bài tập nhóm quy hoạch thực nghiệm

8/3/2019 bi tp nhm quy hoch thc nghim

1/44

Gio vin hng dn: Nguyn Dn

Sinh vin thc hin:1: Nguyn Hn 20A2: Bi Thanh Li 20A3: Phan Vn Trung 20A4: L Hong Anh 20B5: Nguyn Tn Lc 20B

6: inh Vn Soan 20B

Nng 12/2011

I HC NNGI HC BCH KHOA

KHOA HA


2/44

Th nghim u 1 2 3 4 5Nhit tu,0C 25 28 30 32 35Kt qu yu, % 22 24 27 30 33mu 3 4 3 5 5

Bi gii:

p dng cng thc:

Khi :

p dng phng php bnh phng nh nht cho hai bin b0 v b1 ta c h phng trnh

5b0 muu= 1

+5

b1 mu tuu= 1

=5

mu yuu= 1

5b0 mu tuu= 1

+

5

b1 mu t2

u

u= 1

=5

mu yu tuu= 1

Bi tp 1: Lp phng trnh tuyn tnh y = b0+b1t m t nh hng ca nhit nkt qu ca qu trnh

theo cc s liu th nghim sau y:

N

S = m(u Y u )2 min

u= 1

N

S = m(b0+b1t Y u )2 min

u= 1

Trong : u - kt qu th nghim tnh theo phng trnhcho th nghim th uYu - kt qu th nghim th u


3/44

Thay vo ta c:5

mu =

u= 1

205 mu tu =

u= 1

558

5 mu tu =u= 1

612

5

mu t2

u =u= 1

18956

5 mu yu =u= 1

5585 mu yu tu =u= 1

17343

20b0 + 612b1 = 558

612b0 + 18956b1 = 17343

b0 = -7.9694

b1 = 1.1722

V kt qu cui cng c c phng trnh sau:y = -7.9694+1.1722t

Gii h trn tac kt qu sau:

Ta c:


4/44

au:


5/44

Th nghim u 1 2 3 4 5

Nhit tu,0

C 25 28 30 32 35Kt qu yu, % 22 24 27 30 33

cho mu 1 1 1 1 1

Bi gii:

p dngcng thc:

Khi :

p dng phng php bnh phng nh nht cho hai bin b0 v b1 ta c h phng trnh

5b0

u= 1

+5

b1 tuu= 1

+

5

b2 t2

u

u= 1

=5 mu yuu= 1

5b0 tu

u= 1+

5

b1 t2

u

u= 1

+

5

b2 t3

u

u= 1

=5

mu yu tuu= 1

5

b1 t2

u

u= 1

+

5

b1 t3

u

u= 1

+

5

b2t4

u

u= 1

=

5

mu yu t2

u

u= 1

S = (u Y u )2 min

u= 1

N

S = (b0+b1t +b2t2 Y u )

2 minu= 1

Bi tp 2: Lp phng trnh tuyn tnh y = b0+b1t+b2t2 m t nh hng ca nhit

n kt qu ca qu trnh

Trong : u - kt qu th nghim tnh theo phng trnhYu - kt qu th nghim th u


6/44

Thay vo ta c:5

=u= 1

5

5

t3u =u= 1

140220

5 tu =u= 1

1505 t2u =u= 1

4558

5 yu =u= 1

1365

yu tu =u= 1

4147

5

yu t2

u =u= 1

128011

5

t4u =u= 1

4364482

Ta c: 5b0 + 150b1 + 4558b2 = 136

150b0 + 4558b1 + 140220b2 = 4147

4558b0 + 140220b1 + 4354482b2 = 128011

b0 = 12.086

b1 = -0.1646

b2 = 0.022

Kt qu ta thu c phng trnh dng sau:

y = 12.086-0.1646t+0.021996t2

Gii h tac kt qu

sau:


7/44

sau:


8/44

Th nghim u 1 2 3 4 5Nhit tu,

0C 25 28 30 32 35

Kt qu yu, % 22 24 27 30 33

mu 3 4 3 5 5

Bi gii:

p dng cng th

Khi :

p dng phng php bnh phng nh nht cho hai bin b0

v b1

ta c h phng trnh sau:

