04 Giao Trinh Thongke 2008-Latest

1

Chng 1: CC DNG SAI S TRONG HA PHN TCH 1.1. Sai s v cch biu din sai s

Sai s (error) l s sai khc gia cc gi tr thc nghim thu c so vi gi tr mong mun. Tt c cc s liu phn tch thu c t thc nghim u mc sai s. Sai s php o dn n khng chc chn ( khng m bo o) ca s liu phn tch. C hai loi sai s c biu din ch yu trong Ha phn tch l sai s tuyt i v sai s tng i.

1.1.1.Sai s tuyt i (EA) (Absolute error) L s sai khc gia gi tr o c (xi) vi gi tr tht hay gi tr qui chiu c

chp nhn (k hiu l ). EA = xi -

Sai s tuyt i c gi tr m hoc dng, cng th nguyn vi i lng o v khng cho bit chnh xc ca phng php.

* Gi tr qui chiu c chp nhn: (accepted refrence value): l gi tr c chp nhn lm mc so snh, nhn c t:

a) gi tr l thuyt hoc gi tr c thit lp trn c s cc nguyn l khoa hc; b) gi tr c n nh hoc chng nhn trn c s th nghim ca mt s t chc

quc gia hoc quc t; c) gi tr tho thun hoc c chng nhn trn c s th nghim phi hp di s bo tr ca mt nhm cc nh khoa hc hoc k thut; d) k vng ca i lng (o c), ngha l trung bnh ca mt tp hp nht nh cc php o khi cha c a), b) v c). 1.1.2. Sai s tng i (ER) (Relative error) L t s gia sai s tuyt i v gi tr tht hay gi tr bit trc, c chp

nhn.

ER = ix

hay ER % = AE

. 100%

* Sai s tng i cng c th biu din di dng phn nghn (parts per thousand-ppt)

ER = AE

. 1000 (ppt)

Sai s tng i cng c gi tr m hoc dng v khng c th nguyn, c dng biu din chnh xc ca phng php phn tch.

http://hhud.tvu.edu.vn

2

Th d 1.1: Kt qu xc nh hm lng aspirin trong mt mu chun c biu din hnh 1.1. Hm lng ng ca aspirin trong mu chun l 200 mg. Nh vy, php o mc sai s tuyt i t -4mg n +10mg v sai s tng i t -2% n +5% (hay 20ppt n 50ppt). 195 200 205 210

Sai s tuyt i (EA : mg) -5 0 5 10 Sai s tng i (Er : % ) -2,5 0 2,5 5 Hnh 1.1: Sai s tuyt i v sai s tng i khi phn tch aspirin trong mu chun.

1.2. Phn loi sai s 1.2.1. Sai s h thng hay sai s xc nh (Systematic or determinate error):

L loi sai s do nhng nguyn nhn c nh gy ra, lm cho kt qu phn tch cao hn gi tr thc (sai s h thng dng -positive bias) hoc thp hn gi tr tht (sai s h thng mnegative bias).

Sai s h thng gm: - Sai s h thng khng i (constant determinate error): loi sai s ny khng ph

thuc vo kch thc mu (lng mu nhiu hay t). Do , khi kch thc mu tng th nh hng ca sai s ny hu nh khng ng k v c loi tr bng th nghim vi mu trng (blank sample).

- Sai s h thng bin i (proportional determinate error): loi sai s ny t l vi kch thc mu phn tch, khong cch gia cc tr o lun bin i theo hm lng (nng ), do rt kh pht hin. Sai s h thng bin i rt kh pht hin tr khi bit r thnh phn ho hc ca mu v c cch loi tr ion cn.

Sai s h thng khng i v bin i c biu din trn hnh 1.2.


3

Sai s h thng phn nh chnh xc ca phng php phn tch. Hu ht cc sai s h thng c th nhn bit c v c loi tr bng s hiu chnh nh phn tch mu chun hay loi tr nguyn nhn gy ra sai s.

Cc nguyn nhn gy sai s h thng c th gm: - Sai s do phng php hay quy trnh phn tch nh: Phn ng ho hc khng

hon ton, ch th i mu cha n im tng ng, do ion cn tr php xc nh

- Sai s do dng c nh: dng c cha c chun ho, thit b phn tch sai, mi trng phng th nghim khng sch.

- Sai s do ngi phn tch nh: mt nhn khng chnh xc, cu th trong thc nghim, thiu hiu bit, s dng khong nng phn tch khng ph hp, cch ly mu phin din, dng dung dch chun sai, ho cht khng tinh khit, do nh kin c nhn (nh phn tch kt qu sau da trn kt qu trc) ... * Cch loi tr sai s h thng:

- Tin hnh th nghim vi mu trng: Mu trng l mu khng c cht phn tch nhng c thnh phn nn ging nh dung dch mu phn tch.

- Phn tch theo phng php thm chun loi tr nh hng ca cc cht cn tr.

- Phn tch mu chun (hay mu chun c chng nhn- mu CRM: Mu chun l mu thc c hm lng cht cn phn tch bit trc, c dng nh gi chnh xc ca phng php.

Khi lng mu (g)

Khi lng cht phn tch (mg)

Gi tr ng

Sai s h thng khng

i Sai s h

thng bin i

Hnh 1.2: Biu din sai s h thng khng i v bin i


4

- Phn tch c lp: khi khng c mu chun th phi gi mu phn tch n phng th nghim (PTN) khc, tin hnh phn tch c lp loi nhng sai s do ngi phn tch v thit b phn tch, i khi c phng php gy nn.

- Thay i kch thc mu: pht hin sai s h thng khng i v bin i. 1.2.2. Sai s ngu nhin hay sai s khng xc nh (random error or

indeterminate): L nhng sai s gy nn bi nhng nguyn nhn khng c nh, khng bit trc. Sai s ngu nhin thng gy ra do: - Khch quan: nhit tng t ngt, thay i kh quyn, i lng o c

chnh xc gii hn - Ch quan: thao tc th nghim khng chun xc (c th gy ra gi tr bt

thng); thnh phn cht nghin cu khng ng nht Do sai s ngu nhin khng th bit trc c nn loi tr n cn phi lm

nhiu th nghim v tin hnh x l thng k s liu phn tch. Sai s ngu nhin lm cho kt qu phn tch khng chc chn, cn sai s h

thng lm cho kt qu phn tch sai. 1.2.3. Gi tr bt thng (outliers): Gi tr bt thng l nhng gi tr thu c thng rt cao hoc rt thp so vi gi

tr trung bnh. Gi tr bt thng dn n nhng kt qu thu c sai khc nhiu so vi tt c cc s liu lp li ca tp s liu.

Gi tr bt thng do nhng nguyn nhn bt thng xy ra trong qu trnh phn tch gy nn. Do , trc khi x l s liu cn phi loi tr gi tr bt thng.

1.2.4. Sai s tch lu (accumulated error): Trong mt phng php phn tch, sai s ca s liu phn tch thu c thng

bao gm sai s do cc giai on trong qu trnh phn tch ng gp nn. sai s chung l nh th khi phn tch cn phi tm iu kin ti u theo nh lut lan truyn sai s.

Sai s tch lu hay s lan truyn sai s h thng c x l tng t nh sai s h thng. V sai s h thng c du (+) hay (-) nn s dn n s trit tiu sai s v trong mt s trng hp sai s tch lu c th bng khng. - Khi ch c kt hp tuyn tnh ca php o ngu nhin ( kt qu cui cng ca php

cng v tr) th sai s xc nh tuyt i ET l tng cc sai s tuyt i ca php o ring r. Nu m= A+B +C th Em = EA +EB + EC

- Khi biu din nguyn nhn cc kt qu ( kt qu cui cng l php nhn hoc chia), ngi ta dng sai s xc nh tng i ETR

Nu m= A.B/C th CE

BE

AE

m

E CBA RRRRm ++=

Th d 1.2:


5

a) Khi cn mu trn cn phn tch c chnh xc 0,0002 gam c kt qu nh sau:

mchn +mu= (21,1184 0,0002) gam ; mchn= (15,8465 0,0002) gam vy khi lng mu s l mmu= (21,1184 0,002) - (15,8465 0,002) = (5,2719 0,004) gam b) Khi lng dung dch c tnh theo cng thc m=V.d= (3,430,01).(5,660,01)=? Ta c: ERV= 0,01/3,43 ; ERd= 0,01/5,66; ERm= (0,01/3,43)+(0,01/5,66) Do m=(3,43.5,66) [(0,01/3,43)+(0,01/5,66)]. (3,43.5,66)= 19,41380,0909 Nn m= (19,41 0,09)

1.3. lp li, trng, hi t, phn tn * lp li (repeatability): Trong phn tch, khi thc hin cc php th nghim thc hin trn nhng vt liu v trong nhng tnh hung c xem l y ht nhau

thng khng cho cc kt qu ging nhau. iu ny do cc sai s ngu nhin khng th trnh c vn c trong mi quy trnh phn tch gy ra v khng th kim sot

c hon ton tt c cc yu t nh hng n u ra ca mt php o. Khi bo co

cc d liu o, cn xem xt n nguyn nhn v kt qu s thay i ny.

Nhiu yu t khc nhau (khng k s thay i gia cc mu th c xem l ging nhau) c th ng gp vo s thay i cc kt qu ca mt phng php o, bao gm:

a) ngi thao tc; b) thit b c s dng; c) vic hiu chun thit b; d) mi trng (nhit , m, s nhim ca khng kh ...);

e) khong thi gian gia cc php o S thay i gia cc php o do c thc hin bi nhng ngi thao tc khc nhau v/hoc vi cc thit b khc nhau s thng ln hn s thay i gia cc php o do cng mt ngi thc hin vi cc thit b nh nhau trong khong thi gian ngn. * trng (reproducibility): c trng cho mc gn nhau gia gi tr ring l xi ca cng mt mu phn tch, c tin hnh bng mt phng php phn tch, trong iu kin th nghim khc nhau (khc ngi phn tch, trang thit b, phng th nghim, thi gian) (between laboratory precision) . Vi cng mt phng php phn tch, thng xt n lp li hn l trng.

* hi t (convergence): ch s phn b s liu thc nghim xung quanh gi tr trung bnh. Nu lp li tt th hi t tt.


6

* phn tn (dispersion): ch mc phn tn ca kt qu th nghim sau nhiu ln o lp li. phn tn l nghch o ca lp li. Nu kt qu c lp li cao tc l phn tn cc gi tr xung quanh gi tr trung bnh thp. 1.4. chm v chnh xc * chm (precision): dng ch mc gn nhau ca cc gi tr ring l xi ca cc php o lp li. Ni cch khc, chm c dng ch s sai khc gia cc gi tr xi so vi gi tr trung bnh x . Ba khi nim thng k c dng m t chm ca mt tp s liu l lch chun, phng sai v h s bin thin (s xt sau). Tt c cc khi nim ny c lin quan n lch ca s liu phn tch khi gi tr trung bnh: di = xxi

* ng (trurness): ch mc gn nhau gia gi tr trung bnh ca dy ln cc kt qu th nghim v gi tr qui chiu c chp nhn.

Do , thc o ng thng k hiu bng chch. * chnh xc (accuracy): l mc gn nhau ca gi tr phn tch (thng l

gi tr trung bnh x ) vi gi tr thc hay gi tr c chp nhn xt hay . Khi khng c sai s h thng th gi tr trung bnh tin ti gi tr thc nu s php

o rt ln ( N). V vy, c th ni chnh xc tu thuc vo s php o. chnh xc c biu din di dng sai s tuyt i hoc sai s tng i. Trong Ho phn tch, nh gi chnh xc ngi ta pha cc mu t to

(synthetic sample) bit trc hm lng (tc l c gi tr bit trc ) v lm th nghim tm ra gi tr trung bnh sau kim tra xem c s sai khc c ngha thng k gia gi tr trung bnh v gi tr thc hay khng. Vn ny s c xt chng 4.

chm v chnh xc l nhng ch tiu quan trng nh gi cht lng ca s liu phn tch. Thng thng, cn nh gi chm trc v nu phng php phn tch mc sai s h thng th ch c dng nh lng khi sai s ngu nhin nh.


7

Chng 2 CC I LNG THNG K (Descriptive statistics)

2.1. Cc i lng trung bnh * Trung bnh s hc ( x ) (mean, arithmetic mean, average) l i lng dng ch gi tr t c khi chia tng cc kt qu th nghim lp li cho s th nghim lp li. Gi s c tp s liu th nghim lp li x1, x2,, xN th gi tr trung bnh s hc ca tp s liu gm N th nghim lp li l:

x = N

xxx n+++ ...21 =

N

xN

ii

=1 (2.1)

Gi tr trung bnh c tnh cht sau: - Tng lch gia cc gi tr ring r v gi tr trung bnh bng khng.

= 0)( xxi - Tng cc bnh phng lch nh hn tng bnh phng ca bt c lch

no gia gi tr n l v gi tr a no khng phi gi tr trung bnh.

