quy hoach thuc nghiem_châm
TRANSCRIPT
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Bi 1:. Nghin cunh hng ca mt s yut cng nghn qu trnh
chit tch anthocyanin
1.1.nh hng ca t l dung mi
Nghin cu nh hng ca t l dung mi n hm lng v mu
anthocyanin thu c. Ngi N/C lm 6 th nghim trong cc iu kin nhsau:
-Nhit chit: 300C
-Thi gian chit: 45 pht
-Chit trong h dung mi c t l dung mi nc: ethanol thay i nh
bng 1.
Bng1. Cc thng sban u v ktqu th nghim
STT
%Vnc
%Vethanol
T lnc / ethanol
Hm lng %anthocyanin,
mu
18
020 4/1 0,827 3,53
27
030 7/3 0.890 3,50
36
0
40 3/2 0.870 3,46
45
050 1/1 0,857 3,40
54
060 2/3 0,845 3,37
63
070 3/7 0,840 3.34
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1 4
0.93.5
3
0.8 2.5
20.7
1.5
0.6 1
%a
nthocyanin
%a
nthocyanin
30
pht
45
pht
75
pht
90
pht
m
u
mu
1 4
0.9
3.5
3
0.8 2.5
0.7
2
1.5
% anthocya
mu
0.6 1
1 2 3 4 5 6
Hnh 1: Biu biu din nh hng ca t l dung1.2.nh hng cmaithni hgiamnlchnigtv mu anthocyanin
Tin hnh 5 th nghim trong cc iu kin sau:
- Nhit chit: 300C
- Chit trong dung mi c t l nc: ethanol l 7/3
- Thi gian thay i t 30- 90 pht.
% anthoc yanin
mu
Hnh 2: Biu biu din nh hng ca thi gian chit n hmlng v mu anthocyanin thu c
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%a
nth
ocyanin
30oC
40oC
50oC
60oC
70oC
mu
1.3.nh hng ca nhit
Nghin cu nh hng ca nhit, tin hnh 5 th nghim trong cng
iu kin:
- Chit trong h dung mi c t l nc: ethanol l 7/3
- Thi gian chit: 45 pht.- Nhit chit thay i t 300C 700C. Cc thng sc th v kt qu
c th hin trn biu hnh (3).
1 4
0.9
0.8
0.7
0.6
3.5
3
2.5
2
1.5
1
0.5
% anthocyanin
mu
0.5 0
Hnh 3: Biu biu din nh hng ca nhit n hm
lng v mu anthocyaninQua nghin cu nh hng ca mt s yu t cng nghn kh nng thu
nhn anthocyanin chng ti nhn thy: Dung mi, t l dung mi, thi gian
chit, nhit chitunh hng n kh nng chit tch anthocyanin t bp
ci tm. ng vi mi iu kin khc nhau chng ti thu c anthocyanin c
hm lng v mu khc nhau. T cc kt qu nghin cu chng ti chn
c min kho st thch hp ca cc yu t cng ngh cho cc nghin cu tip
theo nh sau:- Chit trong h dung mi c t l nc: ethanol dao ng t 7/3 1/1
- Nhit chit t 30400C
- Thi gian chit trong khong 4575 pht
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2. Ti u ho iu kin chit tch anthocyanin c mu cao t bp ci
tm
Vi mc ch ca ti l thu nhn v s dng cht mu anthocyanin,
chng ti tin hnh ti u ho iu kin chit tch trong khun kh bi ton
ti u a mc tiu thu nhn cht mu anthocyanin c hm lng v mucao nht.
2.1 Chn cc yu tnh hng
Trong qu trnh chit tch anthocyanin phi chu tc ng ca nhiu yu
t cng ngh, song y chng ti chn 3 yu t c thm d phn trn:
- Z1: Nhit chit,0C
- Z2: Thi gian chit, pht
- Z3: T l nc trong h dung mi,%
- Y1: Hm lng anthocyanin, %
- Y2: mu
Phng trnh biu din mi quan h c dng:
Y1 = f ( Z1 , Z2, Z3 ) Y2 = (Z1, Z2, Z3)
Y1 Max Y2 Max
Y1 l hm mc tiu hm lng Y2 Hm mc tiu mu.
