Vi_TCVN9330-1-2012

TIU CHUN VIT NAM

TIU CHUN QUC GIATCVN 9330-1 : 2012ISO 14461-1 : 2005SA V SN PHM SA - KIM SOT CHT LNG TRONG PHNG TH NGHIM VI SINH VT - PHN 1: NH GI NNG LC THC HIN M KHUN LCMilk and milk products - Quality control in microbiological laboratories - Part 1: Analyst performance assessment for colony countsLi ni u

TCVN 9330-1:2012 hon ton tng ng vi ISO 14461-1:2005;TCVN 9330-1:2012 do Vin Kim nghim an ton v sinh thc phm quc gia bin son, B Y t ngh, Tng cc Tiu chun o lng Cht lng thm nh, B Khoa hc v Cng ngh cng b;

B tiu chun TCVN 9330:2012 Sa v sn phm sa - Kim sot cht lng trong phng th nghim vi sinh vt gm c cc phn sau y:

- TCVN 9330-1:2012 Sa v sn phm sa - Kim sot cht lng trong phng th nghim vi sinh vt - Phn 1: nh gi nng lc thc hin m khun lc;

- TCVN 9330-2:2012 Sa v sn phm sa - Kim sot cht lng trong phng th nghim vi sinh vt - Phn 2: Xc nh tin cy s m khun lc ca cc a song song v cc bc pha long lin tip.

Li gii thiu

Mi phng php vi sinh vt bao gm mt s bc k tip theo trnh t c th (ly mu con, pha long mu, a v m khun lc). khng m bo o ca kt qu cui cng c xc nh bi tnh bin thin ca tt c cc bc lin quan trong quy trnh phn tch. thu c cc kt qu c khng m bo o khng qu ln so vi kt qu d kin khi thc hin ng phng php, ngi phn tch cn tun th cc quy nh ca Thc hnh phng th nghim tt (GLP).

C ba yu t quan trng nht t c s m a chnh xc l:

- tnh ng nht ca vt liu mu,

- tnh chnh xc ca cc bc pha long,

- k thut cy v/hoc m khun lc cc a nui cy.

C th nh gi kh nng thc hin k thut m khun lc ca phng th nghim v a ra bin thin mong mun ca phng php bng cch ng ha k vt liu mu, to nhiu dy dch pha long v cy mt s a t cng mt dch pha long.

bin thin qu ln cho thy t nht mt trong cc bc khi thc hin phng php nm ngoi tm kim sot. Xc nh cc bc ny bng cch so snh cc a nui cy lp li, cc dch pha long khc nhau v cc dy pha long khc nhau. Khi pht hin cc bc ca phng php phn tch c s bin thin qu ln, phi tin hnh cc bin php cn thit kim sot cc bc trn.

SA V SN PHM SA - KIM SOT CHT LNG TRONG PHNG TH NGHIM VI SINH VT - PHN 1: NH GI NNG LC THC HIN M KHUN LC

Milk and milk products - Quality control in microbiological laboratories - Part 1: Analyst performance assessment for colony counts

1. Phm vi p dngTiu chun ny a ra quy trnh kim tra nng lc thc hin m khun lc trong phng th nghim bng cch thit lp bin thin ni b phng th nghim v xc nh cc bc c lin quan n bin thin qu mc ca cc kt qu.Quy trnh ny thch hp kim tra vic tun th chnh xc Thc hnh phng th nghim tt (GLP) v cng l mt iu kin tin quyt tham gia kim tra phng php m khun lc trong php th lin phng.V D: Cc mu kim tra thch hp l sa ti nguyn liu, sa thanh trng v sa bt.2. Ti liu vin dnCc ti liu vin dn sau rt cn thit cho vic p dng tiu chun ny. i vi cc ti liu vin dn ghi nm cng b th p dng phin bn c nu. i vi cc ti liu vin dn khng ghi nm cng b th p dng phin bn mi nht, bao gm c cc sa i, b sung (nu c).

TCVN 6263 (ISO 8261) Sa v sn phm sa - Hng dn chung v chun b mu th, huyn ph ban u v dung dch pha long thp phn kim tra vi sinh vt

TCVN 6404 (ISO 7218) Vi sinh vt trong thc phm v thc n chn nui - Yu cu chung v hng dn kim tra vi sinh vt.

TCVN 7150-4 (ISO 835-4) Dng c th nghim bng thy tinh - Pipet chia - Phn 4: Pipet kiu thi ra1).TCVN 7151:2002 (ISO 648:1977) Dng c th nghim bng thy tinh - Pipet mt mc2)TCVN 8488 (ISO 4788) Dng c th nghim bng thy tinh - ng ong chia

3. Thut ng v nh ngha

Trong tiu chun ny s dng thut ng v nh ngha sau:

3.1. K thut m khun lc (colony-count technique)

Vic m s lng vi sinh vt c xc nh bng quy trnh quy nh trong tiu chun ny.

CH THCH: S lng vi sinh vt c biu th theo gam hoc mililit mu th.

4. Nguyn tc (xem Hnh 1)

4.1. Mu phn tch c ng ha sau c pha long n nng lm vic thch hp (v d: 500 CFU/ml n 10 000 CFU/ml) thu c huyn ph.4.2. T dung dch pha long u tin, chun b 4 dy pha long, mi dy gm 12 bc pha long nh phn.

CH THCH: S dng cc bc pha long nh phn (hai ln), khng phi cc bc pha long thp phn (10 ln) nh tin hnh thng thng. Vi cc dung dch pha long nh phn, c th m khun lc trn cc a vi 5 hoc 6 dung dch pha long, mt s ln khun lc s ci thin ng k vic kim tra cc bc pha long.

4.3. Mi dung dch pha long ca mi dy pha long c ra 3 a song song.

4.4. m cc a.

4.5. M ha cc a v m s khun lc trn mi a.

4.6. Lp bng cc s m khun lc v tnh ton "tnh ng nht c ngha thng k" theo hai bc.

4.7. Nu cc gi tr thu c l ng nht c ngha thng k th cht lng ca vic ng dng phng php l tha mn v khng cn tin hnh cc nh gi tip theo.

4.8. Nu cc kt qu khng ng nht c ngha thng k th tin hnh k thut phn tch phng sai (ANOVA) xc nh s bin thin ca kt qu trong iu kin mt hoc nhiu yu t b thay i (ngha l: dy pha long, mc pha long, vic a). Tin hnh cc nh gi tip theo v iu chnh cc yu t c nhn din.

CH THCH: Ngi s dng phi thit k cc ngun pht sinh li ch yu trong khi thc hin phng php.

Hnh 1 - m bo cht lng phng th nghim vi sinh vt:

Thit k nghin cu thm d cho phng php m khun lc

5. Dung dch pha long, mi trng nui cy v thuc th

Cc thao tc c m t trong iu ny v trong iu 9 phi c tin hnh bi mt nhn vin chun b mi trng ring bit hoc c chia u trong nhm vi s phn cng nhim v r rng cho mi thnh vin.

Ch s dng cc thuc th t cht lng phn tch v nc ct hoc nc t nht c tinh khit tng ng, tr khi c quy nh khc. Cc thuc th v nc phi khng cha cc c cht c th tc ng ngc n s pht trin ca vi sinh vt trong cc iu kin phn tch. Mi trng nui cy phi c cng nhn v cht lng vi khun hc. Mi trng kh nc phi c chun b theo hng dn ca nh sn xut.5.1. Dung dch natri hydroxit hoc axit clohydric (khong 0,1 mol/l), chnh pH ca cht pha long v mi trng nui cy.

5.2. Mi trng nui cy: thch trypton-glucoza-cht chit nm men, c b sung sa bt gy.

5.2.1. Thnh phn

Cht chit nm men2,5 g

Dch thy phn tryptic ca casein (trypton)5,0 g

Glucoza ngm mt phn t nc (C6H12O6.H2O)1,0 g

Sa bt gy1,0 g

Thch10 g n 15 ga

Nc1 000 ml

a Ph thuc vo sc ng ca thch

Trong tt c cc trng hp, cn phi thm sa bt gy cho d l mi trng hon chnh kh thng mi v nh cung cp khuyn ngh khng cn phi b sung.5.2.2. Chun b

Cn chun b 2 lt mi trng ca cng mt l. Nu s dng mi trng hon chnh kh thng mi th tun th hng dn ca nh sn xut nhng phi b sung sa bt gy. Chnh pH sao cho sau khi kh trng t 7,0 ( 0,2 khong 450C.