2)( xxi < 2)( axi ( vi a x ) * Trung bnh bnh phng ( x bp): vi tp s liu gm N s liu lp li x1, x2,,xn ta c:

x bp = Nxxx n

222

21 ...+++

(2.2)

* Trung bnh hnh hc hay trung bnh nhn (geometric average) vi cc php o c hm lng cn tm di dng logarit th:

lg x hh= )lg...lg(lg1 21 NnxxxN +++

Do x hh= N Nxxx ..... 21 ( 2.3) * Trung v (median) : Nu sp xp N gi tr lp li trong tp s liu theo th t tng n hoc gim dn t x1, x2, , xN th s nm gia tp s liu c gi l trung v.

- Nu N l th trung v chnh l s gia dy s. - Nu N chn th trung v l trung bnh cng ca 2 gi tr nm gia dy s.

Ch : Gi tr trung bnh hay trung v ca tp s liu c gi l cc gi tr trung tm ca tp s liu. Cc tp s liu khc nhau c cng gi tr trung bnh c th rt khc nhau v ga tr ring l v s th nghim. V vy, trung bnh v trung v khng cho ta ci nhn tng qut v s phn b cc s trong tp s liu. Trong trng hp cn xt n phn tn ( lch khi ga tr trung bnh).

* im t phn v (quartile): Nu sp xp cc s liu trong tp s liu t nh n ln th mi tp s liu c 3 im t phn v: 25 % cc s trong tp s liu sp xp


8

c gi tr nh hn hoc bng im t phn v th nht, 75 % cc s trong tp s liu sp xp c gi tr nh hn hoc bng im t phn v th ba, 50% cc s trong tp s liu sp xp c gi tr nh hn hoc bng trung v (im t phn v th hai). Khong gia im t phn v (interquartile) biu th s khc nhau gia im t phn v th nht v th ba.

C th hnh dung im t phn v theo s sau: Trung v

gi tr 0% 25% 50% 75% 100% gi tr cao thp im t phn v th nht im t phn v th ba.

* S tri (mode): l s c tn s xut hin l ln nht trong tp s liu lp li. Ch : Gi tr bt thng c nh hng ng k ti gi tr trung bnh nhng khng nh hng n s trung v. Do vy, vi nhng tp s liu rt nh, (thng N

9

th ( )

kN

xx

S

m

j

k

iiij

=

= =1 1

2

2 (2.5)

vi N l tng tt c cc th nghim N=m.k (Khi nim ny t dng trong ho hc)

Nu phng sai cng ln th tn mn ca cc gi tr o lp li cng ln hay lp km. * lch chun (Standard deviation) - Mu thng k v mu tng th (statistical sample and population). Trong thng k, mt s xc nh cc quan st thc nghim (hay kt qu php o cc mu phn tch ring r) c gi l mu thng k. Gp tt c nhng mu thng k gi l mu tng th. Nh vy c th xem phn tch mu tng th l nhng php o c th c v v cng ln (N).

Th d: Cn iu tra mc thiu iot trong hc sinh tiu hc thnh ph A. Tin hnh ly mu nc tiu hc sinh mt s trng tiu hc trong thnh ph phn tch hm lng it. Nh vy nc tiu ca mt s hc sinh tiu hc mi trng c ly mu l cc mu thng k. Mu tng th y s l mu nc tiu ca hc sinh tiu hc thnh ph A ni chung.

- Trung bnh mu x v trung bnh tng th . + Trung bnh mu ( sampling fluctuation) ( x ) l gi tr trung bnh ca mt mu thng k gii hn c rt ra t tp hp cc s liu v c xc nh theo cng thc:

N

x

x

N

ii

=

=1

.

+ Trung bnh tng th (population average) () l gi tr trung bnh ca tp hp cc s liu, cng c xc nh theo phng trnh (2.1) nhng vi N rt ln, gn t ti . Khi khng c sai s h thng th trung bnh tng th cng l gi tr tht ca php o.

N

xN

ii

=

=1 khi N . Thng thng khi N > 30 c th xem nh x

- lch chun tng th (Population standard deviation): () c trng cho phn tn cc s liu trong tp hp vi gi tr trung bnh v c xc nh theo phng trnh:

( )N

xxN

ii

=

=1

2

hay 2 = (2.6)

vi N l s th nghim lp li ca tp hp, thc t thng xem cc tp s liu c N>30 l tp hp.

- lch chun mu c on (Sample estimate standard deviation): (S)


10

( )1

1

2

=

=

N

xx

S

N

ii

hay 2SS = (2.7)

vi N l s th nghim trong mu thng k c rt ra t tp hp. S bc t do trong trng hp ny l f =N-1. (Bc t do c th coi l s php o kim tra cn thit c th xc nh c kt qu trong mt tp s liu. Mt cch khc bc t do c hiu l s cc quan st trong mt mu thng k c th t do thay i do bng tng kch thc mu tr i 1 bc t do cho mi trung bnh. Thut ng bc t do cn c dng ch s lch ( )xxi ) c lp dng trong php tnh lch chun) Nh vy, khi N th x v S . Ni cch khc khi N>30 c th xem S .

So vi phng sai, lch chun thng c dng o lp li hn do c cng th nguyn vi i lng o. Khi tnh ton ch khng lm trn s liu ca lch chun cho n khi kt thc php tnh ton v ch ghi gi tr cui cng di dng s c ngha. Nu trng hp c m mu thng k, mi mu lm n th nghim song song th:

( )mnm

xx

S

m n

ij

=

.

1 1

2

bc t do f=m(n-1) (gi thit Sj khc nhau khng ng k). i vi tp s liu nh ( N

11

Nu c nhiu dy s liu lp li (nhiu mu thng k), mi dy c N s liu c ly ngu nhin t tp hp s liu th s phn tn ca trung bnh mu c c trng bng sai chun m thay cho lch chun trong tp hp. S phn tn ny gim khi N tng.

m l lch chun trung bnh hay sai chun v c tnh nh sau :

m=N

Dng sai chun m c trng cho sai s ngu nhin cu phng php phn tch. Tuy nhin, i vi tp s liu hu hn (N

12

Th d 2.1 :Cho kt qu phn tch lp li 35 ln hm lng nitrat (g/ml) nh sau : 0.51 0.51 0.49 0.51 0.51 0.51 0.52 0.48 0.51 0.50 0.51 0.53 0.46 0.51 0.50 0.50 0.48 0.49 0.48 0.53 0.51 0.49 0.49 0.50 0.52 0.49 0.50 0.50 0.50 0.53 0.49 0.49 0.51 0.50 0.49 Sv t tnh cc i lng thng k theo cng thc v so snh vi kt qu tnh theo phn mm MINITAB di y, gii thch ngha cc kt qu .

ham luong nitrat ( micogam/ml)0.530.520.510.500.490.480.470.46

Phan bo cac gia tri thuc nghiem theo tan suat

ham luong n

itra

t ( micro

gam/ml)

0.53

0.52

0.51

0.50

0.49

0.48

0.47

0.46

Do thi khoi cac gia tri thuc nghiem

2.3. Bo co kt qu phn tch 2.3.1. S c ngha v cch ly gi tr gn ng Mt gi tr s hc dng biu din kt qu phn tch s khng c ngha nu khng bit chnh xc ca n. Do vy, khi biu din cn phi ghi r tin cy ca s liu v cc s liu cn c lm trn ch mc khng chc chn ca n (uncertanty). Ni cch khc, s liu ch c cha cc s c ngha. 2.3.1.1. Khi nim s c ngha S c ngha trong mt dy s l tt c cc s chc chn ng v s khng chc chn ng u tin.

Th d 2.2 : Khi c th tch dung dch ng trong buret 50 ml, chng ta c th thy vch cht lng v tr ln hn 30,2 ml v nh hn 30,3 ml. Nu c th c on v tr vch cht lng cp chia khong + 0,02 ml th c th bo co th tch l

Descriptive Statistics for nitrate concentration Total Count : 35 Mean: 0.50413 SE Mean: 0.00260 StDev : 0.01537 Variance : 0.000236 CoefVar : 3.06 Sum of Squares: 8.80810 Minimum: 0.46 Q1: 0.49 Median : 0.50 Q3 : 0.51 Maximum : 0.53 Range: 0.07 Skewness : -0.20 Kurtosis: 0.50

ham luong nitrat ( microgam/ml)

tan xuat

0.530.520.510.500.490.480.470.46

10

8

6

4

2

0

Bieu do phan bo tan xuat ham luong nitrat


13

30,24 ml (4 s c ngha). Trong th d ny 3 con s u tin l s chc chn ng, s cui cng l s khng chc chn ng. Nh vy c th vit 30,24 ml hoc 0,03024 lit (4 s c ngha). S c ngha c qui c nh sau : + Gm cc ch s t nhin 1,2,. 9 + S khng c th l s c ngha hoc khng phi l s c ngha tu thuc vo v tr ca n trong dy s.

- Nu s khng nm gia cc s khc l s c ngha. - Nu s khng nm cui dy s th ch l s c ngha nu ng sau du

phy. - Nu s khng nm trc du thp phn th khng phi l s c ngha.

* Lm trn s: l loi b cc s khng c ngha trong kt qu. Nu b cc s 6,7,8,9, th tng ga tr trc n ln 1 n v. Nu loi b cc s 1,2,3,4, th khng thay i con s ng trc n. Nu loi b s 5 th lm trn s trc v s chn gn nht. V d: 2,25 lm trn thnh 2,2; 2,35 thnh 2,4.

Th d 2.3 : 25,24 c 4 s c ngha 0,15 c 2 s c ngha 15,00 c 4 s c ngha 1,36 c 3 s c ngha 0,0241 c 3 s c ngha 150,00 c 5 s c ngha Khi ly V=5,00 ml c ngha l khi tnh nng phi ly 3 s c ngha. (Nh vy c th ghi gi tr nng l 0,0215; 2,15.10-2 hoc 21,5.10-3 hoc 215.10-4M) Nu ghi th tch bnh l V= 2,0 lit th khi chuyn sang n v ml khng th ghi l 2000 ml (v y ch ghi 1 s c ngha) m phi ghi l 2,0.103ml. 2.3.1.2. Cch ly gi tr gn ng * i lng o trc tip: gi tr o c phi c hoc o, m c. S liu th nghim c ghi theo nguyn tc s cui cng l s gn ng v s trc s cui cng l s chnh xc. * i lng o gin tip. - Php tnh cng v tr : lm trn s thnh s chnh xc v ghi s c ngha theo g tr no c t s c ngha nht. - Php nhn v chia: kt qu ca php nhn v php chia c lm trn s sao cho n cha s c ngha nh gi tr c t s c ngha nht. (Khi tnh bt n tuyt i khgn tnh n du thp phn)

- Php tnh logrit v ngc logrit: + logrit: ly cc ch s sau du phy bng tng cc s c ngha trong s ban u + ngc logarit: ly cc s c ngha bng s cc ch s sau du phy.

Th d 2.4: a) 3,4+0,020+7,31=10,73=10,7 y v 3,4 l s ch c 1 s c ngha sau du phy nn trong kt qu ch ghi 1 s c ngha sau du phy.


14

b) %5470578,88%100.1689,1

05300.0.5481,0.63,35=

Trong dy s trn, khng chc chn ca mi s l 1/3563; 1/5481; 1/5300 v 111689/ Nh vy khng chc chn ca s th nht ln hn so vi khng chc chn ca s th hai v th ba. Do , gi tr c t s c ngha nht l 35,63 nn kt qu cui cng phi c ghi l 88,55% c) log(9,57.104)=4- log 9,57= 4,981 (gi tr 4 c 1 s c ngha; gi tr9,57 c 3

s c ngha ) log(4,000.10-5)=5- log4,000=-4,397940=-4,3479

Antilog(12,5)=3,162277.1012=3.1012 2.4. Quy lut lan truyn sai s ngu nhin - lch chun ca i lng o gin tip

Tt c cc kt qu phn tch nh lng thu c t thc nghim u c cha sai s ngu nhin. V vy, cc gi tr c bo co thng l gi tr trung bnh vit ng s c ngha km theo sai s ngu nhin ca gi tr . Thng thng chng c vit l Sx , vi S l lch chun. Th d: Trong tp s liu th tch dung dch chun dng cho qu trnh chun , cc gi tr th tch thu c l 10,09; 10,11; 10,09; 10,10; 10,12 ml. Nh vy, th tch dung dch chun dng s l Sx = 10,10+0,01 (vi N=5 th nghim lp li).

Ngoi ra, khi s th nghim lp li ln, kt qu phn tch cn c trnh by di dng

NSt

x. v s c xt n trong chng 3.

Tuy nhin, kt qu nh lng thu c t thc nghim trong rt nhiu php o khng phi l kt qu ca php o trc tip m c th c tnh ton t mt hay nhiu php o trc tip. Mt khc, mi s liu thu c trong cc php tnh u c lch chun ring, v vy phi xt n lan truyn sai s gy ra cho kt qu cui cng.

Gi s cc kt qu thc nghim a, b, c, .. l cc s liu thu c t cc php o trc tip M1, M2 , M3. Gi x l gi tr cui cng tnh ton c t cc kt qu ring r a, b, cKhi x l hm ph thuc vo cc tham s a, b, c

Gi cba ,, l lch chun ca cc php o trc tip xc nh a, b, c.. v gi thit l sai s trong cc php o ny c lp ln nhau th lch chun ca i lng x l :

2/122

22

...][ +

+

= b

b

x

a

a

x

x

(theo nh lut lan truyn sai s, biu thc

ny ng khi x l hm tuyn tnh ca cc php o a, b,c). Cch tnh lch chun ca i lng x ny tu thuc vo dng cng thc tnh

em s dng. * lch chun ca tng v hiu: x = a1. a( Sa) + b 1.b( Sb) c1 .c( Sc) vi a1,b1, c1 l cc hng s th

lch chun ca x l


15

......