2.2 Cc bc thchinbi ton quy hoch
2.2.1 Chn phng n quy hoch xc nh hng i ca ti v nhanh chng tin ti min ti u chng
ti chn phng n quy hoch trc giao cp I (TYT 2k) thc nghim yu t ton
phn 2 mc, k yu tnh hng.
Phng trnh hi qui c dng:
Y = b0 + b1x1 + b2x2 + b3x3 + b12x1x2 + b13x1x3 + b23x2x3 + b123 x1x2x3 (1 )
Trong :
b0: H s hi qui.b1, b2, b3 : H s tuyn tnh
b12, b23, b13: H s tng tc i
b123: H s tng tc ba
Mi h s b c trng cho nh hng ca cc yu tn qu trnh chit
tch.
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1
1
a) Chn phng trnh hi qui:
Phng trnh hi qui c chn theo phng trnh (1) mc 2.2.1.
Cc h s b1, b2, b3...b123 c tnh theo s liu thc nghim hm mc
tiu hm lng anthocyanin (Y1).
b) Tnh h sb:V phng n c chn l quy hoch trc giao, theo [1] cc h s bj trong
phng trnh hi qui (1) c xc nh theo cng thc sau:
bj =1
N
(xjuyu ) vi : j = (1, k )Nu 1
bij =N
(xiuxju )yu i j = (1, k ) (2)Nu 1
bijk =N
(xiuxjuxku )yu i j k = (1, k )Nu 1
T s liu thc nghim bng (3), p dng cc cng thc (2) ta tnh c
cc h s b:
b0 = 0,9208 b12 = 0,017
b1 = -0.07 b13 = -0,0255
b2 = 0,0488 b23 = -0,019
b3 = 0,088 b123 = 0,0155
c) Kimnh mc ngha ca cc h sb trong phng trnh 3.1
Cc h sc kimnh theo tiu chun Student (t)
bjtj
Sb j(3)
So snh tj vi tp(f) . Trong : - tp(f) l chun student tra bng ng
vi xc sut tin cy p v bc t do f, f = n01.
bj : l h s trong phng trnh hi quy chn.
Sbj l lch ca cc h s bj
Nu tj > tp(f) th h s bj c ngha.
Nu tj < tp(f) th h s bj b loi khi phng trnh.
kimnh theo chun Student (t) ta thay h s bj, Sbj vo cng thc
(3.3) ta c cc gi tr tj:
t0 = 130,21 t12 = 2,510
t1 = 10,01 t13 = 3,606
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S
S
3
F 2
3
t2 = 6,894 t23 = 2,687
t3 = 12,445 t123 = 2,192
Tra bng tiu chun Student ta c tp(fth) = t0,05(2) = 4,3
Do t12 < tp(fth), t13 < tp(fth), t23 < tp(fth), t123< tp(fth) nn cc h s b12, b13, b23,
b123
loi ra khi phng trnh. Phng trnh ng hc c dng:
y1 = 0,9208 - 0,07x1 + 0,04875x2 + 0,088x3 (4)
d) Kimnh sph hp ca phng trnh hi qui vi thc nghim
S tng thch ca phng trnh vi thc nghimc kimnh
theo tiu chun Fisher (F).2d = 7,8406th
Tra bng tiu chun Fisher ta c F1-p( f1, f2) = F0,95 (4,2) = 19,3
Ta c F < F 1-p. Vy m hnh ton hc chn ph hp vi thc nghim.
2.2.4 Ti u ho thc nghimthu c hm lng anthocyanin cao nht
a) Tnh cc bc chuynng j :
T mc csZ0j, v phng trnh hi qui tuyn tnh i vi hm mc
tiu hm lng chng ti tnh bc chuynng j (j = 1, 2, 3) cho mi yut.
Kt qu c ghi bng 4.