Nu mi trng c chun b t cc thnh phn c bn kh th ha tan v phn phi vo nc c lm nng trc theo th t sau: cht chit nm men, trypton, glucoza v cui cng l sa bt gy. un nng ha tan v phn phi mi trng. B sung thch v un n si, va un va khuy n khi ha tan hon ton. Hoc c th un si hn hp trong 30 min. Lc mi trng qua giy lc, nu cn. Chnh pH sao cho sau khi kh trng t 7,0 ( 0,2 khong 450C.

Phn phi 250 ml mi trng nui cy vo cc chai (6.10). Kh trng cc chai cng lc trong ni hp p lc (6.1) 1210C trong 15 min.

Bo qun mi trng chun b ni ti nhit t 00C n 50C v s dng trong vng 1 thng.

5.3. Cc cht pha long: dung dch mui/pepton hoc dung dch Ringer mt phn t cho mi l.

5.3.1. Dung dch mui/pepton

Cht pha long ny c dng cho cc mc ch chung.

5.3.1.1. Thnh phn

Pepton1,0 g

Natri clorua (NaCl)8,5 g

Thm nc n1000 ml

5.3.1.2. Chun b

Ha tan cc thnh phn trong nc, un nng nu cn. Chnh pH sao cho sau khi kh trng t 7,0 ( 0,2 250C.

5.3.2. Dung dch Ringer mt phn t

5.3.2.1. Thnh phn

Natri clorua (NaCl)2,25 g

Kali clorua (KCl)0,105 g

Canxi clorua khan (CaCl2)0,06 g

Natri hydrocacbonat (NaHCO3)0,05 g

Thm nc n1000 ml

5.3.2.2. Chun b

Ha tan cc mui trong nc. Chnh pH sao cho sau khi kh trng t 6,9 ( 0,2 250C.

5.3.3. Chun b cht pha long

Kh trng cht pha long bng cch hp p lc vi tng lng khng ln hn 500 ml. Sau dng ng ong chia v trng hoc cc dng c phn phi khc (6.11) phn phi cc phn 90 ml cht pha long nhit phng vo cc chai pha long mu (6.8) v trng v dng pipet mt vch dung tch 5 ml hoc pipet chia hoc cc dng c phn phi khc (6.12) phn phi cc phn 5 ml cht pha long vo cc ng nghim v trng (6.9). Khi s dng pipet, chm u pipet vo mt nghing ca bnh cha m bo phn phi chnh xc lng cht pha long.CH THCH: Vic phn phi cc phn cht pha long trc khi kh trng c th dn n s bay hi khng ng u, kt qu l cc phn cht pha long c nng cui cng khc nhau.

Lm mt v bo qun cht pha long v cc phn cht pha long phn phi nhit t 00C n 50C. S dng cht pha long v cc phn cht pha long phn phi chm nht trong ngy tip theo.

6. Thit b v dng c thy tinh

Kh trng cc thit b c kh nng tip xc vi mu cn phn tch, cht pha long, cc nng pha long hoc mi trng nui cy theo TCVN 6404 (ISO 7218) v TCVN 6263 (ISO 8261).

S dng cc thit b, dng c ca phng th nghim vi sinh thng thng v c th sau:

6.1. Ni hp p lc, c th vn hnh 121 0C ( 3 0C.

6.2. T sy, c th vn hnh nhit ln hn 180 0C.

6.3. T m, c th vn hnh 30 0C ( 1 0C v tt c cc im trong di nhit cho php hot ng ca t.

6.4. My o pH, hiu chnh theo nhit , chnh xc n ( 0,1 n v pH.

6.5. Ni cch thy, c th vn hnh 20 0C ( 1 0C, 45 0C ( 1 0C v t 44 0C n 47 0C.

6.6. Knh lp, loi c phng i t 2x n 4x v loi c phng i t nht 8x.

6.7. Bi thy tinh, c ng knh khong 6 mm.

6.8. Chai pha long, dung tch danh nh t 150 ml n 250 ml, c np y kn, cha t 5 n 10 vin bi thy tinh (6.7). Cho bi thy tinh vo trc khi kh trng chai pha long.

6.9. ng nghim, di 150 mm, ng knh 15 mm, c np y.

6.10. Chai, dung tch danh nh 500 ml, c np y, bo qun cc phn 250 ml mi trng nui cy.

6.11. ng ong chia , vi cc chia chnh, ph hp vi TCVN 8488 (ISO 4788) hoc cc dng c phn phi khc c chnh xc tng ng.

6.12. Pipet mt vch hoc pipet chia , c hiu chun, c kh nng phn phi 1 ml, 5 ml v 10 ml ph hp vi loi A ca TCVN 7151 : 2002 (ISO 648:1977), hoc TCVN 7150-4(ISO 835-4) hoc cc dng c phn phi khc c chnh xc tng ng.6.13. a Petri, lm bng thy tinh khng mu trong sut hoc vt liu nha, ng knh trong ca y l 90 mm v khng gy nh hng n kt qu m khun lc.6.14. My khuy trn c hc, c kh nng trn lng cha trong ng nghim, lm vic theo nguyn tc lch tm (v d: my trn vortex).

6.15. Cn, c phm vi o thch hp, c th cn chnh xc n 0,05 g.

7. Ly mu

Mu gi n phng th nghim phi l mu i din v khng b h hng hoc b bin i cht lng trong sut qu trnh vn chuyn v bo qun.

Vic ly mu khng quy nh trong tiu chun ny. Nn ly mu theo TCVN 6400 (ISO 707).

8. Chun b mu th

8.1. Sa

Lc mnh mu th bng cch o chiu vt cha mu 25 ln sao cho vi sinh vt c phn tn ng u. Trnh to bt hoc bt lan rng. Khong thi gian xen k gia trn mu v ly cc phn mu th khng qu 3 min.

8.2. Sa bt

Trn u lng cha trong vt cha kn bng cch lc v o nhiu ln. Nu mu th ng trong vt cha kn nguyn vn v qu y c th trn u th chuyn mu sang vt cha ln hn sau trn u.

9. Cch tin hnh

9.1. Yu cu chung

Trong phng php m khun lc, cc a cy c th thng xuyn khng m c hon ton hoc ch m c mt phn do nhiu nguyn nhn khc nhau (khun lc mc lan, nm mc pht trin, v.v). i vi phng php hin hnh, ch chp nhn thiu mt s gii hn kt qu m (xem 10.1). Nu thiu qu nhiu kt qu th hoc l do vt liu mu khng thch hp i vi php th hoc c sai st k thut. Trong trng hp nh vy, lp li quy trnh vi vt liu mu khc thch hp hn hoc tun th nghim ngt quy trnh phn tch.

9.2. S lng cc bc pha long thp phn

Xc nh s bc pha long thp phn da trn mt vi sinh vt d kin c trong mu nh sau:

a) khi s m d kin nh hn 100 000 trong 1 ml hoc trong 1 g mu th chun b dung dch pha long thp phn n 0,1 (mt bc pha long thp phn);b) khi s m d kin t 100 000 n 1 000 000 trong 1 ml hoc 1 g mu th chun b dung dch pha long thp phn n 10-2 (hai bc pha long thp phn);

c) khi s m d kin ln hn 1 000 000 trong 1 ml hoc trong 1 g mu th chun b dung dch pha long thp phn n 10-3 (ba bc pha long thp phn).9.3. Chun b dung dch pha long thp phn u tin

9.3.1. Sa

Dng pipet v trng (6.12) ly 1 ml mu th (8.1) v thm vo 9 ml cht pha long (5.3) (hoc 10 ml mu th trong 90 ml cht pha long hoc 11 ml mu th trong 99 ml cht pha long). Lc u dung dch pha long ban u [v d: lc bng tay 25 ln chuyn ng ngang 300 mm trong 7 s hoc dng my khuy trn c hc (6.14) t 5 s n 10 s] t c pha long 10-1.

9.3.2. Sa bt

9.3.2.1. M vt cha (8.2), dng dao trn ly lng phn mu th cn thit v tin hnh theo 9.3.2.2. y ngay vt cha.

9.3.2.2. Cn 10 g mu th vo dng c thy tinh thch hp (v d: cc c m) v sau cho sa bt vo chai pha long cha cht pha long thch hp (5.3). Cch khc, cn 10 g mu trc tip vo chai pha long cha cht pha long thch hp. ha tan mu th, xoay nh nhng lm t sa bt sau lc chai (v d: 25 ln vi chuyn ng ngang 300 mm trong khong 7 s). C th s dng my trn kiu nhu ng lm thit b lc. yn trong 5 min, thnh thong lc.

9.4. Chun b cc dung dch pha long thp phn tip theo

Chun b dung dch pha long tip theo theo TCVN 6263 (ISO 8261).

