bài tập nhóm quy hoạch thực nghiệm
TRANSCRIPT
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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
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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
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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:
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au:
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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
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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:
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sau:
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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
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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:
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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
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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
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*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
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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
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*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
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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
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*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
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*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
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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
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.
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..........
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:
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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)
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im
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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
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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:
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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
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y:
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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
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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
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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|
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Vy vi xc xut tin cy p=95% c th kt lun rng phng trnh lp ph hpvi s liu thc nghim
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1
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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)
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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
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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:
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* 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,
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** 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.
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