5b0 muu= 1

+5

b1 mu tuu= 1

+

5

b2 mu t2

u

u= 1

=5 mu yuu= 1

5b0 mu tuu= 1

+

5

b1 mu t2

u

u= 1

+

5

b2 mu t3

u

u= 1

=5 mu yu tuu= 1

5

b1 mu t2

u

u= 1

+

5

b1 mu t3

u

u= 1

+

5

b2 mu t4

u

u= 1

=

5

mu yu t2u

u= 1

N

S = m(u Y u )2 min

u= 1

Bi tp 3: Lp phng trnh tuyn tnh y = b0+b1t+b2t2 m t nh hng ca nhit n

kt qu ca qu trnhtheo cc s liu th nghim sau y:

Yu - kt qu th nghim th uTrong : u - kt qu th nghim tnh theo phng trnh

N

S = m(b0+b1t +b2t2 Y u )

2 minu= 1


9/44

Thay vo ta c:5 mu =u= 1

20

5

mu t3

u =u= 1

593898

5 mu tu =u= 1

612

5

mu t2

u =u= 1

18956

5 mu yu =u= 1

5585

mu yu tu =u= 1

17343

5

mu yu t2

u =u= 1

545139

5

mu t4

u =u= 1

18806504

Ta c:20b0 + 612b1 + 18956b2 = 558

612b0 + 18956b1 + 593898b2 = 17343

18956b0 + 593898b1 + 18806504b2 = 17343

b0 = 7.02366

b1 = 0.16839

b2 = 0.01659Kt qu ta thu c phng trnh

dng sau:

y = 7.02366+0.16839t+0.01659t2

Gii h trn tac kt qu

sau:


10/44

Bi tp 4

BI LM

I. T VN

S th nghim N= 210-n'. Nh vy, gim s th nghim th n' phi l ln nht.

* Khng chn cc h thc sinh l tch ca 2 bin m:Nu chn xi=xj.xkTo tng phn xc nh: 1= xi.xj.xkNh vy: (hiu ng tuyn tnh hn hp vi cc tng tc cp)* Khng chn cc h thc sinh c s tha s lin k nhau :Chn

Ti tng phn xc nh :

To tng phn xc nh tng hp :

Nh vy+ Nu n' = 6 ta c K1:

Nhng theo k th 2

u khng tha mn, vy vi n'= 6 ta khng th chn c h thc tha mn yu c+ Nu n' =5 ta c:

K 1:K 2: b qua

tha mntha mntha mn

Lp quy hoch thc nghim nghin cu nh hng ca 10 yu t n qu trnh sao cho cac h s c trngcho cc hiu ng tuyn tnh khng hn hp vi cc hiu ng tng tc cp. Bit rng cc tng tc b bn, bba v tng tc cp ca yu t th nhtv th hai khng ng k.

+ Nu n' = 7 th phi lp TY 23,nh vy s yu t b sung l 7trong khi cc hiu ngtng tc cp, tng tc b ba c th b qua( xp x bng 0) l: C32+C33=4


11/44

II. XT KH NNG GIIVi n'= 5 ta c TPB1. Trng hp 1:

*Lp TY 25

*Chn hi thc sinh:

*To tng phn xc nh :

*To tng phn xc nh tng hp:

5102

3216 xxxx 4217

xxxx

5218 xxxx

5319 xxxx

54110 xxxx

)1(1 6321 xxxx)2(1 7421 xxxx

)3(1 8521 xxxx)4(1 9531 xxxx

)5(1 10541 xxxx

7621)2)(1(1 xxxx

8631)3)(1(1 xxxx

9641)4)(1(1 xxxx

10651)5)(1(1 xxxx

8732)3)(2(1 xxxx

9742)4)(2(1 xxxx

10752)5)(2(1 xxxx

9843)4)(3(1 xxxx

10853)5)(3(1 xxxx

10954)5)(4(1 xxxx

87654)3)(2)(1(1 xxxxx

97653)4)(2)(1(1 xxxxx

107643)5)(2)(1(1 xxxxx

98652)4)(3)(1(1 xxxxx

108642)5)(3)(1(1 xxxxx

109632)5)(4)(1(1 xxxxx

98751)4)(3)(2(1 xxxxx

108741)5)(3)(2(1 xxxxx

109731)5)(4)(2(1 xxxxx

109821)5)(4)(3(1 xxxxx


12/44

*Xc nh iu kin hn hp cc hiu ng :