221

221

221 +++= cbax ScSbSaS

* lch chun ca php nhn v chia:

x = 1

11.

c

ba

c

ba th ......

22

1

22

1

22

1 +

+

+

=

c

Sc

bSbb

a

Sa

x

S cax

Khi , kt qu s c biu din di dng x = 1

11.

c

ba

c

ba Sx .

* lch chun ca php tnh logarit:

x= k.lna th Sx=

a

Sk a.

x= k.loga th Sx=

a

Sk a.

30,2

Cc gi tr lch chun trong php o trn c gi l sai s tuyt i ca php

o. i lng

a

Sa gi l sai s tng i.

Th d 2.5: a) Tnh gi tr biu thc: (65,060,07) +(16,130,01)-(22,680,02)= 58,51? ta c xS 073,002,001,007,0 222 =++= V biu din x = 58,51 0,07

lch chun tng i ca php o l %1,0%100.51,5807,0 =

b) ?0,356006,0623,4

)2,04,120).(02,067,13(.

=

==

cba

x

ta c 222

623,4006,0

4,1202,0

67,1302,0

+

+

=

x

S x =0,0026 do vy

Sx=356,0.0,0026=0,93 kt qu cui cng s l x = 356,00,9 Th d 2.6 : Tnh lch chun s mmol Cl- trong 250,0 ml dung dch mu, nu ly 25,00 ml dung dch mu ny chun bng dung dch chun AgNO3 c nng ( 0,11670,0002) M. Th tch dung dch AgNO3 tiu tn sau 3 ln o lp li l 36,78; 36,82 v 36,75 ml. HD gii : - th tch dung dch chun AgNO3 trung bnh l: 36,78 ml

- p dng cng thc tnh lch chun th tch chun ta c S= 0,035 - Vy 04,078,36

3=AgNOV (ml)

- S mmol Cl-

c chun trong 250 ml mu : X= (0,11670,0002).((36,780,04).10= 42,92 ?


16

Ta c : 019,010.78,3604,0

1167,00002,0 2

22

=

+

=

x

S x

Do Sx= 42,92.0,019=0,082 Kt qu s mmol Cl- trong 250 ml mu l (42,920,08) mmol

Ch : Trong qu trnh tnh ton v c s lan truyn sai s nn cn trnh lm trn s khi vic tnh ton cha kt thc.


17

Chng 3 HM PHN B V CHUN PHN B 3.1. Biu din s liu nh lng

Trong phn tch nh lng, s liu thc nghim l cc s liu thu c khi tin hnh cc php phn tch nh lng. h thng ho nhng s liu ny nhm thu c ci nhn tng qut hn hoc phc v cho nhng nghin cu tip theo, ngi ta biu din chng di dng biu hoc th. Cc dng biu thng gp l biu ct hay biu hnh ch nht (bar chart), biu hnh qut (pie chart), biu tn sut (historgram) hay biu ng gp khc (pylogon). Nu cn biu din gi tr thc nghim ca cc tp s liu khc nhau, th s dng ln ca cc s liu. Trong trng hp cn biu din cc s liu trong cng tp s liu th thng dng tn sut ca gi tr trong tp s liu.

Trong phn trnh by di y ch xt n biu biu din tn s xut hin ca gi tr trong tp s liu di hai dng biu tn sut v biu ng gp khc .

Cch tin hnh: Cc gi tr trong tp s liu c chia thnh cc nhm khc nhau (category) v kim tra tn sut ca gi tr biu din kt qu o di dng im ring bit trn trc s (c chia tuyn tnh 1 chiu) v nhn nh v mt cc im (trng hp ny gi l phn b 1 chiu) hoc biu din dng bc thang (ct) bng cch tp hp cc gi tr ring r thnh k cp c b rng d (5 < k < 20) (k cn bc hai tng cc gi tr o c).

Th d 3.1: Ngi ta xc nh ng thi Al trong mt mu thp 12 phng th nghim (PTN). Mi PTN cho 5 gi tr phn tch thu c trong nhng ngy khc nhau. Cc gi tr ny c h thng ha nh bng 3.1:

Bng 3.1: Kt qu phn tch hm lng Al (%) trong mu thp

STT PTN X1 X2 X3 X4 X5

1 A 0,016 0,015 0,017 0,016 0,019

2 B 0,017 0,016 0,016 0,016 0,018

3 C 0,015 0,014 0,014 0,014 0,015

4 D 0,011 0,007 0,008 0,010 0,009

5 E 0,011 0,011 0,013 0,012 0,012

6 F 0,012 0,014 0,013 0,013 0,015

7 G 0,011 0,009 0,012 0,010 0,012

8 H 0,011 0,011 0,012 0,014 0,013

9 I 0,012 0,014 0,015 0,013 0,014

10 K 0,015 0,018 0,016 0,017 0,016

11 L 0,015 0,014 0,013 0,014 0,014

12 M 0,012 0,014 0,012 0,013 0,012

Gii hn 8 10 12 14 16 18 20 .10-3%

trn ca cp

ca Hnh 3.1: Phn phi tn sut khi xc nh ng

thi hm lng Al trong mu thp ti 12 PTN.

M M M L M L M L L I L K H I K H I K H I I G H F G H C G F C F F B E F B K G E E B K G E C A B D D E C A B D D D C A A A


18

Nh vy c tt c N=60 gi tr. Gi tr thp nht l ca PTN D c 2D

X =0,007%.

Gi tr cao nht ca PTN A l 5AX = 0,019%. Sau khi tp hp cc s liu thnh k= 7

cp vi rng ca cp l d= 0,002 %Al ta c k N . Cp th nht gm cc gi tr 0,007 v 0,008 % Al, cp th hai l 0,009 v 0,010 % Al.... Nh vy ta c phn b tn sut thc nghim c trnh by hnh 3.1 v biu tn sut phn trm hnh 3.2.

Tan

x

uat

(%

)

2018161412108

35

30

25

20

15

10

5

0

Hnh 3.2. Biu phn trm tn sut hm lng Al trong kt qu phn tch cc PTN

T dng phn b tn sut c th thy c nh tnh v s xut hin sai s ngu nhin. Khi sai s ngu nhin ln th phn b rng, sai s ngu nhin nh th phn b hp v nhn, nhng trong trng hp ny khng cho bit v sai s h thng v sai s h thng khng lm thay i dng phn b.

3.2. Phn b l thuyt Khi h thng ho cc gi tr o v biu din chng trn th bng cch v tn

sut ca gi tr no vi mt trc l gi tr , ta lun thu c cc phn b dng ct nh trn, c bit khi ch c sai s ngu nhin. Do , cho php gi thit c nhng qui lut ton hc lm c s ca nhng phn b .

3.2.1. Phn b chun (Phn b Gauss) Gi s tin hnh rt nhiu th nghim lp li v thu c rt nhiu cc gi tr (N

) trong c mt s yu t ngu nhin nh hng n cc gi tr ny v cc nguyn nhn gy nh hng c tnh cng tnh, nh hn gi tr o.

Khi rng ca lp nh (d 0) th phn b tn sut c biu din bng hm

mt xc sut sau: 2)(

21

21)(

pi

=

x

exy (3.1)

trong : pi 3,1416 e 2,7183; l tham s v l lch chun, c trng cho phn tn ca php o (measure of dispersion); l tham s v l gi tr


19

tht hoc gi tr trung bnh, c trng cho php o v tr phn b (measure of location) ; x l to hoc gi tr trn trc honh; Y: tung d, chiu cao ca ng biu din tung ng vi gi tr x.

V tr v dng ng cong c xc nh bi v . Cc i ca ng cong ti y' = 0, tc l im x= . Cc im un l x1= - v x2 = + . Nu cho . th y = f(x). Khi y = 0 th x = . Tuy nhin, trn thc t c th b qua cc gi tr ca trc tung khi x ngoi khong 3 .

Hnh 3.5: Phn b chun vi cc gi tr trung bnh cng khc nhau.

Hnh 3.6 : Biu din hnh hc ca lch chun

Nu k hiu

=

xZ th Z l mt bin ngu nhin v hm phn b c dng

2.

21

21)( ZezY =

pi (3.2) khi Z=1 v z=0

Hm phn b Z ny c gi l phn b chun hay phn b Gauss. Phng trnh (3.2) m t mt xc sut ca phn b, l tng din tch gia ng cong v trc x l 1 n v. ng biu din cn c gi l ng cong sai s (error curve).

Nu ly tch phn ca hm phn b chun t - n + th ton b phn din tch gii hn bi ng cong biu din xc sut xut hin cc gi tr xi. Gi tr xc sut ny gn lin vi tin cy thng k P. Ni cch khc, phn din tch gii hn bi ng cong l tin cy thng k xut hin xi trong khong tch phn.

i vi cc tp s liu c cng gi tr thc s c cng din tch ng cong Gauss nhng nu cng nh th ng cong cng hp v cng nhn, chnh xc cng ln. Xc sut gi tr o nm ngoi gii hn trn ca tch phn l =1-P. Phn din tch P cng c biu din theo % so vi tng din tch v gi l tin cy thng k.

Trong khong th mt xc sut chim 68 % din tch ca ng cong. Trong khong 2 th mt xc sut chim 95 % din tch ng cong. C

ngha l c 95 % gi tr trung bnh mu nm trong khong:

mt xc sut

lch chun


20

- 1,96(n

)< x < +1,96 (n

). Do khong bin thin gi tr thc l:

x - 1,96(n

)< < x +1,96 (n

) (y l khong tin cy c on ca gi tr trung bnh).

Trong khong 3 th mt xc sut chim 99,7 % din tch ca ng cong. Tc l x - 2,97(

n

)< < x +2,97 (n

)

a s cc kt qu o trong phng php phn tch thng thng u tun theo phn b chun (tr cc php m). Tuy nhin, khi x l thng k, c bit trong cc php phn tch a bin khng c gi thit trc l c phn b chun trong cc tp s liu thu c t cc phng php phn tch (nh phn tch lng vt, phn tch bn nh lng... ) m phi kim tra xem tp s liu c tun theo phn b chun hay khng.

Nu k hiu tin cy thng k xut hin ga tr xi nm trong vng (-, xi) l P(xi). T hm phn b chun, khi cho gi tr ui(x) ta tnh c tin cy thng k Pi (ng vi din tch Pi v ngc li. Thay cho tnh ton, ngi ta lp sn bng s tra gi tr u khi bit P hoc ngc li (xem ph lc 1 )

Ch : -Trong thc nghim c nhng tp s liu tun theo phn b chun (gi tr trung bnh, trung v v s tri trng nhau). Tuy nhin cng c mt s tp s liu khng theo phn b ny m theo phn b lch (skewed distribution) (tn xut ca s tri>trung v>trung bnh). Khi gi tr skewed tin ti khng th phn b lch tr thnh phn b chun. Nhng dng phn b lch ny c th t c gn phn b chun nu chuyn cc kt qu sang dng logarit ri tnh gi tr trung bnh v lch chun . Phn phi ny gi l phn b log-chun (log-normal distribution). 3.2.2. Phn b Poiison:

Trong mt s phng php phn tch hin i, kt qu php o l cc i lng nguyn ri rc, nh m xung vi phn trong Ho phng x, m lng t trong phn tch ph Rn ghenS liu thc nghim trong cc phng php ny c c im nh sau:

- Kt qu trong tp s liu l nhng s m cc s kin xy ra trong mt khong thi gian.

- Xc sut xy ra s kin trong mt n v thi gian l nh nhau vi cc khong thi gian khc nhau.

- S s kin xy ra trong khong thi gian ny c lp vi khong thi gian khc. Nu lp li nhiu ln cng mt th nghim th mi quan h gia gi tr o v tn

xut c biu din bng hm phn b xc sut nh sau:

.

!.

x

eyx

= vi x= 0,1, 2, 3 v l trung bnh ca s cc s kin

trong khong thi gian xt. Phn b ny c gi l phn b Poisson, cc i lng c trng thng k l:

- Gi tr trung bnh = .


21

- Phng sai 2 = - Gia v c quan h: = 1/2 vi l s thc v >0

Hnh 3.6. Phn b Poisson vi cc gi tr khc nhau ca trung bnh cng.

Phn b Poisson l phn b ri rc. Khi nh th phn b c dng bt i xng. S bt i xng gim nhanh khi tng v dng ng phn b tin ti phn b chun.

Thc t khi n > 15 th c th coi nh xp x phn b chun. ng vi bng phn b chun s c 68,3 % cc gi tr trong gii hn - 1/2 v +1/2.

3.2.3. Cc phn b c bit. 3.2.3.1. Phn b Student (t) Phn b chun xt trn ch thch hp vi trng hp s php o ln (N).

Khi s php o nh, mt phn b c th lch khi qui lut ca phn b chun, do cn loi tr khng tin cy bng phn b i xng bin dng gi l phn b student (t).

