tin hoc trong hoa hoc
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TRNG I HC CN THKHOA SPHM
THNG K HA HCV TIN HC TRONG HA HC
ThS. Hunh Kim Lin
2006
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THNG TIN V TC GI
PHM VI V I TNG SDNGCA GIO TRNH
1. THNG TIN V TC GI
H v tn: Hunh Kim LinSinh nm: 1955Cquan cng tc:B Mn: Ha Hc Khoa: S Phm
Trng: i hc Cn Tha ch Email lin h: huynhkimlien@ctu.edu.vn
2. PHM VI V I TNG SDNGGio trnh c th dng tham kho cho cc ngnh : C nhn Ha hc, S Phm
Ha hc, Cng ngh Ha HcC th dng cho cc trng: i hc S Phm, i hc Khoa Hc T Nhin, Cao
ng S Phm
Cc t kha: Phng sai, lch chun, Sai s ngu nhin, Sai s h thng,Chun thng k, MS Excel, Chem win, Chem office, MS flash.Yu cu kin thc trc khi hc mn hc ny: Xc sut thng k v tin hc cn
bn (trnh A)
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MC LCBA.......................................................................................................................................1
THNG TIN V TC GI ................................................................................................2
MC LC ...........................................................................................................................3
PHN I: THNG K HA HC.......................................................................................8
Chng 1: I CNG V THNG K .........................................................................8
I. SAI S NGU NHIN V SAI S H THNG. .....................................................8
1. Cc khi nim thng dng: ....................................................................................8
2. Sai s ngu nhin:.....................................................................................................9
3. Sai s h thng: ......................................................................................................10
4. Lan truyn sai s h thng v sai s ngu nhin: ...................................................12
II. HM PHN B (DISTRIBUTION FUNCTION) ..................................................121. Cc khi nim cbn: ............................................................................................12
2. Hm phn b chun (Normal distribution function): .............................................13
3. Hm phn b mu:..................................................................................................18
III. CC CHUN (TEST) THNG K........................................................................24
1. Khi qut v phng php kim nh thng k: ....................................................24
2. Chun Dixon (Zlt = n,PQ ) .......................................................................................26
3. Chun (t) (Zlt =p,n ).........................................................................................284. Cc chun : .......................................................................................................30
5. Chun Fisher. (Zlt =III f,f,P
F ).....................................................................................33
6. Chun Cochran . (Zlt= GP,f,n) ..................................................................................34
7. Chun Student (t-Test): ..........................................................................................35
8. Chun Gauss (Zlt = Up)...........................................................................................38
9. Chun Duncan. (Zlt = thf,R,Pq ) ...............................................................................39
CU HI N TP........................................................................................................45TI LIU THAM KHO..............................................................................................45
Chng 2: PHN TCH PHNG SAI...........................................................................46
I. KHI QUT V PHN TCH PHNG SAI (ANALYSIS OF VARIANCE).....46
1. Mc ch v ngha: ..............................................................................................46
2. Nguyn tc v thut ton: .......................................................................................46
II. PHN TCH PHNG SAI MT YU T (SINGLE FACTOR) ........................47
III. BI TP NG DNG............................................................................................50
1. Bi tp 1:.................................................................................................................50
2. Bi tp 2:.................................................................................................................52
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BI TP ........................................................................................................................56
TI LIU THAM KHO..............................................................................................56
Chng 3: PHN TCH HI QUY ..................................................................................57
I. KHI QUT V PHN TCH HI QUY................................................................57
1. Mc ch v ngha : .............................................................................................572. iu kin thc hin: ...............................................................................................57
II. PHNG TRNH HI QUY TUYN TNH N GIN (Y=ax + b). .................57
1. Nguyn tc tm cc h s ca phng trnh hi quy: .............................................57
2. Tnh cc h s a , b v cc thng s cn thit: .......................................................58
3. Xt ngha ca h s hi quy (chun Student):.....................................................59
4. Kim nh s tuyn tnh gia x v y ca phng trnh hi quy ( chun Fisher): .60
5. Trnh by phng trnh hi quy km vi cc c trng cn thit:.........................60
6. ng dng phng trnh hi quy: ............................................................................61III. PHNG TRNH HI QUY TUYN TNH NHIU BIN.................................62
IV. BI TP NG DNG............................................................................................62
1. Bi tp 1:.................................................................................................................62
2. Bi tp 2:.................................................................................................................65
BI TP ........................................................................................................................66
TI LIU THAM KHO..............................................................................................67
PHN II: TIN HC NG DNG TRONG HA HC ..................................................68Chng 1: PHN TCH DLIU BNG MICROSOFT EXCEL .................................68
I. CNG C PHN TCH DLIU TRONG EXCEL. .............................................68
II. NG DNG PHN TCH DLIU. .....................................................................70
1. Loi gi tr bt thng (aberrant observation): ......................................................70
2. Thng k m t:......................................................................................................71
3. So snh phng sai:................................................................................................74
4. So snh gi tr trung bnh vi hai phng sai ng nht:.......................................76
5. Phn tch phng sai mt yu t: ...........................................................................796. Hi quy tuyn tnh n gin:..................................................................................82
7. Hi quy tuyn tnh a tham s: ..............................................................................85
BI TP ........................................................................................................................88
TI LIU THAM KHO..............................................................................................88
Chng 2: CHNG TRNH MS EQUATION ..............................................................89
I. CA SNG DNG. ..............................................................................................89
1. Cch mca s: .....................................................................................................892. c im ca ca s:..............................................................................................90
3. Cch ng ca s: ..................................................................................................90
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II. THANH MENU. .......................................................................................................90
1. Menu File: ..............................................................................................................90
2. Menu Edit: ..............................................................................................................90
3. Menu View: ............................................................................................................91
4. Menu Format: .........................................................................................................915. Menu Style: ............................................................................................................91
6. Menu Size:..............................................................................................................92
7. Menu Help: .............................................................................................................92
III. TNH NNG K THUT. .....................................................................................93
1. Thanh k hiu: ........................................................................................................93
2. Thanh khung mu: ..................................................................................................94
IV. BI TP NG DNG............................................................................................95
1. Bi tp 1:.................................................................................................................952. Bi tp 2:.................................................................................................................96
3. Bii tp 3: ...............................................................................................................96
4. Bi tp 4:.................................................................................................................96
5. Bi tp 5:.................................................................................................................96
TI LIU THAM KHO..............................................................................................97
Chng 3: CHNG TRNH CHEMWIN ......................................................................98
A. CHNG TRNH CHEMWIN 3.............................................................................98I. CA SNG DNG............................................................................................98
II. THANH MENU.....................................................................................................99
III. TNH N NG K THUT................................................................................104
B. CHNG TRNH CHEMWIN 6...........................................................................107
I. CA SNG DNG..........................................................................................107
II. THANH MENU...................................................................................................108
III. CC THANH CNG C. .................................................................................109
IV. CCH M THVIN V NP TRANG MU. ...........................................111V. BI TP NG DNG. ......................................................................................112
BI TP...................................................................................................................115
TI LIU THAM KHO............................................................................................116
Chng 4: CHNG TRNH CHEMOFFICE ..............................................................117
A. CHNG TRNH CHEMDRAW..........................................................................117
I. CA SNG DNG..........................................................................................117
II. THANH MENU...................................................................................................118III. BI TP NG DNG. .....................................................................................121
B. CHNG TRNH CHEM3D.....................................................................................130
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I. CA SNG DNG: ............................................................................................130
II. THANH MENU: .....................................................................................................131
III. THANH CNG C. ..............................................................................................134
III. TNH NNG K THUT: ...................................................................................136
IV. BI TP P DNG .............................................................................................137BI TP...................................................................................................................141
TI LIU THAM KHO........................................................................................141
Chng 5: CHNG TRNH MICROSOFT POWERPOINT 2003.............................142
I. CA SNG DNG. ............................................................................................143
II. THANH MENU. .....................................................................................................143
1. Menu File: ............................................................................................................143
2. Menu Edit: ............................................................................................................144
3. Menu View: ..........................................................................................................1444. Menu Insert:..........................................................................................................145
5. Menu Format: .......................................................................................................145
6. Menu Tools:..........................................................................................................145
7. Menu Slide Show: ................................................................................................146
III. XY DNG CC SLIDE.....................................................................................148
1. Qun l cc slide: .................................................................................................148
2. a thng tin ln slide: ........................................................................................1493. nh dng tng th cc slide: ...............................................................................151
IV. SDNG CC HIU NG NG. ..................................................................155
1. p dng cho cc thnh phn ca mt trang slide (dng Custom Animation): ....155
V. K THUT TRNH DIN.....................................................................................159
1. Cch bt u v kt thc trnh din: .....................................................................159
2. Bt u cc hiu ng v chuyn slide, quay li hiu ng trc:..........................159
3. Cc hot ng khc khi trnh din:.......................................................................160
VI. BI TP NG DNG..........................................................................................1601. Bi tp 1:...............................................................................................................160
2. Bi tp 2:...............................................................................................................163
BI TP ......................................................................................................................164
TI LIU THAM KHO............................................................................................164
Chng 6: CHNG TRNH MACROMEDIA FLASH (FLASH)..............................165
I. CA SNG DNG V MT S KHI NIM C BN................................165
1. Ca s chng trnh: ............................................................................................1652. Cc khi nim cbn: ..........................................................................................166
II. THANH MENU. .....................................................................................................166
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1. Menu File : ...........................................................................................................166
2. Menu Edit : ...........................................................................................................167
3. Menu View : .........................................................................................................167
4. Menu Insert:..........................................................................................................167
5. Menu Modify:.......................................................................................................1686. Menu Text: ...........................................................................................................171
7. Menu Control: ......................................................................................................171
8. Menu Window:.....................................................................................................171
III. THANH CNG C (TOOLS). .............................................................................173
IV. BI TP NG DNG..........................................................................................175
1. Bi tp 1:...............................................................................................................175
2. Bi tp 2:...............................................................................................................180
3. Bi tp 3:...............................................................................................................1834. Bi tp 4:...............................................................................................................187
5. Bi tp 5:...............................................................................................................196
6. Bi tp 6:...............................................................................................................197
7. Bi tp 7:...............................................................................................................198
8. Bi tp 8:...............................................................................................................199
9. Bi tp 9:...............................................................................................................200
BI TP ......................................................................................................................201TI LIU THAM KHO............................................................................................202
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PHN I: THNG K HA HC
Chng 1: I CNG V THNG K
I. SAI S NGU NHIN V SAI S H THNG.
1. Cc khi nim thng dng:
Trong thc nghim ha hc khi o i lng X nhiu ln lp li cng cc iu kinging nhau, thu c mt dy cc gi tr xi vi i = 1, 2, ..., n.
Mi gi tr xi gi l mt yu t ca tp h p, n l dung lng ca tp hp(observations).
K hiu tp hp {xi}
a) Tp hp mu (samples)
- Nu n hu hn, dy xi to thnh mt tp hp mu
b) Tp hp tng qut (populations)
- Nu n , tp hp mu trthnh tp hp tng qut .