Bng 4. Kt qu tnh bc chuynng j ca cc yu t
Cc mc Cc yu tnh hng
Z1,0C
Z2,
pht
Z3,
%
Mc cs 35 60 60
Khong bin thin ( j ) 5 15 10
H s bj -0,070 0,048 0.088
bj j -0,350 0,731 0,88
Bc chuynng ( j ) -1,980 4,150 5Lm trn -2 4 5
Theo bng s liu (4) ta c : b = 0,88, theo ti liu [1]max
Chn bc chuynng 3 0,5 . 3 = 0,5.10 =5.
Cc bc chuynng ca yu t x1, x2 c tnh:
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b
b
b1 11 3
3 3
b2 22 3
3 3
= -1,98
= 4,1
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b) Tchc th nghim leo dc:
T kt qu cc bc chuynng j bng (4), chng ti t chc th
nghim leo dc v im xut pht t tm thc nghim.
Th nghim theo hng chn, kt qu c biu dinbng 5.Bng 5: Ktqu th nghim theo hng leo dc
Yu t
TN
Z1,0C
Z2,
pht
Z3,
%y1,% y2
1( Tn ti
tm )
35 60 60 0,927
2 33 64 65 0.962
3 31 68 70 0,985
4 29 72 75 1,113 4,720
5 27 76 80 0,997
Nhn vo bng 5, kt qu th nghimtt nht th nghim th t. Ti nhit
chit 290C, thi gian chit 72 pht, t l nc trong h dung mi l 75%
chng ti thu c hm lng anthocyanin cao nht l 1,113 %. Ti th nghim
ny, chng ti xc nh mu ca anthocyanin l: 4,720. y cha phi l
mu thu c cao nht. V th, chng ti tin hnh tm iu kin chit tch ti
u thu c anthocyanin c mu cao.
2.2.5 Xy dng m t ton hc vi hm mc tiu mu.
a) Chn phng trnh hi qui:
Phng trnh hi qui c chn theo phng trnh (1) mc 2.2.1. Cc h
s b1, b2, b3...b123 c tnh theo s liu thc nghim hm mc tiu mu
(Y2).b) Kim tra mc ngha ca h sb trong phng trnh hi qui:
Sau khi kim tra mc ngha ca cc h s b ta c: t13 < tp(fth), t23 < tp(fth),
t123< tp (fth) nn cc h s b13, b23, b123 b loi ra khi phng trnh.
Phng trnh hi qui c dng:
y 3,7709 - 0,55x10,2826x2 + 0,5291x30,2224x1x2 (5)
c) Kimnh sph hp ca phng trnh hi qui vi thc nghim:
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Cc bc kim tra c trnh by ph lc 6.
Sau khi kim tra ta c phng trnh hi qui (5) ph hp vi thc nghim.
2.2.6: Ti u ho thc nghim thu c anthocyanin c mu cao.
Sau khi kim tra phng trnh hi quy ph hp vi thc nghim,
chng ti tin hnh ti u ho thc nghim bng phng php leo dc thu
c anthocyanin c mu cao.
* Tnh bc chuynng ca cc yu t
Cng t mc cs Zj v phng trnh hi qui i vi hm mc tiu
mu. Chng ti tnhbc chuynng j (j = 1, 2, 3) tng t nh mc 2.2.3.
Kt qu c th hin bng 6.
Bng 6: Tnh bc chuyn ng j ca cc yu t
Cc mcCc yu t
Z1,0C
Z2,
phtZ3, %
Mc cs 35 60 60
Khong bin thin ( j ) 5 15 10
H s bj -0,55-
0,2826
0,5291
bj j -2,57-
4,2395,29
Bc chuynng ( j ) -2,07 -3,2 4
Lm trn -2 -3 4
*T chc th nghim leo dc cho hm mc tiu mu:
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cng ngh thc tin tt cho c hai hm mc tiu, ng thi nng cao tnh ton
din v tnh thuyt phc cho kt qu thu c.