Hm ca phn b t c dng:

212

)1(),(+

+=f

ftBftY vi B l hng s v f l bc t do.

Hm phn b ny ph thuc bin t mt cch ngu nhin. th ca hm t c dng ca hm phn b chun v c y tnh cht nh hm

phn b chun nhng nhn ca th hm phn b t ph thuc vo bc t do (hnh 3.7).


22

Hnh 3.7: Phn b Student vi f=1; f=3, f=5, f=100 v phn phi chun.

Chiu cao v rng ca cc ng cong ca phn b t 8 chun ho ph thuc vo bc t do f ca lch chun. Bc t do f cng nh th ng cong cng t. Khi N th S v phn b t chuyn thnh phn b chun Z (thc t ch cn xt vi N>30). Cc gii hn tch phn ca phn b t ph thuc vo xc sut P v bc t do f c cho trong ph lc 2. Khi bit hai gi tr f v P c th tra bng t tm gi tr tch phn ca phn b t. Hai loi bng tra gi tr t tng ng vi phn b t mt pha hoc hai pha (hnh 3.8).

Chun t (Student-test) c dng tnh khong tin cy ca s liu thc nghim, so snh gi tr trung bnh thc nghim v gi tr tht, so snh 2 gi tr trung bnh hoc tnh khng m bo o ca lch chun mu khi s mu nh.

3.2.3.2. Phn b Fisher (F)

Gi s c 2 tp s liu vi kch thc mu N1 v N2, phng sai tng ng l S12

v S22 vi cc bc t do f1= N1-1 v f2= N2-1 v lp t s :

PP chun

Phn phi chun

/2 /2

Hnh 3.8 : Phn b Student 1 pha (1 sided) v hai pha (2 sided).

xc sut P


23

22

21

SS

F = (F>1)

Th hm mt xc sut c dng: 2

2

1

22

),,( 21

1

21

)1(ff

f

ffx

ff

xAY+

+

=

trong , x l bin ngu nhin v A l hng s ph thuc f1 v f2; 0 x +.

ng cong thu c mang c tnh ca mt pha, c v trong gc phn t th nht gia x=0 v x= (hnh 3.9).

Hnh 3.9. Phn b F vi hai bc t do f1 v f2.

Nu ly tch phn hm phn b trong gii hn 0...Fp ( Fp

24

Hm phn b vi 2 nm trong gc phn t th nht trong min t 2=0 n 2= c dng ph thuc vo bc t do f (hnh 3.10).

Nu f nh, ng cong bt i xng, nu f tng s bt i xng gim v f ta c ng cong Gauss vi >0. Ly tch phn hm phn b trong gii hn t 0 n 2P (2P15

Phn phi Poisson

F=S12/S22

Nsx

t/

= 2/22 fS=

=

xZ

!/xeP x =

f=2 f=10

Hnh 3.10: Phn b 2 vi f bc t do.


25

3.4. Khong tin cy, gii hn tin cy v khng m bo ca i lng o

Khong tin cy (confidence interval- CI) ca i lng o l gi tr thc biu th khong tn ti gi tr trung bnh hay cn gi l khong bt n ca s liu thc nghim trung bnh.

Gii hn tin cy (CL: confidence limit) l gi tr ln nht v nh nht ca khong tin cy.

Vic tnh ton khong tin cy ca gi tr trung bnh ch c thc hin khi sai s h thng xut hin khng ng k.

Vi mt tp s liu tun theo phn b chun, khi bit lch chun , th s sai

khc gia gi tr thc v gi tr trung bnh x khng ln hn Z ln sai chun ca tp hp. Ni cch khc

NZx 30 v tun theo phn b chun.

i vi cc tp s liu nh (tc l cc mu thng k c N

26

RtRxCL .+=

Gi tr tR tra tin cy thng k P=0,95 v P=0,99 nh bng 3.2.

Bng 3.2. Gi tr t tra theo khong bin thin R tin cy thng k 95% v 99%

N 2 3 4 5 6 7 8 9 10

tR 0,95 6,4 1,3 0,72 0,51 0,40 0,33 0,29 0,26 0,23 0,00

tR 0,99 31,83 3,01 1,32 0,84 0,63 0,51 0,43 0,37 0,33 0,00

3.5. Mt s bi ton lin quan n khong tin cy

3.5.1. X l s liu thc nghim tm khong tin cy ca gi tr thc

- Khi cha bit lch chun S hay khong bin thin CV

Gi s c tp s liu thc nghim : x1, x2, ...xN. T d8y s ny ta tm c gi tr trung bnh, phng sai S2 v lch chun S.

Nh vy, vi tin cy P=0,95, tra bng ta c t(P,f) v xc nh c gi tr cn tm

nm trong khong N

Stx =)(

Th d 3.2: Kt qu phn tch hm lng it trong mt mu nc bin Thanh Ho theo phng php ng hc xc tc -trc quang ln lt l: 24,75; 25,12; 24,76;

26,28; 25,15 g/l. Tm khong xc nh ca hm lng thc it trong mu nc ny. (SV t gii)

- Khi bit lch chun S hay khong bin thin CV

Gi s c tp s liu thc nghim : x1, x2, ...xN.

* Nu N30: c th xem nh tp s liu ca mu thng k l tp hp v tp s liu tun theo phn phi chun. Do vy, tin cy thng k 95% ta c Z=1,96, nn khong tin cy s l:

NS

x 96,1)( =


27

3.5.2. Xc nh s th nghim cn tin hnh thu c chnh xc mong mun:

Theo cng thc: NS

tx =)(

Gi tr - x = NS

t c gi l khng chc chn, hay khng m

bo o ca kt qu thc nghim. Khi s th nghim ln th gi tr ny gim c n bt k gi tr no mong mun x . mt mc hm lng cht cn phn tch c th, gi tr - x v lch chun S c cho trc (theo ISO), t ta s tnh c i lng

NS

t . Tra bng vi t(P=0,95; n ) =1,96 s tm c N

kt qu thc nghim c tin cy cho trc.

3.5.3. Chn phng php phn tch thch hp c sai s nh hn gii hn cho trc .

Mi phng php 8 bit u mc sai s tng i cho trc. Bi ton t ra l cn chn phng php no sau N ln th nghim th t chnh xc CV(%) mong mun.

Theo cng thc = NS

t nu 8 bit v N, s tnh c S sau , thay

vo cng thc trn xem c tho m8n iu kin CV a % cho trc hay khng. Theo ISO, vi cc mu c nn phc tp, quan h gia CV(%) v nng

cht phn tch c cho bng 3.3.

Bng 3.3: Quan h gia nng cht phn tch v CV cho php

Hm lng

100

g/kg

10

g/kg

1

g/kg

100

mg/kg

10

mg/kg

1

mg/kg

100

g/kg 10

g/kg 1

g/kg 0,1

g/kg

CV(%) 2 3 4 5 7 11 15 21 30 43

Cng theo ISO, sai s tng i c nh gi qua chnh xc ca phng php l :

1 ppb sai s tng i cho php t -50 % n +30 % > 1 ppb n 10 ppb, sai s tng i cho php -30% n +10%

> 10 ppb, sai s tng i cho php -20% n +10%.


28

Chng 4: CC PHNG PHP KIM TRA THNG K

4.1. Nguyn tc php kim tra thng k (significant tests) Mc ch ca cc php kim tra thng k l lm cho kt qu phn tch c din

gii mt cch khch quan nhm gii p cu hi c s khc nhau gia cc kt qu thu c hay khng. Ni cch khc, cn kim tra xem gi thit thng k cc kt qu o cng tp hp l ng hay sai?

Trong thc t phn tch, nh ho hc thng t ra gi thit v phn tch thng k s liu a ra xc sut v gi thit . Ni cch khc ta gi thit l ng (gi thit o- null hypothesis) v tnh ra xc sut l gi thit ng.

Cch tin hnh: T kt qu cn kim tra ca mu, tnh gi tr ca mt i lng cn kim tra , xc nh min trong tn ti vi xc sut P nh trc. Nu nm ngoi min th gi thit chn (hai i lng ging nhau) b bc b v s khc nhau gia cc i lng thu c gi l s khc nhau c ngha.

Khi kt lun ngi ta tun theo 3 qui tc sau: - Gi thit cn kim tra b bc b nu sai lm loi mt (b ci ng) xut hin t

hn 100 (1% tng trng hp) (P 0,99 hay tr s P tc l Pvalue 0,05) th kt lun s khc nhau khng c ngha, tc l c xem nh ging nhau mc tin cy 5%.

- Nu sai lm loi mt nm trong khong 5% v 1% (0,95 < P < 0,99 hay 0,01

29

Nguyn tc: Sp xp cc s liu thu c theo chiu tng hoc gim dn v dng Q-test nh gi kt qu nghi ng khc xa bao nhiu so vi s cn li trong tp s liu. Tnh gi tr Q theo biu thc (1) v so snh vi gi tr Q chun trong bng 4.1:

Qtnh= minmax xx

xx canlanngonghi

So snh Qtnh v Qchun (P=0,90; N). Gi tr nghi ng s chnh l gi tr bt thng nu Qtnh > Qchun (P,N).

Bng 4.1 : Gi tr chun Q dng loi b gi tr bt thng.

N Mc tin cy

90% 95% 99%

3 0,89 0,94 0,99

4 0,68 0,77 0,89

5 0,56 0,64 0,76

6 0,48 0,56 0,70

7 0,43 0,51 0,64

8 0,47

9 0,44

10 0,41

Ch : Nu s php o ln (N >10) th cch pht hin theo chun Q khng nhy, do trong php kim tra ny ch c gi tr nghi ng v hai gi tr khc ca php o c s dng. Khi , kim tra s tn ti ca gi tr bt thng, ngi ta dng tiu chun 2.

Th d 4.1 : Kt qu xc nh hm lng CaCO3 (%) trong mt mu olomit thu c nh sau: 54,31;54,36; 54,40; 54,44 ; 54,59 %.

Hjy kim tra xem gi tr nghi ng 54,99 c phi l gi tr bt thngkhng?

Gii: S gn nht ca 54,99 l 54,44.

Ta c: Q= 8,031,5499,5444,5499,54

=

Vi 5 lVi 5 ln th nghim v P=0,90 tra bng chun Q ta c Qchun=0,56. vy Qthc nghim >Qchun hay ga tr 54,59 l gi tr bt thng.

* Tiu chun 2: (p dng cho tp s liu c N>10)

Da trn khong gii hn tin cy: x 2 cha 95 % s liu o c vi x l gi tr trung bnh ca tp s liu (8 loi b s liu nghi ng) v l lch chun tp hp. Nhng gi tr no ngoi khong trn s c loi b.

*Tiu chun 3: Gi s tp s liu thc nghim c sp xp theo th t tng dn

xL , x2, , xH . Tnh gi tr trung bnh x v lch chun S v kim tra cc gi tr nghi ng theo cch sau:

Trc tin tnh S

xxT H = i vi gi tr cao nghi ng.


30

V S

xxT L= vi cc gi tr thp nghi ng.

Sau so snh gi tr T tnh dc vi gi tr Tchun (s php o: N) trong bng 4.2 mc ngha 5% v 1%:

Nu Ttnh>Tchun th xL v xH l sai s th cn loi b mc ngha thng k 8 cho.

Bng 4.2: Gi tr T chun 5% v 1 % ca s khng ph hp vi gi tr bt thng trong mu chun.

Gi tr chun Gi tr chun S php o (N)

5% 1%

S php o (N)

5% 1%

3 1,15 1,15 15 2,41 2,71

4 1,46 1,49 16 2,44 2,75

5 1,67 1,75 18 2,50 2,82

6 1,82 1,94 20 2,56 2,88

7 1,94 2,10 30 2,74 3,10

8 2,03 2,22 40 2,87 3,24

9 2,11 2,32 50 2,96 3,34

10 2,18 2,41 60 3,03 3,41

12 2,29 2,55 100 3,21 3,60

14 2,37 2,66 120 3,27 3,66

Ngoi ra, cc gi tr bt thng c th c nhn bit bng cch dng th boxplot trong phn mm thng k MINITAB.

4.3. S dng chun thng k trong cc php so snh 4.3.1. So snh trong mt tp s liu (1 sample)

4.3.1.1. Kim tra s tun theo phn b chun

Trong rt nhiu php tnh thng k, tp s liu cn phi tho m8n iu kin tun theo phn phi chun, tc l phi tho m8n cc iu kin ca phn phi chun t ra. Vic s dng cc phn mm thng k cho php n gin hn th tc tnh ton bng cch xt ga tr lch (skewness) trong thng k m t hoc dng cc chun thng k nh Kolmononov- Smirnov.


31

Th d 4.2. Kt qu phn tch hm lng Ni( mg/kg) trong mu t nh sau: 22 15 18 25 21 12 23 20 20 42 22 31 22 13 8 33 12 23 12 16 30 36 15 17 28 26 26 16 23 26 15 17 17 14 14 18 12 35 30 15 13 14 14 14 13 7 43 59 25 37 7 10 8 13 2 14 11 19 5 12 19 11 15 2 15 31 9 11 26 33 27 13 12 20 26 16 15 22 6 10 . Hjy kim tra xem cc s liu trong tp s liu trn c tun theo phn phi chun khng.

