 74,512 +1 +1 +1 -1 +1 0,2 0,2 0,2 0,2 29,613 +1 -1 -1 +1 +1 0,2 0,2 0,2 0,2 94,814 +1 +1 -1 +1 +1 0,2 0,2 0,2 0,2 49,615 +1 -1 +1 +1 +1 0,2 0,2 0,2 0,2 68,616 +1 +1 +1 +1 +1 0,2 0,2 0,2 0,2 51,817 +1 0 0 0 0 -0,8 -0,8 -0,8 -0,8 61,818 +1 +1,4 0 0 0 1,2 -0,8 -0,8 -0,8 95,419 +1 -1,4 0 0 0 1,2 -0,8 -0,8 -0,8 41,720 +1 0 +1,4 0 0 -0,8 1,2 -0,8 -0,8 7921 +1 0 -1,4 0 0 -0,8 1,2 -0,8 -0,8 42,422 +1 0 0 +1,4 0 -0,8 -0,8 1,2 -0,8 77,623 +1 0 0 -1,4 0 -0,8 -0,8 1,2 -0,8 5824 +1 0 0 0 +1,4 -0,8 -0,8 -0,8 1,2 45,625 +1 0 0 0 -1,4 -0,8 -0,8 -0,8 1,2 52,3

Tnh bj, bjs theo cc cng thc cho:


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0 11 13

1 22 14

2 33 23

3 44 24

4 12 34

61,54; 4,5; 0, 2

17,34; 1,3; 1, 2

6,4; 4,09; 0,56

4,7; 5,34; 0,76

4,37; 2,18; 1,9

b b b

b b b

b b b

b b b

b b b

* Kim nh D=2

Dng 2ts

s - mun tnh phng sai ti sinh ta lm bn th nghim ti tm0 01 2

0 03 4

40

01

0

242 0

01

61,5%; 59,3%

58,7%; 69%

160,9%

4

4

1( ) 5,95

3

i

i

ts i

i

y y

y y

y y

n

s y y

* Kim nh cc h s:2

20 0

2 22

22uj

1

2 22

2ui uj

1

22

2' 2uj

1

5,950, 238 0, 238 0, 4878

25

5,950, 297 0, 297 0,545

202 2( 2)

5,950,371 0,371 0,61

2 16( )

5,95

16(0, 2) 7(0,( )

tsb b

ts tsbj bjN k

u

ts tsbij bijN k

u

tsbjj N

u

ss s

N

s ss s

x

s ss s

x x

ss

x

2 2

0,743 0,743 0,8648) 2(1, 2) bjj

j

bj

bj

s

bt

s

Suy ra:t0= 126,15; t1=31,9; t2=11,7; t3= 8,64; t4=8,04; t12=3,57; t13=0,328; t14=1,97; t11=5,2;t22=1,5; t33=4,73; t44=6,22; t23=0,91; t24=1,25; t34=3,8

Chn =0,05 bc t do n0-1=4-1=3. Tra bng t=2,35. So snh ta thy:22 13 14 23 24; ; ; ; 0

Vy phng trnh hi quy c dng: 2 2 2

1 2 3 4 1 2 3 4 1 3 4

2 2 2

1 2 3 4 1 2 3 4 1 3 4

61,54 17,37 6,4 4,7 4,37 2,18 1,9 4,5( 0,8) 4,09( 0,8) 5,34( 0,8)

58,9 17,37 6,4 4,7 4,37 2,18 1,9 4,5 4,09 5,34

y x x x x x x x x x x x

x x x x xx x x x x x

Kim nh s ph hp ca m hnh:Ta tnh 2 5,95tss bc t do l 3


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252 2

1

2

2

1 396,2( )

( 1) 25 10

26,44, 4

5,95

du i i

i

du

ts

s y yN k

sF

s

Bc t (Bc t do ca phng sai d, V1, f1, N-l): 15

Bc mu (Bc t do ca phng sai lp, V2, f2, n0-1): 3Chn =0,05. Tra bng Fisher ta c F=8,6.F F m hnh ph hp.a v bin Z vi cng thc:

0

1 21 2

3 43 4

55 17,5; ; ;

25 22,5

80 32,8;

20 18,8

j j

j

j

z z z zx x x

z

z zx x

Thay vo ta c:2 21 3 4 1 2 3 4 1 390,64 0, 242 0,07 0,35 0,00388 0,00506 0,0072 0, 0120 0,015y Z Z Z Z Z Z Z Z Z Z