Vy mt tp hp tng qut cha ng v s yu t v v s tp hp mu. Mt khc,khi c 2 tp hp mu no , chng c th thuc v cng mt tp hp tng qut hocthuc v hai tp hp tng qut khc nhau.
c) Gi tr trung bnh (mean, average)
Vi tp hp mu:
n
x i= (trung tm phx n b)
Vi tp hp tng qut:
=x (tr sng, k vng)
d) Ph n, variance)
-Ph sai mu:
ng sai (dispersio
ng
f
d 2i
f: bc t do ca phng sai
-Phng sai tng qu
1n
)xx(S
2i2 ==
di: lch ngu nhin
t
n
)x( 2i2 =
e) on)lch chun (standard deviati
- lch chun mu : S
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- l chun tng qut:
tandard erro of the mean)
ch
- lch chun tngi (s
n
SS = x
f) Khong bin ng R (range)R = xmax-xmin
- H s bin ng CV (Coefficient of variation): 100x
SCV =
2. Sai sngu nhin:
Sai s ngu nhin pht sinh do hng lot nguyn nhn khng kim sot c vlun
u nhin
i hon ton ngu nhin. Khi n tng th s du (+) cng xpx s
sngu nhin. unhin
s ngu nhin. N biu tho cng c ngha l lp li ca php o. N thay i ngu
nhin
tc) Tr
hp l mt yu t no ca tp hp y m tt cccy i tp hp u tn ti mt trung tm phn b.. Tphp {
lun c mt trong bt c php o no
a) lch ng lch ngu nhin di c cc tnh cht sau :
- Du (-) hay (+) thaydu (-).
- Gi tr tuyt i |di| cng thay i hon ton ngu nhin nhng gi tr cng nh sc tn s xut hin cng ln, ngc li gi tr cng ln s c tn s xut hin cng nh.
= 0d i - Tng i s
Nhng tnh cht trn cho thy lch ngu nhin di
l du hiu tn ti ca saiTuy nhin, mt gi tr d ring l khng th coi l i din cho sai s ngi
. i din cho sai s ngu nhin phi l ton b tp hp {di}.
b) phn tn
- Phng sai : l i din cho sai s ngu nhin (khng cng th nguyn vi xi)
- lch chun (mu hoc tng qut) l thc o ca sai phn tn ca kt qu
ty thuc phng php o lng, iu kin o lng, ln ca i lng o vvo c nhn ngi o lng. Chnh v th m lch chun l mt thng s thng k
quan trng c s dng rng ri rong nhiu ngnh khoa hc.ung tm phn b:
Trung tm phn b ca mt tpu t khc quy t xung quanh. Mxi} c trung tm phn b l x
t i lng ngu nhin X c biu din bng hai thng s :Tm li, m
- x : biu th trung tm phn b
- S: biu th phn tn
Ch :
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- S c dng biu dinsai sngu nhin ca php o
ng c th gim thiu ti mc ty mun
thng:
a) P
Th d : Cc qu cn chun, dung dch m pH chun dng cho my o pH.
ia gi tro c so vi gi trng cai l
- Khng th loi bc sai s ngu nhin nhbng cch tng ln s ln o n mt cch tng ng.
3. Sai sh
hn bit sai s h thng v sai s ngu nhin.Gi s x l gi trng ca i lng X, gi tr ny cn c theo mu chun hoc
chtchun.
Sai s h thng ca php o l hiu s gng o.
= x x
Sai s h thng c cc tnh cht sau :
Sai s thng c xem xt khi | | > S
ring l:
n hng nh, vvy t
- C du hngnh :
- Khi < 0 : gi l sai s tha.
- Khi > 0 : gi l sai s thiu.
- C ln || cng hngnh cho mii lngo.
h
Php o coi nh khng mc sai s h thng khi | | < S.
- l tngi sca nhng sai shthng
= i
Mi i pht sinh t ngun sai s ring, mi ngun c du v lng i s cng c du v l n hng nh.
- Sai s h thng tng ix
biu thng (accuracy).
- Sai s ngu nhin tng i
x
b) Phn bit
Sbi u th chnh xc (prescision).
ng v chnh xc :
cao khi- Mt php o c ng x cng gn x
o nhng gitr xi
- Mt php o c chnh xc cao khi s ln o lp li in ht nhau chphn b st gn gi tr x . Tuy nhin khng phi c ng cao th nht thit c
chnh xc cao.
Phn bit 4 trng hp :
o c chnh xc cao, nhng ng km : S nh v || > S.
+ Php o c chnh xc km, nhng ng cao : S ln v || < S.
+ Php
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+ Php o c chnh xc v ng u km : S ln v || > S.
Php o c chnh xc v ng cao : S nh v || < S.
khng hon ho ca nh ch to dng co lng hoc dngco
: Cc vch chia ca buret khng u nhau, qu cn b mi mn...
t cc tp cht trong ha cht em s dng phn tch hahc.
ch nn gi l sai s t l.
t phn trong dung dch lm thp kt quphn
n php loi b sai s h thng :
th php o phi gm hai giai on:
iai on 2 : Tin hnh o trn mu so snh.
+
c) Phn loi sai s h thng :
- Sai sdng c :
L sai s gy ra do sxung cp trong qu trnh s dng.
Th d
- Sai sha cht :
L sai s gy ra do c m
Th d : Lng nh SiO2 trong NaOH, lng nh Fe3+ trong HCl...
- Sai sc th:
L sai s thuc v nguyn l ca phng php phn tch.
Th d : Phng php phn tch th tch c hai sai s phng php quan trng :
- Sai s ch th.
- Sai s t l : gy ra do xc nh khng ng nng dung dch chun.
V vy nu cht phn tch c nng cng cao th phi tiu tn nhiu th tch dungdch chun, do s mc sai s h thng cng ln. Sai s ny t l vi hm lng cacht phn t
Trong phng php phn tch trng lng, c hai loi sai s tri chiu nhau :- Sai s thiu : gy ra do kt ta tan mtch.
- Sai s tha : gy ra do s cng kt ca kt qu lm cho tng kt qu phn tch.
d) Cc bi
- Nguyn l ly so theo hiu s.
Theo nguyn l ny, c c mt so ng
- Giai on 1 : Tin hnh o trn mu nghin cu.
- G
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Kt quo ly theo hiu s ca cc so thu c mi giai on.
php phn tch, tin hnh phn tch vi mu nghincu, n hnh vi mu trng l mu khng c mt chtnghi
php thm chun :
y mu so snhc c chtchun
- ng vi hm lng x1 ca mu, o c tn hiu phn tch l y1.
- ng vi hm lng x2 = x1 + a (thm vo), o c tn hiu phn tch l y2.
Mu so snh c la chn thch hp cn c theo ngun gc pht sinh sai s hthng.
* Th nghim trng :
loi tr sai s ha cht trongthu c kt qu x . Sau ti1n cu nhng c thc hn trong cng iu kin vi mu nghin cu, thu c kt
qu x2. Hm lng cht em phn tch c tnh : x = x1 - x2
* Phng
Cn gi l phng php thm. Khc vi th nghim trng, ch to bng cch ly mu nghin cu v cho thm mt lng chnh x. Vy :
Nu gia tn hiu phn tch y v hm lng x c quan h tuyn tnh th :
x1 y - y2 1
Phng php thm c s dng rng ri khi phn tch cc hm lng vt nhm
=y1
lo b sai s h thng gy ra bi thnh phn th 3 m nhiu khi khng bit r.
iu kin p dng thnh cng phng php thm l quan h gia x v y phi
i b sai s ha cht ln y1.
thng v sai sngu nhin:
a cc so gin tip. Bnng v sai s ngu nhin dn n cc thut ton lan truyn
II. H1. C
hin lin tc :
c th ca X lp y mt hay mt khong ca trc s, hoc lpy t
{X = a} = 0.
i
tuyn tnh v ngoi ra cn phi lm th nghim trng lo
4. Lan truyn sai sh
Sai s ca so trc tip c lan truyn sang sai s ccht khc nhau ca sai s h thsai s cng khc nhau.
M PHN B (DISTRIBUTION FUNCTION)c khi nim cbn:
a) i lng ngu n
Mt LNN (i lng ngu nhin )X c gi l LNN lin tc nu:
- Tp hp cc gi tran b trc s.
- Xc sut X nhn mt gi tr c th no lun lun bng khng, ngha l vimi s a : P
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Nh vy i vi LNN lin tc, xc sut n nhn gi tr trong mt khong no r
b) Hm
Hm nh trn ton b trc sc gi l hm mt ca LNN lin tcX nu :
P{a < X < b }
P{a < X < b } l din tch hnh thang cong gii hn bi th hm s y = (x) v 2
t c quan tm. Xc sut ny c quyt nh bi mt hm gi l hm mt xcsutca X
mt xc sut :
(x) xc
(x) 0 vi mi x
= 1dx)x( +
Vi mi a < b
=b
a
dx)x(
ng thng x = a v x = b
x
y
a b
2. Hm phn bchun (Normal distribution function):
a) Hm Gauss
Hm Gauss (x) (t tp hp tng qut) vi bin s x v cc thng s, :2-x1
2e.1
)x(
=2.
Hm (x) mang y mi tnh cht ca mt hm mt xc sut.
th:
th(x) theo x c dng i xng hnh chung.
* Cc i : 0dx
)x(d = khi x = .
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ng (x) c cc i :
=
= 0,399/2.
1)x(
0dx
)x(d 2=
* im un : khi x = .
ng (x) c hai im un i xng qua trc thng ng x = v cch trc .
Ti cc im un :
( + ) = ( - ) = 0,242/
Bng 1. Cc gi trng lu ca hm phn b chun
x
(x) (x)
x
x
(x) (x)
x
x
(x) (x)
x
x
(x) (x)
x
x
(x) (x)
x
x
(x) (x)
x
x
(x) (x)
x
x
(x) (x)
x
x
(x) (x)
x
x
(x) (x)
x
x-
(x) (x)
x -2-3
php gii tch Ton hc, tch phn xc nh dx)x(f c gi tr bng din tch S
bao hm gi f(x) l mt
hm mt xc sut, ngha l khi f(x) = (x) th tch phn f
tin cy cc gi ng l x ca tp hp {x} ri vo khong (a , b). Vy din tch S
x (x)
2
0,399/
0,242/
0,054/ 3 0,0044/
-2-3 - 2 23 3
b
Ta
a ng f(x), trc x v hai ng thng ng x = a v x = b. Khi
b
a
dx)x( = P biu th xc sut
cho tr ri
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c gi g bng xc sut. Mi quan h gia din tch S v P mi hmmt t , trong n b n.
M y P phi lun lun gn lin vi khong y (a , b)l kho cy c sut tin cy
Khi (a , b) n nh (- , + ) th xc sut P = 1 : s ki ng l xnm trong khong ) l mt s kin chc chn xy ra, xc n ny
phi = 1.
Phn bit hai loi khong tin cy : khong i xng v khong bt i xng.
- Khi a i xng vi b qua im x = th (a , b) l khong i xng.
khong bt i xng.
trn ny ng choxc su c hm ph chu
t khc, xc sung tin
t tin cng vi x
(a , b).VP.
i rng th n gi tr ri(- , + sut ca s ki
- Khi khng tha iu kin trn (th du a, b ng cng mt pha so vi hoc a, bkhng cch u ( t hai pha th (a , b) l
Bng 2. Mt s khong tin cy v xc sut tin cy ng lu trn ng phn b chun
Khong tin cy
x = a x = bP =
b
dx)x( Loi khong tin cya
i xng
i xng
-
- 2
+ 0,682
0,954
- 3 + 3
+ 2
0,997
-
-
+ 2
+ 20,977
0,9540,5
0,8142
0,954
2
682,0
=+
=+
bt i xng
bt i xng
2
i xng
Nh
Th d : P = 0,682 c ngha l c 1000 gi tr ring l x trong tp hp {x} th c 682gi tr x nm trong khong (- ; + )
n xt :* Bt lun l bao nhiu, din tch S bao hm gia ng (x) v ton b trc x c
gi tr = 1; ngha l P = 1.