98764321)4)(3)(2)(1(1 xxxxxxxx

108765321)5)(3)(2)(1(1 xxxxxxxx

109765421)5)(4)(2)(1(1 xxxxxxxx109865431)5)(4)(3)(1(1 xxxxxxxx

109875432)5)(4)(3)(2(1 xxxxxxxx

109876)5)(4)(3)(2)(1(1 xxxxx

1457812345610125691356813467451035925824723611 b246910346810689356710456792345678134910124810123891257101234579

25678910135789101456891016791012367810124678923458910237910781034789

2457834561056923568234671245101235915814713622 b

1469101234681012689123456710124567913456782349104810389571034579

15678910235789102456891026791036781046789134589101379101278101234789

..........................................................................33 b

457810234562569103568403467101451359101258101247101236101010 b


13/44

2.Trng hp 2:

*Lp TY 25

*Chn h thc sinh

*To tng phn xc nh:

*To tng phn xc nh tng hp:

54326 xxxxx 54317 xxxxx

54218 xxxxx 53219 xxxxx

432110 xxxxx

)1(165432

xxxxx )2(1 75431 xxxxx

)3(185421

xxxxx )4(1 95321 xxxxx

)5(1 104321 xxxxx

12469134681689101356714567910123456781034924823891025723457910

12567893578945689679236782467891012345891237917813478910

7621)2)(1(1 xxxx

8631)3)(1(1 xxxx

9641)4)(1(1 xxxx

87654)3)(2)(1(1 xxxxx

97653)4)(2)(1(1 xxxxx

107643)5)(2)(1(1 xxxxx


14/44

*Xc nh iu kin hn hp ca hiu ng:

10651)5)(1(1 xxxx

8732

)3)(2(1 xxxx

9742)4)(2(1 xxxx

10752)5)(2(1 xxxx

9843)4)(3(1 xxxx

10853)5)(3(1 xxxx

10954)5)(4(1 xxxx

98652)4)(3)(1(1 xxxxx

108642

)5)(3)(1(1 xxxxx

109632)5)(4)(1(1 xxxxx

98751)4)(3)(2(1 xxxxx

108741)5)(3)(2(1 xxxxx

109731)5)(4)(2(1 xxxxx

109821)5)(4)(3(1 xxxxx

98764321)4)(3)(2)(1(1 xxxxxxxx

108765321)5)(3)(2)(1(1 xxxxxxxx

109765421)5)(4)(2)(1(1 xxxxxxxx

109865431)5)(4)(3)(1(1 xxxxxxxx

109875432)5)(4)(3)(2(1 xxxxxxxx

109876)5)(4)(3)(2)(1(1 xxxxx

1237856104693682672341023592458345712345611 b

1236910124681012568913467101356791456781459101358101348912571012479

1678910123457891034568910245679102356781023467892891037910478105789


15/44

3. Trng hp 3

*Lp: TY 25

*Chn h thc sinh :

*Trong trng hp ny th cc h s c trng cho hiu ng tuyn tnh khng hn hp vi cchiu ng tng tc cp nhng cc hiu ng c bc lin k lai hn hp vi nhau.