5.5. K hoch thc nghim bc 2 (114)5.5.1. M t vng phi tuyn (vng hu nh n nh)(114)5.5.2. Cc k hoch bc hai trc giao (115)5.5.3. Cc k hoch bc hai tm xoay (119)5.5.4. Cc v d (131)


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CHNG 6. CC K HOCH THC NGHIM C BIT (177)6.1. Phng php n hnh k hoch ha thc nghim v ti u ha(177)6.2. K hoch ha tin ha cc thc nghim (181)6.3. K hoch thc nghim khi nghin cu biu thnh phn-tnh cht(184)6.3.1. Phng php mng n hnh6.3.2. K hoch mnh n hnh Scheffe

6.3.3. K hoch trung tm n hnh6.3.4. K hoch thc nghim khi nghi n cu mt phn biu 6.3.5. K hoch ti u D6.3.6. K hoch vi s cc tiu ha sai s c tnh h thng6.3.7. K hoch ha thc nghim khi nghi n cu quan h ph thuc ca tnh cht v ot l cc cu t


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CHNG 7. TI U HA THC NGHIM

7.1. Cc bc thc hin ti u ha thc nghimBc 1- Xc nh mt im xut pht nm trong min gii hn tng th ca cc bin uvo. Chn im lm mc cbn, chn khong bin thin ca tng bin xc

nh min gii hn ca quy hoch thc nghim trc giao cp mt.Bc 2- Lm cc th nghim theo quy hoch trc giao cp mt- Xy dng phng trnh hi quy bc nht .Nu phng trnh hi quy bc nht khng tng thch th chuyn ti thc hin bc 4.Nu phng trnh hi quy bc nht tng thch th thc hin bc 3.Bc 3- Xc nh vect gradient ca hm mc tiu ti mc cbn v xut pht t mc c

bn xc nh ta cc im thc nghim nm cch u nhau trn hng ca vectgradient vi khong cch t chn ph hp vi i tng nghin cu. Lm thcnghim xc nh mt im c gi tr hm mc tiu tt nht trn hng gradient.Chn im tm c lm im xut pht mi v quay v bc 2 .Bc 4- Lm cc th nghim theo quy hoch cp hai (trc giao hoc quay).Bc 5- Xy dng phng trnh hi quy bc hai.- Nu phng trnh hi quy bc hai khng tng thch th chuyn ti thc hin bc6.

- Nu phng trnh hi quy bc hai tng thch th thc hin bc 7.Bc 6- Thu hp khong bin thin ca cc bin u vo ri quay v bc 5.Bc 7- Tm cc tr ca hm mc tiu thu c dng phng trnh hi quy bc hai thuc bc 5 v lm li thc nghim kim chng v nh gi kt qu.

7.2. Quy hoch thc nghim tm cc tr7.2.1. t bi ton

Ta dng cc phng php bnh phng cc tiu, QHTG cp 1 v QHTGcp 2 tm mi quan h gia y v cc xi (i=1,k) di dng a thc bc nht hoc bchai. Lu rng cc a thc biu din gn ng i t ng tht.By gi hy tm * * *1 2, ,..., kx x x sao cho

* * *1 2 1 2 1 2

* * *1 2 1 2 1 2

( , ,..., ) ( , ,..., ) ( , ,..., )

: ( , ,..., ) ( , ,..., ) ( , ,..., )k k k

k k k

y x x x y x x x x x x D

Hay y x x x y x x x x x x D

Lc ny c th chn mt trong hai cch sau:


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- Nu chng tha mn vi chnh xc ca m hnh thng k thu c th gii bi ton trn chng ta ch vic dng cc phng php ca quy hoch ton hctng ng tm cc tr. Nu m hnh l tuyn tnh th dng cc phng php caquy hoch tuyn tnh, nu m hnh l phi tuyn tnh th dng cc phng php caquy hoch phi tuyn tnh.