Thc t khng th c mt nghim chung cho c hai qu trnh tc
y1max, y2max m ch tm c nghim tho hip (x1, x2, x3) cc gi tr y1, y2nm gn y
1max, y
2max. tm c nghim tho hip chng ti sdng phng
php chp tuyn tnh :
yL = 1 y1 + 2 y2
Trong :
- 1 l h s quan trng ng vi hm mc tiu hm lng (y1)
- 2 l h s quan trng ng vi hm mc tiu mu (y2)
Vi mc ch thu nhn cht mu anthocyanin ng dng lm cht ch th
trong ho phn tch. chng ti u tin cho hm mc tiu hm lng.
Chn: - 1 = 0,6, 2 = 0,4
Ta c phng trnh hm a mc tiu : yL = 0,6y1 + 0,4y2
Cc h s ca phng trnh hi quy c tnh theo bng 8
Bng 8. Tnh h sca phng trnh hi quy
H s b y1 y2 yL
b0 0,928 3,7709 2,06
b1 -0,07 -0,55 -0,262b2 0,04875 -0,2826 -0,0837
b3 0,088 0,5291 0,2644
b12 -0,2223 -0,22
Ta c phng trnh hi quy:
yL = 2,060,262x1- 0,08379x2 + 0,2644x30,222x1x2 ( 6 )
Tin hnh ti u ha hm a mc tiu tm gii php cng ngh thc tin
ph hp.
*Tnh cc bc chuynng j : cho HMT yL
Cng tng t nh mc 2.2.3, chng ti tnh bc chuyn ng j (
1 , 2 , 3 ) cho cc yu tnh hng. Kt qu c biu din bng 9.
Bng 9. Tnh bc chuyn ng ca cc mc yu t
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Cc mc
Cc yu t
Z1,0C
Z2,
phtZ3, %
Mc cs 35 60 60Khong bin thin
( j )5 15 10
H s bj-
0,262
-
0,083970,2644
bj j -1,31-
1,4273
Bc chuynng( j )
-1,48 -1,427
3
Lm trn -1,5 -1,5 3
* T chc th nghim leo dc cho hm mc tiu YL:
Bng 10: Kt qu th nghim theo hng leo dc ca hm chp YL
KTN
Cc yu tnh hng
Y1,% Y2 YLZ1,
0CZ2,
phtZ3,%
1 ( TN
ti tm)35 60 60 0,927 3,991 2,149
2 33,5 58,5 63 0,943 4,403 2,287
3 32 57 66 0,972 4,911 2,548
4 30,5 55,5 69 0,983 4,952 2,553
5 29 54 72 1,110 4,967 2,6566 28,5 52,5 75 0,915 4,500 2,397
Nhn vo bng 10, ti th nghim th 5 hm chp yL t gi tr ln nht
yLmax= 2,656. So snh vi cc th nghim leo dc vi tng hm mc tiu ta c
PA.TN Z1,0C Z2, Z3, % y1,% y2 yL
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pht
Theo
% Anth
29 75 75 1,113 4,72 2,556
Theo
mu
29 51 72 0,975 5,001 2,585
Theo
hm chp
29 54 72 1,110 4,967 2,656
T bng trn chng ti tm c iu kin tt chit cht mu
anthocyanin t bp ci tm trong mi trng trung tnh
l: Nhit chit 29 0C
Thi gian chit l 54 pht
H dung mi nc -ethanol l 72-28
Bi ton 2
KT QU NGHIN CUNH HNG CA NHIT V
NNG KIMN QU TRNH TCH TP CHT RA KHI X
SI XENLULO
(Trch mt phn ti lun vn Thc sca CN NguynB Trung
khoa Ha, trngi hc SphmHN)
I. KT QU NGHIN CU T TI
1. Nghin cu nh hng ca nhit v nng kimn qu trnh tch
tp cht ra khi xsi xenlulo.