Gii: S dng phn mm Minitab 14 tnh cc i lng thng k trong thng k m t.Kkt qu thu c nh sau:

Variable Mean StDev CoefVar Minimum Median Maximum Skewness Kurtosis

Ni 18.99 9.91 52.17 2.00 16.00 59.00 1.22 2.51

Biu tn sut xut hin cc gi tr trong tp s liu c dng:

Ni

Frequency

60483624120

25

20

15

10

5

0

Mean 18.99

StDev 9.907

N 80

Histogram (with Normal Curve) of Ni

Gi tr skewness kh nh, ng biu din tn sut gn vi phn phi chun.

Nu s dng thut ton kim tra phn phi chun (Normality test) vi chun Kolmogorov- Smirnov ta c cc gi tr: KS=0,119, P-value 0 hay

32

- Quyt nh da trn mc tin cy thng k s dng trong trng hp phn

phi chun: Nxz

)( 0=

-Tm phn phi mu ca gi tr thng k nu khng nh n ng.

y phi gi nh rng Nxz

)( 0= c phn phi chun vi gi tr trung

bnh bng "khng" v phng sai bng "mt".

- Tnh gi tr Z v so snh vi g tr Zchun trong bng 4.3.

Bng 4.3: Gi tr Z cc mc tin cy thng k khc nhau.

Mc tin cy (%) 50 68 90 95 99 99,9

Z (2 sided 0,67 1,000 1,645 1,960 2,576 3,29

Z (1 sided) 0,0 0,407 1,282 1,645 2,326 3,08

Nu Z 1,96 th loi b gi thit o (vi =0,05). Nu chn = 0,01 th xt khong -2,58 n +2,58.

Phng php ny ch p dng cho tp s liu tun theo phn phi chun. Nu Z< Zbng th chp nhn gi thit o hay ni cch khc v o khc nhau khng c ngha thng k. Nu s dng phn mm thng k th gi thit o c chp nhn nu Pvalue P ( thng chn l 0,05 tc l khi gi thit ng m loi b th s mc sai lm loi mt vi xc sut l ).

Khi cn so snh s khc nhau gia hai i lng th phn b xc sut c dng l phn b 2 pha (2 sided). Trung hp hai i lng khc nhau th c th dng phn phi xc sut 1 pha (1 sided) so snh gi tr no ln hn.

Th d nu gi tr Pvalue=0,027 th c ngha l ch c 2,7% c hi o. Do vy, cn kt lun l o .

4.3.1.3. So snh gi rt trung bnh mu v gi tr c chp nhn (chun t).

Chun student c dng so snh xem c s khc nhau c ngha gia gi tr thc nghim x v gi tr thc hay khng. Phng php ny cng c dng so snh kt qu thc nghim vi gi tr chun trong mu kim tra cht lng (quality control standard) v mu chun so snh (standard reference materials- SRM).

Php so snh ny da trn khong tin cy ca gi tr trung bnh. Nu s khc nhau gia gi tr tm c v gi tr thc ln hn khng m bo o ca php o th th chng t c s khc nhau c ngha gia hai gi tr ny tin cy thng k d8 cho.

Vi tp s liu c N >20 hoc khi bit lch chun tp hp th N

Zx

.

Vi tp s liu c N

33

Mt cch khc, so snh v x ngi ta tnh gi tr tthcnghim = .x N /S sau so snh vi gi tr tchun(P,f) (tra chun Student 2 ui.

Nu tthcnghim> tchun hoc Pvalue P th gi thit o b bc b tc l khng c s khc nhau c ngha thng k gia gi tr trung bnh v gi tr thc.

Phng php ny cng c dng nh gi sai s h thng ca phng php phn tch bng cch tin hnh phn tch lp li N th nghim t mu chun (8 c gi tr thc hoc gi tr c chp nhn ) v nh gi s sai khc gia gi tr x vi gi tr thc .

Tnh gi tr t theo biu thc NS

xt

= v so snh vi t(P,f) vi f=N-1

Nu ttnh < tbng c th kt lun x khng khc hay phng php ch mc sai s ngu nhin tc l phng php c ng chp nhn c.

Nu ttnh > tbng th phng php phn tch mc sai s h thng.

Cch so snh ny cn c p dng :

- So snh phng php nghin cu vi phng php chun bng cch so snh gi tr trung bnh ca tp s liu trong phng php nghin cu vi kt qu c phn tch bng phng php chun.

- Xt nh hng ca nguyn t l (so snh khi c nguyn t l v khi khng c nguyn t l)

- nh gi nh hng ca dung mi chun khi thm 1 dung mi khc.

Th d 4.3: Khi nghin cu phng php trc quang xc nh As(III) bng vi thuc th bc ietyl ithio cacbamat sau khi hyrua ho bng k thut kh in ho, cc tc gi j phn tch As(III) trong mu t to (c mt As(V) sau 5 ln lp li. Kt qu thu c (trung bnh lch chun) nh sau:

Mu As thm vo(g) As(III) tm thy(g) As(III) As(V)

Nc my 10 50 9,60,4

20 50 19,70,3

Nc bin nhn to 10 50 10,20,4

20 50 20 0,3

Hjy kim tra xem phng php nghin cu c mc sai s h thng hay khng v c nn p dng phn tch asen trong nc bin khng?

Ngun: M.H. Arbab-Zavar, M. Hashemi :Talanta 52 (2000) 10071014.

4.3.2. So snh hai tp s liu (2 samples)

4.3.2.1. So snh phng sai ca hai tp s liu (chun Fisher : 2 2 )

Chun Fisher c dng so snh chm (precision) ca hai tp s liu hoc hai phng php khc nhau. Gi s c hai tp hp kt qu phn tch thu c t hai


34

ngi phn tch, hai PTN phn tch hoc hai phng php vi hai gi tr phng sai l 21 v

22 , bc t do tng ng f1 v f2. Nh vy, cn gii p cu hi

21 v

22

c phi l phng sai ca cng tp hp khng?

Vy gi thit thng k trong trng hp ny l 2222

1 == .

Vi cc tp s liu ca mu thng k c s th nghim xc nh v khng ln th bi ton tr thnh so snh hai gi tr 21S v

22S .

Nu "gi thit o" tho m8n th t s 22

21

SS

phi tun theo phn phi chun Fisher

vi cc bc t do l f1 v f2 v gi tr F dc tnh theo cng thc:

Ftnh= 22

21

SS

>1

Khi , s bc b gi thit kim tra nu Ftnh > Fbng (P, f1, f2) (chun 2 ui: 2-tailed-test) hoc P > P. Ni cch khc, hai phng sai

21S v

22S c s khc nhau c

ngha hay chnh xc cc s liu thc nghim gia hai mu thng k (hoc hai phng php) l khc nhau.

Nu lp li hai phng php khc nhau th c th kim tra xem phng php A chnh xc hn hay km chnh xc hn phng php B (kim tra chun 1 ui: one-tailed-test). Nu Fthc nghim > Fchun (P,f1, f2) th c th kt lun phng php A km chnh xc hn phng php B.

Th d 4.5: nghin cu phng php, cn so snh lp li ca hai php o khi xc nh Na theo phng php quang ph pht x ngn la. Cc ga tr lch chun thu c ( tnh theo phn trm tng i) nh sau:

Phng php 1: S1= 3%; f1= 12

Phng php 2: S2 =2,1%; f2=12

Ta c F= 19,41,23,4

2

2

22

21

==

SS

Tra bng chun F ta c F(0,95; 12;12)=2,79

F(0,95; 12;12)=4,16

Vy F= 4,19 > 4,16 nn c th kt lun rng lp li ca hai php o khc nhau c ngha, hay lp li ca hai phng php khng ging nhau.

Khi cn so snh chnh xc ca phng php nghin cu c ln hn c ngha so vi phng php tiu chun khng?

Th d 4.6: nh gi mt phng php mi c xut xc nh SO42-

trong nc thi cng nghip, ngi ta so snh ca phng php ny vi phng php tiu chun qua th nghim sau:

Phng php

Gi tr trung bnh

S th nghim lp li

Bc t do lch chun (mg/)l

Phng php tiu chun 72 8 7 3,38


35

Phng php xut 70 8 7 1,50

Hi c s khc nhau v ng ca hai phng php hay khng.

( SV t gii)

4.3.2 2. So snh 2 gi tr trung bnh thc nghim (Chun Student: 2t).

Gi s c hai gi tr trung bnh Ax v Bx thu c t hai d8y php o vi s th

nghim lp li l nA v nB c lp nhau. Gi thit o cn kim tra l Ax v Bx ging

nhau hay s khc nhau gia Ax v Bx c phi do sai s ngu nhin hay khng? iu

c ngha l cn kim tra xem c s khc nhau c ngha gia hiu ( Ax - Bx ) v gi tr 0 hay khng.

Cch lm:

Bc 1: Kim tra xem lp li ca hai tp s liu (qua phng sai 2AS v 2

BS ) c ng nht khng hay c khc nhau c ngha thng k hay khng? (chun F).

- Nu 2AS v 2

BS ng nht ( khc nhau khng c ngha) th tnh Spooled theo bc 2.

-Nu hai phng sai khng ng nht th tin hnh bc 3, s dng phng sai ca A v B.

Bc 2: Nu 2AS v 2

BS ng nht

Tnh lch chun hp nht Spooled ca hiu 2 gi tr trung bnh Ax v Bx vi s th nghim nA v nB tchun(P,f) (tra chun t 2-pha) th s khc nhau gia Ax v Bx l c ngha thng k.

Nu tthcnghim > tchun(P,f) (tra chun t 1-pha) th s khc nhau gia Ax > Bx l c

ngha thng k. Hoc Pvalue

36

2

22

1

21

21

ns

ns

xxtcalc

+

=

So snh vi ga tr tchun tra bng vi P=0,95 v bc t do f tnh theo cng thc

2

11 2

2

2

22

1

2

1

21

2

2

22

1

21

+

++

+

=

n

ns

n

ns

ns

ns

f

Trong mt s trng hp, phng php trn khng thch hp so snh hai gi tr trung bnh thc nghim v s mu hn ch, mi phng php so snh ch phn tch mt mc hm lng, lm lp li n ln, do khng thch hp cho ton b vng nng kho st. Vic so snh nh gi phng php phn tch s c trnh by trong phn 4.4.

Th d 4.6: so snh 2 phng php xc nh hirocacbon a vng thm (phng php hunh quang v phng php UV) trong t, ngi ta tin hnh cc php phn tch vi 10 th nghim ca mi phng php. Gi tr trung bnh thu c ca phng php hunh quang l 28,00 mg/kg , lch chun S = 0,30 mg/kg; ca phng php UV l 26,25 mg/kg; S= 0,23 mg/kg. Hi gi tr trung bnh ca hai phng php c khc nhau c ngha hay khng?.

( Sv t gii)

4.3.2.4. H s tng quan (coefficient of corelation:

COR)

Cng thc tnh h s tng quan tuyn tnh Pearson s c trnh by trong chng 6.

Trong a s trng hp, h s tng quan Pearson (R) gia tng cp bin thng c dng. i lng ny c trng cho mc quan h tuyn tnh gia hai bin.

R nm trong khong t -1 n +1. Nu R>0 th hai bin c tng quan ng bin cn R

37

4.4. So snh 2 phng php Gi s chng ta nghin cu phng php A phn tch cht cha bit no .

Sau khi tm c cc iu kin ti u cho php xc nh cn tin hnh nh gi phng php phn tch vi phng php tiu chun. Nu s dng phng php so snh hai gi tr trung bnh s khng thch hp v kt qu ph thuc vo nh hng ca lng cht nn khc nhau c trong mu phn tch. Khi , cn tin hnh th nghim theo tng cp. Vi mi mu phn tch cn lm ng thi hai phng php: Phng php ang nghin cu v phng php tiu chun v tin hnh vi cc kch thc mu khc nhau . Cc gi tr thu c ln lt l x1A, x1B; x2A, x2B..xiA v xiB.. Cc kt qu thu c c th so snh theo phng php tng cp hoc phng php th.

4.4.1. So snh tng cp

nh gi phng php phn tch ang nghin cu vi phng php chun, cn phi so snh tng cp kt qu (mi kt qu ca mi phng php mt mc nng nht nh) v s dng chun t so snh tng cp (a paired- t- test).

Gi thit o trong trng hp ny l khng c s khc nhau c ngha v kt qu phn tch cng hm lng cht phn tch trong cng mu ca hai phng php. Ni cch khc, cn so snh hiu s trung bnh ca hai tp s liu c khc khng c ngha hay khng.

Gi tr t c tnh theo cng thc: t= NSx dd . ; Vi BABAd xxN

xxx ii =

= )(

dx l trung bnh s sai khc gia cc cp gi tr.

V Sd lch chun c on ca s sai khc.

gi tr tchun c tra trong bng chun vi mc ngha P=0,95 v (n -1 ) bc t do.

Nu ttinhP=0,05 th gi thit "khng" c chp nhn, c ngha l hai phng php khng c s khc nhau c ngha. Phng php ny cn gi l phng php hiu s.