- Nu khng tha mn vi chnh xc ca m hnh thng k thu c

(chng hn cn nghi ng min xc nh cn qu rng nn chnh xc qu th) thtrc khi dng cc phng php ca quy hoch ton hc ta phi tm cch thu hpvng cha im cc tr, hay ni mt cch vn tt ta phi d ng thc nghim tmvng cc tr.Ni dung ch yu ca chng ny l xt cch gii quyt th hai. Theo cch n y, ccphng php quy ho ch thc nghim tm cc tr chia ra lm hai giai on:

G1:Tm vng cha im cc tr bt u t m hnh cp 1.G2:Sau khi tm c vng cha im cc tr ta dng QHTG cp 2 xy

dng m hnh bc hai. Cui cng ta dng cc phng php quy hoch phi tuyn

tm cc tr.Trong giai on 1 li c th dng mt trong hai hng :+ Hng dng o hm (PP leo dc Box-Winson)+ Hng khng dng o hm (tm kim) (PP n hnh u)

7.2.2. Phng php leo dc Box-Wilson7.2.2.1. Ni dung ca phng php:

Ta bit rng, vect gradien ca hm y=f(x) ti im x0(k hiu grad f(x0)) chra lng tng nhanh nht ca h m y ti im x.

0 0 00 1 2

1 2

( ) ( ) ( )( ) ...

k

k

f x f x f xgradf x i i i

x x x

Trong : 0( )

j

f x

x

o hm ring theo bin xj; i1, i2,,ikl cc vect n v theo

cc trc ta ti x0.Ni dung ca phng php Box-Winson nh sau:Xt mt min con D0D. Tm ca D0l Z0v Z0 ng vi x0.QHTG cp 1 tm phng trnh hi quy bc nht:

0

1

( )k

j j

j

y b b x f x

Kim nh s ph hp ca y nu y ph hp vi m hnh ngha l mt cong yxp x c mt phng th D0khng phi l im cc tr. Bc chuyn sang vng conD1c thc hin theo hng grad f(x 0). Theo hng ny ta c i cho ti khi hm ykhng tng c na. Ta c immi- vi im mi ta lp li qu trnh nh trn.C lm nh vy cho ti khi phng trnh bc nht khng ph hp na Ta tm cvng cha im cc tr. Ta chuyn sang giai on hai.7.2.2.2. Thut ton leo dc:a. Tnh cc thnh phn ca gradien:


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Gi s ta ang im x 0ta mun chuyn ti im x theo hng gradien ti x 0.Theo khai trin Taylo, ta c:

2 220 0 0

0 21 1 1 1

( ) ( ) ( )1 1( ) ( ) 2 ... ...

2! 2!

N k k k

j i j j

j i j jj i j j

f x f x f xf x f x x x x x

x x x x

Nu ta ly gn ng n s hng bc nht th t x0n x hm f tng cmt lng l :

0 0 01 2

1 2

( ) ( ) ( )...

k

k

f x f x f xx x x

x x x

Nu min D0m hnh l tuyn tnh th:0( ) ( 1, )

j

j

f xb j k

x

V vy, hm f tng c mt i lng l:1 1 2 2 ... k kb x b x b x

b. Chn nhn t c s v tm bc i: tm bc i hiu qu nht ta s ti s hng no trong hm trn tng ln

nhiu nht. Bin tng ng l nhn t c s.Ta gi cc khong bin i tng ng theo bin tht l ; 1,jz j k . Tm ch s j

*t:

* * max j jj jb z b z

Vy j*l ch s nhn t c s.- Ta chn di bc chonhn t c s l hj*-Cc di bc hj*ca cc nhn t khc c tnh theo hj*:

** * * * *

j j j j j

j j

j j j j j

h b z b zh h

h b z b z

c. Tin hnh th nghim- Gi Z0l tm min D0trong ta tht, k hiu l im M0. Vi di bc trn. Ta n M1c cc ta :

1 0 2 ; 1,j j jz z h j k

Lm th nghim o kt qu ta c y1. C lm nh vy ta c mt dy cc y0,y1,,yn. Va lm va o va so snh:

0 1y y tip tc lm y2

1 2y y tip tc lm y3.

1p py y mt cong bt u gim. Ta dng li M p+1vi trung tm min nghin cumi Dp. Kim nh s ph hp ca phng trnh bc nht. Nu tha mn th ta li tm di bc v lm th nghim. C nh vy cho n khi m hnh bc nht khng phhp na Ta n vng cc tr.7.2.2.3. Th d: Tip tc th d nghin cu qu trnh bin tnh nhm bng Mo.Bng QHTG cp 1 ta c m hnh:y=83,1+20x1+11,9x2-5,1x3-9,4x4V kim nh m hnh thy ph hp.