1.1.nh hng ca thi gian v nng NaOHn lng tp cht tch
ra.Gai b sau khi c tch sb, dng tay tc gai b kh ra thnh
nhng si mng, ct li thnh b, cc b c khi lng xp x nhau khong 15
gam. Ngm cc b si vo dung dch NaOH cc nng nghin cu trong
thi gian tng ng. Kt qu thu c bng sau:
Bng 1.1: Kt qu kho st nh hng ca thi gian v nng
n lng tp cht tch ra
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Thi gian
C%
5 gi 10 gi 15 gi 20 gi
1% NaOH 10.098 10.978 12.46 13.01
3% NaOH 12.92 14.1 15.4 14.815% NaOH 14.01 14.67 15.598 14.46
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1%3%5%
%tachduoc
18
16
14
12
10
8
6
0 5 10 15 20 25
thoi gian (gio)
ra.
Hnh 1.1. nh hng thi gian v nng NaOH n lng tp cht tch
Nhn xt: T 3 ng biu din nng 1%, 3% v 5% ta nhn
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thyrng nng kim 1% th hiu qu tch l khng cao. Khi tng nng
kim ln 3% v 5% th hiu qu tch tng ln nhiu. Tuy nhin hiu qu tch
trong khong nng t 3 5% l khng khc nhau lm trong khong thi
gian t 10 15h.
iu ny c thc gii thch nh sau: nng qu long 1%, thigian ngn ban u cha ho tan cc tp cht bao bc bn ngoi nn hiu
qu ca qu trnh tch khng cao. Sau thi gian t 10 15 gi, cc lp bn
ngoi b ho tan nn toiu kin thun li cho NaOH thm nhp vo bn
trong ho tan hemixenlulo, lignin v cc cht c phn t lng thp khc c
trong cc b si.
Nh vy qua th cho ta thyrng c thi gian v nng u c nh
hng r nt n qu trnh tch tp cht ra khi xsi gai.
1.1.2. Ti uho thc nghim qu trnh tch tp cht ra khi si gai.
tin ti min ti u, chng ti chnphng n thc nghim yu t ton
phn. Hai yu tnh hng n qu trnh l nng (Z1) v thi gian ngm
(Z2). Hm mc tiu cntc l lng tp cht tch ra khi si l ln
nht hay ni cch khc hiu qu tch l cao nht.
quy hoch thc nghim ton phn, chng ti tin hnh b tr th
nghim thay i ng thi cc yu t, mi yu tc tin hnh 3 mc:
mc trn, mc di v mc cs th nghimtm phng n
Mc trn, mc di, khong bin thin c trnh by bng 1.2, ma
trn quy hoch thc nghim c trnh by bng 1.3.
Bng 1.2. Cc mc thnghim.
Mc di Mc cs Mc trn Khong bin
thin ( )
Z1 (C% NaOH) 3 4 5 1
Z2 (thi gian ngm) 10 12.5 15 2.5
Lp ma trn quy hoch:
Vi 2 yu t nhit v nng (k = 2), mi yu t c hai mc l
mc trn v mc di. Vy s th nghimc tin hnh l N = 22 = 4 th
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th
nghim.
Phng n tin hnh trnh by bng sau:
Bng 1.3. Ma trn quy hoch thc nghim.
N
Cc yu t theo t
l thc
Cc yu t theo t l
m ho
Gi tro
c
Z1 (C%) Z2 (t) X0 X1 X2 X1X2 Y
1 5 15 1 1 1 1 15.598
2 3 10 1 -1 -1 1 14.1
3 5 10 1 1 -1 -1 14.67
4 3 15 1 -1 1 -1 15.4
5 4 12.5 1 0 0 0 14.82
6 4 12.5 1 0 0 0 14.8
7 4 12.5 1 0 0 0 14.75
Thitlp phng trnh hi quy:
Tnh h s hi quy: Cc h s hi quy c tnh theo cng thc ton hc
(3), (4), (5). T s liu thc nghim ta xc nh c cc gi tr b0, b1, b2 nh
sau:
b0 = 14,982 ; b1 = 0,192 ; b2 = 0,557 ; b12 = - 0,093
Vi kt qu trn ta c phng trnh hi quy theo ton hc:
Y = 14,982 + 0,192 X1 + 0,557 X20,093 X1X2
kimnh ngha ca h s hi quy v s tng thch ca phng
trnh hi quy vi thc nghim, ta phi tm phng sai ti hin S2 . Do vy
chng ta phi lm thm 3 th nghimtm phng n v thu c gi tr Y0 .