9.5. Lm tan chy mi trng

Trc khi bt u cc thao tc theo 9.6, lm tan chy mi trng (5.2) v ngui trong ni cch thy (6.5) t nhit t 44 0C n 47 0C. Kim tra nhit ca mi trng bng cch t nhit k vo 250 ml thch (v d: thch nc) ca mt bnh ring bit ging vi bnh ng mi trng. Rt thch lng trong vng 2 h sau khi lm tan chy.

9.6. Chun b cc dung dch pha long nh phn v cy mi trng

9.6.1. Dy pha long u tin (S1)

Ly 12 ng nghim pha long (6.9) cha 5 ml dung dch pha long c bo qun lnh (5.3.3).

Chun b cc dung dch pha long nh phn (D1, D2,) bng cch dng pipet mi dung tch 5 ml ly 5 ml huyn ph ca dung dch pha long trc (9.4) vo ng nghim cha 5 ml cht pha long. Trn huyn ph 5 ln trong 5 s bng my khuy (6.14) trc mi ln ly huyn ph a vo ng nghim. t chai pha long thp phn cui cng (9.4) tr li t lnh ngay sau khi ly dch cy u tin.

Trc khi bt u tin hnh dy pha long nh phn tip theo, cy 3 a Petri (P1, P2 v P3) t mi dung dch pha long trong s 12 dung dch pha long, dng pipet chia hoc pipet mt vch dung tch 1 ml (6.12). S dng pipet v trng mi cho mi dung dch pha long.

Sau khi cy tt c cc a ca cc dy (S1), rt t 12 ml n 15 ml mi trng nui cy (9.5) c lm nng chy v c lm m (t 44 0C n 47 0C) vo mi a Petri theo th t ng nh lc cy mu. Cn thn trn mi trng vi dch cy bng cch xoay u cc a Petri cc khun lc phn b ng u sau khi . cho hn hp ng li bng cch t cc a Petri trn b mt phng v ngui.

9.6.2. Cc dy pha long tip theo (S2, S3 v S4)

Sau khi hon thnh dy pha long u tin v a, chun b cc dy pha long th hai, th ba v th t (S2, S3 v S4) tng t, bt u mt dy pha long bng cch trn u lng cha trong chai pha long thp phn cui cng (9.4) c bo qun trong t lnh trong thi gian ch i trc . Dng 2 phn hoc 3 phn, mi phn gm 250 ml mi trng c lm tan chy a cho mi dy pha long v loi b phn mi trng cn d.

9.7. m

Lt p cc a chun b v t trong t m (6.3) nhit 30 0C trong 72 h ( 2 h. Khng chng cao qu 3 a. nh du v tr mi a trong chng (di-gia-trn).

CH THCH: Thng tin ny c th c ch nu xut hin bin thin qu ln gia cc a cy v c th xem xt nh hng ca vic chng a.

Khng cc chng a st nhau v st thnh t m cng nh trn t m. Khng cc khay trong t m.

9.8. M ha ngu nhin cc a v m khun lc

9.8.1. M ha ngu nhin

Khng m cc a theo th t pha long hoc nhm theo cc bc pha long v iu ny c th dn n kt qu c lng bin thin thp do ngi m khun lc c th c mong mun ch quan ti kt qu. V vy, trc tin phi kim tra v m ha cc a nh m t trong iu ny bi mt ngi khng tham gia vo qu trnh m 9.8.2.Bt u t dy pha long nht, la chn cc dung dch pha long c th m c, ngha l cc dung dch pha long c s m d kin trung bnh t 5 n 300 khun lc mi a.

M ha tt c cc a ca dung dch pha long m c, s dng bng s ngu nhin m ha ngu nhin cc a ca tt c cc dy v cc dung dch pha long. Xem Bng 1 v mt bng s ngu nhin nh vy; cng c th s dng cc s ngu nhin t 1 n 144. Xa cc du hiu ban u trn cc a, nn s dng cc nhn dn c th tho ri. Ch r cc a khng m c bng du tr trong quy trnh (xem Bng 2).

9.8.2. m khun lc

Kim tra cc a di nh sng du. h tr qu trnh m, nn dng knh lp (6.6) v/hoc my m thch hp. Ch trnh m nhm cc thnh phn mu khng ha tan hoc cc cht kt ta trong cc a vi cc khun lc nh. Kim tra cc khun lc nghi ng ny, s dng knh lp c phng i cao hn (6.6) phn bit khun lc vi cc cht ngoi lai, nu cn.

in cc s m khun lc vo Bng 2.

lch gia cc s m c th nh hng n kt qu nh gi, nh cc a b mc lan trn ton b hoc mt a cn c ch r trong bng (Hnh 2).

Vic m ha v m khun lc phi thc hin trong ngy kt thc m.

10. c lng thng k

10.1. S ph hp ca b d liu

CH THCH: Trong cc bng v cc cng thc, k hiu i, j v k tng ng l ch s ca dy pha long S, cc dung dch pha long D v cc a P.

Khi khng th m c tt c 3 a song song ca bt k mt pha long no th loi b tt c cc s m ca dung dch pha long 3 dy pha long khc.

Thc nghim c th c lng thng k khi c t nht d liu ca 5 mc pha long nh phn. C th chp nhn thiu 5% s m [ngha l: 3 a cy trong tng s 60 a (5 mc pha long) hoc 4 a cy trong tng s 72 a (6 mc pha long)] tr trng hp s thiu nm trn trong b ba a song song. Trong trng hp ny, phi loi b tt c cc a ca dung dch pha long tng ng.

Thm vo , s m trung bnh mong mun ca 5 hoc 6 pha long chp nhn c phi nm trong khong 5 n 300 khun lc trong mi a.

Nu tt c cc iu kin trn khng c p ng th b d liu coi nh khng y . Lp li quy trnh, chn vt liu mu thch hp hn (nu c qu nhiu a khng m c) hoc tun th nghim ngt hn cc hng dn s dng, hoc c hai........................................

Bng 2 - V d v bng ghi cc s m khun lc, s dng trong qu trnh m a c m ha trong Bng 1 theo s th t a

aS khun lcaS khun lcaS khun lcaS khun lc

13773109

23874110

33975111

44076112

54177113

64278114

74379115

84480116

94581117

104682118

114783119

124884120

134985121

145086122

155187123

165288124

175389125

185490126

195591127

205692128

215793129

225894130

235995131

246096132

256197133

266298134

276399135

2864100136

2965101137

3066102138

3167103139

3268104140

3369105141

3470106142

3571107143

3672108144

CH THCH: Cc a khng c m hoc khng th m c th ghi "-", cc a khng c khun lc th ghi "O".

10.2. c lng b d liu y (xem Hnh 2)10.2.1. Lp bng cc s m

Mi s m thuc v mt dy pha long c th Si (S1, S2, S3 hoc S4), mt dung dch pha long c th Dj (D1, D2, D6, trong D1 l dung dch pha long t nht cho cc a c th c kt qu) v mt a c th Pk (P1, P2 hoc P3). khi qut hn v cc kt qu, sp xp cc s m Bng 2 theo dy pha long Sj, nng pha long Dj v cc a cy Pk nh trong Bng 3.

Bng 3 - Lp bng cc s m

SiDjPk

P1P2P3

S1D1C111C112C113

D2C121C122C123

DjC1j1

S2D1C211

D2

Dj

SiDjCij2

..

CH THCH: C123 l s m ca a th ba, dung dch pha long th hai ca dy pha long u tin.

10.2.2. Kim tra tnh ng nht ca vic a: php th G2PBc u tin trong c lng thng k cc s m (xem Hnh 2) bao gm xc nh ln ca tnh ng nht c ngha thng k ca cc a cy lp li. Php th ny c tin hnh pht hin cc kt qu l "qu tt", ngha l: pht hin cc trng hp trong bin thin cc s m gia cc a song song P1, P2 v P3 nh hn d kin ("phn tn thp"). iu ny, v d, c th c nguyn nhn t vic m ha cc a khng hp l trc khi m.Tnh ng nht c ngha thng k l s t c nh sau (xem Ph lc A c thm thng tin chi tit v php th G2P v v d Bng 6).

a) Xc nh gi tr trung bnh Cij ca tng nhm 3 a song song (v d: C12 = C121 + C122 + C123)/3, trung bnh cc kt qu ca mi hng c ch ra trong bng).

b) Tnh ton mi a vi s m Cijk, gi tr Cijk.ln(Cijk/Cij), vi Cij l gi tr trung bnh cc kt qu ca b a cy song song.

c) Tnh tng 3 gi tr trong mi hng (cc a cy song song) v nhn vi 2 thu c gi tr G2 ring bit cho mi b a song song. Vit cc thng s vo trong bng. Tnh tng cc gi tr ca tt c cc pha long ca cc dy pha long thu c gi tr th:

G2P = 2 [ eq \i \su(i,s, )

eq \i \su(j,d, )

eq \i \su(k,p, )(Cijk ln eq \f(Cijk, Cij)) ]

Trong s l s dy pha long, d l s bc pha long v p l s a.