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1 1

2 2

3 3

20.0,15 3

11,9.100 1190

5,1.60 306

b z

b z

b z

Ta c:* * 2 2j j

b z b z ngha l j*=2. Vy nhn t c s l Z2 . Ta chn bc hj*=h2=100C (tng

nhit qu nung).1 1

1 22 2

3 33 2

2 2

310 0,03

1190

30610 3

1190

b zh h x

b z

b zh h x

b z

Yu t Z4c tnh cht nh tnh v ch ly gi tr +1 v -1. V s x4m nntrong tt c cc bc ta ly Z4m.

Cc gi tr chuyn ng leo dc cho trong bng. Th nghim 12 cho kt qu ttnht (366 ht/1cm2). ln ca tham s ti u tha mn yu cu ca ngi nghin

cu nn phng php quy hoch dng li.TT Danh mc %Mo(Z1) Nhit nung (Z2)

Thi giannung (Z3)

Tc ngui (Z4)

(y) Sht/1cm2

bj 20 11,9 -5,1 -9,4bjZj 3 1190 -360 -9,4Bc hj 0,03 10 -3 -9,4TN tng ng 0,43 850 57 ChmNt 0,46 860 54 Chm

10 TN thc 0,49 870 51 Chm 108TN tng ng 0,52 880 48 ChmNt 0,55 890 45 Chm

11 TN thc 0,58 900 42 Chm 19612 Nt 0,61 910 39 Chm 36613 Nt 0,64 920 36 Chm 313

TN tng ng 0,67 930 33 Chm14 TN thc 0,70 940 30 Chm 142Thc ra y ta mi ti im:Z1=0,61; Z2=910; Z3=39; Z4= -1 l tm min D1nhng ngi nhn li gii thamn nn ta dng li.

7.2.3. Phng php n hnh u tm cc tr7.2.3.1. ng li chung:

Mt n hnh u trong khng gian k chiu c tm gc ta l mt a dinc ng k+1 nh cch u gc ta v c di cc cnh bng nhau.Trong khng gian mt chiu, n hnh u l mt on thng. Trong khng gian haichiu, n hnh u l mt tam gic u. Trong khng gian ba chiu, n hnh u lmt hnh chung tam gic u.


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-u tin ta tin hnh k+1 th nghim xut pht sao cho cc im th nghim lcc nh ca n hnh u ni trn. Ta c k+1 kt qu ra: y 1, y2,,yk+1. S c mtim th nghim ng vi kt qu ra km nht.

- Ta thay im bng im phn chiu ca n qua tm ca mt i din.im nh cng vi cc im cn li ca n hnh c li to thnh mt n hnh umi.

-i vi n hnh u mi, ta ch lm thm mt th nghim im nh. Ta liso snh cc kt qu ra v s tm c im th nghim ng vi kt qu ra km nht. y c ba vn cn gii quyt:

-Xy dng n hnh xut pht th no?- Chuyn t ta gi xjsang ta tht nh th no?- Tm cc ta ca im nh th no?

7.2.3.2. Xy dng n hnh xut pht:Ngi ta chng minh c rng, cc ta ca k+1 nh ca n hnh u

trong khng gian Rkl cc ta ca k+1 vc t hng ca ma trn sau:

1 2 3 1

1 2 3 1

2 3 1

3 1

( )( )( ).......( )( )......( )( )( )( ).......( )( )......( )

(0)( 2 )( ).......( )( )......( )

(0)(0)( 3 ).......( )( )......( )

........................................

j k k

j k k

j k k

j k k

x x x x x xx x x x x x

x x x x x

x x x xX

1

1

..............