* ng phn b chun c nh cng cao khi cng nh (. l thc o ca phn
chun ca hai i lng sai s ngu nhin c coi l trng nhaukhi chng c v . ng phn b chun s khc nhau khi hai thng s
ny k
tn). Khi cng nh th chnh xc cng cao, cc gi tr x ring l cng tp trungli xung quanh trung tm phn b.
* ng phn b cng thng s
hc nhau.Quy tc 3 (ba xch ma) :
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T bng 2, khong (a , b) vi a = - 3 v b = + 3ng vi xc sut P rt ln,= 0,9 cho khong ny rt nh, bng 1 -0,997 n m ngoi khong (a , b) ny rthim
vi ln m gp
mt gi tr ring l x* * c th l mt gi tr bt thng cnc xt xem c loi l ni dung ca quy
Quy tc 3 c th huyn thnh quy tc 2, 4... ty thuc vo xc sut c chn.Khi dng quy tc 3, ch c 2 th xcsut cc gi tr b loi
Cch p dng quy
quy tc ny l phi bit trc ca php o.
97.Vy xc sut gi tr ring l x i ra ngoi= 0,003 (tc l 3 ph nghn). Nhng gi tr ring n
gp.
Vy vi mt php o bit trc , nu ch mi o lp li c
> + 3 hoc x* < - 3 , xb ra khi cc gi tr ring l khc khng.
tc 3.
cp nhn 0,3% cc gi tr b loi b ; khi dng quy t
b cao hn, = 1 - 0,954 = 0,046, tc l 4,6%.
tc 3 trong thc hnh :Mc ch ca quy tc ny l loi b cc so c gi tr bt thng. iu kin p
dngCch tin hnh :
Gi s nghi nggi tr x* trong tp hp mu {x} dung lng n. Tin hnh loi bx* v dung lng cn li l n - 1. Tnh 1nx v coi 1nx = .
- Nu tm thy |x* - 1nx | > 3 loi b x* .
- Nu tm thy |x* - 1nx | < 3 khng loi b x*.
Vy s loi b hay chp nhn x* rt ph thuc vo xc sut P.
Th d : Mt php o hm lng nguyn t X cho cc gi tr sau :
3,45; 3,48; 3,47; 3,57* (%)
C loi b gi tr x* khng, nu theo quy tc 3 v 2 ? ( php o c = 0,04%)
3,473,46754
3,473,473,483,45x 1n =
+++=
|3,57* - 3,47| = 0,10 < 3.0,04 = 0,12 (quy tc 3)
|3,57* - 3,47| 0 2.0,04 = 0,08 (quy tc 2)
b.
R nhikhc a c
= 0,1 >
Theo quy tc 3 khng nn loi gi tr 3,57; nu theo quy tc 2 th c th loi
b) Hm Gauss chun ha
t nhiu i lng ngu nhin gp trong t n tun theo hm phn b Gauss. Snhau gia chng th hin s khc nhau c c thng s v . Tuy nhin, khi p
dng hm Gauss trong thc t, xc sut P cng vi khong (a , b) no rt c ch . tin cho vic tnh ton P, tp hp {x} c bin i thnh tp hp {u} :
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.dudx-x
u =
=
du.e.2
2= 1
22u.
1
du..e.
1dx.e.
1(x)dx
2u.2
1-x
2
1
== 2
2
t :2u
11 2e.2
)u(
=
(x)dx = (u)du.
==b
a
)b(u
u(a)
(u)du(x)dxP vi
-a
=
=
-b)b(u
)a(u
Bin ngu nhin x t l tuyn tnh vi bin ngu nhin u; nhng khc u ch l x li lng c th nguyn ca i lng o v cn ph thuc cc thng s v , trongkhi u khng c hai tnh cht trn.
Nu lch d = x - c th nguyn th
=d
u khng th nguyn ( lch rt gn)
Hm c thng s = 0 v ng t nh hm Gauss vtrn v thay = 0 v =
1.ng cch tra bng tch phn
:
u nhin Z , k hiu Z (x) = P{Z < x})
P{Z Z } = (Z) = P{Z < Z } = 1-
= 1- P : Mc ngha hay xc sut ngvc
Xc sut tin cy mt pha (one tail)
X ti y hai pha (two tail) i xng (Px) hoc bt i xng (
(u) gi l hm Gauss chun ha, y l mt hm Gauss c bit khi c = 1. th biu din t
Xc sut P theo khong (a , b) c tnh d dng bLaplace .
- ng dng ca hm phn b chun:
Cc khi nim
im phn v ca i lng ng
(Hm phn b
>
P = 1- : Xc sut tin cy
P )c sut n c
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Z/2 Z1-/2Z
P = 1- P = 1-
ng dng 1: Tnh gii hn tin cy (GHTC, confidence limits) v khong tin cy
(KTC, confidence level) vi xc sut P cho trc :
Khi bit xc sut Px, tra bng tm gi tr uP (Bng tch phn Laplace).
* i vi gi tr ring l x :
T
=
xu gii hn tin cy ca ng vi xc sut P :
GHTC() = x uP.Khong tin cy ca xung quanh x ng vi xc sut P l :
KTC(x)
Gi tr u P.
* Vi gi tr
= uP.ty thuc vo xc sut
x :
V n.x
un
x
==
GHTC c ut P l :
a ng vi xc s
GHTC()n
.ux P
=
x )n
.u P
= KTC(
Khong (x - uP. ; x + uP.) rng hn khong (n
ux;
n
.u-x PP
+
) nn c lng
theo x c hiu qu hn theo x.3. Hm phn bmu:
a) Hm phn b Student:
Hm phn b chun thch hp cho tp hp tng qut {x} vi dung lng n rt ln ( n> 30). Tp hp mu {x} vi dung lng nh (n 2) tun theo hm phn b Student. HmStudent c vai tr thay th hm phn b chun khi n nh v trc ht c s dng lng . Tng t hm (u), hm Student c cho dng hm mt xc sut(
ct) vi bin ngu nhin t thay cho u.
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+
+f
1.
+
=
2
1f
f
1f
..f
)t(
f : s bc
221 t
2
vi : - < t < +
t do = n -1
S
xt
= hoc n.
xt
=
S
Bin ngu nhin t c gi l lch rt gn mu
=
t1-x
dte.t)x( (hm Gamma)
ng vi mi f m ng.
(t) l m t vi mi gi tr ca f
0
t hm (t) tng
t hm mt xc su
P = 1-
0 /2t/- t 2
H o xc sut Px bngnhn 5 ; 0,99
tp,f : Student (tra bng h s Student phn ph
ng ca hm phn b Studen
ng dng 1 :Tnh gii hn tin cy
x tp,f.S
i vi gi nh
m phn b Student i xng , vi t trong khong (-t, +t ) sao chg gi tr thng dng : 0,90 ; 0,9
h s lc)
dng t
i vi gi tr ring l x :
GHTC() =
tr trung b x :
GHTC() = x nS.tf,p
Th d : Php xc nh Ni trong thp cho kt qu :
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x = 1,76% vi S = 0,08%
Tnh GHTC( 95.
Gii :
) xung quanh gi tr trung bnh ng vi Px = 0,
Khi Px = 0,95; f = 5 - 1 = 4 t0,95;4 = 2,78
Ta c :
GHTC() = 1,76 4
.78,2 = (1,76 0,11) %08,0
ng dng 2: Tnh P ng vi KTC cho trc v f cho trc :
Php o pH sau 6 ln o cho kt qu :
Biu din kt quy :
% Ni = (1,76 0,11) % ng vi n = 5; P = 0,95.
Th d :
x = 2,87 vi S = 0,019
Tnh P cho KTC( x ) = 0,03 d( ng bng h s Student y ).
Gii :
KTC( x ) =n
t f,p = 0,03S
.
|tp,f| =
S
n. 0,03 =
019,0
6. 0,03 = 3,78
5.
tp,5 2,57 3,37 4,03 4,77
Tra ngc bng h s Student tnh P ng vi f = 6 - 1 =
T bng h s Student, ta c :
Px 0,95 0,98 0,99 0,995
t 3,37 < 3,87 < 4,03
0,98 < ? < 0,99
P = 0,98 +3,37)-03,4(
3,37)-0,98)(3,87-99,0(# 0,988
0,988 v n = 6.
t mt gi tr CV cho trchoc khong tin cy
Biu din kt qu :
pH = 2,87 0,03 ng vi P =
ng dng 3: Tnh s ln th nghim song song x cho trc :
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(Dng bng h s Student y )
Th d : Php xc nh C (3 ln) trong mt cht hu cmi tng hp cho kt qu x =44,3%
c ca php o cha thit lp cng thc ha hc v cntng
vi S = 0,4%.
Tuy nhin chnh xs ln th nghim song song n sao cho KTC ( x ) 0,25% ng vi P = 0,95. Hy tm
n.
Gii :
T cng thc :
x ) = n
S.t f,p KTC(
x
S
t
n=
iu kin : KTC( ) 0,25%x
0,25
S
t
n
.
Ngi ta chp nhn Sn # S3 = 0,4%, do :
V ch bit S (n =3) nn php tnh n y ch l gn ng
1,60,2525,0t
f,p
Tm cp gi tr n, t
0,4
S
n n =
p,fbng h s Student :
11 12 13n
t0,95;f 2,20 2,18 2,16
f;95,0t
n
1,51 1,59 1,67
f,Vi n = 13 th pt
n
= 1,67.
Vy n 13.
Vy mun nng cao chnh xc u phi tr gi : tng t 3 ln 13 ln. V th ccdng c c cp chnh xc cao thng rt t tin.
ng dng 4: Loi b so c gi tr bt thng :
ax). Ta tnhGi s nghi ngx* trong dy o lp li n ln (x* c th l xmin hoc xmn-1 Sn-1 (v loi b x* khi tnh ton). Nu tm thy :
|x* -vx
x n-1| > 4.Sn-1
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th c th loi b x*.
l quy tc Graf - Henning c p dng cho 4 < n < 1000.
b) H
. Hm phn b2 cho php
c l ng t S khi n nh
m phn b2Hm phn b Gauss v Student cho php c lng
2
2
2
22 S)1n( ==
Sf
Khong bin thin : 0 +
u (t) ch bin s ngu nhin 2 tn ti
(0 , + ).(2) c y tnh cht ca mt hm mt xc sut :
2
( ) . ( )2 2=1
Vy hm mu ( ) khc vi h
trong khong
2 m m
.2
2 2ef
.2
2 f / 2 f
2
2( )
P = 1- = 1- P
2
2
2 2 2
/2/2 1-
Hm phn b(2) , ni chung l bt i xng, nhng bt i xng s cnggim khi f tng ln
ng dng:
- Tnh GHTC ca t S ng vi xc sut P i xng hoc bt i xng
- Kim nh mt gi tr cho trc no c cn l lch chu ng qut cho Shay khng (s cp trong chun 2 )
c)
c ccph ut ths hng sai ny phi mang tnh cht ngu nhin.
sai khc ngu nhin ny theo t s F v bin ngu nhinmi:
( )
n t
Hm phn b Fisher (F)
Gi s c hai tp hp mu {x1} c dung lng nI v {x2} c dung lng nII,ng sai mu 2S v 2S . Nu hai tp mu ny thuc v cng mt tp hp tng qIsai khc gia 2 p
II
Fisher ngh biu th s
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2IIS
2ISF = vi khong bin thi n : 0 F +
Fisher tm ra hm phn b ((F), mt hm phn b mu c dng sau y :
ong : fI = nI - 1, fII = nII - 1.
t :
0
ng vi khong (F(a) , F(b))
Xc sut mt pha :
Hm phn b Fisher l mt cng c hu hiu so snh cc loi phng sai rt
ln dng ng cong cng i
Tr
(F) c y tnh cht ca mt hm mt xc su
+
= 1dF)F(
- Xc sut hai pha :
-
ng vi khong (0 , F(b))
hay gp trong thc nghim ha hc.