4.4.2. Phng php th

V s liu trn th hai chiu mt trc l phng php phn tch (gi s l phng php M) v mt trc l phng php chun (gi s phng php N)

Phng php N

Gi s theo phng php M s sai khc l M v sai s tuyt i l M cn theo phng php N s sai khc l N , sai s tuyt i l N .

Phng php M

A

B

M

N


38

Mun so snh hai phng php ngi ta so snh hai t s M

M

v

N

N

bng cch

ly t s hai i lng ny (hay chnh l dc cu ng biu din).

NMNM

NM

NN

MM tg

///

//

=

=

Kt qu ny cng ln (cng gn 1) th s chnh xc ca phng php M ang nghin cu cng cao. ng biu din khi s tin ti ng thng. Chng ta s xt gi tr ny di dng h s tng quan ca phng trnh hi qui tuyn tnh bc mt (R 1) chng 6.

Th d 4.8: Kt qa phn tch Hg (g/l) trong mu nc bt bng phng php FIA (Phng php A) v phng php thng thng (Phng php B) trong 20 mu thu c nh sau:

STT 1 2 3 4 5 6 7 8 9 10

PP A 47,5 29,5 74,4 5,5 30,9 9,8 25,5 2,9 8,6 23,8

PP B 51,8 27,4 71,6 6,0 29,2 8,0 23,2 3,2 8,8 23,5

STT 11 12 13 14 15 16 17 18 19 20

PP A 84,4 147,0 30,6 19,9 33,9 25,0 107,6 18,0 125,3 84,9

PP B 87,9 150,5 29,8 19,8 29,0 25,3 107,5 15,1 134,6 81,4

(Ngun: T.Gao, J. Baasner, M.gradl, A. Kistner, Analytical Chimica Acta, 320, (1996), 171-176.)

Hjy dng phng php so snh tng cp xem cc kt qu xc nh ca hai phng php c trng nhau khng?

( Sinh vin t gii)

(Cho kt qu tnh theo phn mm MINITAB 14.0 nh sau:

N Trung bnh lch chun sai chun

ppA 20 46.7500 42.2895 9.4562

ppB 20 46.6800 43.9953 9.8376

Khc nhau: 20 0.070000 3.235836 0.723555

95% CI for mean difference: (-1.444418, 1.584418)

T-Test of mean difference = 0 (vs not = 0): T-Value = 0.10 P-Value = 0.924

Hjy nh ga kt qu trn v a ra kt lun v s ging hay khc nhau ga hai phng php.


39

Chng 5: PHN TCH PHNG SAI

Trong chng trc chng ta xt bi ton so snh gi tr trung bnh ca hai tp s liu trong tp hp bng cch dng chun t. Vic so snh s chnh xc hn nu cng nhiu tp s liu trong tp hp c xt n nu. Tuy nhin, nu cn so snh nhiu hn hai gi tr trung bnh th chun t khng cn ph hp. Do vy cn xt n nh hng ca yu t trong nhm v giu cc nhm qua nh gi phng sai. Phng php ny thng c gi l phn tch phng sai (analysis of variance- ANOVA) hn l thut ng phn tch trung bnh a nhm (multi-group means analysis).

Nh vy, c th ni, phn tch phng sai l phn tch tc ng ca mt hay nhiu yu t n kt qu th nghim qua tham s phng sai. c th l nh hng ca mt hay nhiu yu t hay nh hng tng h ca nhng yu t . Ngoi vic dng so snh nhiu ga tr trung bnh, ANOVA cn c dng nh gi nh hng ca nhng ngun sai s khc nhau n dy kt qu th nghim t nh gi c nh hng ca cc ngun sai s n s phn b mu .

Ngun sai s c chia thnh hai dng: - nh hng ngu nhin ca yu t thm vo. - nh hng c nh hay c kim sot ca th nghim. Ni cch khc, phn tch phng sai l lm th nghim theo qui hoch nh trc

nhm kho st nh hng c ngha ca cc yu t n kt qu th nghim qua vic nh gi phng sai theo chun Fisher.

Nu ch so snh hai gi tr trung bnh th phn tch phng sai tr thnh php so snh s dng chun t.

Cc bi ton v phn tch phng sai c 3 dng ch yu: - So snh nhiu ga tr trung bnh: thc cht l bi ton mt yu t, k mc th

nghim, mi mc nghin cu lp li n ln (one way ANOVA or one - factor ANOVA). - Bi ton hai yu t A v B, yu t A c k mc th nghim, yu t B c m mc

th nghim, mi mc ca A v B lm lp li n ln (two-way ANOVA). - Bi ton 3 yu t tr ln (Latin squares).

5.1. So snh mt s gi tr trung bnh Gi s cn so snh s khc nhau c ngha thng k hay khng ca cc gi tr

trung bnh mu ,1x ,2x ,3x ,kx trong cng tp hp. Cc trung bnh mu ny thu c t n th nghim trong mi mu thng k.

Mu thng k 1: x11 , x12 , ., x1n v c gi tr trung bnh l 1x

Mu thng k 2: x21 , x22 , ., x2n v c ga tr trung bnh l 2x

Mu thng k th i : xi1, xi2 ,., xij v c ga tr trung bnh l ix


40

Mu thng k k : xk1 , xk2 , , xkn v c ga tr trung bnh l kx

Gi thit o trong trng hp ny l cc mu c ly t cng tp hp c trung bnh mu l v phng sai tp hp l 2 . Ni cch khc cn kim tra gi thit o l = 1x = 2x == kx . Khi cc mu thng k thuc cng tp hp th phng sai trong m mu (within-sample) phi chnh l phng sai gia cc mu (between sample). Vic so snh ny c thc hin qua chun F bng cch tnh t s hai phng sai gia cc mu thng k v trong cng mu thng k ri so snh vi gi tr trong bng F (hoc so snh ga tr P value vi ) a ra kt lun thng k.

* Phng sai trong cng mu thng k:

1

)(1

211

21

=

=

n

xx

S

n

jj

1

)(1

2

2

=

=

n

xx

S

n

jkkj

k 1

)(1

2

21

=

=

n

xx

S

n

jiij

i

Mi mu c n th nghim lp li, do c n-1 bc t do. Tng s mu thng k l k mu. Vy bc t do i din cho tt cc cc mu l f0 =k(n-1).

Do vy, phng sai trong cng mu (within-sample estimation of variance/ within-sample mean square) s l:

)1(

)(1 1

2

1

2

==

= ==

nk

xx

k

SMS

k

i

n

jiij

k

ii

within

*Phng sai gia cc mu: (between-sample estimation of variance)

Trung bnh tp hp : k

x

X

k

ii

=

=1

phng sai gia cc mu:

1

)(1

2

=

=

k

XxkMS

k

ii

between bc t do f1=k-1

Nu gi thit o l ng th hai phng sai phi khng khc nhau hay nh nhau. Cn nu ga thit o l sai th phng sai gia cc mu phi ln hn phng sai trong cng mu thng k.

Ni cch khc ta tnh bi thc:

wwithin

betweencalculate MS

MSF = v so snh vi ga tr Fbng(P=0,95; f1=k-1; f0=k(n-1)

Nh vy nu Ftnh >Fbng th gi thit o b loi b tc l cc ga tr trung bnh ca cc mu thng k l khc nhau c ngha. iu ny c th do c mt gi tr trung bnh khc vi cc gi tr trung bnh khc, hoc cc gi tr trung bnh khc ln nhau hoc cc gi tr trung bnh phn thnh hai nhm ring bit. Mt cch n gin tm


41

ra nguyn nhn s khc nhau gia cc gi tr trung bnh l sp xp cc ga tr trung bnh theo th t tng dn v so snh s khc nhau ca hai gi tr trung bnh cnh nhau vi i lng biu th s khc nhau c ngha ti thiu (A). Nu hiu hai ga tr trung bnh cnh nhau ln hn A th c ngha chng gy ra s khc nhau trong tp hp.

i lng A c tnh theo cng thc sau:

),(.2

fptkSA = Vi S l lch chun c on trong cc mu

withinMSS =

t l gi t chun student tra bng vi tin cy thng k P=0,95 v bc t do f= k(n-1).

Th d nu c 4 gi tr trung bnh cu 4 mu A, B, C, D ln lt l 92 , 97, 99 v 102. tnh ton trn cho thy chng khc nhau c ngha . S th nghim lp li trong mi mu l 3 v S= 3 th gi tr A= 3,26.

Hiu ca hai ga tr trung bnh gia hai mu A v B l 5 >3,26. Vy nguyn nhn cc gi tr trung bnh mu ny khc nhau l do hai mu A v B khc nhau c ngha gy ra.

So snh cc ga tr trung bnh cng c th p dng cho bi ton c hai yu t v nh ga c nh hng tng h ca hai yu t ny.

Th d 5.1: Mt PTN A cn ch to mu chun xi mng xc nh hm lng cc kim loi theo phng php hunh quang tia X (XRF). Mu chun c ly ngu nhin t cc bao xi mng, sau nghin nh v trn tht u ri gi i phn tch cc PTN. nh gi ng u ca mu ngi ta chia mu chun ban u (c xem nh tp hp) thnh 8 mu nh (mu thng k). Tin hnh phn tch hm lng Al (tnh theo phn trm Al2O3) trong mi, lm lp li 6 ln. Kt qu thu c nh sau:

Hjy dng phng php ANOVA kim tra xem gi tr trung bnh gia cc mu c ging nhau khng v kt lun thnh phn mu c p ng yu cu ng nht khng.

PTN/M laps

1 2 3 4 5 6

1 4,5 3,9 4,9 5,3 5,1 4,9

2 5,3 5,1 4,8 5,0 4,6 4,9

3 5,5 5,2 5,0 4,7 4,6 5,2

4 4,9 5,2 5,2 4,7 4,3 4,5

5 5,3 5,6 5,7 5,1 4,9 5,1

6 4,9 4,6 5,1 5,3 4,8 5,0

7 5,2 4,7 4,9 5,1 5,7 5,3


42

8 4,9 5,0 5,2 5,4 5,6 5,7

Gii: Nhp s liu vo phn mm MINITAB 14 di dng ct l % Al2O3 v yu t l cc mu t 1 n 8 .

Vo Stat->ANOVA-> Analysis of Means, nhp response l %Al2O3. Trong Distribution of data chn Normal, factor 1 l ct cha Mu, alpha-level l 0.05, v tick vo OK. Kt qu thu c nh sau:

mau

Mean

87654321

5.50

5.25

5.00

4.75

4.50

4.657

5.393

5.025

One-Way ANOM for Al2O3(%) by mauAlpha = 0.05

Trong th trn, ng trung tm chnh l trung bnh chung (grand mean), hai ng pha ngoi l gii hn quyt nh (decision limit). Nu cc im ch gi tr trung bnh ca cc mu nm trong gii hn quyt nh th kt lun l khng c bng chng ni rng cc gi tr trung bnh mu l khc nhau. Ni cch khc cc mu ny u thuc cng tp hp hay mu chun tho mjn tnh ng nht.

Sinh vin t kim tra bng cch tnh ton theo cng thc.

5.2. Phn tch phng sai mt yu t (one-way ANOVA) Gi s vic thay i yu t A (c th l nng ion cn, phng th nghim trong

sn xut, iu kin t nhin) c nh hng n kt qu thc nghim (nh hp th quang, chiu cao pic, bn sn phm, nng ). Mc th nghim c th l cc mc nng , cc phng sn xut khc nhau, cc cng on khc nhau).

nghin cu nh hng ca yu t A, ngi ta tin hnh k mc th nghim, mi mc nghin cu lp li n ln, kt qu th nghim l cc gi tr yij ( vi i=1k v j= 1 n nh bng 5.1. Bng 5.1: Qui hoch thc nghim phn tch phng sai 1 yu t k mc th nghim, mi mc th nghim lp li n ln.

Mc a1 a2 a3 ... ai ... ak


43

S ln TN

1 y11 y21 y31 yi1 yk1

2 y12 y22 y32 yi2 yk2

3 y13 y23 y33 yi3 yk3

...

j y1j y2j y3j yij ykj

...

N y1n y2n y3n yin ykn

Tng ct

so snh s sai khc gia cc kt qu khi thay i cc mc ca A, ngi ta so snh phng sai do s thay i cc mc nghin cu gy nn vi phng sai chung ca th nghim xem chng c khc nhau ng tin cy hay khng. Nu s khc nhau khng ng tin cy th c th kt lun yu t A s nh khng ng k n kt qu th nghim v ngc li.

Vic so snh phng sai c thc hin qua chun F.

2

2

TN

Atinh S

SF = >1 v so snh vi Fchun (P, fA, fTN)

trong SA2 l phng sai ca th nghim khi thay i cc mc khc nhau ca

yu t A.

STN2: l phng sai chung ca th nghim v lm th nghim bao gi cng mc sai

s.

fA: bc t do ca cc mc nghin cu ca yu t A 8 lm; f= k-1

fTN: bc t do ca s nghin cu 8 tin hnh trong qui hoch nghin cu : f2= k(n-1)

Gi thit thng k l: H0 : SA2STN

2 v Ha: SA2 STN

2

V F>1 nn :

- Nu Ftinh Fbng th Ftnh ng tin cy, tc l SA2 v STN

2 khc nhau c ngha hay yu t A c nh hng n kt qu nghin cu.