CH THCH: Cng thc thay th, (A.6) v (A.7), c nu trong Ph lc A.

d) So snh gi tr ca G2P vi cc gi tr ("khi" bnh phng) trong Bng 4 vi 40 bc t do (df = 40) trong trng hp 5 dung dch pha long v 48 bc t do (df = 48) trong trng hp 6 dung dch pha long. Nu mt s kt qu b thiu (xem 10.1) th ly bc t do df tr i s kt qu b thiu trc khi so vi cc gi tr trong Bng 4.

Nu gi tr G2P thp hn gi tr xc sut 0,995 trong bng th cc s m ca cc a song song c bin thin thp hn d kin: cc gi tr rt ng nht v mt thng k. Ngi phn tch phi xc nhn iu ny bng cch m ha ngu nhin cc a mt ln na v m li t u. Trong trng hp xc nhn, phi lp li vic thc nghim v tun th nghim ngt quy trnh.Nu gi tr G2P ln hn gi tr xc sut (P) 0,995 % trong bng th bin thin gia cc a song song khng qu thp. Tip tc php th theo 10.2.3.

Nu gi tr G2P ln hn ng k so vi gi tr xc sut 0,01 th cc a song song c bin thin ln hn d kin. Lu n iu ny v c th s dng din gii v bin thin tng th trong cc giai on sau ny.

Bng 4 - Cc gi tr cho (P = 0,01 v P = 0,995)df

df

3759,8918,594468,7123,58

3861,1619,294569,9624,31

3962,4320,004671,2025,05

4063,6920,714772,4425,78

4873,6826,51

Hnh 2 - Lu phn tch thng k i vi nghin cu thnh tho m khun lc ti phng th nghim

10.2.3. Th nghim tnh ng nht chung ca cc s m

10.2.3.1. Php th G2ABc th hai (xem Hnh 2) bao gm xc nh ln ca tnh ng nht c ngha thng k ca ton b d liu. Trong php th ny, cc s m c so snh vi cc gi tr d kin, c tnh n nh hng ca cc dung dch pha long.

Tnh ng nht c ngha thng k chung l mt s n thu c nh sau (xem thm V d).

a) Xc nh tng cc s m khun lc trong bng ((Cijk) v trong tng th tch mu ((Vijk) ca cc khun lc c m.

b) Th tch mu ca bc pha long cao nht Dd cha cc khun lc c m (D5 hoc D6) c coi l mt n v th tch (Vd = 1). i vi mi bc pha long thp hn Di, th tch mu tng ng Vi c th xc nh nh sau:

Vi = Vd.2aTrong a l s dung dch pha long nh phn gia Di v dung dch pha long cao nht Dd.

V D: Gi s dung dch pha long cao nht c cc s m hp l l D5, tng ng vi pha long 2-10, th tch mu ca dung dch pha long ny c coi l 1 n v th tch (V5 = 1). Th tch tng ng V3 ca dung dch pha long D3 (2-8), vi hai bc pha long nh phn t D5, c tnh: V3 = V5 x 22 = 4 n v th tch V5.

1) Tnh s khun lc d kin trong mt n v th tch mu e = (Cijk/(Vijk v t tnh s khun lc d kin E(Cijk) = e.Vijk i vi mi pha long.2) Tnh gi tr Cijk.ln[Cijk/E(Cijk)] i vi s m Cijk ca mi a.

3) Nu Cijk = 0 th gi tr nu trn cng tng ng vi 0.4) Tnh gi tr G2(2) i vi mi hng (mt b cc a song song) (ngha l: G2 vi 2 bc t do nu khng b thiu d liu) bng cch cng gi tr ca b ba a v nhn tng ny vi 2.

G2(2) = 2 [ eq \i \su(k,p, )(Cijk . ln eq \f(Cijk, E(Cijk))) ]

Ghi cc gi tr ny vo bng.

5) Tnh tng cc gi tr G2(2) ca ton b cc dy pha long v dung dch pha long trong bng. Bng cch ny, gi tr ca php th thu c nh sau:

G=

So snh gi tr G2A vi cc gi tr trong Bng 5, trong df bng s lng s m s dng trong tnh ton tr i 1. Nu gi tr Gthp hn gi tr xc sut 0,01 trong bng th cc s m l ng nht c ngha thng k. Tt c cc bc trong khi tin hnh phng php (ly mu con, pha long, a, m khun lc) coi nh c chp nhn. Cc c lng tip theo l khng cn thit v c th kt lun v php phn tch ti y.

Bng 5 - Cc gi tr cho (P = 0,01)df

df

5683,536796,82

5784,756898,02

5885,966999,22

5987,1770100,42

71101,62

Nu gi tr G ln hn gi tr xc sut 0,01 trong bng th tin hnh phn tch phng sai (ANOVA) ca d liu xc nh v nh lng ngun cc bin thin qu mc (ly mu con, pha long, a, m khun lc).C 2 l do c th lm tng gi tr G:a) Khi phn tng thm mc trung bnh th l do c th l c mt s bc trong phng php c bin thin ln hn d kin mt cht nu phng php c p dng ng. Tuy nhin, bin thin vt mc ny khng qu ln khng th s dng kt qu trong tnh ton. Trong trng hp ny, hm G2A l du hiu cnh bo: phng php vn c th c p dng nhng cn phi kim sot thng xuyn m bo bin thin khng tr nn qu ti t.

b) Khi phn tng thm tr nn ng k (G2A ln hn nhiu so vi gi tr P = 0,01) th c mt bc hoc nhiu bc c bin thin cao mc khng th chp nhn c.

Xem xt k lng v hiu chnh cc yu t c bin thin qu cao.

V D: Bn dy pha long S1, S2, S3 v S4 c chun b t mt mu ng nht tt v c pha long thch hp. Chun b mt dy pha long nh phn t mi mu con v cy 3 a song song P1, P2 v P3 t mi dung dch pha long Dj. Su dung dch pha long cho kt qu m c. Cc s m c trnh by trong Bng 6. Tt c cc gi tr c tnh ton bng my tnh in t c y cc chc nng sau kt qu c lm trn n 2 hoc 3 ch s thp phn.Bng 6 - D liu gc

SiDjPkCijE(Cijk)G2

P1P2P3

S1D12-684113109102,00205,804,997

D22-774827075,33102,900,984

D32-835433337,0051,451,483

D42-910131613,0025,721,397

D52-1071199,0012,860,896

D62-110231,676,434,256

S2D12-6238236226233,33205,800,356

D22-7154153153153,33102,900,004

D32-8154126111130,3351,457,226

D42-933343835,0025,720,395

D52-1016152117,3312,861,161

D62-114565,006,430,403

S3D12-6154151136147,00205,801,280

D22-784687274,67102,901,831

D32-844656357,3351,454,899

D42-925352528,3325,722,275

D52-1013131313,0012,860,000

D62-115032,676,436,993

S4D12-6238236224232,67205,800,496

D22-7154156146152,00102,900,371

D32-863615660,0051,450,437

D42-917322825,6725,724,980

D52-1011101211,0012,860,182

D62-111744,006,435,062

Cc ch s: i = 4, j = 6, k = 3.Tng = G = 52,364

Trong Bng 6, ct Cij gm cc gi tr trung bnh ca 3 a song song, ct E(Cijk) gm cc gi tr trung bnh d kin ca cc dung dch pha long.Cc gi tr E(Cijk) c tnh nh sau (cc bc pha long c s dng: t D1 = 2-6 n D6 = 2-11).Cc th tch c biu th theo n v tng i ( pha long cao nht 2-11 c coi l n v th tch):

- V1 (tng ng vi D1) = 25V6 = 32

- V2 (tng ng vi D2) = 24V6 = 16

- V3 (tng ng vi D3) = 23V6 = 8

- V4 (tng ng vi D4) = 22V6 = 4

- V5 (tng ng vi D5) = 21V6 = 2

- V6 (tng ng vi D6) = 1

- Tng s khun lc: 84 + 113 + 109 + 74+ + 1 + 7 + 4 = 4862

- Tng th tch (theo n v V6): (12 x 32) + (12 x 16) + (12 x 8) + (12 x 4) + (12 x 2) + (12 x 1) = 756