(0)(0)(0).......( )( )......( )

(0)(0)(0).......(0)... ( 1) ......( )

(0)(0)(0).......(0)(0).................( )

j k k

k k

k

jx x x

k x x

kx

Trong x1ty chn, cc xjkhc (j=2,k) c tnh theo cng thc sau:1 2

( 1)j

xx

j j

Tm ca n hnh l (0,0,0,..,0) di cnh l a=2x. Khi k=6, x1=0,5 tc l a=1 tac:

0,5 K=2 K=3 K=4 K=5 K=6-0,5 0,289 0,204 0,518 0,129 0,109

0 0,289 0,204 0,518 0,129 0,1090 -0,578 0,204 0,518 0,129 0,1090 0 -0,612 0,518 0,129 0,1090 0 0 -0,632 0,129 0,1090 0 0 0 -0,645 0,1090 0 0 0 0 -0,654

Chuyn ta xjsang ta Zjvn theo cng thc:


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0

; 1,j jj

j

Z Zx j k

Z

7.2.3.3. Tm ta ca im nh:Xt n hnh u trong khng gian Rkvi k+1 nh. Gi s nh Z 1 ng vi kt

qu ra b nht. Mt i din nh Z1to thnh bi cc nh cn li Zi(i1).Trc ht ta tnh cc ta im trng tm ca t din:

1

1

1; 1,

kc i

j j

ii j

Z Z j kk

nh mi Z1i xng vi nh b i Z1qua trng tm Zc. Vy:1

1 1' 1' 1 1' 1

1

1 2( ) 2 ( ); 1,

2

kc i

c j j j

ii j

Z Z Z Z Z Z Z Z Z j kk

7.2.3.4. Th d: Ngi ta nghin cu phn ng xy ra theo s A B C D trong dung dch ru. Cht lng v s lng sn phm D (k hiu y) ph thuc v o

cc yu t:Z1: Thi gian phn ng (gi)Z2: Nng ru trong dung dchZ3: Nng cht CZ4: Nng cht DZ5: Nng [B/A]Bng yu t cho nh sau:TT Z1 Z2 Z3 Z4 Z5

0

jZ 2,0 0,65 0,1 0,25 1,2

jZ 0,2 0,15 0,025 0,05 0,2Dng phng php n hnh u vi k=5, ta c ma trn X:

0,500 0, 289 0, 2040 0,1580 0,129

0,5 0, 289 0, 204 0,158 0,129

0 0,578 0, 204 0,158 0,129

0 0 0,612 0,158 0,129

0 0 0 0,632 0,129

0,645

X

Bin i sang cc bin Z bng cc cn g thc:31 2

1 2 3

544 5

0,102,0 0,65; ; ;

0, 2 0,15 0,025

1,200,25;

0,05 0, 20

ZZ Zx x x

ZZx x

Khi , ma trn ca n hnh ban u vi kch thc thtc dng (ta c su nh can hnh u):


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TT N Z1 Z2 Z3 Z4 Z5 yZ1 1 2,10 0,693 0,105 0,258 1,225 0,760Z2 2 1,90 0,693 0,105 0,258 1,225 0,491Z3 3 2,00 0,564 0,105 0,258 1,225 0,513Z4 4 2,00 0,650 0,085 0,258 1,225 0,675Z5 5 2,00 0,650 0,100 0,218 1,225 0,693Z6 6 2,00 0,650 0,100 0,250 1,075 0,666

Y l gi tr o c.Theo bng y2=min(y1, y2,,yt)=0,491Ta b nh Z2thay n theo nh x phn chiu, ta c nh Z7i xng vi Z2quamt to bi cc im 1,3,4,5,6.Tnh ta trng tm C ca mt to bi cc im 1,3,4,5,6

( )1

( )2

( )3

( )4

( )5

4.2, 0 2,12,02

5

3.0, 65 0, 564 0, 693 0,6415

2.0,105 0,085 2.0,10,099

53.0, 285 0, 218 0, 250

0,2985

4.1,225 0,0751,195

5

C

C

C

C

C

Z

Z

Z

Z

Z

Ta :(7) ( ) (2) (7) ( ) (2)

(7 )1

(7 )2

(7 )3

(7 )4

(7 )5

2 2 ; 1,5

2.202 19 2,14

2.0, 641 0, 693 0, 589

2.0, 99 0,105 0, 093

2.0, 298 0, 258 0, 238

2.1,195 1,225 1,165

C C

j j j

Z Z Z Z Z Z j

Z

Z

Z

Z

Z

im mi th 7 cng vi cc im cn li to nn mt n hnh 1,3,4,5,6,7. Ta s cbng sau:

N Z1 Z2 Z3 Z4 Z5 y1 2,10 0,693 0,105 0,258 1,225 0,7603 2,00 0,569 0,105 0,258 1,225 0,5134 2,00 0,650 0,085 0,258 1,225 0,6755 2,00 0,650 0,100 0,218 1,225 0,6436 2,00 0,650 0,100 0,250 1,075 0,6667 2,14 0,589 0,093 0,238 1,165 0,810