Dng ng biu din ca hm F (Nu fI , fII cngxng)
0,8
0,
0,4
0,2
6
1 2 3 4
10 ; 50( )
10 4( );
(F) If = fII =
(F) If = fII =
( )
.
( )/F
f
fFI
II
f fI II=
f
2
f
2+ 1I II
+ 2
) .( / )
f FI
II
fI
(f + f I II-1
2
/f
fII 2
P F dFF a
F b
= ( )( )
( )
P F dFF b
= ( )( )
0
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ng dng: Chun thng k F :
xem c s khc bit h thng hay ngu nhin :
ng sai nh k hiu , fII.
So snh hai phng sai mu
Cch tin hnh:
- Phng sai ln k hiu 2IS , fI.
- Ph 2IIS
Tnh2
2I
tn
SF = v so snh vi Flt =
III f,f,PF
SII
- Nu Ftn < Flt : S khc bit gia hai phng sai mang tnh ngu nhin (khngngk).
ng (ng k).
so snh tay ngh gia hai k thut vin A v B, ngi ta ly mt muphn hso
S
- Nu Ftn > Flt : S sai khc gia hai phng sai mang tnh h th
Cch kim nh thng k ny gi l kim nh theo chun F.
Th d : tch ng nht ri phn chia thnh nhiu mu mang s hi u khc nhau ln vo
ng lot mu phn tch khc (mc ch l khng bit c l mu th nghim songng).
Kt qu phn tch c x l thng k tnh ra S :
KTV A :A
S = S5 = 0,4%
KTV B : BS = S6 = 0,9%
o snh tay ngh ca A v B, chn P = 0,95.
Gii :
Tra bng tm Flt = F0,95;5;4 = 6,26
V Ftn < Flt nn c th kt lun l tay ngh ca cc k thut vin l tng ngnhau. Kt lun ny c ngvc (mc ngha ) = 0,5%.
N (TEST) THNG K.
1. Khi qut vphng php kim nh thng k:
a) Gi thit thng k:
mt cch khch quan cckt q
F =0,9
= 5,062
tn 0,4 2
III. CC CHU
Cc phng php kim nh thng k cho php gii thchu th nghim. Th d, c hai kt qu trung bnh Ix v IIx ca hai k thut vin khi
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phn tch cng mt mu ng nht. Mun bit s sai khc gia Ix v IIx mang bn chtngu nhin hay h thng, cn phi dng phng php kim nh thng k.
Nu cho rng Ix v IIg phi mang bn cht ngu nhin. Mt gi thit thng k nh vy c gi l gi
thit H
x thuc v cng mt tp hp tng qut th s sai khc cachn
0 (Null Hypothesis). Ngc li, nu cho rng Ix v IIx khng thuc cng mt tp
hp t g ph i mang bn cht h thng. Gi thit ny cgi l H a l bc b H1 v ngc li.
b) M
ngha), k hiu l ty thuc vo s dng xc sut hai pha(two tail) hay mt pha (one tail).
te :
nh thng k. cn phi dng cc chun thng k u tin chn mc ngha thc
ng qut th s sai khc gia chn1.(Alternative Hypthesis) Nu chp nhn H0 c ngh
c ngha :S chp nhn hay bc b mt gi thit thng k bao gicng phi gn vi mt xc
sut tin cy xc nh v gn lin vi mt xc sut ngvc nht nh ( trong kim nhthng k cn gi l mc
c) Chun thng k Z(Z st)
kim h hp, sau phi chn mt bin ngu nhin Z thch hp cho bi ton thng k.
Bin ngu nhin Z c hm mt (Z) v c sn cc im phn vP
Z hay ZP ghi bngthng k.
Th d : Z c th l bin ngu nhin hi t nh u, t, 2, F... Chn bin no th chunthng .
Ngo n c theo xc sut mt pha hay hai pha th gitng n thng k mt pha hay hai pha.
Th d ai pha, chun F mt pha...
i tr Z tra bng thng k gi l gi tr l thuyt, k hiu Zlt.
ng k mt pha, ch cn tra mt trong hai gi tr Zlt, ly Zlt(a)hoc l lt
- Khi dng chun th ng k hai pha, cn tra hai gi tr Zlt : Zlt(a) v Zlt(b) nu Zlt l
k mang tn bin y : chun u, chun t, chun F..
i ra, nu chun thng k cng l chu
: Chun t h
G
- Khi dng chun thy Z (b).
PZ . Khi : Zlt(a) = Z v Zlt(b) = 1Z .
Zx th ch cn tra mt gi tr Zlt l .
Gi tr Z tnh c t s liu thc nghim (rt ra t tp hp mu {x}) gi l gi trthc .
i th t H0 p nhn khi Ztn < ZP hoc Ztn nm trongkhong (Z (a), Z (b))
Gi c chp nhn khi Ztn > Zlt(a) hoc Ztn < Zlt(b).
Nu cc iu kin H0 khng tha mn, c ngha l chp nhn H1.
Tuy nhin, nu Zlt l
nghim v k hiu Ztn
Sau , so snh Zlt vi Ztn, v kt lun :
G i theo chun hai pha c chlt lt
thit H theo chun mt pha0
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/2Z- /2ZZZ-
Bc b H0Chp nhn H0
ng k:
thit H0 khi gi thit ny ng mc ngha ca kim nh , ngha l tin cy ca kim nh l (1-). Th d : =5% c nh sai lm ca kim nh ny 5%, v vy tin cy l 95%.
Erro): Ngc li vi sai lm loi I, Sai lm loi II l loisai l khi gi thit ny sai mc ngha no .
tc :
b H0 h chn = 0,01, tc l P = 0,99.
Khi chp nhn c l P = 0,95.
* Khi nm gia Zlt;0,99 v Zlt;0,95 th cn thn, tt hn ht l lm thm th nghimb su
Cc loi sai lm trong trong kim nh gi thit th
- Sai lm loi 1 (Type I Erro): Bc b gi no ngha l gi
- Sai lm loi II (Type IIm ca vic chp nhn gi thit H0
Cn phi tun th nguyn
* Khi bc t
* H0 th chn = 0,05, t
ng ri hy k t lun.
2. Chun Dixon (Zlt= n,PQ )
a) M
ng trong mt tp hp mudung
b) C
- Sp xp cc so theo trnh t t nhn ln :
x1 < x2 < ... < xn
R
- Nu nghi ng x1 :
c ch :
Chun Dixon dng loi b so c gi tr bt thlng 3 n 8.
ch thc hin :
- Tnh R :
= |x1 - xn|
RQ tn =
- Nu ngh
x-x 2*1
i ngx :n
Chp nhn H0 Chp nhn H0Bc b H0 Bc b H0 Bc b H0
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R
- Gi tr Q
x-xQ
1-n*n
tn =
lt tra bng n,PQ .
Gi thit thng k : H0 : khng nn loi b x1 hay xn.
H
+ Nu Qtn < Qlt : Ch p nhn H0
H1
Bng cc im phn v
1: loi b x1 hay xn.
+ Nu Qtn > Qlt : Chp nh n
n,PQ
n P = 0,90 P = 0,95 P = 0,99
3
8
0,89
0,40
0,94
0,77
4
0,56
0,51
0,48
0,99
0,89
0,76
0,70
0,64
0,58
4 0,68
5 0,56 0,6
6 0,48
7 0,43
Th d : C 4 so : 8,26 8,28 8,29 v 8,42.
C nn loi b so 8,42 hay khng ?
Gii :
t gi thit thng k
H : khng loi b so 8,42
1
Tn
R 6 - 8,42| = 0,1
0
H : Loi b so 8,42
h:
= |8,2 6
0,8116,0
8,29-42Q tn ==
Nu c
8,
hn P = 0,95 ; Q0,95;4 = 0,77
Qtn > lt : bc b t H0, c th b so 8,42 ng theo qui tc trn,khi bc b n
Q gi thi loi . NhH0 nn ch P = 0,99. Khi , Q 0,89 Q lt .
khng nn lo tr 8,42 v Q Q < Q0,99.
Theo quy tc trn th nn lm thm th m b sung.Gi s m thm th im thu c s 8,32 :
0,99;4 = tn < Q
i b gi 0,95 ,n : chp nh
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M vun loi b so tip theo th cn tnh li tn vi Sn-1 1nx , sau so snh
vi . n-1.
m phn v
P = 0,95 P = 0,99
p
Bng cc i p,n
n P = 0,90
3
4 1,65 1,69 1,72
5
6
1,79
1,89
1,87
2,00
1,96
7 1,97 2,09 2,27
8
9
2,04
2,10
2,17
2,24
2,37
2,
10
1,41
2,15
1,41
2,29
1,41
2,13
46
2,54
111 2,19 2,34 2,6
Nhn xt :
So snh v Q :
m nh ch
dng 3 gi tr x1, x2, x3 hoc x1, xn-1, xn, v vy khi n cng ln th chun Q cng trnnkhn
Bin tn dng ht tt c s liu ca tp hp mu nn chun c th thch hpcho d
Bin Q khng tn dng ht cc s liu ca tp hp mu, mi ln ki
g thch h p.
ung lng n nh v ln.
Th d 1 : Ly th d trong chun Q :
n = 4 S = 0,0774 x = 8,3125
tn =
414. 07274,0
8,3125-42,8= 1,706
> = 1,69 v
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Gi gi tr hm lng phi tm l xmax.
Gi x = 11,0 ppm.
n = + .S.
n
1n.S
xx max x = xtn
max tn1-n
Cho tn = 0,95;5 = 1,87 (tra bng)
5
1-5= 12,5 ppmxmax = 11,0 +1,87.0,9
Vy t lun l h cha bt u b nhim.khi xi > 12,5 ppm/l th c th k
Th d 3 : Hiu sut thu hi alcaloid t mt nguyn liu thc vt sau 5 ln xc nhl x = 85% vi S = S5 = 2 %. Trong mt ln thu hi khc c hiu sut x = 92%.