- S khun lc d kin E(Cijk) i vi th tch V6 = 4862/756 = 6,43- S khun lc d kin E(Cijk) i vi th tch V5 = 4862/756 x 2 = 12,86

- S khun lc d kin E(Cijk) i vi th tch V4 = 4862/756 x 4 = 25,72




D liu ca mi dy pha long c nu trong s ca Hnh 3. ng thng ca mi biu trong s bn biu nu ln gi tr d kin nu cc dung dch pha long trong dy pha long c tin hnh hon ho. Mi gi tr trong mt dy pha long c k hiu bng du "+".Hnh 4 a) biu din bn ng thng trong Hnh 3 trong cng mt biu . Cc lch tng h ca chng cho bit cht lng ca vic ly mu con. Hnh 4 b) ch ra trung bnh ca cc a song song (ct Cij trong Bng 6) i vi mi dy pha long. Trong trng hp l tng, tt c cc im nm trn cng mt ng.

a) Dy pha long 1

b) Dy pha long 2

c) Dy pha long 3

d) Dy pha long 4CH DN

X pha long 2-x (thang nh phn)

Y s m khun lc (thang nh phn)

Hnh 3 - Cc dy pha long

a) Cc trung bnh theo khi lng ca mi dy pha long

b) Cc trung bnh ca cc a cy song song ca mi dy pha long

CH DN

X pha long 2-x (thang nh phn)

Y s m khun lc (thang nh phn)Hnh 4 - Cc trung bnh theo khi lng v cc trung bnh ca cc a cy song song

10.2.3.2. Tnh gi tr ca php th G

i vi php tnh ny, s dng cc gi tr trong ct Cij.

Tnh gi tr G (ct cui cng ca Bng 6) ca cc a song song trong mi hng.

2 [ 84ln (84/102) + 113ln (113/102) + 109ln (109/102)] = 4,9972 [ 74ln (74/75,33) + 82ln (82/75,33) + 70ln (70/75,33)] = 0,984

2 [ 1ln (1/4) + 7ln (7/4) + 4ln (4/4)] = 5,062

Cng tt c cc gi tr thu c gi tr tnh ng nht chung ca vic cc a song song:

G= 4,997 + 0,984 + + 5,062 = 52,364

S bc t do ca G c tnh bng tng tt c cc bc t do ca cc G2. Trong v d nu trn khng thiu s m no, do mi hng c 2 bc t do v tng s bc t do df = 48.

Gi tr G(52,364) ln hn gi tr 26,51 v nh hn gi tr 73,68 trong Bng 4, do bin thin gia cc a song song nm trong gii hn cho php.

10.2.3.3. Tnh gi tr ca php th G

Cc gi tr E(Cijk) trong ct c dng tnh ton:

G= 2 [ 84ln (84/205,80) + 113ln (113/205,80) + . + 4ln (4/6,43)] = 2 (420,35) = 840,70

S bc t do = (72 -1) = 71

Gi tr G (840,70) ln hn nhiu so vi gi tr 101,62 trong Bng 5. Do bin thin tng th vt qu gii hn; phn tch thng k nn c tip tc tm ra ngun gc ca bin thin ln ny.

10.2.4. Ngun gc ca cc sai s v phn tch phng sai (xem Hnh 2)

10.2.4.1. Mi yu t trong quy trnh u lin quan n bin thin xc nh (thut ng thng k gi l phng sai). Tr cc thay i l tng gia cc a song song, khng th bit ln ca phng sai i vi mi yu t nu phng php c p dng ng. Tuy nhin, s thay i ln hn nhiu so vi gi tr d kin ch ra cc vn trong mi bc c th ca phng php.

V mi kt qu lin quan n mt s ngun bin thin (ly mu con, pha long, a hoc c th kt hp cc yu t ny), cn phi tch cc bin thin lin quan n mi ngun t d liu chung. iu ny c thc hin bng cch dng k thut thng k gi l "phn tch phng sai".CH THCH 1: i vi c s l thuyt ca k thut nu trn, xem cc sch v thng k.

CH THCH 2: Nu thiu d liu (xem 10.1) th cc gi tr tng ng c th c c lng bng cch ly trung bnh ca cc a song song trong cng dy pha long v cng nng pha long. Sai s trong thao tc ny nh v c nh hng khng ng k n kt lun.

10.2.4.2. Thc hin tch cc phng sai nh sau.

Cc s m Cijk trong bng (xem 10.2.2) c chuyn thnh cc gi tr Tijk theo cng thc:

Tijk = eq \r(,Cijk) - eq \r(,E(Cijk))Trong E(Cijk) l gi tr d kin ca mi dung dch pha long (xem 10.2.3).

CH THCH: Nu mt hoc nhiu d liu b thiu c c lng (xem Ch thch 2 trong 10.2.4.1) th phi tnh li cc gi tr d kin E(Cijk) ca b d liu tng th trc khi thc hin chuyn i.

Cc s m vi sinh vt tun theo phn b thng k gi l phn b Poisson. Tuy nhin, vic phn tch phng sai li tun th theo phn b Gauss (cn gi l phn b chun), do cc d liu th khng c s dng trc tip trong tnh ton. Chuyn dng khai cn ca d liu s cho phng sai ca cc s m hp l hn vi gi nh l thuyt ca k thut phn tch phng sai.

10.2.4.3. T d liu chuyn dng, Tijk s tnh c gi tr tng s.

Tng s tt c cc gi tr chuyn dng:

(v) =

Trong :

s l s dy pha long Si (= 4);

d l s cc bc pha long Dj (=5 hoc 6);

p l s a Pk (=3).

Tng bnh phng ca tt c cc gi tr chuyn dng:

(w) =

Tng bnh phng ca ton b cc nhm a cy lp li, Pk:

(x) =

Tng bnh phng ca ton b cc dy pha long Si:

(y) =

Tng bnh phng ca ton b cc dung dch pha long Dj:

(z) =

10.2.4.4. Cc tng ny c dng tnh cc tng bnh phng:Ngun bin thinTng bnh phng

Gia cc dy pha long S((1) = eq \f(s x (y) - (v)2, s x d x p)

Cc bc pha long D trong dy pha long S((2) = eq \f(d x (x) - (y), d x p)

Gia cc a song song((3) = (w) - eq \f((x), p)

Tng s((4) = (w) - eq \f((v)2, s x d x p)

Cc tng bnh phng c dng tnh cc bnh phng trung bnh, cc bnh phng trung bnh ny cng vi cc bnh phng trung bnh d kin trn l thuyt c dng xc nh cc phng sai. Cc bc ny c trnh by trong Bng 7.Bng 7 - ANOVA: Cc bnh phng trung bnh v cc bnh phng trung bnh d kin

Ngun bin thinTng bnh phngdfBnh phng trung bnhBnh phng trung bnh d kin

Gia cc dy pha long S((1)s - 1= eq \f(((1) , s - 1) + p+ dp

Gia cc bc pha long D trong cc dy pha long((2)s(d - 1)= eq \f(((2) , s(d - 1)) + p

Gia cc a song song((3)s x d(p - 1)= eq \f(((3) , s x d(p - 1))

Tng((4)s x d x p -1

Phng sai gia cc a, c c lng t S. Phng sai gia cc bc pha long trong dy pha long, , c c lng t (- )/p v phng sai gia cc dy pha long, , c c lng t (- )/dp.Tng phng sai = + + .

10.2.5. c lng cc phng sai

Gi tr l tng cho nn l khong 0,25; kinh nghim cho thy nu tng phng sai ln hn 1 th c sai li nghim trng.Nu tng phng sai nh hn 1 th phng php coi nh trong iu kin kim sot thng k. Nu tng phng sai ln hn 1 th mt hoc nhiu yu t mt kim sot v mt thng k. Trong trng hp ny, phi m rng phn tch phng sai xc nh nhng khuyt im khi tin hnh phng php.

Vic m rng phn tch thng k bao gm hai bc. u tin, phng sai lin quan n "cc bc pha long trong dy pha long" c phn thnh mt phng sai lin quan n cc bc pha long v mt phng sai sinh ra bi s tng tc c th xy ra gia cc bc pha long v cc dy pha long. Sau cc phng sai ny c kim tra v ngha thng k phn loi cc yu t theo mc quan trng. Trc ht nn tnh cc gi tr b sung sau y:Tng bnh phng ca cc dung dch pha long D:

((5) = eq \f(d(z) - (v)2, s x d x p)Tng bnh phng ca s tng tc gia cc bc pha long v cc dy pha long c tnh bng:

((6) = ((2) - ((5)

Cc kt qu phn tch phng sai hon chnh c nu trong Bng 8.