Trong phn mm thng k c th s dng tr s P (Pvalue) so snh vi P. nu Pvalue< P=0,05 th khng nh rng khng phi tt c cc gi tr trung bnh cc mc th nghim khc nhau u ging nhau. Ni cch khc l yu t A c nh hng n kt qu th nghim

=

n

jjy

12

=

n

jkjy

1=

n

jijy

1

=

n

jjy

13

=

n

jjy

12


44

Trong qu trnh tnh ton trnh nhm ln, ngi ta lp bng cc cng on tnh phng sai so snh cho bi ton mt yu t, k mc nghin cu v n ln lp li nh sau:

Bng 5.2: Bng tnh phng sai khi nghin cu nh hng ca yu t A.

Ngun bin thin (Source of variation)

Bc t do

( Degree of freedom) f

Tng cc bnh phng

( Sum of squares)

2)( xxi

Trung bnh bnh phng

( mean of square)

S2

A k-1 SSA= SS2-SS3

12

=

kSSS AA

Sai s th nghim

( residue error)

k(n-1) SSTN= SS1- SS3

)1(2

=

nkSSS TNTN

Total kn-1 SStotal=SS1-SS3

Cc k hiu trn c tnh nh sau:

=

=

1jiji ynA (tng cc ga tr trong mt ct).

n

AA ii

= ( trung bnh ct)

=

=

n

iiAk

Y1

1 (trung bnh chung) ( overall average)

=

=

k

iiASS

1

22

=

=

k

iiA

nSS

1

23 )(

1 ( SS: Sum of squares);

( S2: mean of squares)

= =

=

k

i

n

jijySS

1 1

21 )( vi

=

=

n

jiji yA

1( tng cc gi tr trong mt ct )

Ftinh= 2

2

TN

A

SS

So snh Ftinh vi Fbang(P,f1,f2) vi P=0,95; f1=k-1; f2=k(n-1).

Nu Ftinh < Fbang th kt lun rng yu t A gy nh hng khng ng k n kt qu th nghim v ngc li.


45

Th d 5.2: Kt qu phn tch tng hm lng Hg (g/g) bng phng php HPLC trong 3 loi ng vt thn mm (Rap., Nev., Sca.) 8 im ven b bin Bohai - Trung Quc thu c nh sau:

a im

Loi

1 2 3 4 5 6 7 8

Rap. 0.042 0.063 0.059 0.038 0.053 0.199 0.060 0.038

Nev. 0.033 0.062 0.096 0.027 0.044 0.077 0.039 0.031

Sca. 0.005 0.044 0.068 0.016 0.014 0.099 0.021 0.026

Hjy dng phng php phn tch phng sai mt yu t nh gi xem loi v a im c nh hng n s tch lu Hg trong ng vt thn mm hay khng.

Ngun: W. Yawei et al. / Environmental Pollution 135 (2005) 457 - 467

( Nu s dng phn mm MINITAB 14 th kt qu vn tt thu c nh sau:

Ngun phng sai DF SS MS F P

Loi ( gia cc loi) 3 0.00476 0.00159 0.76 0.524

Sai s( trong mt loi) 28 0.05816 0.00208

Tng 31 0.06292

S = 0.04558 R-Sq = 7.57% R-Sq(adj) = 0.00%

Individual 95% CIs For Mean Based on

Pooled StDev

Level N Mean StDev ---------+---------+---------+---------+

1 8 0.06900 0.05348 (-------------*------------)

2 8 0.05113 0.02483 (------------*-------------)

3 8 0.03663 0.03212 (-------------*------------)

4 8 0.04263 0.06164 (------------*------------)

---------+---------+---------+---------+

0.025 0.050 0.075 0.100

Pooled StDev = 0.04558

Fisher 95% Individual Confidence Intervals

All Pairwise Comparisons among Levels of Muc

Simultaneous confidence level = 80.51%


46

Hjy gii thch kt qu trn

5.3. Phn tch phng sai hai yu t (two-way ANOVA) Gi s c hai yu t nh hng n kt qu th nghim A v B. Yu t A c k

mc nghin c, yu t B c m mc nghin cu, mi mc th nghim lp li n ln. Lp bng qui hoch nghin cu tc ng ca hai yu t n kt qu th nghim nh bng 5.3:

Bng 5.3. Qui hoch thc nghim phn tch phng sai 2, yu t A c k mc th nghim, yu t B c m mc; mi mc th nghim lp li n ln.

Yu t A

a1 a2 ... ai ... ak

b1 y111, y112, ..., y11n y211, y212, .. y21n yi11, yi12, .. yi1n yk11, yk12, .. yk1n

b2 y121, y122, .. y12n y221, y222, .. y22n yi21, yi22, .. yi2n yk21, yk22, .. yk2n

B ...

bj y1j1, y1j2, .. y1jn y2j1, y2j2, .. y2jn yj1, yij2, .. yijn ykj1, ykj2, .. ykjn

...

bm y1m1, y1m2, .. y1mn y2m1, y2m2, .. y2mn yim1, yim2, .. yimn ykm1, ykm2, .. ykmn

Tng ct A1 A2 Ai Ak

Cc bc tnh phng sai theo bng trn ln lt nh sau:

=

=

n

u

ijuij yY1

(tng cc kt qu nghin cu trong 1 ) =

=

n

u

ijuij yY1

22 )(

== =

==

m

jij

m

j

n

u

ijui YyA11 1

(tng cc kt qu nghin cu trong 1 ct)

== =

==

n

iij

k

i

n

u

ijuj YyB11 1

(tng cc kt qu nghin cu trong mt hng)

= == = =

===

k

i

m

jji

k

i

m

j

n

u

ijuiju BAyY1 11 1 1

(tng cc ct = tng cc hng)


47

= = =

=

k

i

m

j

n

u

ijuySS1 1 1

21

=

=

k

iiA

nmSS

1

22

.

1

=

=

m

jjB

nkSS

1

23

.

1

2

1 1 1 1

2

1

24 )(

..

1)(..

1)(..

1 = = = ==

===

k

i

m

j

n

u

m

jj

k

iiiju B

nmkA

nmky

nmkSS

Mu kt qu tnh ton ANOVA c trnh by trong bng 5.4

Bng 5.4: Bng phn tch phng sai hai yu t

Ngun bin thin

(Source of variation)

Bc t do

(Degrees of freedom)

f

Tng cc bnh phng

(Sum of squares)

2)( xxi

Trung bnh bnh phng

(Mean of square)

S2

A k-1 SSA=SS2-SS4

12

=

kSSS AA

B m-1 SSB= SS3-SS4

12

=

m

SSS BB

AB (k-1).(m-1) SSAB= SStotal-SSA-SSB-SSe

SSAB =SS1-SS2-SS3+SS4 )1)(1(2

=

mkSSS ABAB

Sai s th nghim

(Residue error)

mk(n-1) SSe=SStotal-SSA-SSB

SSe= SS1 - = =

k

i

m

jijY

n 1 1

21

)1(2

=

nmkSSS ee

Tng mk(n-1) SStotal= SS1-SS4

Trong :

SA2 v SB

2: phng sai c trng cho nh hng ca yu t A v B n kt qu th nghim.

2

2

2

2

2

2

;;e

ABAB

e

BB

e

AA S

SFSSF

SSF ===

SAB2: phng sai c trng cho nh hng tng h ca c hai yu t A v B n

kt qu th nghim.

S e2 : phng sai c trng cho sai s th nghim.

Bc t do:

fA=k-1: fB = m-1 ; fAB= (k- 1).(m-1) ; fe = m.k.(n-1)


48

So snh FA, FB, FAB vi gi tr Fbang vi P=0,95; f1= fA hoc fB hoc fAB v f2= fe v kt lun v mc nh hng ca tng yu t n kt qu th nghim nh phn 5.1.

Th d 5.3: Trong thc nghim so snh kh nng tch loi Cu2+ trong nc ca nha vng cng (% Cu2+) mt ngi phn tch lm th nghim phn tch phng sai 2 yu t l 5 ngy lm th nghim v 4 loi nha. Mi th nghim lm lp li hai ln. Kt qu thu c bng di y.

Hjy nh gi xem c s khc nhau c ngha ca cc loi nha theo thi gian hay khng cng nh c s tng tc cu hai yu t nghin cu hay khng. Biu din kt qu tnh c vo bng ANOVA. Ly P=0,95.

Loi 1 Loi 2 Loi 3 Loi 4

1 20,2

20,2

6,8

7,2

45,5

47,0

20,1

20,9

Ngy 2 28,1

29,6

22,6

23,5

15,5

16,0

7,5

8,6

3 8,7

9,0

38,7

38,2

6,7

7,1

52,7

53,0

4 30,4

30,9

50,6

51,1

18,9

17,6

60,4

61,2

5 50,7

50,5

18,8

18,5

30,5

30,9

67,6

67,2

Ngun:

( Sinh vin t gii theo cng thc tnh ton j nu ).

Hng dn: S dng phn mm MINITAB, s liu c nhp vo dng sau:

Ct th nht l kt qu % Cu t trn xung di theo th t tng ngy v tng loi

Ct th hai l ngy phn tch theo th t 8 s liu l 1( ngy th 1) sau n 8 s liu l 2 ( ngy th hai) Ct th ba l loi nhp theo th t 1, 1; 2, 2; 3, 3 ; 4, 4 ; 5, 5, ln lt t ngy 1 n ngy 5.

Vo Stat->ANOVA->2-way.

Chn response l %Cu

Row factor l ngy

Column factor l loi

V nh du vo dislay mean

Kt qu thu c nh sau:

Source DF SS MS F P

Ngay 4 3359.0 839.761 3205.20 0.000


49

Loai 3 1922.9 640.962 2446.42 0.000

Interaction 12 8267.9 688.992 2629.74 0.000

Error (within) 20 5.2 0.262

Total 39 13555.1

S = 0.5119 R-Sq = 99.96% R-Sq(adj) = 99.92%


Pooled StDev

Ngay Mean ---------+---------+---------+---------+

1 23.4875 *)

2 18.8500 *)

3 26.7625 (*

4 40.1375 (*)

5 41.8375 (*

---------+---------+---------+---------+

24.0 30.0 36.0 42.0


Pooled StDev

Loai Mean ----+---------+---------+---------+-----

1 27.83 (*

2 27.60 *)

3 23.57 (*)

4 41.86 (*

----+---------+---------+---------+-----

25.0 30.0 35.0 40.0

Kt qu trn cho thy tt c cc tr s P ca ngy, loi v nh hng tng h

ca chng (interaction) u bng 0.000 tc l nh hn =0,05 chng t c nh hng c ngha n kh nng loi Cu trong nc. S phn b gi tr trung bnh ca % Cu theo ngy v theo loi nha u cho thy cng tng thi gian th kh nng hp th Cu cng ln v tt nht loi nha th 5.


50

5.4. Bi ton phn tch phng sai 3 yu t tr ln- phng php vung Latinh

Trong trng hp cn nghin cu nh hng ca 3 yu t tr ln, xy dng bng qui hoch thc nghim, ngi ta s dng phng php vung La tinh (Latin square).

Nguyn tc: khng mt iu kin nghin cu xc nh lp li trong mt hng hay mt ct. Ni cch khc trong bng qui hoch thc nghim khng c hai ging nhau.

Gi thit c 3 yu t A, B, C, mi yu t c 4 mc nghin cu. Mi m t mt iu kin nghin cu l t hp cc mc nghin cu ca 3 yu t. Th d: s 1 khi lm th nghim ly mc a1, b1 v c1.

Ta c bng qui hoch thc nghim nh sau:

Bng 5.5: Qui hoch thc nghim phn tch phng sai 3 yu t, mi yu t 4 mc th nghim.

b1 b2 b3 b4 Tng hng

a1 c1

y1111; y1112; y1113

c2

y2121; y2122; y2123

c3

y3131 y3132; y3133;

c4

y4141 y4132; y4143

A1

a2 c2

y1221; y1222; y1223

c3

y2231; y2232; y2233

c4

y3241; y3242; y3243

c1

y4211; y4212; y4213

A2

a3 c3

y1331; y1332; y1333

c4

y2341; y2342; y2343

c1

y3311; y3312; y3313

c2

y4321; y4322; y4323

A3

a4 c4

y1441; y1442; y1443

c1

y2411; y2412; y2413

c2

y3421; y3422; y3423

c3

y4431; y4432; y4433

A4

Tng ct

B1 B2 B3 B4 Y*

Cch tnh cc gi tr trong bng trn nh sau:

A1= y111+y122+y133+y144

A1: tng cc gi tr y ( y l gi tr trung bnh ca 3 ln th nghim lp li ca cng 1 trong cng iu kin), c mc a1 ( tc l tng trung bnh ca cc kt qu ca cc trong hng a1).

Tng t ta c cc gi tr khc:

A2, A3, A4 l tng cc kt qu c mc a2, a3, a4.

B1...B4 l tng cc kt qu c mc b1, b2, b3, b4.


51

C1...C4 l tng cc kt qu c mc c1, c2, c3, c4.