Phi chng c mt bin ng ng k v nguyn liu trong ln ny ? Cho P = 0,95.
tn =
5
4.2
= 3,9
0,95;5 =
lt
4. C
85-92
tn = 4,96
tn > c s bin ng ng k v nguyn liu.
c chun 2 :Chun 2, chun Bartlet ( Zlt = 2 f,p )
a. M
ng co lng, ca phng php phntch, i chnh xc quy nh (chun 2).
h t t dy phng sai mu rt ra t mt tp hp mu
tun theo u n Bartlet).
b.Ki ng) :
c ch :
Kim nh chnh xc thc t (ca dca tay ngh ngi phn tch) so v
Kim n nh ng nht ca m
nh l t phn b chun (chu
m nh chnh xc thc t (chun 2 thng th chnh xc quy nh l cho sn bi nh ch to dng c o lng hoc
phng php phn tch em s dng.. chnh xc thc t l S :
2 =tn 2
2Sf
Dng chun hai pha vi xc sut P v tra bng 2 tm gi tr 22
P1 v2
2
P1+
+Nu 2
2
P1 < 2
P1+ K2
t lu : t khng t yu cu.
Nu
n chnh xc thc
+ 2tn 0 13,3t lu : Chic cn ny ng hn n s i.
2P1 +
2
2
tn
2
4;99, =K n b xu cp c h xc, c a cha l
Gi s : Sau khi sa cha, S5 = 0,0003g.
2( 0002)
22tn 0,
4.(0,003)= = 9 < 13,3
K i phc.
c. Kim ai mu (chun 2 theo Bartlet)Gi s g sai mu nh s fj = 1, 2, ..., k vi fj = nj - 1.
sa ti hin (cn gi l phng sai mu c trng s k hiu: )
:
t lun : chnh xc cn c kh
nh tnh ng nht ca dy phng s
c k phn 2jS
+Tnh phng i 2thS2
k,nS
=2thS
j
2jj
f
S.f(fth=fj = nj-k)
2jj S.ff
t g fj.log =
2thth S.
B = 2,303(fth.lo 2thS -2thS )
C =
+ 111
1 thj
Theo Bartlet bin ngu nhin B i vi dy ph sai ng2
ff)1k(3
, /C ng nht s tun theonh lut vi bc s t do f = k - 1 nu tt c fj > 2.
Theo Bartlet :
C
B2tn =
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2lt : tra bng
2
f,
Pvi f = k - 1.
< : dy phng sai mu S thng nht. Ngha l cc phng sai
c dy phng saiu phng sai tng qut khc nhau.
g phng sai ng nht.
Chun Bartlet l mt cng c quan trng ca php phn tch phng sai.
:
+Nu 2 tn lt jmu 2jS cng mt phng sai tng qut.
2 2
+Nu2tn >
2lt : dy phng sai khng ng nht. Ngha l c
2jS
thuc v hai hoc nhi
* Bartlet khng cho bit trong bao m my nhm
Ch
V C lun un > 1, kl im nh nhanh :
u tin tnh B v so snh vi 2f,P
:
- Nu B < 2f,P
: khng cn tnh C.
- Nu B > 2 : tnh thm C, lm nh trn.f,P
C ong 4 mu thp khc nhau bng cch o th tch khu khc nhau. Hy kim nh tnh ng nht ca
mu, bin lun vnh hng ca cc thnh phn trong thp n chnhxc c
Th d : Khi xc nh % tr CO2, ta thu c cc lch chun mcc phng sai
a php xc nh % C.
jj
x (%) Sj (%) fj Loi thp
1
3
8
32
32
C pha thm 1,2% Si v 1,2% Cr
Loi thp khng pha thm
1,03 0,005 24 C pha 14% Cr
2 1,23 0,007
4
1,30
1,38
0,010
0,00
28 Loi thp Ferro mangan
Gii :
t Si = 1000Sj.(kt qu khng thay i)
i Si 2iS fj fj.2iS log
2iS fj.logS
2i
1
2
4
9
100 28
600
2.800
1,3979
2,0000
33,5496
56,0000
5 25 24
3
7
8
4 32 1.568 1,6802 54,0864
10 64 32 2.048 1,0062 57,7984
116 7016 201,4344
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60,487016
S2 ==116th
= 116 x 1,7816 = 206,6656
03(fth. lo f )
(206,6656 - 201,4344) = 12,0475
: B >
C =
log 2thS = 1,7816
2thSfth. log
B = 2,3 g 2thS - j. log2iS
= 2,303
2lt =
23;99,0 = 11,3
2lt So snh
Tnh thm :
1 thf)1k(3
+ j 1f11
+++
116
1
32
1
28
1
32
1
24
1
)14(3
1= 1,0146= 1 +
0146,1
0475,12 2Btn == = 11,8740 11,87
Kt l ng sai mu l khng ng
nht.Phng a
loi b v = = 5,99.
Vy cc phng sa n
Phng php xc nh % C trong mu thp Ferro mangan c chnh xc kmhn so vi i.
5. Chun
C
2iSun : V
2tn = 11,87 >
299,0 = 11,3 nn cc ph
on : C l tnh khng ng nht do S = 0,010 ln nht trong dy ny. T3tnh li 2tn . Kt qu thu c
2tn = 5,63 v
2lt
22;95,0
i c li l ng nht.
cc mu thp cn l
Fisher. (Zlt=III f,f,P
F )
a) Mc :m nh tnh ng nht ca hai phng sai mu v rt
u {xI} v {xII}.
ch .
ng sai em kim nh S >S .
chChun Fisher dng ki 2IS
2IIS
ra t hai tp hp m
iu kin : Cc tp hp ny tun theo nh lut phn b un
b) Cch thc hin :
Trong hai ph 2I II2
Ftn
=2
I
S
S
2
II
(lun lun ln hn 1)
So snh Ftn vi Flt =III f,f,P
F :
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Nu Ftn III f,f,P
F th hai phng sai khng ng nht.
Chun F l cng c quan trng ca php gii tch phng sai.
6. Ch G P,f,n)
a) M c ch :
Chun Cochran dng kim nh trong dy phng sai mu c cng dung
lng
j n nht c ng nht vi cc phng sai cn li khng.
b) Cch th
Gi s , k,
un Cochran . (Zlt=
2jS
n = n, phng sai l 2maxS
c hin :
c k phng sai mu 2S dung lng n bng nhau v nh s j = 1, 2,j
phng sai ln nht2 xmaS l
Tnh Gtn theo cng thc :
=tnG 2
jS
p,f,n vi f = k - 1.
So snh gi tr G i tr Glt :- Nu Gtn < Glt : sai bit khng ng k so vi cc phng sai cn li;dy
phng sa
ng sai cn li.
L v p vi th hai trong dy phng saicho n khi thu c dy phng sai ng nht.
2maxS
Tra bng gi tr Glt trong bng im phn v G
tn vi g2
xmaS
i 2jS l ng nht.
- Nu G > G : 2 c sai s h thng vi cc phxmaStn lt
oi 2 xmaS a xem xt v c th th ti2
xmaS
Lu :Khi th vi xmaS th2 hai , so snh Gtn 2 vi Gp,f,n , trong f = k -2.
Th d : Php xc nh % Cl- trong 4 mu khc nhau cho kt qu sau :
1) 11 1
2) 11,26 14,32 14,27
3) 18,60 62
m nh tnh ng nht ca cc phng sai trc khi tnh Sn,k).
ii :
,28 1,30 11,31
18,72 18,
4) 16,45 16,42 16,50
Hy tnh lch chun c trng s Sn,k ca php xc nh ny. Cho P = 0,95.(Lu : Cn phi ki
G
im nh tnh ng nht ca cc phng sai mu theo chun Cochran :K
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21S = 0,0002333 = 3,071033
22S
23S = 0,004133
2S = 0,0016334
2maS =
22S x
G =tn 2jS
2maxS =
077032,3071033,3 = 0,5877
> Gtn = 0,5877.
t.
3 phng sai cn li ng nht vi nhau
Glt = G0,95;3;3 = 0,7977
Cc phng sai mu l khng ng nh
Loi b 22S ra khi dy phng sai trn .Xem xt 3 phng sai cn li.
k,nf
2
jj2 k,n
S.fS = = 3900599.2 = 0,001966 vi fn,k= nj-k
443
7. Ch
a) Mc
trung bnh
Sn,k= 0,0
Sn,k 0,04%
un Student (t-Test):
ch :
- Kim nh s sai khc gia hai gi tr Ix v IIx trong iu kinv mang tnh ngu nhin hoc hthng
i hn tin cy - nh gi kt qu phn tch.
b) C
i
2IS
2IIS (sau khi kim nh bng chun F) Sai s
.
- Tnh ton gi
ch thc hin :
* Kim nh ha gi trtrung bnh :
Tnh ttn theo cng thc :
III
IIIIII
2IIIIII S)1n(S)1n( +
2
IIItn
xxt =nn
)2nn(nn.+
+
I = nII = n th :* Nu n
nxx
t IIItn
= SS 2II
2I +
- Tra tlt = tp trong b,f ng im phn v (vi f = nII + nII - 2)
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- So snh t lt :
Nu ttn < tlt : S sai khc gia hai gi tr trung bnh mang tnh ngu nhin.
N u t t mang tnh h thng.
Ch
tn v t
n > tlt : S sai khc gia hai gi tr trung bnh
: Nu c snh vi xgi tr xnn so no ng hn.
Th d : Hm lng % N tm thy trong cc mu phn tch bi hai nhm sn xutcho kt qu sau
bit gi tr
: Ix = 9,36 vi SI = 0,09 v IIx = 9,57 vi SII = 0,034 , nI = nII = 4.Hy so snh hai k
Gii :
t qu trung bnh ? (P = 0,95)
Kim h ng nht ginh tn a hai phng sai :
2tn 0,034F = = 7,0
20,09
Ftn < Flt : Hai phng sai ng nht.p d n t so snh hai gi tr trung bnh :
F = F = 9,28lt 0,95;3;3
ng chu
4.034,009,0
9,57-9,36t
2tn +=
2= 4,36
= t0,95;6 = t0,99;6 = 3,71
ung bnh sai khc rt ng k.
* Tn
tlt 2,45
Hai gi tr tr
h gii hn tin cy :
n.x
(vi f = n -1)tP,f=S
n
Stx f,P hay GHTC () =
n
Stx f,P =
Th d 1 : Kt qu phn tch nguyn t X l 53,2; 53,6; 4,9; 52,3; 53.6; 53.1 mg.y phng php phn tch c mcV sai s h thng khng nu gi tr thc ca X l 56,3
Gii
mg ? (P = 0,95)
:
- Kim tra so coi gi tr no.
gi tr bt thng trong dy s liu thu c theo chun Q :khng l
= 53,45.- Tnh : x
- Tnh : S = 0,85.
- Tnh : t = 8,2.tn
Tra bng : t = t = 2,57lt 0,95;5
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t mc sai s h thng.
Th d ch Al2O3, thu c cc kt qu (%) : 2,25; 2,19; 2,11;2,38; ca Al2O3 bng bao nhiu, vi P = 0,95 ?
tn > tlt : Phng php
2 : Sau 5 ln phn t2,32. Vy hm lng
Gii :
- Kim tra ch khng b g .nh :
un Q : i tr no- T x = 2,25.
- Tnh : ,11.
ra b lt = t0,95;4 = 2,7
S = 0
- T ng : t 8.
5
S.t 4;95,0 = 0,14
= (2,25 0,14) %
*So snh gi tr
Hm lng thc ca Al2O3 :
Ngha l trong khong 2,11 - 2,39 %.
x vi gi trtht (bit trc)Tnh ttn = n.