Bng 8 - Cc kt qu phn tch phng sai

Ngun bin thinTng bnh phngdfBnh phng trung bnhGi tr F

Gia cc dy pha long S((1)s - 1= eq \f(((1) , s - 1)

Gia cc bc pha long D((5)d - 1= eq \f(((5) , d - 1)

Tng tc((6)(s -1)(d - 1)= eq \f(((6) , (s - 1)(d - 1))

Gia cc a song song((3)s x d(p - 1)= eq \f(((3) , s x d(p - 1))

Tng((4)(s x d x p) - 1

Xc nh cc gi tr F kim tra ngha ca cc phng sai tng ng. Cc gi tr F l cc t s gia hai phng sai trong t s l phng sai tnh c lin quan n yu t cn xem xt v mu s l phng sai d kin nu yu t cn xem xt khng to nn tng phng sai. Nu yu t cn xem xt khng c nh hng th hai phng sai c lp v t s ca chng nm trong di xc nh trc . Tuy nhin, nu h s lm tng bin thin th phng sai t s tr nn qu ln khin cho t l F nm ngoi di xc nh trc.So snh cc gi tr F thu c vi F(f1, f2) tng ng trong Bng 9 mc xc sut P = 0,01. Cc k hiu f1 v f2 l s bc t do (ct df trong Bng 8) ca phng sai t s v phng sai mu s tng ng.

Bng 9 - Gi tr F cho cc kt hp la chn ca f1/f2 (P = 0,01)f1f2F

3125,95

3155,42

4125,41

5154,56

12402,66

15482,44

10.2.6. Kt lun v hnh ng tip theoCn xem xt k cc yu t lm cho gi tr F vt qu gi tr ti hn trong Bng 9. Chng cho thy cc kha cnh (tnh ng nht mu, thao tc) nn c a v iu kin kim sot trc khi s dng cc s m khun lc. Gii thch v gi tr c ngha ca cc yu t khc nhau nh sau:

a) gia cc dy pha long S: sai s h thng trong vic chun b cc dy pha long khc nhau (ng ha vt liu mu, phn phi);

b) gia cc bc pha long D: sai s trong cch chun b cc bc pha long;

c) tng tc: sai s chung khi tin hnh cng vic;

d) gia cc a: nu phng sai gia cc a ln hn nhiu so vi 0,25 th phi gii thch r; nu gi tr cao khng phi do cc b a cy song song thay i qu mc (xem cc thnh phn ca gi tr G trong Bng 6) th xem xt nh hng ca vic chng a bng cch tnh tng s m khun lc ca ton b d liu theo v tr ca chng trong chng a (trn-gia-di);

e) c lng nh hng ca phng sai cao trong phn tch phng sai.

Sau khi xem xt v hiu chnh cc bc khim khuyt trong p dng phng php, phi lp li ton b quy trnh xc nhn rng cc vn c hiu chnh.

V D: D liu trong Bng 6 c thay th bi cc gi tr chuyn dng: hai ct bn phi a ra tng ca cc gi tr ca mi hng v tng cc gi tr bnh phng ca mi hng.

Bng 10 - Cc d liu c chuyn dngSiDjPkTngTng bnh phng

P1P2P3

S1D12-6-5,181-3,716-3,905-12,80155,895

D22-7-1,542-1,089-1,777-4,4086,720

D32-8-1,257-0,615-1,428-3,3003,998

D42-9-1,910-1,466-1,072-4,4486,946

D52-10-0,941-0,270-0,586-1,7971,302

D62-11-2,536-1,122-0,804-4,4628,336

S2D12-61,0821,0170,6882,7862,676

D22-72,2662,2252,2256,71615,038

D32-85,2374,0523,36312,65255,153

D42-90,6730,7591,0922,5242,222

D52-100,4140,2870,9961,6961,245

D62-11-0,536-0,300-0,086-0,9220,385

S3D12-6-1,936-2,057-2,684-6,67715,184

D22-7-0,979-1,898-1,659-4,5357,311

D32-8-0,5400,8890,7641,1141,667

D42-9-0,0720,844-0,0720,7000,723

D52-100,0190,0190,0190,0570,001

D62-11-0,300-2,536- 0,804-3,6407,167

S4D12-61,0821,0170,6212,7192,589

D22-72,2662,3461,9396,55114,398

D32-80,7640,6370,3101,7121,087

D42-9-0,9490,5850,220-0,1441,291

D52-10-0,270-0,424-0,122-0,8160,268

D62-11-1,5360,110-0,536-1,9622,659

Tng ca tng (v)-10,685

Tng ca tng bnh phng (w)214,260

(x) = tng ca [cc bnh phng ca cc tng i vi mi nhm a cy lp li]= tng ca cc bnh phng ca cc gi tr trong ct "tng" Bng 10

= (-12,801)2 + (-4,408)2 + + (-1,962)2= 598,070

(z) = tng ca [ cc bnh phng ca cc tng ca cc dy pha long]

Cc dy pha longTng sCc bnh phng

S1-31,216974,445

S225,452647,800

S3-12,981168,494

S48,05964,953

Tng(v)1855,693 (y)

Cc tng (v), (w), (x) v (y) c s dng tnh cc tng bnh phng:Ngun bin thinTng bnh phng

Gia cc dy pha long S((1) = = 101,508

Cc bc pha long D trong dy pha long S((2) = = 96,263

Gia cc a cy song song((3) = 214,260 - =14,903

Tng((4) = 214,260 - =212,674

T cc tng ca cc bnh phng thu c, tnh cc bnh phng trung bnh sau coi nh chng l cc bnh phng trung bnh d kin theo l thuyt c lng cc phng sai.Ngun bin thinTng bnh phngdfCc bnh phng trung bnh

Gia cc dy pha long S((1) 3= = 33,836

Cc bc pha long D trong dy pha long((2) 20= = 4,813

Gia cc a song song((3) 48= = 0,310

c lng cc thnh phn phng sai (xem Bng 7):

c c lng bi = 0,310;

c c lng bi (- )/p = (4,813 - 0,310)/3 = 1,510;

c c lng bi (- )/dp = (33,836 - 4,813)/18 = 1,612;

c c lng bi 0,310 + 1,501 + 1,612 = 3,424.

Khi gi tr cui cng ny ln hn gi tr cho php cc i l 1 th phi tin hnh xem xt tip theo v ngun gc ca sai s.

Cc gi tr sau y phi c tnh ton trc tin:

(z) = tng bnh phng ca cc tng cho mi bc pha long

Cc bc pha longTngCc bnh phng

D1-13,974195,271

D24,32518,703

D312,178148,299

D4-1,3681,872

D5-0,8590,739

D6-10,986120,695

Tng(v)485,579(z)

Tng bnh phng i vi cc bc pha long ((5) = =38,879Tng bnh phng ca tng tc ((6) = 96,263 - 38,879 = 57,384

S dng cc gi tr ny hon thnh bng phn tch phng sai (xem Bng 8):

Bng 11 - Tnh ton

Ngun bin thinTng bnh phngdfBnh phng trung bnhGi tr F

Gia cc dy pha long S101,5083s12 = 33,8368,845a

Gia cc bc pha long D38,8795s52 = 7,7762,033 (n.s.)b

Tng tc57,38415s62 = 3,82612,321a

Gia cc a song song14,90348s32 = 0,310

Tng212,67471

a C ngha mc ngha P = 0,01.

b n.s.: khng c ngha thng k.

So snh cc gi tr F mc xc sut 0,01 (Bng 9) vi cc gi tr tnh c di y:- gia cc dy pha long S: F0,01 (3;15) = 5,42 < 8,845 c ngha;- gia cc bc pha long D: F0,01(5;15) = 4,56 > 2,033 khng c ngha;

- tng tc: F0,01 (15;48) = 2,44 < 12,321 c ngha.

Gi tr "gia cc dy pha long S" cao cho thy sai s h thng trong vic chun b cc dy pha long (tnh ng nht ca vt liu mu, phn phi mu) hoc tnh khng n nh ca qun th vi sinh vt.

Gi tr "tng tc" cao cho thy v mt thng k, c s khc nhau v tuyn tnh (hoc phi tuyn tnh) gia cc s m khun lc ca bn dy pha long (xem Hnh 3 v Hnh 4). iu ny khng nh rng c sai st trong cc thao tc k thut, nhng cng c th cho thy cc tng tc sinh hc hoc cc pht trin bt thng m ngi m khun lc c th nhm.

