Tin Hoc Trong Hoa Hoc

8/9/2019 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: [email protected]

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} :

16


17/202

.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

17


18/202

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.

18


19/202

+

+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 :

19


20/202

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 :

20


21/202

(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

21


22/202

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

22


23/202

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

23


24/202

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

24


25/202

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

25


26/202

/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

26


27/202

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

28


29/202

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


30/202

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 =

31


32/202

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

32


33/202

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 :

33


34/202

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

34


35/202

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)

35


36/202

- 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

36


37/202

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

37


38/202

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)

38


39/202

- 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

39


40/202

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

40


41/202

* 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

41


42/202

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 )

42


43/202

= 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 :

43


44/202

'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 :

44


45/202

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

45


46/202

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

46


47/202

- 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= =

47


48/202

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

48


49/202

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

49


50/202

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

50


51/202

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

51


52/202

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

52


53/202

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

53


54/202

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

54


55/202

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.

55


56/202

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

56


57/202

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)


58/202

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 =

58


59/202

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

59


60/202

- 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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