S
x
Nu ttn < tlt : x # (s t gi l ng
ttn > t
khc bi a 2 gi tr u nhin)
lt : x do sai s thng no )
0,02% cht X .Hai phng php phn tch cho ccgi tr
; ,42
( c th h
Th d : Mt mu cha 49,06o:
PPA : 49,01 ; 49,21 ; 49,08
PPB : 49,40 ; 49,44 49
Tnh x , GHTC v nh gi 2 kt qu (P = 0,95)
Gii :
- Kim tra cc gi tr bng chun Q: khng b gi tr no
- Ax = 49,10% SA = 0,10
- Bx = 49,42% SB = 0,02
* So snh v x
310,4906,49
ttnA= = 0,69 < t0,95;2 = 4,310,0
Ax # : s khc bit ch do sai s ngu nhin
ttnB = 302,0
42,4906,49 = 31 > t0,95;2 = 4,3
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Bx : s khc bit do sai s h thng
* So snh vng:
3.02,01,0
42,4910,49 ttn =
22 += 5,43 > t0,95; 4 = 2,78
ng h s ai s h thng)
tn =
Hai gi tr tru bn c s ai khc ng k (s
* So snh lp li:
F 2502,0
10,02
== > FS
S 2
2B
2A 19
l i ca hai th nghim cng sai khc nhau mt cch h thng.
* Tnh gii hn tin cy:
0,95;2;2 =
p l
nS.t 95,0 A2; = 0,25
n
S.t B2;95,0 = 0,05
A
B=(49,42 0,05)% nm ngoi khong tin cy
8. Ch n Gauss (Zlt= Up)
a) Mc dng kim nh s sai khc gia hai gi tr trung bnh
=(49,10 0,25)% nm trong khong tin cy
u
c ch :Chun Gauss Ix v
IIx c ng sai tng qut 2
b) Cch th n ngu nhin x tun theo hm phn b chun :
cng ph
c hin i vi bi
- Tnh Utn theo cng thc :
Utn =III nn +
IIIIII n.nxx
*Nu nI = nII = n :
Utn =2nx III
- Tra bng U = U .
Vi gi trng nh:
0,95 6 (P = 0,9)
0,99
x
lt p
U = 1,64 (P = 0,90)0,90
U = 1,9
U = 2,52 (P = 0,99)
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- So snh Utn v Ult.
Th d em phn tch hai mu king, thu c kt qu :
Mu B ca phng php phn tch
:
Mu A
A
Ce
L
Sb
Th
ppm
2,7
0,6
1,5
0,81
0,17
1,5
0,08
s 1290 1090 ppm 95 ppm
0,45
a 3,93 3,61 0,09
0,61
C th coi h king ny thuc c ng m t loi khng ? Cho P = 0,95.ai mu
Gii :
Tnh Utn ca cc nguyn t theo cng thc :
2
1.
x-xu BAtn
=
As Ce La* Sb Th
Utn 1,49 0,62 2,51 0,57 1,77
- Tra bng Ult = U0,95 = 1,96.
V U a La ln t Ult nn hai mu k ng ny khng cng mt i.9. Chun Duncan. (Zlt= q )
a) Mc ch
Chun Duncan ki h s sai khc gia m tr trun h viln t cc gi tr trung n li, t s t p s sai khc h th ngunhin gia cc gi tr trung bnh v nh gi tc dng nh hng ca cc yu t gy ra skhc bit c r trung bnh.
iu kin thc hin kim nh Duncan :
- Phi oan chc rng cc phng sai mu l ng nht (kim nh chun Bartlet).
- Phng sai ti hin v phng sai i snh l khng ng nht (kim nh bngchun Fisher).
Ch : Kim nh chun Bartlet v Fisher c thc hin trc khi kim nhchun Duncan.
b) Cch thc hin :
Gi s c k mu nh s i = 1, 2, 3, ..., k, mi mu i c tin hnh ni th nghim
song song, t tnh c gi tr trung bnh
tn c hn r r i lo
thf,R,P
:
c dngbnh c
m n t gi g bnl rn c hit l ng v
a gi t
ix v* Kim nh tnh ng nht ca theo Bartlet :
2iS .2iS
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C
B2tn =
Kim nghim : 2tn F ia tr trun s sai s g, tin hng chun D
tn < Flt h th trung bnh..
nh the an :
tn v qlt :
* Tm q
ix theo th t t ln n nh v nh s bc r = 1, 2, 3, ..., k.
Gi s n so snh c ix v ix , vi ix > ix .
Tm s bc r v r tng ng r < r.
R l s a ix v ix bc t ng i gi :
R = r- r + 1Gi th lt ng Duncan.tr bc R v f dng tnh q trong b
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* Tnh qtn :
"n'nS iithtn +
"n'.n2.
"x-'xq ii
ii
=
* So s :
Nu qtn < : s sai khc gia
nh qtn v qlt
thf;R;95,0q i vx ix khng mang tnh h thng
Ghi: ix ix
Nu q : s sai k tn >thf;R;95,0
q hc gia ix v ix mang tnh h thng ng k.
Ghi : 'x i > "x i
N kh c giau qtn >thf;R;99,0
q :s sai ix v ix mang tnh h thng rt ng k.
Ghi: ix >> ix Th d : ch to mu chun dng cho phng php phn tch bng ph pht x
nguy ng nht ri ln lt ca thnh cc ming nh c3x3 cm i ming, ngi ta tin hnh xc nh % Cr4 ln 6 mu nhvy, ring mu th hai th c phn tch 3 ln v st , phi lo .
H y k tnh t c u c n c ng s sau (xpln l di c t) :
i
ni
n t, ngi ta chn mt tm st 2. kim tra tnh ng nht ca m
. C 5 ming th chn ming th nm lm mu phn tch. Chn ch b r i b
t the
im trao chiu
ng nha tm s
a cc m hun c theo b liu
1 2 3 4 5 6
1
2
3
4
42
1,41
1,44
,
1,39
1,41
1,38
1
1,41
1,42
1,37
1,34
1,38
1,36
1,37
1,32
1,33
1,32
1,
1,421 42
1,4 1,34
1,38 1,37 1,34
1,423 1 07 1 5 1,3ix ,4 ,40 58 1,370 1,328
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G i :i
t X = 100x - 140 : chuyn thnh bng :
i 2 3 4 5 6
i
n 1
1
2
+
+
+
+
+ 2
- 1
+ 1
-
+
+
-
-
-
-
- 2
- 3
- 4
- 3
- 8
- 7
- 6
- 8
3
4
2
2
1
4
2
1
1
+ 2
3
6
2
6
X + + 2 + - - 12 - 299 2 17
X 45-
2iS 1,583 2,333 3,000 4,25 0,667 0,917
iX +
2,+ 0,50 - 4,25 - 3,0
25+ 0,667 - 7,25
* Kim nh tnh ng nht ca 2iS theo chun Bartlet :
Lp bng sau :
I S fi fi. log f i.log2iS2iS
2iS
2i
1
2
3
1,583
2,333
3,000
3
2
3
4,749
4,666
9,000
0,19948
0,36791
0,47712
0,59844
0,73583
1,43136
4
5
4,25
0,667
3
3
12,75
2,001
0,62839
- 0,17587
6 0,917 3 2,751 - 0,03763 - 0,11289
1,88517
- 0,52797
1 3 917 4,010017 5,
Tnh :
2,1128=17
35,917
f
.f
i
i =
S2iS2th =
= fj(fth = 17)
log 2thS = 0,3248
fth.log2thS = 5,52247
B = 2,303(fth.log2thS - fi.log
2iS )
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= 2,303(5,52247 - 4,01001) = 3,438
= = 11,1 > 32lt2
5;95,0 ,483 =2tn
Vy cc phng sai l ng nht.
* Kim nh tnh khng ng nht gia phng sai theo chun Fisher :
2thS = 2,1128 fth = 17
( )
= 221
xn1
S
2
iiiids xnN1k
+++ 45
-29
12
17
2
22222
++=
2344443
2
4
9
1-6
1
22
2thS = 50,60 fs = 6 - 1 = 5
23,952,112850,60
S
SF
2
2ds
tn ===th
Flt = F0,95;5;17 = 2,81 F0,99;5;17 = 4,34
Vy cc phng sai l khng ng nht v : Ftn > Flt
* Kim nh theo chun Duncan :
Sp xp li cc gi tr trung bnh t ln n nh, ta c bng nh sau :
r 1 2 3 4 5 6
iX
i
ni
+ 2,25
1
4
+ 0,667
2
3
+ 0,50
3
4
- 3,0
4
4
- 4,25
5
4
- 7,25
6
4
1,4532,11S th ==
Tnh :
- So snh hai gi tr ix = + 2,25 v ix = + 0,667 :
2,02134
3.4.2.
453,1
0,667-25,2q tn =+
= (R = r- r + 1 = 2 - 1 + 1= 2)
- So snh hai gi tr ix = 2,25 v ix = + 0,50 :
2,4144
4.4.2.
453,1
0,50-25,2q tn =+
= (R = 3)
Lp bng :
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'x i "i R qx Kt luntn q0,95;R;17 q0,99;R;17
+ 2,25
+
+ 0,667
- 3,00
2
4
5
2,021
7,24
8,97
2,98
3,22
3,28
4,10
4,41
4,50
>>
>>
>>
>>
+ 0,50 3 2,41 3,13 4,30
- 4,25
-7,25 6 13,10 3,33 4,56 >>
0,667
+ 0,50
+ 0,50
- 3,00
- 4,25
-7,25
- 3,00
2
3
4
5
2
0,21
4,68
6,28
10,10
4,83
2,98
3,13
3,22
3,28
2,98
4,10
4,30
4,41
4,50
4,10
>>
>>
>>
- 4,25
-7,25
3
4
6,55
10,69
3,13
3,22
4,30
4,50
>>
- 3,00
- 4,25
- 4,25
-7,25
-7,25
2
3
2
1,72
5,86
4,14
2,98
3,13
2,98
4,10
4,30
4,10
>>
>>
Phng php lp bng ny ca Doerffel tuy khi qut nhng khng tin cho vic
bin lun kt qu. Gio s C Thnh Long ngh mt phng php khc :
Nguyn tc :
Vic so snh gi tr trung bnh cng mt lc ging nh vic phn hng nhiu ibng trong cch thi u vng trn. Trong trn ha , mi i c 1 im; trong trnthn
imtng
g (> hoc >>), i thng c 2 im, i thua 0 im. S ln thng m (tng ng>>) c ghi di dng ch s di bn phi ca im tng kt.
Gi tr trung bnh cng ln th c im tng kt cng cao.
Cc gi tr trung bnh c coi l hon ton tng ng nhau khi c cngkt v cng ch s.
i 1 2 3 4 5 6
i (% Cr) 1,423 1,407 1,405 1,358 1,370 1,328x
im tng kt 83 83 83 31 31 0
T bng trn, c th kt lun :
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Hm lng % Crnhng phn u ca tm st (3 mu u tin) l hon ton ngnht ,hm l
um cc sai s trn trong thc nghim ha hc.
li
3- t mc ch s dng ca cc chun thng k: Bartlet, Fisher,
1- Doerffel Thng k trong ha hc phn tch NXB H&THCN 1983
2- C Th ng k trong thc nghim ha hc H
nhau v c th dng lm mu chun. Dc theo chiu di ca tm st, k t mu s 4ng % Cr cng trnn km ng nht. Do khng nn dng lm mu chun.
CU HI N TP
1- Phn bit sai s ng nhin v sai s h thng. Cho bit cch loi tr hoc lmgi
2- Cch loi b cc s u bt thng thu c trong thc nghim ha hc.