Trong trng hp ny, khng c thng tin v gi tr c th chp nhn "gia cc bc pha long D" khi c s tng tc ng k. Kt qu ny ch cho bit rng khng c sai s h thng trong vic chun b cc dung dch pha long.

Ph lc A

(Tham kho)

Kim tra trung bnh theo khi lng ca mu th v tnh ng nht ca cc s m khun lc

A.1. Gii thiu

Khi c sn cc s m khun lc t cc a cy song song, t cc th tch huyn ph khc nhau hoc t nhiu hn mt pha long th tt nht l s dng mi thng tin c lng mt tt nht. Tnh s khun lc trung bnh. Nguyn tc ny c xut bn ln u trong ti liu v vi sinh vt (xem Ti liu tham kho [2]).

Thng thng ch nhng a c s m khun lc t 30 n 300 hoc t 25 n 250 l "ng tin cy". Nu tun th cc quy tc ny th thng ch cn li mt dung dch pha long c s khun lc m c v khng cn tnh s khun lc trung bnh. Tuy nhin, c th l lng ph khi b qua cc a t hn 25 khun lc, c bit v cc a c s khun lc thp l ng tin cy nht v mt sinh hc. Gii hn di, trung bnh khong 5 khun lc trong mi a c th l gii hn di ph hp v mt thng k.

S khun lc trung bnh l tng bng tng s tt c cc s m khun lc quan st c chia cho tng cc th tch mu lin quan (c biu th theo mu ban u). Cch biu th chung nht ca nguyn tc chung ny nh sau:M = eq \f((Ci, (Vi) = eq \f(C1 + C2 + ... + Cn, V1 + V2 + ... + Vn) (A.1)

Trong :

M l s m khun lc trung bnh trn mililit mu ban u;

Ci l s cc khun lc m c trn a th i;

Vi l th tch c s dng m khun lc trn a th I;

i l s a c s dng (i = 1, 2, n);

n l tng s a c s dng.

thu thp d liu vi tin cy thch hp, trc ht cn kim tra xem cc khun lc ng nht hay khng (bin thin ngu nhin). lm c iu ny, mt ch s ng nht chung l cn thit kim tra cc s m khun lc c ngun gc t V1, V2, Vn c phi c ly t mt huyn ph ng nht hay khng. Php th tt nht l G2, c m t trong A.2.Nu php th cho thy d liu khng ng nht th khng nn s dng tt c cc s m khun lc tnh trung bnh trn my tnh. ng thi, kt qu c th l du hiu xem xt thm v l do ca s khng ng nht ny.

C hai thnh phn chnh cu thnh ch s ng nht G2 trong vic m khun lc. N c s dng xc nhn s ng nht ca d liu trc khi tnh s khun lc trung bnh v c th c p dng trong phn tch " lch" tm ra s khng ng nht trong b d liu ca cc s m khun lc. Trong tiu chun ny, ch s G2 c s dng kim tra tnh ng nht chung, ngoi ra cn kim tra tnh ng nht chung ca vic cc a song song.

A.2. Php th G2 v tnh ngu nhin (tnh ng nht)

Ch s G2 l ch s ng nht thch hp kim tra d liu m khun lc. Ch s ny c biu th dng n gin nht l:

G2n-1 = 2

(A.2)

Trong :

Oi l s m khun lc quan st c th i;

Ei l s m d kin th i;

i l s ln m (i = 1, 2, n);

n l tng s ln m.

Trong php th "tnh ng nht ni b", gi s cc s m khun lc quan st c tun th cc th tch nghin cu. Do , cc s m khun lc d kin Ei thu c bng cch tnh phn ca tng ca tt c cc khun lc quan st c m v l thuyt mi th tch phi cha:Ei = eq \f(Vi, (Vi) (Ci(A.3)

Trong thc t, cng thc chung (A.2) bao gm vic tnh tng gi tr Ei v khng cn dng cho cc mc ch khc (trong trng hp ny, Cng thc A.2 c s dng bi v ANOVA cn cc gi tr Ei). Trong trng hp , a cc s m d kin trong Cng thc (A.3) v cu trc li cng thc, thu c dng thch hp hn:

G2n-1 = 2

(A.4)

Hin nhin l trong kim tra tnh ng nht ni b, th tch thc v bt k s m t l vi chng c th thay th Vi. Trong hu ht cc trng hp, t l ny c th c la chn theo cch sao cho cc th tch tng i Ri l cc s nguyn n. S thay i ny khin cng thc tr nn kh n gin tnh ton bng my tnh cm tay.

G2n-1 = 2

(A.5)

Trong :

Ci l s cc khun lc m c t a th i;

Vi l th tch thc c s dng cho a th i;

Ri l th tch tng i tng ng c s dng trn a th i;

i l s cc a c s dng (i = 1, 2, , n);

n l tng s cc a c s dng.

Cc ch s c gi tr cao cho thy s bin thin cao hn mc ngu nhin, gi l phn tn rng. Cc ch s c gi tr thp cho thy bin thin thp bt thng (phn tn hp).C th xem xt ngha thng k ca phn tn rng hoc phn tn hp bng cch so snh cc gi tr tnh c vi phn b x2 l thuyt mc xc sut la chn vi n - 1 bc t do. Cc gi tr cn thit trong cc v d c m phng trong Bng A.1. Cc tnh ton nu trn s cho ch s ng nht chung GA2 c s dng trong ni dung chnh.

Chng trnh BASIC tnh ch s ng nht theo Cng thc (A.5) c nu trong Ph lc B.

Bng A.1 - Cc gi tr ti hn la chn cho phn b X2dfXc sut

Phn tn hpPhn tn rng

0,990,950,100,050,010,001

1-0,0042,713,846,6310,38

20,0200,1034,615,999,2113,81

30,1150,3526,257,8111,3416,27

40,2970,7117,789,4913,2818,47

50,5541,1459,2411,0715,0920,52

A.3. S ph hp chung ca vic cc a song songCch s dng th hai ca php th G2 trong tiu chun ny l kim tra tnh ng nht ca vic cc a song song. C th thc hin iu ny bng cch tnh gi tr ca ch s ng nht i vi mi b a song song. Vic ny c n gin ha bi thc t l tt c cc th tch (hoc th tch tng i) trong cc b a cy song song l nh nhau v c th c thay th bi gi tr 1. Do , tng ca cc th tch tng i bng vi tng s a song song (n) v Cng thc (A.5) c th c vit li nh sau:

G2n-1 = 2

(A.6)

Trong :

Ci l s cc khun lc c m trn a th i; i l s a c s dng (i = 1, 2, n);

n l s cc a cy song song.

Gi tr cng ca ch s ng nht c s dng kh hn ch. Tuy nhin, nu c m b a cy vi s lng a song song bng nhau th c th thy u im ca vic b sung G2. Cng cc gi tr G2 v cc bc t do tng ng t tt c cc b a cy thu c gi tr ca tnh ng nht chung. Trong tiu chun ny, i lng nu trn c gi l GP2:G2P = G2m(n-1) = eq \i\su(j=1,m, G2(n-1))

(A.7)

Mt vi phn tn rng phi nm trong khong cho php trong tnh ton i vi cc phn tch vi sinh, bi v cc php o th tch vn ph hp vi cc chun lm vic chp nhn c v cc php o ny khng bao gi chnh xc tuyt i. Do , dng nh l mc xc sut 1% (P = 0,01) thay cho mc 5% thng thng cn c coi l "gii hn cnh bo".

A.4. Cc v d

A.4.1.1. V d 1: Php th tnh ng nht chung

Mt b cc s m khun lc thu c t hai pha long thp phn lin tip. Chun b hai dy a song song t mi pha long. Cc s m c nu trong Bng 2, vi cc th tch tng i.

Bng A.2 - m s khun lc hai dy pha long song song v cc th tch tng i pha longCc s m khun lcTngTh tch tng iTng

10-4251305556101020

10-5313667112

Tng62322

Tnh ngu nhin ca ton b a c tnh bng ch s G2 nh sau:G2 = 2[251ln(251/10) + 305ln(305/10) + 31ln(31/1) + 36ln(36/1) - 623ln(623/22)] = 7,607Khi c bn s hng trong tng, s c 4 - 1 = 3 bc t do. Phi so snh gi tr tnh c vi cc gi tr trong hng th ba ca phn b Bng A.1.