58/62

Sau khi tin hnh th nghim 7, im km nht ca n hnh 1,3,4,5,6,7 l imth ba. Hnh chiu ca n qua mt 1,4,5,6,7. Loi nh Z 3, thm vo nh Z8i xngvi Z3qua mt 1,4,5,6,7.Ta tm C ca mt 1,4,5,6,7 l :

( )

1

( )2

( )3

( )4

( )5

2,1 3.2 2,142,048

50, 693 3.0, 65 0, 589

0,6465

0,105 0, 085 2.0,1 0, 0930,096

52.0, 285 0, 218 0, 250 0, 338

0,2645

3.1,225 1,075 1,1651,183

5

C

C

C

C

C

Z

Z

Z

Z

Z

T y ta tnh c Z

8

(8) (8) (8)1 2 3

(8) (8)4 5

2,04; 0,633; 0,098;

0, 247; 0,190;

Z Z Z

Z Z

N Z1 Z2 Z3 Z4 Z5 y1 2,10 0,693 0,105 0,258 1,225 0,7604 2,00 0,650 0,685 0,258 1,225 0,6755 2,00 0,650 0,100 0,218 1,225 0,6936 2,00 0,650 0,100 0,250 1,075 0,6667 2,14 0,589 0,093 0,238 1,165 0,810

8 2,04 0,633 0,098 0,247 1,190 0,792o c y8=0,792 nn loi Z6 ng vi:

Y6=min(1,4,5,6,7,8). Thm Z9. C tip tc nh vy n khi t c im ti u.

7.2.4. Quy hoch th c nghim gii bi ton nhiu mc tiu7.2.4.1. t bi ton

Trong thc t thng xut hin bi ton: Trong mt h thngcn nghin cumi qua h ca m bin ra y 1, y2,,ymi vi k bin vo x1, x2,,xkv sau khi nhnc m hnh biu din cc mi quan h , cn tm mt phng n trong min rngbuc cho trc sao cho t c cc tr ca m mc tiu.Bi ton cc tr nu trn c th pht biu nh sau:Xt m hm, k bin

1 1 1 2

2 2 1 2

1 2

( , ,..., )

( , ,..., )

................................

( , ,..., )

k

k

m m k

y f x x x

y f x x x

y f x x x

Hy tm im * * *1 2( , ,..., )kx x x sao cho:


59/62

* * *1 1 2 1 1 2

* * *2 1 2 2 1 2

* * *1 2 1 2

( , ,..., ) ( , ,..., )

( , ,..., ) ( , ,..., )

.....................................................

( , ,..., ) ( , ,..., )

k k

k k

m k m k

f x x x f x x x

f x x x f x x x

f x x x f x x x

1 2,...,( , )kx x x D

l bi ton quy hoch a mc tiu. Trong thc t, rt kh c mt im* * *1 2( , ,..., )kx x x lm cc i m mc tiu cng mt lc nh vy. Do im

* * *1 2( , ,..., )kx x x

thng gi l im l tng v thay vo vic tm im l tng ngi ta tm imtiu theo mt ngha no .

Nh vy, QHTN gii bi ton nhiu mc tiu chia ra lm hai giai on:G1:Xy dng m hnh gm m phng trnh hi quy. lm iu ny ta vn tinhnh N th nghim nhng ti mi im th nghim ta khng ch o gi tr ca mtbin ra m ca m bin ra y1, y2,,ym. Khi ta c bng:N X1 X2 Xk Z1 Z2 Zk y1 y2 .ym

12...N

y11 y12y1my21 y22y2m.

.

.

YN1 yN2yNm

i vi mi bin y1, y2,,ymcch xy dng m hnh v kim nh vn nh c.G2: Tm im ti u chung cho m m hnh. n y ta c hai cch:

- Nu ta tha mn vi chnh xc ca cc m hnh th ta ch vic s dng ccphng php quy ho ch a mc tiu tm cc tr.