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b
T cng thc tnh phng sai ti hin, ta c: S2 th = 0,0013 ; Sth =
0.03605551
Kimnh cc h s c ngha ca phng trnh hi quy:
bS c ngha ca h s hi quy c theo tiu chun Student: ti i
Si
Bng cch tnh nh trn ta thu c cc gi tr ti nh sau:
t0 = 828,833 t2 = 30,896
t1 = 10,650 t12 = 5,158
Tra bng tiu chun Student ta c t0,05 (2) = 4,303.
Qua bng s liu trn ta thy cc c h s b0, b1, b2 , b12 l c ngha vi
tin cy P
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b
tng, n mt gii hn no th gi tr phn trm tp cht tch ra l cc i v
qu gii hn th gim xung.
Ti u ho c thc hin nh sau:
Chn bc nhy ca yu t x1 l 1 = 0,05. Da vo 1 ta tnh c gi tr
b
2 theo cng thc sau : 1 11 3 b3 3
b2 22 3
3 3
c: 2 0,05.0,557.2,50,192.1
0,36 T cc thng s tnh c, chng ti tin
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hnh thc nghimti u ho cc mc csv bc nhy nh sau:
Bng 1.4. Kt qu ti u phng trnh hi quy Y1 m tnh hng
ca nng v thi gian ln hm lng cc cht tch ra.
Ccyu t X1 (nng
kim C%)
X2 (thi gian ngm)
Mc cs 4 % 12,5 gi = 750
pht
Bc nhy 1 = 0,05 2 = 0,36 gi
20 pht
Nng
NaOH
Thi gian ngm Y (lng cht
tch ra)
4,05 770 pht 14,3
4,10 790 pht 14,81
4,15
(TNTT)
810 pht -
4,20
(TNTT)
830 pht -
4,25 850 pht 15,34
4,30 870 pht 15,62
4,35 890 pht 15,2
Nh vy hiu qu tch tp cht ra khi xsi gai th phi tin hnh
cc thng s sau nhit phng.
Cc thng sti u Gi tr
4,3%
Nng kim C%Thi gian ngm 870 pht (14 gi30
pht)
II.NHN XT V NH GI V BI TON TI U HA TRONG
TI
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1. V phng php nghin cu
3.1.1. Phng php quy hoch thcnghim.
Trong ti trn, CN NBT s dng phng php quy hoch thc
nghim tan phn. Ni dung ca phng php nh sau:
- Kho st nh hng ca cc yu tn qu trnh tch tp cht ra khi sixenlulo, chn ra cc yu tnh hng nht.
- Xy dng ma trn quy hoch thc nghim.
- T chc th nghim v ly mu thng k.
- T s liu thu c, xy dng m hnh thng k thc nghim m t
qu trnh.
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ZZ
- Kimnh thng k.
- Phn tch s tng tc gia cc yu tnh hng n qu trnh x l si
trong phm vi chn.
quy hoch thc nghim ton phn, tc gi tin hnh b tr th nghimthay i ng thi cc yu t. Mi yu tc tin hnh 3 mc: Mc trn;
mc di; mc cs th nghimtm phng n. Xy dng hm mc tiu
cntc (max hay min).
Lp ma trn quy hoch:
Vi 2 yu t l nhit v nng (k = 2), mi yu t c hai mc l mc
trn v mc di. Vy s th nghimc tin hnh l N = 22 = 4 th nghim.
tin cho vic tnh ton, ta chuyn t h trc t nhin Z1, Z2 c th
nguyn sang h trc khng th nguyn m ho. Vic m ho c thc hin ddng nh chn tm ca minc nghin cu lm gc to.