Gi tr 7,607 tnh c cho thy bin thin cao nhng so vi phn b l thuyt (gi tr xc sut 1% = 7,81) th ton b d liu c th c xem nh l mt phn b ngu nhin t mt huyn ph ng nht. T s ca cc s m gia cc dung dch pha long (556/67 = 8,3) khc nhiu so vi gi tr l tng 10 : 1, nhng c th xut hin do tnh c. Khng c l do thng k thch hp coi b d liu l khng ng nht. Do , s khun lc trung bnh c th c tnh theo Cng thc (A.1) s dng cc s m khun lc sn c.A.4.1.2. V d 2: Php th tnh ng nht chung bng cch phn tch lch

Vi b d liu tng t khc, nhng c ba a song song, c cc gi tr c nu trong Bng A.3.

Bng A.3 - S m khun lc ca ba dy a cy song song v cc th tch tng i

pha longCc s m khun lcTngTh tch tng iTng

10-5122749228810101030

10-6121510371113

Tng32533

B d liu dng nh phn tn kh rng, nh c ch ra bi ch s:G2 = 2[122ln(122/10 + 74ln(74/10)+ + 10ln(10/1) - 325ln(325/33)] = 15,077

Gi tr ny gn nh bng gi tr xc sut 1% ca 5 bc t do. Kt qu ny c ngha l vic tnh s khun lc trung bnh da trn d liu ton b l khng thn trng.

Do , cn phi kim tra trng hp tip theo. Cng thc (A.3) c th c s dng lp li nghin cu cc chi tit ca b d liu.

Tng bin thin bao gm ba thnh phn: chnh lch gia cc a song song ca pha long 10-5, chnh lch gia cc a song song ca pha long 10-6 v chnh lch gia cc mc pha long:G(2)2 = 2[122ln(122/10) + 74ln(74/10) + 92ln(92/10) - 288ln(288/30)] = 12,127

G(2)2 = 2[12ln12 + 15ln15 + 10ln10 - 37ln(37/3)] = 1,020

G(1)2 = 2[288ln(288/30) + 37ln(37/3) - 325ln(325/33)] = 1,930

Cc ch s di ca G2 biu th s bc t do km theo.Vic chia nh tng bin thin dn n php phn tch lch v c th c nu trong Bng A.4.

Ch tng 12,127 + 1,020 + 1,930 = 15,077 ng bng tng cc ch s c tnh trn.

Bng A.4 - Tng bin thin - Phn tch lch

Ngun bin thinG2dfP

Gia cc pha long1,9301> 0,10

pha long song song 10-51,0202> 0,30

pha long song song 10-612,1272> 0,01

Tng15,0775< 0,02

Khi so snh cc gi tr tnh c vi phn s l thuyt theo s bc t do thch hp, dng nh nguyn nhn ng ch ca phn tn rng trong d liu l s khng ng nht trong b a song song c s khun lc cao (t pha long hn). Khng nn s dng cc s m ny. iu ny c ngha l nn tnh gi tr trung bnh t duy nht mt b s m khun lc thp ( pha long cao hn). Do , khng nn coi b d liu ny b thiu hn cho d d liu ban u l khng ng nht.

A.4.1.3. V d 3: Tnh ng nht ca cc s m song song

Cng mt nhn vin th nghim s dng 2 a kp to nn 5 b s m khun lc. Thc hin nh gi thng nht trn cc a song song. Cc s m ban u c sp xp theo gi tr s m trung bnh tng dn nh trong Bng A.5. i vi mi cp s m, ch s ng nht c tnh bng Cng thc (A.6).Bng A.5 - Cp s m: ch s ng nht

S bCc s m song songG2df

122180,4011

235410,4741

380992,0211

41911642,0561

53402972,9051

Tng7,8575

Tng (7,857) ca ch s G2 ring r l mt kiu thng k ng dng trong ni dung chnh tnh s thng nht chung ca vic cc a song song (GP2). Theo cc gi tr c ch ra trong hng th nm ca Bng 5, s ph hp chung ca cc a song song trong v d dng nh tha mn.C th s dng cc d liu ny hoc cc d liu tng t theo cch khc. C th so snh mi ch s ring r vi cc gi tr ca phn b theo bc t do tng ng. Tuy nhin, vi cc trng hp c xc nh l km ph hp hoc cn nghi ng th khng cn ch qu nhiu, bi v G2 thay i ngu nhin v him khi c th sinh ra cc gi tr lch mt cch tnh c.

Tuy nhin, khi hng chc hay hng trm cc gi tr ch s c dng trn biu kim sot th kt qu c th cho cc thng tin hu ch khng ch cho ngi tin hnh phn tch m cn cho cc gii hn ca tin hnh phng php.

Khi d liu t c trong cc kha o to, c th dng cc gi tr ch s ring bit theo ng th t. Thng c gi tr thng tin nhiu hn khi v cc gi tr so vi s m khun lc trung bnh.

Thng xuyn c khuynh hng gi tr ch s cao (ph hp km, phn tn rng) do s m trung bnh tng ln. Xu hng ny dng nh xut hin trong v d trn. Cui cng, cc kt qu phi ch ra gii hn trn (v c th l gii hn di) ca phng php khi mt ngi phn tch thc hin (cc gi tr ch s lun c dng trn mt biu kim sot ring r).

Ph lc B

(Tham kho)

Chng trnh BASIC tnh ch s G210PRINT "LIKELIHOOD RATIO-INDEX G^2"

20INPUT "NUMBER OF TERMS, n-"; N

30C = 0; R = 0; S = 0; T = 0; D = 0

40FOR I=1 TO N

50PRINT "/=";/

60INPUT "COLONY COUNT="; C

70INPUT "RELATIVE VOLUME ="; R

80IF (C=0) THEN W = 0: GOTO 100

90W = C*LOG(C/R)

91REM IN THIS BASIC LOG = NATURAL LOGARITHM

100S = S + W

110T = T + R

120D = D + C

130NEXT /

140Y = 2 * (S - D*LOG(D/T)

150REM IN THIS BASIC LOG = NATURAL LOGARITHM

160PRINT

170PRINT "INDEX G^2="; Y

180PRINT

190PRINT

200INPUT "ANOTHER SET?(Y/N)"; H$

210IF H$ = Y" GOTO 20, ELSE TO 220

220END

TH MC TI LIU THAM KHO[1] TCVN 6400 (ISO 707), Sa v sn phm sa - Hng dn ly mu[2] FARMILOE, F.J., CORNFORD, S.J., COPPOCK, J.B.M and INGRAM, M. The survival of Bacillus subtilis spores in the baking of bread. J. Sci. Food. Agric., 5, 1954, pp. 294-404

1) B tiu chun TCVN 7150:2002 (ISO 835:1981) (gm 4 phn) c hy b v c thay th bng TCVN 7150:2007 (ISO 835:2007) Dng c th nghim bng thy tinh - Pipet chia .

2) TCVN 7151:2002 (ISO 648:1977) c hy b v c thay th bng TCVN 7151:2010 (ISO 648:2008) Dng c th nghim bng thy tinh - Pipet mt mc.

_1435082751.unknown

_1435391777.unknown

_1435392298.unknown

_1435393786.unknown

_1435394621.unknown

_1435402815.unknown

_1435403219.unknown

_1435403477.unknown

_1435400294.unknown

_1435393916.unknown

_1435393934.unknown

_1435393799.unknown

_1435393186.unknown

_1435393673.unknown

_1435392371.unknown

_1435391803.unknown

_1435392282.unknown

_1435391781.unknown

_1435084972.unknown

_1435391581.unknown

_1435391753.unknown

_1435085182.unknown

_1435085248.unknown

_1435085289.unknown

_1435391518.unknown

_1435085384.unknown

_1435085284.unknown

_1435085220.unknown

_1435085019.unknown

_1435085143.unknown

_1435084982.unknown

_1435085011.unknown

_1435084538.unknown

_1435084670.unknown

_1435084928.unknown

_1435084600.unknown

_1435082815.unknown

_1435084471.unknown

_1435082802.unknown

_1435081628.unknown

_1435081918.unknown

_1435081956.unknown

_1435082626.unknown

_1435082682.unknown

_1435082605.unknown

_1435081937.unknown

_1435081866.unknown

_1435081881.unknown

_1435081875.unknown

_1435081826.unknown

_1435080870.unknown

_1435081553.unknown

_1435081463.unknown

_1435081496.unknown

_1435081415.unknown

_1435081333.unknown

_1435080639.unknown

_1435080735.unknown

_1435080771.unknown

_1435080700.unknown

_1435072788.unknown

_1435073391.unknown

_1435074110.unknown

_1435073138.unknown

_1435071192.unknown

Vi_TCVN9330-1-2012

Documents

Transcript of Vi_TCVN9330-1-2012