37812561012469123681672341013591458123457345622 b

369104681056892346710235679245678245910235810234895710479

2678910345789101234568910145679101356781013467891891012379101247810125789

....................................................................................33 b

237810156146910136810126710123412359101245810134571023456101010 b

236924682568910346735679104567810459358348910257247910

6789234578913456891245679123567812346789101289137914781578910

3216xxxx

5217 xxxx

4218 xxxx

5439 xxxx

5432110 xxxxxx


16/44

*To tng phn xc nh:

*Tao tng phn xc nh tng hp:

)1(1 6321 xxxx )2(1 7521 xxxx

)3(1 8421 xxxx )4(1 9543 xxxx

)5(1 1054321 xxxxxx

7653)2)(1(1 xxxx

8643)3)(1(1 xxxx

965421)4)(1(1 xxxxxx

10654)5)(1(1 xxxx

8754)3)(2(1 xxxx

974321)4)(2(1 xxxxxx

10743)5)(2(1 xxxx

985321)4)(3(1 xxxxxx

10853)5)(3(1 xxxx

10921)5)(4(1 xxxx

87654321)3)(2)(1(1 xxxxxxxx

9764)4)(2)(1(1xxxx

1076421)5)(2)(1(1 xxxxxx

9865)4)(3)(1(1 xxxx

1086521)5)(3)(1(1 xxxxxx

10963)5)(4)(1(1 xxxx

9873)4)(3)(2(1 xxxx

1087321)5)(3)(2(1 xxxxxx

10975)5)(4)(2(1 xxxx

10984)5)(4)(3(1 xxxx

987621)4)(3)(2)(1(1 xxxxxx

10876)5)(3)(2)(1(1 xxxx

109765321)5)(4)(2)(1(1 xxxxxxxx


17/44

*Xc nh iu kin hn hp cc hiu ng:

109864321)5)(4)(3)(1(1 xxxxxxxx

109875421)5)(4)(3)(2(1 xxxxxxxx

109876543)5)(4)(3)(2)(1(1 xxxxxxxx

145781456102456913468135672345101345924835723611 b

1369102568101568924671014679234567829101358102358913471023479

134567891024578910234689102356791016781026789148910157910237810

245782456101456923468235671345102345914815713622 b

2369101568102568914671024679134567819102358101358923471013479

2345678911457891013468910345689135679102678101678924891025791013781023789

....................................................................................33 b

45781045612456910346810356710123453459101248101257101236101010 b

3789103691256856891012467467910123456781012935812358910347

34567891245789123468912356796781267891048912378


18/44

III. NHN XT

Nh Vy nu lp TBP 2n-n' ta nn chn h thc theo iu kin sau tha mn c 2 iu kin tr* Chn n'

Vi n-n' l

Vi n-n' ch

*Chn h thc sinh

Vi k =2i + 1i = 1m Vi n-n' l

Vi n-n' ch

T bi ton trn ta nhn thy rng: Nu chn h thc sinh l tich ca ca k tha s th ta c 2kh nng sau y vi k >=2 ta c:

+ k chn : Th cc hiu ng bc lin nhau s hn hp vi nhau v khi k = 2 th hiu ng tuyntnh s hn hp vi cc hiu ng tng tc cp.+ k l : Cc h s c trng cho cc hiu ng tuy n tnh khng h n hp vi cc hiu ng tng

tc cp v cc hiu ng lin k khng hn hp vi nhau.

2

1'

nnm

12

'1

'

i

nn

m

i

Cn

12

'

nn

m

kmlji xxxxx ...

2

1'

nnm

12

'

nnm


19/44

.


20/44


21/44

..........


22/44


23/44


24/44


25/44


26/44


27/44

:


28/44

Bi gii:

Bng s liu: Nhit (0c)Nng (M)

Mc di 20 1Mc trn 40 5

X10 = (X+

1 +X-1)/2 = 30

1/Tm thc nghim: X20 = (X+

2 +X-2)/2 = 3

1 = (X+

1 -X-1)/2 = 10

2/Khong bin thin: 2 = (X+2 -X-2)/2 = 2

3/Lp bng quy hoch ta c:

u X1u(0c) X2u(M) u x1u(0c) x2u(M)

1 20 1 1 -1 -12 40 1 2 1 -13 20 5 3 -1 14 40 5 4 1 1

4/Tng hp cc kt qu tnh ton trn v bng quy hoch ta c:Cc ch tiu u X1 X2Mc c s 30 3

Khong bin thin 10 2Mc trn 40 5Mc di 20 1Bin m

Th nghim u x1 x2 y1 - - y1

2 + - y23 - + y34 + + y4

Bi tp 5: Hy cho bit tm thc nghim, khong bin thin, mc trn, mc di ca s liu thc ngh

chu nh hng ca hai yu t nhit trong khong (20-400c) v t (1-5M)