Trong h m ho khng th nguyn ta c c:
Mc trn: - k hiu +1
Mc cs: - k hiu 0
Mc di: - k hiu1
Cng thc chuyn t h n v thc qua n v m ho khng th nguyn:
0j jXj
Zj
Zmax
; j = 1, ..., k (1)
ZminZj j j
2; j = 1, ..., k (2)
Thitlp phng trnh hi quy m tnh hng ca cc yu
tn qu trnh nghin cu:
Tnh h s hi quy: Cc h s hi quy c tnh theo cng thc ton hc
nh sau: nYi
b i 10N
n
XijYib i 1i
N
(3)
(4)
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th
Y
2 0
0
b
th
S
S
i
2
bj1
n
(Xj .X1 )i .Yii 1
N(5)
T s liu thc nghim trn, p dng cc cng thc (3), (4) v (5) xc nh
c gi tr b0 , b1 , b2 v b12 .T kt qu trn c phng trnh theo ton hc: Y = b0 + b1X1 + b2X2+
b12X1X2Kimnh ngha cah s hi quy v stng thch ca
phng trnh hi quy vi thc nghim. Tm phng sai ti hin S2 . Do vy,
phi lm thm 3 th nghimtm phng n v thu c ba gi tr 0 v gi
tr Y0 ti tm.
Phng sai ti hin c tnh theo cng thc:1 m
S th (Yim 1 i 1Y0 )2 (6)
m
(Yi Y0 )2S i 1 (7)th m 1
trong m l s th nghimtm phng n.
- Sc ngha ca h shi quy c kim nh theo tiu chun Student:
ti biS
i
bi: l h s th i trong phng trnh hi quy.
Sbi: lch qun phng ca h s th i.
SSbi (8)N
Tra bng tiu chun Student ta c t0,05 (2) = 4,303.
Nu gi tr tj
> t0,05
(2) = 4,303 th cc h s bj
l c ngha, ngc li l
khng c ngha.
Xy dng phng trnh hi quy vi cc h s bj c ngha.
- Kimnh stng thch ca phng trnh hi quy vi thc nghim theo
tiu chun Fisher.2
Ta c F duth
(9)
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N ^
i i(Y Y )2
vi S2 i 1 (10)du N L
(N l s th nghim, L l h s ngha) Yi4
bi Xii 1
(11)
Thay s vo tnh ton ta c cc gi tr Y1; Y2 ; Y3 ;Y4.So snh vi F(1-
p)(f1,f2) vi P = 0,05; f1 l bc t do ca phng sai tng thch (f1= n - l); f2 l
bc t do ca phng sai ti hin (f2 = m 1).Tra bng ta c c gi tr
F(0,95)(f1,f2)Nu F < Ftb th phng trnh hi quy tm c l tng thch vi thc
nghimvi mc ngha 95% , ngc li l khng tng thch.
1.3. Ti u ho thc nghim.
Tin hnh ti u ha bng phng php thc nghim leo dc nht.
Vect grad y l mt vectc chiu biu th s bin thin nhanh nht ca
grad y(x), gi tr ca
gian yu t.
grad y(x) thay i tim ny sang im khc trong khng
Vi m hnh tuyn tnh bi k,
grad y
hoc
y
ix1
grad y
y
x 2
b1 i
j
b2 j
...
...
y
kx k
bk k
(12)
(13)
Chuynng theo grad yl chuynng theo ng ngn nht n im
ti u, bi v hng grad l hng c nghing dc nht dn tim cho
n im cc i.
Trong trng hp k yu t vic tnh ng dc nht trn mtp ng c
thc hin nh sau:Chn bc nhy ca yu t x1 l 1; da vo 1 ta tnh c 2 theo cng
thc:bj j
j i .b
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i i
(14)
trong : i l bc nhy ca yu t th i
bi, bj l h s hi quy ca cc yu t tng quan
74i, j l khong bin thin ca tng yu ttng ng
thay s vo ta tnh c cc gi tr j khc.
T cc thng s tnh c, tin hnh thc nghimti u ho cc mc
csvi cc bc nhy tnh sn trc. Chuynng grad phi bt u t
im 0 (mc csca mi yu t) v dng li khi tm c im ti u nu
nhng hn cht vo cc yu t lm cho chuynng tip tc theo hng
grad khng hp l na.
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