Y* = A1+A2+A3+A4 = B1+B2+B3+B4 = C1+C2+C3+C4

= =

=

n

j

n

iijySS

1 1

21 (tng bnh phng cc ga tr c mt trong bng ).

2

12

2

12

2

125

1

24

1

23

1

22

)(1)(1)(1

1

1

1

===

=

=

=

===

=

=

=

n

qiq

n

jj

n

ii

n

qq

n

jj

n

ii

Cn

Bn

An

SS

Cn

SS

Bn

SS

An

SS

2

2

2

2

2

2

e

CC

e

BB

e

AA

SSF

SSF

SSF

=

=

=

Bng phn tch phng sai trong trng hp ny c dng sau:

Bng 5.5: Bng phn tch phng sai ba yu t

Ngun bin thin

(Source of variation)

Bc t do

(Degrees of freedom) f

Tng cc bnh phng

(Sum of squares)

2)( xxi

Trung bnh bnh phng

(Mean of square)

S2

A n-1 SSA= SS2-SS5

B n-1 SSB= SS3-SS5

C n-1 SSC= SS4-SS5

Sai s th nghim

(Residue error)

(n-1)(n-2) SSe=SStotal-SSA-SSB-SSc

)2)(1(2

=

nn

SSS ee

Tng n2-1 SStotal= SS1-SS5

So snh cc gi tr tnh ton vi gi tr tra bng Fbang (P,f1,f2) vi P=0,95 ; f1=n-1 v f2= (n-1)(n-2) sau kt lun v mc nh hng ca tng yu t n kt qu th nghim nh phn 5.1

Phng php ny c s dng nghin cu nh hng ca nhiu yu t trong nng nghip, y hc sinh hc, xj hi hc...

Th d 5.4: kho st nh hng ca nng thuc th o- phenantrolin (A), pH dung dch (B) v nhit (C) n hp th quang ca dung dch phc mu Fe(II)- o- phenantrolin , ngi ta tin hnh th nghim theo phng php vung la tinh vi 3 yu t nh hng, 4 mc th nghim. Kt qu trung bnh (sau 3 ln lm lp li) nh

12

=

n

SSS AA

12

=

n

SSS BB

12

=

n

SSS CC


52

sau:

b1 b2 b3 b4

a1 c1

0, 351

C2

0,522

c3

0,24 5

c4

0,248

a2 c2

0,356

C3

0,258

c4

0,452

c1

0,526

a3 c3

0,211

C4

0,356

c1

0,456

c2

0,521

a4 c4

0,169

C1

0,254

c2

0,255

c3

0,452

Hjy nh gi xem c nh hng c ngha ca cc yu t n hp th quang ca dung dch phc mu hay khng? Biu din kt qu tnh c vo bng ANOVA. Ly P=0,95.

Hng dn: Nhp cc s liu trong bng trn vo chng trnh MINITAB 14. Ct kt qu c vo theo th t t tri sang phi v t trn xung di trong bng tr n. Ba ct km theo l cc bin A, B, C. Trong ct A c nhp cc s 1,2 3, 4, thay cho a1 a4 , tng t cho cc ct B v C. S liu sau khi nhp vo MINITAB c dng:

Abs A B C

0.351 1 1 1

0.522 1 2 2

0.245 1 3 3

0.248 1 4 4

0.356 2 1 2

0.258 2 2 3

0.452 2 3 4

0.526 2 4 1

0.211 3 1 3

0.350 3 2 4

0.456 3 3 1

0.521 3 4 2


53

0.169 4 1 4

0.254 4 2 1

0.255 4 3 2

0.452 4 4 3

V cc s liu khng phi l ma trn cn bng do vy khng dng balanced ANOVA m phi dng dng General linear Model

Vo STAT->ANOVA->General linear Model. Nhp response l ct hp th quang, Model l ct A, B, C, random factor chn A, B. Trong results chn: In addition, coefficient for covariate terman table of unusal observation. V bm vo OK. Kt qu thu c nh sau:

General Linear Model: Do hap thu quang versus A, B, C

Factor Type Levels Values

A random 4 1, 2, 3, 4

B random 4 1, 2, 3, 4

C fixed 4 1, 2, 3, 4

Analysis of Variance for Do hap thu quang, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P

A 3 0.03245 0.03245 0.01082 0.74 0.568

B 3 0.05463 0.05463 0.01821 1.24 0.375

C 3 0.04671 0.04671 0.01557 1.06 0.434

Error 6 0.08827 0.08827 0.01471

Total 15 0.22206

S = 0.121289 R-Sq = 60.25% R-Sq(adj) = 0.63%

Kt qu tnh ton cho thy c 3 yu t A, B, C u c gi tr P> 0,05 tc l c 3 3 yu t khng nh hng c ngha n kt qu th nghim.

( Sinh vin t gii bng cch tnh ton theo cng thc).


54

Chng 6: PHN TCH TNG QUAN V HI QUI

Trong thc t phn tch, xc nh hm lng cht ngi ta thng s dng phn tch ho hc v phn tch cng c.

- Phn tch ho hc c s dng rng ri do c chnh xc cao, lm t th nghim vi s t mu, v c p dng phn tch nhng mu chun. Tuy nhin, phng php ny c gii hn pht hin thp v tn nhiu thi gian phn tch.

- Phn tch cng c chim hn 90 % cc php phn tch do: + nhy cao, xc nh c ng thi nhiu nguyn t, phn tch c s

lng ln mu cng mt lc, v vy tn t thi gian phn tch, gi thnh phn tch r hn so vi phn tch ho hc .

+ Phn tch cng c kt ni c vi my tnh, do c th t ng ho, x l s liu trn my tnh, nh gi thng k v so snh c cc s liu lu tr trong b nh.

Tuy nhin, cc phng php phn tch cng c lun cn nh chun (c mu chun). T s liu thu c i vi cc mu chun, dng ng biu din tn hiu phn tch theo nng cht nh phn tm phng trnh h qui v chnh xc ca phng php phn tch da trn phn tch hi qui. Mc d vy, khng phi gia 2 bin ngu nhin lun c tng quan, do cn phi kim tra bng phn phi 2 chiu. Ni cch khc, cn phi tr li c cc cu hi sau:

- ng chun c tuyn tnh khng? Nu l ng cong th dng biu din l g? - Mi im trn ng chun u mc sai s khi phn tch. Vy ng biu din

no s i qua tt c cc im thc nghim ny? - Gi thit ng chun l thc s tuyn tnh th sai s v gii hn tin cy ca

nng xc nh c l bao nhiu? -Gii hn pht hin ca phng php l bao nhiu?

6.1. Phn tch tng quan (correlation analysis) Phn tch tng quan c dng nh gi mi quan h gia hai hay nhiu bin

thng qua h s tng quan. Hai loi h s tng quan thng dng nht l h s tng quan Pearson hoc Spearmen.

H s tng quan r biu th mc quan h tuyn tnh gia hai bin v tnh c nu tp s liu tho mn cc iu kin sau:

- Cc tp s liu (cc bin) tun theo phn phi chun. - Gi tr cc bin l c lp nhau. - Phi loi b gi tr bt thng trc khi tnh h s tng quan.Trng hp nu

khng tun theo phn phi chun th nn s dng h s tng quan phn hng Spearmen.


55

6.1.1. Cch tnh h s tng quan Pearson (the product-moment correlation coefficient)

H s tng quan c tnh theo cng thc sau:

yx

XY

SSCOV

r =

vi COV l ng phng sai ca hai tp s liu X v Y v c tnh theo cng thc:

n

yyxxCOV iiXY

=

))(( vi n l s ga tr trong tp s liu X v Y

Do vy

=

=

))()()(()).((

))()()(()(

2222

2

yyxx

yyxx

yyxx

xyxnr

ii

ii

ii

iii =

n

yy

n

xx

yxn

yx

ii

ii

iiii

22

22 )()()((

1

Khi r=1 th tp hp cc im (xi, yi) hu nh nm trn ng thng tc l hai bin c tng quan tuyn tnh tuyt i . Khi r>0 th x v y c quan h ng bin cn r

56

Nhng yu t nh hng ln n h s tng quan l:

+ Khong bin i ca cc s liu trong tp s liu.

+ khng ng nht ca mu.

+ Sai s th.

Th d 6.1:

Phn tch hm lng gluco trong mu theo phng php ng chun. S ph thuc gia hp th quang v nng gluco trong mu nh sau:

Nng gluco mM

0 2 4 6 8 10

hp th quang

0,002 0,150 0,294 0,434 0,570 0,704

SV hjy vn dng cng thc nu trn tnh h s tng quan Pearson r v kt lun mc tng quan tuyn tnh gia hai i lng nu trn.

tr li cu hi v hai bin X v Y ang xt tht s c tng quan tuyn tnh hay khng chng ta c th s dng chun student kim tra bng cch tnh gi tr t v so snh vi gi tr t trong bng cho trc.

)1(2.

2r

nrt

=

vi r l h s tng quan Pearson, n l s th nghim hay s s liu trong mi bin. Sau , so snh ga tr ttinh vi tchun tra bng tin cy thng k mong mun (thng chn P=0,95), s bc t do f=n-2.

Gi thit "khng" l gi thit gia X v Y khng c tng quan t c khi

ttinh < ttra bang . Nu ttinh >ttra bang th x v y c tng quan tuyn tnh.

Nu tnh ton bng cc phn mm thng k, c th s dng gi tr P ( Pvalue) v so snh vi khng tin cy . Thng thng nu Pvalue

57

6.1.2. H s tng quan Spearmen (rs): H s ny cng c dng biu th mc tng quan hai bin nhng khc vi h s tng quan Pearman, n xp th hng mi bin thay v tnh gi tr.

1)-N( Nd6

- 1 = r2

2N

=1is

y d l s khc nhau gia cc th hng trong hai phng php xp hng.

Khi N>=10 th rs c th c dng tnh gi tr t theo phng trnh trn.

6.1.3. H s tng quan Kendall :

H s ny phc tp hn Spearman v ch nn dng khi c nhiu hn 2 tp s liu cn so snh v c tnh nh l hiu s ca cp ph hp tr i hiu s cp khng ph hp.

Cp ph hp l khi (xi-xj)*(yi-yj)>0

Cp khng ph hp khi (xi-xj)*(yi-yj)

58

13 905.8 128.9 451.3 52115 32.2 356.5 * 378.6

14 * 59.6 510.8 39931 38.0 131.3 706.5 312.3

15 898.0 74.7 377.7 26779 34.1 89.6 381.1 108.6

16 558.9 101.8 397.9 21001 24.5 235.9 439.6 183.9

17 1217.0 53.3 528.4 23656 23.8 139.0 990.5 83.6

18 1160.0 71.9 633.5 31204 37.6 143.1 * 98.0

19 955.3 100.7 469.2 26171 30.5 189.7 839.1 207.8

Nhng s liu k hiu du * l nhng s liu th j c loi b. Khi tnh ton, thng thng phn mm thng k s xem nhng s liu ny c gi tr bng gi tr trung bnh ca tp s liu.

S dng phn mm thng k MINITAB 14, vo Stat-> basic Statistics-> Correlation. Nhp Variable l cc ct cha hm lng 8 kim loi v chn mc display P-value.

Kt qu thu c nh sau:

Ti Cr Mn Fe Ni Cu Zn

Cr 0.334

0.205

Mn 0.638 0.184

0.006 0.465

Fe 0.356 0.424 0.345

0.161 0.079 0.147

Ni 0.151 0.434 0.034 0.028

0.578 0.072 0.894 0.911

Cu -0.452 -0.098 -0.365 -0.032 -0.218

0.068 0.697 0.124 0.898 0.385

Zn 0.592 0.087 0.595 0.477 0.173 -0.022

0.026 0.747 0.015 0.062 0.521 0.934

Pb -0.065 0.129 -0.113 0.551 -0.150 0.250 0.040

0.810 0.621 0.655 0.018 0.566 0.316 0.888

Cell Contents: Pearson correlation

P-Value

Gi tr trong mi gm h s tng quan Pearson v gi tr P. Hjy kt lun v chiu hng v mc tng quan gia hm lng cc kim loi nu trn.

6.2. Phng php bnh phng ti thiu Gi s c 2 tp s liu : x: x1 x2 x3 x4 x5 ... (nng ) y: y1 y2 y3 y4 y5 ... (tn hiu phn tch)


59

ng chun s biu din s ph thuc tuyn tnh gia tn hiu o v nng cht nh phn nu phng trnh hi qui c dng y = a + bx. Trong a l im ct trc tung ca ng biu din (ng chun) v b l dc ca ng chun. Trong thc t phn tch, khi h s tng quan r > 0,99 c th xem c tng quan tuyn tnh tt gia x v y v phng trnh hi qui c dng nh lng y theo x.

T cc im trn th ( x1; y1) ( x2; y2)........ (xn; yn) ta s tm c im trng tm (centroid of all points) ( x ; y ).

Khi c quan h tuyn tnh gia bin c lp x (nng ) v bin ph thuc y (tn hiu phn tch ) th vn quan trng l lm th no tm c ng thng ng nht i qua tt c cc im trn ng chun (trong khi mi im

Documents

04 Giao Trinh Thongke 2008-Latest