29/44

im


30/44

u 1 2 3 4 5 6 7Cu (%) 9 11 13 15 17 19 21y, mg/l 18 19 21 26 27 29 32

Bi gii:

Tm thc nghim:

C0 =7 Cu /N =u= 1

15

Khong bin thi = Cu+1 - Cu = 2

xu = (Cu - C0 )/Khi ta c bng s liu sau yu 1 2 3 4 5 6 7

Cu (%) 9 11 13 15 17 19 21y, mg/l 18 19 21 26 27 29 32

xu -3 -2 -1 0 1 2 3

Thc hin php tnh ta c:7

yu =u= 1 172

7

xu yu =u= 1 68

7

x2u =u= 1

28

Khi : b0 =7 yu/N =

u= 124.57

b1 =7 7

xu yu / x2

u =u= 1 u=1

2.43

Bi tp 6: D ng p ng p p n p ng n n t p mt x p x tuy n t ntheo cc s liu thc nghim sau:

Vy phng trnh theo bin m l: y=24.57+2.43xuPhng trnh theo bin c th nguyn l: y=24.57+2.43(Cu -15)/2 = 6.25+1.215Cu

Cc bin khng th nguyn


31/44


32/44

Bi tp 7: Biu din cc s liu thc nghim bng phng trnh bc hai theo bng s liu sauu 1 2 3 4 5 6 7

tu,h 8 10 12 14 16 18 20wu=yu(kg) 15.2 20.5 25.3 30.6 36.1 42.1 59.5

Bi gii:

Tm thc nghim:t0 =

7 tu /N =u= 1

14

Khong bin thin: = tu+1 - tu = 2

xu = (tu - t0 )/

u 1 2 3 4 5 6 7tu,h 8 10 12 14 16 18 20

wu=yu(kg) 15.2 20.5 25.3 30.6 36.1 42.1 59.5

xu -3 -2 -1 0 1 2 3

7

yu =u= 1

229.3

7

xu yu=u= 1

186.9

7

x2u =u= 1

28

7

x4u =u= 1

196

7

x2u yu =

u= 1

984.1

b1=7 7

xu yu / x2u =

u= 1 u=1

6.68

Cc bin khng th nguyn:

Thc hin php tnh ta c:

Khi ta c bng s liu sau y:


33/44

Mt khc ta c h phng trnh sau:

b0N +

7

b2x2uu= 1

=

7

yuu= 1

7b0 xu

u= 1+

5

b2 x4

u= 1

=

7

yu x2

u

u= 1

Gii h phng trnh trn ta c;b0= 29.57

b2= 0.80

Vy phng trnh theo bin m l: y=29.57+6.68xu+0.8x2

u

Phng trnh theo bin c th nguyn l: y=20.7-2.11t+0.194t2 vi

xu=(tu-14)/2


34/44

y:


35/44

x1 x2 x3 x1x2 x1x3 x2x3 x1x2x3 yu1 yu2 yu3 Yu

1 -1 -1 -1 1 1 1 -1 73 69 68 702 -1 1 -1 -1 1 -1 1 58 58 64 603 1 -1 -1 -1 -1 1 1 54 59 52 554 1 1 -1 1 -1 -1 -1 84 94 92 905 -1 -1 1 1 -1 -1 1 100 106 109 1056 -1 1 1 -1 -1 1 -1 98 90 97 957 1 -1 1 -1 1 -1 -1 77 85 78 808 1 1 1 1 1 1 1 105 95 100 1009 0 0 0 89 83 86 86

Bi gii:

b0=8 yu/N =u= 1

81.875

b1=

8

yux1/N =u= 1 -0.625

b2=8 yux2/N =u= 1

4.375

b3=8yux3/N =u= 1

13.125

b12=

8

yux12/N =u= 1 9.375

b13=8 yux13/N =u= 1

-4.375

Bi tp 8: Cho s liu thc nghim sau y.Xc nh phng trnh thc nghimv sau kim tra bng hai tiu chun student v Fisher