So snh v phn biDucan, Cohran, Student.
TI LIU THAM KHO
nh Long Gio trnh x l thTng hp TP HCM 1991
3- ng Hng Thng Thng k v ng dng NXB GD 1999
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Chng 2: PHN TCH PHNG SAI
I. KHI QUT V PHN TCH PH SAI (ANA YSIS OFVARIANCE)
1. Mc ch v ngha:Cn phn bit hai loi yu tnh hng n gi tr ca mt so thc nghim : yu
t cb n v yu t ngu nhin.
NG L
Yu tcbn : Bao gm mt nhm cc iu kin cbn ca th nghim. Mi iukin bn. Trong th nghim Ha hc, yu t cbn thng lyu t ng ha hc hoc lm thay i vn tc phn ng. Th d :nhit , p sut, nng cc cht xc tc, nng tc cht... l cc yu t cbn. Mi
ca th nghi m gi l mc cnh ca yu t cbn. Chng hn, nhc kh 3 mc cnh l pH = 2, pH = 3, pH = 4.
Kh m, vi khong mc cnh chn th yu t cbn c g h thng ca gi tr trung bnh. Nu xt v mt sai s th
l yu c kh nng gy ra sai s hthng ca php o.
Khi c nhiu phng th nghim cng tham gia phn tch mt mu ng nht bng mtquy tr thng gia cc gitr trung bnh thu c bi mi phng th nghim. Tnh hung ny rt hay gp trong thc
ngi ta ch n mt yu t c bn c bit gi l yu tphng th nghim vi s mc cnh bng ng bng s phng th nghim tham gia.
c coi l mt yu t clm dch chuyn cn b
iu kin c th hng ca pH o st
i lp k hoch th nghith y ra s thay i c tnhyu t cbn t
nh phn tch ging ht nhau, thng xy ra c s khc bit h
t kim nghim. Khi p nh
Yu t ngu nhin : Th hin khi lp li th nghim vi cc iu kin c bnc nhng gi tro khc nhau. y l sai s ngu nhin thunty c nh ca yu tbn
i gi tro cha ng thi ca yu t c bn v yu t nguhin.
ch ca phn tch phng sai l tch bit v so snh tng loi yu tn gi tro: nh hng gia cc y n v u t cbn vi cc yu tgu nhin. Hn na g i pht hin mt lot nh hng
c bit ch th hin khi c mn tch phng sai c s dng rng ri trong Ha phn tch pht hin v nh
gi vai tr ca ngun sai s khc nhau. Trong Ha hc ni chung, phn tch phng sai lt cng c tm ra i a nh th nghim.
Ty theo s yu t cbn dnh em kho cu, phn tch phng sai mt yu t,hai y ai mc cnh.
2. Nguyn tc v thut ton:
ng ging o do mi yu t gy ra c c trng bng mtphng sai mu vi b do tng ng. Php so snh nh hng ca cc yu t rtthnh php kim nh tnh ng nht ca cc yu t.
khng h thay i, thu a th nghim. c lng sai s ngu nhin ny vi mi mc c
c cn phi tin hnh mt s th nghim song song.
M ng nh hng n
Mc u t cb i nhau, gia cc yn , phn tch phn sa cn cho php
t ng thi hai hay nhiu yu t cbn.Ph
m cc iu kin t u h trong hoch
u t, nhiu yu t... Thng thng mi yu tc kho cu t nht vi h
S th ca gi trc s t
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- ng nht ca 2 phng sai : chun Fisher.
- ng t chun Bartlet hoc Cochran.
ai :
- Ph n b u nhin thun ty ngi tr
n v yu t cn n gi tro.
+Nu v ng nht (theo Fisher) : yu t cbn khng nh hng n ktq
+ Nu v khng ng nh ln t , c th tch thnh hai phn
ri t ngu thun ty
Thnh phn a yu t cbn A
v c gii quyt a vo s lp li n i mc jc u cho mi mc (th nghim i xng) th:
(n l th nghim song song)
NG SAI MT YU T (SINGLE FACTOR)
Gik, m i 2
1. Trnh tthc hin:
Bc 1:
Kim nh tnh
Kim nh tnh nh ca mt dy phng sai :
Thut ton :
C hai loi phng sai c trng ca phn tch phng s
ng sai ti hi 2thS iu th tc dng ca yu t ng:o.
- Phng sai i snh 2dsS : biu th tc dng chung ca yu t ngu nhib
2thS
2dsS
uo.
t, S2thS2dsS
2ds
2thS
2dsS
ng :Thnh phn 2thS c a yu nhin
2AS c
Mi quan h gia v S 2A 2thS
2ds S d i m
a yu t A , nu ni ng
2dsS =2thS + nS
s ln2A
II. PHN TCH PH
Mc ch : nh gi snh hng ca mt yu t no trn cc gi tr trung bnhca kt quo
s kho st nh hng ca yu t cbn A vi k mc cnh, nh s j = 1, 2,... ,i mc tin hnh th nghim song song nh s = 1, ,... ,n
Lp bng ghi kt quo xji v tnh thm cc cc d liu cn thit
... kj
1 2i
1 x
2
...
n
x
11 x21 xk1
x22
...
x2n
...x12
...
x1n
k2
... =n k
jinN
xkn
1i 1j= =
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i
n
1ii
j
xx
n
== 1x 2x ... kx
=
=k
1jjTT
n
ix T=
T2 ... Tk11i
=
=1i
2i
2j xx 21x
22x ...
2kx
kn
=1j
2jx
2s 21s
22s ...
2ks j
Cc k hi
*Trung b ca mu
u :
nh
i
1ij n
x
n
ix== =
inj
T
chung:* Trung bnh
Nx =
T
* SST: Tng bnh phng chung (Total Sum of Squares)
SST = 1n
2)xx( +=i 1
i1 2n
2)xx( +.+ kn
NT
x2
2ji
=
1i
2)xx( ==1i
2i ki
* SSF : Tng bnh phng do y (Sum of Squares for Factor)
SSF =
u t
N
T
n
T...
n
T
n
T)xx(jn
2
k
2k
2
2k
1j 1
212
j ++==
* SSE : Tng bnh phng do sa
SSE = SST SSF
* M F : Trung bnh bnh phng ca yu t (Mean Square forFactor)
MSF
2 +
i s (Sum of Squares for Erro)
S
=1k
SSF
= (fs = k-1 )
* M E : Trung bnh bnh phng c sai s (Mean Square forErro)
MSE
S2ds
S a
=kN
SSE
= 2thS (fth= N-k)
* M T : Trung bnh bnh ph (Mean Total Sum of quare)S ng chung
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MST =1N
SST
= (fchung = N-1)
* Ftn=
2ChungS
2thSSSE
2dsSSF = (Flt =
Sthds f,f,P
F )
So snh Ftnv F+Nu Ftn < Fl : v S ng nht (theo Fisher) :
Yu t cbn A khng nh hng n kt quo.. Ly biu th sai s
c s t do fchung = N - 1.
+Nu Ftn > Flt
trungbn
Ngun gc phng saic
t do
Tphng cc
lchPhng sai
Thnca phng sai
lt2
t th ds
2
S 2
chungS
ngu nhin ca ton b php o, vi b
: 2S v 2S khng ng nht (theo Fisher)th ds
Yu t cbn A nh hng ng kn kt quo .Trong dy gi trh nht nh c mt hoc vi cp c sai bit h thng (tin hnh bc 2)
S bng bnh
h phn
Tc dng chung cayu t ngu SSF 2dsS = 1k
SSF
2dsS = + n
(th nghim i
2thS
2AS
cbn v k - 1nhin xng)
T SSEc dng ring cayu t ngu nhin N - k2thS = kNSSE
2thS
Ngu nhin ha mi2ChungS = 1N
SST
2chungS tc dn ca yu t c N- 1g
bn v ngu nhinSST
Bc 2 :Kim nh tnh ng nht ca phng sai theo chun Bartlet hocCochran (khi th nghim i xng cc n =n):j
cc phng sai ln cho n khi cc phng sai cn li u ng nhtBcCn loi b
3 : Kim nh tnh khng ng nht ca v heo chS2th2dsS t un Fisher :
2th
2ds
tnS
SF =
thds
i F
f,f,PltFF = vi fs = k - 1 fth = k(n - 1)
So snh v v F :
tn ltn : y A khng c n cc gi tro
rn cc so cn li sau khi loi bbc 2)
tn lt
Nu F < F , k t lu u t nh hng ng k(t
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Nu Ftn > F , k t lun : yu t A c ng ng n gi tro.gi tr trung bnh nht nh c mt hoc vi cp c sai bit h thng., cn tin hnh bc4 nh s t gia cc gi tr trung bnh.
B c 4
lt nh h k Trong dy
kim ai bi
: Ki nh s sai bit h gia cc gi tr trung bnh theo chunDuncan:
p xp l
m thng
Ta s i jx theo trnh t t ln , nh s c r = 1, 2, ..., k; sau tinnh nhphn chun Duncan.
III G DNG
1.
H ng Ca (%) trong mu vi c xc nh bng 3 phng php khcnh hm lng C c c bnh ng bi cc phng php phn
tch khc nhau khng?B
PP1
12 11 13 10
Bc 1
n nh bh
. BI TP N
Bi tp 1:
m lau. Hy cho bit a thu h
ng kt qu:
12 10 11 12 9 12
PP2 12 14 15 16
PP3
: Lp bng v ghi cc d liu cn thit
Gi t thng k
ng Ca khng bnh hng bi phng php phn tch
c c s khc bit )
PP1 PP2 PP3
12
10
11
12
9
12
12
14
15
16
12
11
13
10
nj 6 4 4 N
fj 5 3 3
thi
H0 : Hm l
(Cc gi tr trung bnh thu c xem nh tng ng nhau)
H1 : Hm lng Ca bnh hng bi phng php phn tch
(Cc gi tr trung bnh thu
=14
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jx 11 14,25 11.
T 66 57 46
1,6 2,916667 1,666667
=
5
2jS
NTx2
2jiSST = 48,9286
SSF =N
TTTT 2222
n...
k
k ++ = 27,1786nn 2
2
1
1 +
SSE = SST SSF = 21,75
MSF =1k
= dsS = 13,5893 (fSSF 2
s = k-1 = 3-1 = 2)
MSE = kNSSE
=2
thS = 1,97 73 (fth= N-k = 14-3 = 11)
2th
2ds
tnS
SF = = 6,8727 > F 1 = 3,98
Yu t ng php phn tch c nh g n kt o
B c 2
0,95;2;1
ph hn qu
: Ki tnh n t ca cc p g sai bng n Barlet
PP1 PP2 PP Slog 1,0206 1,3947 0,6655 3,0808
= 2,3 Sg - Sf )
B = 2,30 log1,9773 808) = 0, = 5 f = k-1 =
Cc phng sai ng nhtB
m nh g nh hn chu
3
2
jj Slogf2jjf
B 03( th lof2th
2jj log
3(11. 3,0 4053 < 2 2;95,0 ,99 ( 2)
c 3: Ki nh kh nht h Fishe
m nh t ng ng ca 2thS v2dsS t eo chun r :
2th
2ds
tn S
SF = = 6,8727 > F0,95;2;11 = 3,98
Cc phng sai v l khng ng nht Yu t phng php phn tchc nh thu c
B
2 2thS dsS
hng n cc kt qu
c 4: Kim nh s sai bit h thng gia cc gi tr trung bnh theo chun Duncan :
Lp bng
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r 1 2 3
jx 14,25 11,5 11
n 4 6j 4
Sth = 1,4062
Qtn =jj
jj
th
jj
nn
n.n.2
S
xx
+
Qlt = v i f = 11 v R = r
/ - r// +thf;R;95,0
Qthf;R;99,0
Q ( v th
1)
jx jx R Qtn K t lu n
14,25
11
11,5
11
2
3
3,91
5,06
0,78
3,12
3,26
3,12
4,4
4,64
4,4
>
>>
Bng im tng kt:
Ph g p PP PP2 PP3
H 11,5
i m t 1 41 1
u
t H1.Ngha l hm lng Ca thu c t 3PP phn tch c s khc bit . Trong PP1 v PP3 xem nh cho kt qu tng ngnhau
C ng ca Ca nn khng th kt lun l PP no chokt qu
2. Bi tp
Hy so snh nh hng ca cc halogenur alkyl CH3I (a1), C3H7I (a2), C4H9I (a3),C2H5Br (a 7Br (a5) n hiu sut ng polimer theo cch gc t do,
da vo bng s liu o hiu sut :
thf;R;95,0Q
thf;R;99,0Q
,5 11 2
n ph 1
m l ng Ca 11 14,25(%)
ng k t
Kt l n:
Bc b gi thit H0 , chp nhn gi th
.
h :y cha bit gi trng .