- Nu ta khng tha mn vi chnh xc ca cc m hnh th ta phi lm tipbng quy hoch thc nghim. Ta s xt ph ng php Harring ton.7.2.4.2. Phng php Harrington:a. T hp cc mc tiu:

Harrington dng phng php t hp cc mc ti u thnh mt mc tiu chungbng vic a ra mt hm mong mun dng:

1 2. ..... mQ q q q

Trong : q1, q2,,qm l cc hm ca y1, y2,,ymv y1, y2,,ymli l hm ca x1,

x2,,xk.Q=F(x1, x2,,xk)

Nhng chn qj, i=1,m nh th no? Harrington ngh chn qjsao cho khi yjtt nht trong mt khong mong mun th qj=1. Cn yj xu nht ngha l ngoikhong mong mun th qj=0. Cc gi tr trung gian ca qjcho theo thang gi tr sau:

yj qjRt tt 0,8-1


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Tt 0,63-0,80t 0,37-0,63Xu 0,20-0,37Rt xu 0-0,2

S d chn 0,37 v 0,63 v 1/e=0,37 v 1-1/e=0,63 tm biu thc lin h gia qjv yj(j=1,m) Harrington lm nh sau:Gi s: j jjy y y j=1,m

'

'

| |

| |

0'

1nej

nej

y

j y

j j

j

j

q ee

y yy

y

Cn nel hng s c xc nh nh sau:i vi mi y1, ta ly cc gi tr xc nh ;xd xdj jy q . Thay vo phng trnh trn ta s

gii c ne.1

'1

ln ln(1/ )

ln | |eq

ny

Th d:Gi s ta c: 100yj220Ta chn 200; 0,85xd xdj jy q

Khi :

'

2( )3

2( )3

200 160 40 2

60 60 3

10,85

1 2ln[ln( - )]

0,85 3

n

n

jy

e

e

n

b. T chc th nghim:u tin, ta b tr th nghim nh bng 1. Sau ta tnh cc gi tr Q i(i=1,N)

theo cng thc 1 2. ..... mQ q q q . Trong qj c tnh theo cng

thc'

'

| |

| |

1nej

nej

y

j yq e

e

v

0' j jj

j

y yy

y

. Nh vy ta c bng sau:

N X1 X2 Xk Z1 Z2 Zk y1 y2 .ym Q

12...N

y11 y12y1my21 y22y2m.

.

.

YN1 yN2yNm

Q1Q2.

.

.

QN


61/62

Ta tm cc phng trnh hi quy y1, y2,,ym. T y ta c th xy dng cphng trnh hi quycho 1 2( , ,..., )kQ x x x

tm cc tr ca Q ta dng phng php Box -Winson hoc phng php nhnh u tm ring cc tr ri li dng quy hoch phi tuyn tnh.

Gi s ta tm c im ti u l * * *1 2( , ,..., )kx x x cho 1 2( , ,..., )kQ x x x . Thay* * *

1 2( , ,..., )kx x x vo cc phng trnh hi quy ca y1, y2,,ym ta tm c cc gi trtng ng * * * * * * * * *1 1 2 2 1 2 1 2( , ,..., ); ( , ,..., );....; ( , ,..., )k k m k y x x x y x x x y x x x .7.2. Cc phng php ti u ha(248)7.3. Ti u ha nh hm nguyn vng(251)7.4. Phng php nghin cu b mt p tr7.4.1. Ti u ha bng phng php leo dc theo mt p tr (B mt biu din)(96)7.4.2 Ti u ha bng phng php nghin cu b mt biu din (mt p tr)(124)


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TI LIU THAM KHO

[1]. Nguyn Minh Tuyn. Quy hoch thc nghim. NXB KHKT, 2005.[2]. Bi Cng Cng, Bi Minh Tr. Gio trnh xc su t v thng k ng dng.NXBGTVT, 1997.[3]. Nguyn Don . Gio trnh quy hoch thc nghim. NXBKHKT, 2004.

[4]. Giang Th Kim Lin. Bi ging quy hoch thc nghim (Cc phng php thngk x l s liu thc nghim), H Nng 2009.[5]. L c Ngc. X l s liu v k hoch ha thc nghim. Khoa Ha, HQGHN,2001.[6]. Douglas C. Montgomery. Design and Analysis of Experiment, 5 th, John Wileyand Son, inc, 2001.[7]. Zivorad R.Lazic. Design of Experiments in Chemical Engineering. Wiley -VCH,2004.[8]. Design Expert Software 8.0 (Ph n mm quy hoch thc nghim) .

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