Gi s phng trnh tuyn tnh.Khi ta c : y=b0+b1x1+b2x2+b3x3+b12x12+b13x13+b23x23+b123x1x2x3

ct chnh ct ph kt quu


36/44

b23=

8 yux23/N

=u= 1

-1.875

b123=

yux123/N

=u= 1

-1.875

u (yu1-u)2 (yu2-u)

2 (yu3-u)2 1+2+3 S2yuk

1 9 1 4 14 72 4 4 16 24 123 1 16 9 26 134 36 16 4 56 285 25 1 16 42 216 9 25 4 38 197 9 25 4 38 198 25 25 0 50 259 9 9 0 18 9

S2yuk =

8

S2(yuk)/N =u= 1

18

S2(y)= S2(yk)/ m = 6

S

2

(bi) = S

2

(y)/N = 0.75S(bi) = (S

2(bi) = 0.87

Sai s tin cy: (bi)=t(p;f).S(bi)

Tra bng student ng vi p=0.95 v f=N(m-1)=8.2=16

1/ Phng sai theo hng:

5/ Kim tra tiu chun student:

4/ Phng sai ca h s bi bt k ca phng trnh

Vy y=81.875-0.625x1+4.375x2+13.125x3+9.375x12-4.375x13-1.875x23-1.875x1x2x3

3/ Phng sai ca kt qu trung bnh ca ch tiu c ti u h

2/ Phng sai ti hin tng th nghim ca cuc th nghi


37/44

t(0.95;16)=2.12 v S(bi)=0.87

y (bi)= 1.84

b0 b1 b2 b3 b12 b13 b23 b12382 0 4.375 13.125 9.375 -4.375 -1.88 -1.875

u yu u |u -yu| (u -yu)2

1 70 69.375 0.625 0.390632 60 59.375 0.625 0.390633 55 55.625 0.625 0.390634 90 90.625 0.625 0.390635 105 104.375 0.625 0.390636 95 94.375 0.625 0.390637 80 80.625 0.625 0.39063

8 100 100.625 0.625 0.39063

8

(u-yu)2 =

u= 1

3.125

S2ph = 3.125

* T V S2(y)>S2ph cho nn:

F = S2(y)/S2ph = 1.92

Chun s Fb=248 ng vi p=95% v f1= N(m-1)=8x2=16 v f2=N-N'=8-7=1

Theo phng trnh thu c theo tiu chun student ta c:6/ Kim tra tiu chun Fisher:

N: h s ca phng trnh ban u

* Phng sai ph hp:8

(u-yu)2 /(N-N' ) =

u= 1

N': h s c ngha sau khi kim tra bng tiu chu

i chiu vi cc h s trong phng trnh trn th ta thy rng h s bi= -0,63 l khng ngha(|bi|


38/44

Vy vi xc xut tin cy p=95% c th kt lun rng phng trnh lp ph hpvi s liu thc nghim


39/44

1


40/44

x1 x2 x3Mc c s 0.4 840 60Khong bin thin 0.15 100 60Mc trn(+) 0.55 940 120Mc di(-) 0.25 740 0

Ma trn QHTN v cc kt qu c cho bng sau:

STT x0 x1 x2 x3 x4 y

1 1 1 1 1 1 1002 1 -1 1 1 -1 813 1 1 -1 1 -1 954 1 -1 -1 1 1 365 1 1 1 -1 -1 130

6 1 -1 1 -1 1 697 1 1 -1 -1 1 908 1 -1 -1 -1 -1 64

Cc yu t nh hng n thng s ti ux1 - lng molip en a vo nhm, tnh bng %

x2 - nhit nung nng ,0Cx3 - thi gian nung nng, pht

x4 l yu t nh tnh, c hai gi tr :lm lnh nhanh ( lm lnh bng graphit),lm lnh chm(gch chu la)

-

Ma trn quy hoch thc nghim

Cc mc ca cc yu t cho bng sau:

graphitGch chu lu

Bi tp 9 : Nghin cu s bin i tnh ca nhm nguyn cht bng molip en.