2:
4), C3H (%) ca phn
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j1
(a )
2
(a )
3
(a )
4
(a )
5
(a5)
i
1 2 3 4
1
2
3
4
5
86,3
86,5
69,6
81,8
5,5
42,5
64,3
79,0
61,0
52,5
76,0
83,8
72,8
89,0
93,2
70,7
64,8
38,5
77,0
91,5
80,0
79,8 87,3
92,3 78,0
76,5 83,7 31,3 76,5
6
7
8
87,1
82,5
90,0
64,8
67,3
7
72,9
58,7
87,5
74,5
68,0
38,1
Gii :
Bc 1:: Lp bng v ghi cc d liu cn thit
Gi thit thng k
H0 : Cc halogenur alkyl khng nh hng n hiu sut
(Cc hiu sut thu c xem nh tng ng nhau)
H1 : Cc halogenur alkyl c nh hng n hiu sut
(Cc hiu sut thu c c s khc bit )
j
i
1
(a1)
2
(a2)
3
(a3)
4
(a4)
5
(a5)
1
2
3
4
5
6
7
8
79,8
86,3
86,5
92,3
76,5
87,1
82,5
90,0
87,3
69,6
81,8
78,0
83,7
64,8
67,3
75,5
42,5
64,3
79,0
61,0
31,3
72,9
58,7
52,5
76,0
83,8
72,8
89,0
76,5
87,5
74,5
93,2
70,7
64,8
38,5
77,0
91,5
68,0
38,1
80,0
jx 85,125 76 57,775 81,6625 66,075Tj 681 608 462,2 653,3 528,6
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T= Tj = 2933,12jS 27,4364 66,7086 242,1678 59,1655 361,3421
N = 40
SST = NT
x
22ji = 9379,9397
SSF =N
T
n
T...
n
T
n
T k21 +++ = 4082,1962222
k21
SSE = SST SSF = 5297,7437
MSF =1k
SSF
= 2dsS = 1020,549 (fs = k-1 = 5-1 = 4)
MSE =kN
= 2th
S = 151,3641 (fSSE
th= N-k = 40-5 = 35)
2th
2dsS
tn SF = = 6,7539 > F0,95;4;35 = 2,65
Yu t halogenur alkyl n ng (cc hiu sut thuc c s khc bit nhau)
ng nht
ca
B t ca cc phng sai theo chun Cochran
Gtn =
c nh hng n hiu sut ca ph
Ch : Nu th nghim i xng (nj = n), s dng gi tr2jS kim nh s
cc phng sai theo chun Cochran
c 2: Kim nh sng nh2maxS = 361,3421
2j
2max
S
S=
8204,756
3421,361= 0,4774
Glt = G0,95;k-1,n = G0,95;4;8 = 0,3910 ; G0,99;4;8 = 0,4627 < Gtn
loi b = 361,3421
Xem xt 4 phng sai cn li := 242,1678
Gtn =
2maxS
2maxS
2j
2max
S
S=
3421,3618204,756
1678,242
= 0,6123
Glt = G0,95;3;8 = 0,4377 ; G0,99;3,8 = 0,5209 < Gtn loi b = 242,1678
Xem xt 3 phng sai cn li :
= 66,7086
2maxS
2maxS
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2j
2max
S
S=
7086,663421,3618204,756
7086,66= 0,2029Gtn =
i l ng nht
SST =
Glt = G0,95;2;8 = 0,5157 > Gtn = 0,2029
3 phng sai cn l
Bng s liu bi 2 ct a3 v a5
T = 2933,1 - 462,2 528,6 = 1942,3
N = 40 16 = 24
k=3
Nji
SSF =
T 22x = 1412,6895
N
T
n
T
...n
T
n
T 2
k
2k
2
22
1
21
+++ = 339,5158
SSE = SST SSF = 1073,1737
MSF =1k ds
SSF= 2S = 169,7579 (f = k-1 = 3-1 = 2)s
MSE =kN th
SSE= 2S = 51,1035 (f = N-k = 24-3 = 21)th
Bc 3: Kim nh tnh khng ng nht ca 2thS v 2dsS theo chun Fisher :
2th
tn S
2dsSF = = 3,3218 > F0,95;2;21 = 3,47
11,9898,87
1185,40SF
*2ds
tn = S *2
==
Kt lun: Chp nhn gi thit thng k H1, cc halogenur alkyl c nh hng n hiua2, v
th
Flt = F0,95;3;28 = 4,57 < 11,98
Cc hiu sut ca a1, a2, v a4 khng c s khc bit nhau.
sut ca phn ng polimer ha. Sau khi loi b a3 v a5 th cc hiu sut cn li a1,a4 khng c s khc bit nhau.
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BI TP
1 ng (%) ca H2SO4 do 3 nhm sinh vin thc hin nh
Nhm 1: 79 8
N2 68 70 76
Kim nh xem h ng trung bnh ca cc nhm thu c c ging nhau khng?
2. nh gia hiu sut ca phng php chit thuc tr su Basudin t cc h dungc kt qu sau (%):
CH3CO 3 78,4 76,4 78,4 76,1
CH3CO 8 93,9 98,8 98,8 97,8
C 95,8 94,8 96,8 96,8 94,3 95,8Cho P=0,95
TI LIU THAM KHO
4- Doe k a hc phn tch NXB H&THCN 1983
5- C h Long Gio trnh x l thng k trong thc nghim ha hc H
Tng h p TP HCM 19916- n ng v ng dng NXB GD 1999
. Kt qu phn tch hm lsau:
6 94 89
hm 2: 71 77 81 88 Nhm 3: 8
m l
mi thu
OH: 78,4 72,2 71,6 73,
OH:CCl4 (1:1): 95,9 96,8 97,8 95,
H3COOH:CCl4(1:2): 96,8 95,5
rffel Thng trong h
Thn
g Hng Th Thng k
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Chng 3: PHN TCH HI QUY
I. KHI QUT V HN TCH HI QUY.
1. Mc ch v ngha :
Trong nghin cu khoa hc, thng phi v th ph thuc ca i lng y voa vo cc cp gi tr thc nghim (xi , yi), th biu din s ph thuc ny
c th l ng thng hoc l ng cong. C mt s phng php i tm cc hmng thc nghim, trong c phng php hi quy.
Biu thc ton hc ca hm ph hp ny gi l phng trnh hi quy, cng c tonc h ph hp gi l phn tch hi quy .
Trong ha hc, phn tch hi quy c dng tm cho cc th chun gia cc t chnh xc v tn hiu ph y. i c phng trnh hi quy, c
th s dng ngc phng trnh ny : o tn hiu phn tch y* ca mu phn tch ri tnhra hm lng x* theo phng trnh hi quy, nh vy trnh c nhc im ca phpc chiu theo th
- Php chiu th thng km chnh xc
n ic v mt h i tt c cc im ca thmang tnh ch quan ca ngi v v c th gy ra nhng sai s ln.
n phng trnh h uy c th theo di c s bin ng ca tn hi ng hiu chnh cc thng s ca
phng trnh hi quy cho ph hp vi khch quan. Ngoi ra, phn tch hi quy cho phpng tin c c x* m t cch hch quan.
2.iu kin thc hin:
p nhn S2(x)
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2. Tnh cc hsa , b v cc thng scn thit:
a) Trng hp tng qut :
m ring phn ca SSE theo a vb p ng 0.
Thay Yi = axi + b :
SSE = (yi axi - b)2 minimum
cho a v b tha mn iu kin trn th cc o hhi b
0a
)E= ;
SS(0
b
)SSE(=
axi - b)2 = 0 (1)
i( - axi -b)2 = 0 (2)
ng trnh (1) v (2) :
Do :
2 (yi -
2 x yi
Gii h ph
a =( ) i2i xk
2
iiii yxyxk x
b =k
xay ii
L p kh u
4. yi
.
xi.yI
c nghim (xi , yi) ;
ng bnh ph a hi quy
o d li :
1. xi
2. 2ix 52 iy
3. ( xi)2 6.
=
=k
1i
k : s cc cp th
Cc k hiu
SST: T ng c cc sai s trong phn tch
SST = ( ) yy2
2i k
SSE: Tng bnh phng do sai s
SSE = 2 by ii yxay
ng bnh phng do hi quy
ii
SSR: T
+ 2SSR = SST SSE = i )ybax(
MSR = SSR
E2k
SSE
(vi Y = ax + b)MS =
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R2 =SST
SSR x
b
u b = 0 (ng hi quy qua gc ta ) :Y = a.x
: H s c nh
) Trng hp c bit :
N
=
2
ii
x
yxa
i
'
2i aySSE = ii yx'
MSE =1k
SSE
* Cch tnh 2yS ,2
y /S , 2aS ,
2bS , :
=
2
a /S
2YS 2kSSE
=
y yxayb iii2i 2k
2
Y /S =
1k
xay ii2i Vi f = k-1y
=2aS
( ) 2
i
2
i
2Y
xxk
kS Vi f = k-2
2bS =
( ) 2i2i xxk V i f = k-2
2i
2Y xS
=2a /
S 2i
2
Y
x
S / Vi f = k-1
3. Xt ngha ca hshi quy (chun Student):
t gi thit thng k
H0 : H s hi quy khng c ngha
H1 : H s hi quy c ngha
Gi tr thng k:
Xt h s a : ttn=2aS
a
Xt h s b: ttn=2
bS
Bin ln:
b
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- ttn < tlt = tP, k-2 : chp nhn gi thit H0
P, k-2 : chp nhn gi thit H1
Ch
- ttn > tlt = t
: Nu h s b khng c ngha (b = 0) Chn ng hi quy Y/ , tnh a/ v ccthng s cn thit
4 i p g h y nFis
ikim n tch
ph th nghim song song i) l m
H hng trnh hi quy khng thch hprnh hi qu
Gi tr thng k
. Kim nh stuyn tnh g a x v y ca hn trnh i qu ( chuher):
Khi tnh c cc h s a, b cha chc l x v y tuyn tnh vi nhau, do cn phnh xem gia x
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