Thng s ti u ha y c chn l s ht nhm trn b mt 1cm2

* M hnh c chn l m hnh tuyn tnh. Khi ta c : y = b0+b1x1+b2x2+b3x3+b4x4

Vi biu thc sinh : x4=x1x2x3. Cc th nghim khng lp li

Thc nghim c thc hin l TYP : 24-1 suy ra N=8(TN)

Cc mc

Cc yu t

x4-

x4 - tc lm lnh ()

x1,x2,x3 l nhng yu t nh lng


41/44

STT y0u y0u -y0 (y

0u -y0)2

1 80 0 02 82 2 43 78 2 4

3

y0u/3 =u= 1

80

3

(y0u -y0)2 =u= 1

8

S2th=

3

(y0u -y0)2/(n0-1) =u= 1

4

H s trong m hnh tnh ton :

b0=8

y/N =u= 1

83.125

b1=8

yx1/N =u= 1

20.625

b2=8

yx2/N =u= 1

11.875

b3=8

yx3/N =u= 1

-5.125

b4=8

yx3/N =u= 1

-9.375

Kt qu ta thu c phng trnh c dng sau : y=83.125+20.625x1+11.875x2-5.125x3-9.375x4

Phng sai ti hin:


42/44

* Kim tra tiu chun student:

Sbj = (S2

th/ N) = 0.71

Khi :t1 = 29.17

t2= 16.79

t3= -7.25

t4= -13.26

Tra bng tp(f) vi p=0.05 v f=n0-1= 3-1=2 ta c tp(f)=4.3

* Kim tra tiu chun Fisher:

b0 b1 b2 b3 b483.125 20.625 11.875 -5.125 -9.375

STT y u y - u (y -u )2

1 100 101.125 -1.125 1.265632 81 78.625 2.375 5.640633 95 96.125 -1.125 1.265634 36 36.125 -0.125 0.015635 130 130.125 -0.125 0.015636 69 70.125 -1.125 1.26563

7 90 87.625 2.375 5.640638 64 65.125 -1.125 1.26563

S2t t =8(y -u )

2/(N -L) =u= 1

5.46

F= S2t t/S2

t h = 1.365

Tra bng F1-p(f1,f2); p=0.05, f1=3,f2=2 ta c F1- p(3,2) =19.2

Ta c bng s liu sau:

V F1-p(f1,f2)>F do phng trnh tng thch vi thc nghim

Theo phng trnh thu c trn ta c:

R rng ta thy | tj | >tp(f) do cc h s hi quy u c ngha v phng trnh hi quy

tj = |bj| / Sbj,


43/44

** Ti u ha thc nghim bng phng php ng dc ng

1 = 2.b1.1/2.b2 = 0.0261

3= 2.b3.1/2.b2 = -2.5895

Bng thc nghim ti u ha

Tn x1 x2 x3 x4 y

Mc c s 0.4 840 60 - -H s bj 20.625 11.875 -5.125 -9.4 -

Khong bin thin 0.15 100 60 - -bj. j 3.0938 1187.5 -307.5 - -

Bc j 0.0261 10 -2.5895 - -

Bc lm trn 0.03 10 -2.6 - -

Th nghim tng tng 0.43 850 57.4Gch

chu la -Th nghim tng tng 0.46 860 54.8 " -

Th nghim th 9 0.49 870 52.2 " 108Th nghim tng tng 0.52 880 49.6 " -Th nghim tng tng 0.55 890 47 " -

Th nghim th 10 0.58 900 44.4 " 196

Th nghim th 11 0.61 910 41.8 " 366Th nghim th 12 0.64 920 39.2 " 313

Da vo bng ti u ny th ta thy nhn c kt qu tt nht th nghim 11.gi tr ca thng s ti u tha mn ngi nghin cu

Mc c s x1=0.4, x2=840, x3 =60,x4 lm lnh chm(lm lnh bng gch chu la)

Bc chuyn ng ca yu t x1, x3 vi bc chuyn ng ca x2 l 2

im bt u l im khng.


44/44

Documents

bài tập nhóm quy hoạch thực nghiệm