Xu ly so lieu

26/04/2012

X L THNG K V QUY HOCH HO THC NGHIM

MN HCNI DUNG1. PHN B CHNG TRNH 2. K HOCH THI V PHN B IM 3. K HOCH GING DY

X L S LIU

Nm hc: 2011 2012 Nguyn Quc Thng ST: 0983177314

4. CNG BI GING

X L THNG K

X L THNG K V QUY HOCH HO THC NGHIM

1. PHN B CHNG TRNH (15 tit)Ni dung S tit

2. K HOCH THI V PHN B IM6 3 10 5 IM IM THNG K (20%) IM GIA K (20%)

Chng 1: Cc khi nim trong x l thng k Chng 2: Cc hm phn b Chng 3: X l kt qu thc nghim bng thng k Chng 4: Php phn tch tng quan v phn tch hi quy tuyn tnh Chng 5: nh lut lan truyn sai s ng dng trong biu din kt qu thc nghim

IM CUI K (60%)

6

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Chng 1: CC KHI NIM TRONG X L THNG K


1.1.S o, ch s c ngha 1.2. Sai s, phn loi sai s Khi nim 1.3. Cc i lng thng k c trng

1.1.S o, ch s c nghaMt i lng vt l trc tip c c S o gin tip tnh c

C th nguyn

Khng c th nguyn

1.4. Phn bit ng, chnh xc q=m+p

n v o lng Ch s c ngha

1.5. lp li

Q = q .Q^



1.1.S o, ch s c nghaCSCN tin cy, CSCN khng tin cy Xc nh s CSCN Bo ton s CSCN khi chuyn n v

1.1.S o, ch s c ngha

CSCN tin cy, CSCN khng tin cyQui tc 1a Mt s o c th c nhiu CSCN tin cy nhng duy nht ch c mt CSCN khng tin cy ng hng sau cht k t tri sang phi. V d: 14,54 8,755 0,00072 0,1086 3,08

Quy tc Ch s c ngha

Lm trn s Php cng, tr Php nhn chia Php ly logarit Php lu tha

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1.1.S o, ch s c ngha Bo ton s CSCN khi chuyn n v Qui tc 2 S CSCN trong mt s o bt k (trc tip, gin tip) phi gi nguyn trong mi php chuyn i n v o lng. V d: 0,56L = 0,56 x 103 mL = 560 mL

Xc nh s CSCN Qui tc 1bS CSCN ca mt s o c tnh t ch s u tin khc 0 k t tri sang phi, mi ch s 0 sau CSCN u tin, bt k ng v tr no, u l CSCN. V d: 14,53 3,07 0,1080 4 CSCN 3 CSCN 4 CSCN 0,00074 8,750 2 CSCN 4 CSCN



1.1.S o, ch s c ngha Lm trn s Qui tc 3 Trong s o gin tip - CSCN sau cht c tng ln 1 n v nu ng sau n l ch s v ngha ln hn 5 -CSCN sau cht gi nguyn gi tr nu ng sau n l ch s v ngha nh hn 5 V d: 18,176 18,18 18,174 18,17

1.1.S o, ch s c ngha Lm trn s

Trong s o gin tip, ch s v ngha bng 5 th: CSCN sau cht c tng ln 1 n v nu n l mt ch s l CSCN sau cht gi nguyn nu n l mt ch s chn k c s 0 V d: 18,175 18,18 18,165 18,16 18,205 18,20

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Chng 1: CC KHI NIM TRONG X L THNG K 1.1.S o, ch s c ngha Php cng, tr


Qui tc 4 Chn s hng cht: l s hng c ln nht. S CSCN sau du phy trong kt qu cui cng c ly bng vi s CSCN sau du phy ca s hng cht. 137.34 = 0.01+ 15.9994 + 15.9994 169.3388

169.34



1.1.S o, ch s c ngha Php nhn chia

1.1.S o, ch s c ngha Php ly logarit Qui tc 6 S CSCN trong s lgarit w c tnh t ch s khc 0 u tin k t tri sang phi ca phn nh tr, mi ch s 0 sau CSCN u tin, bt k v tr no ca phn nh tr, u l CSCN. Trong php ly lgarit: w = log x, s CSCN ca w phi bng s CSCN ca x. V d: log 134 = 2.1271048 2.127

Qui tc 5 Chn tha s cht: l s hng c ln nht. S CSCN trong kt qu cui cng c ly bng vi s CSCN ca tha s cht. V d: Ht 10.00 mL dung dch HCl em chun bng dung dch NaOH 0.10195N th tiu tn ht 9.35 mL. Tnh CN HCl. CN HCl = 9.35 x 0.10195 : 10.00 = 0.09532325 S CSCN: 3 5 4 = 0.0953

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1.1.S o, ch s c ngha Php lu tha Qui tc 7 S CSCN ca phn m c xc nh ty thuc xut x ca n l s t nhin hay s lgarit Trong php ly anti-lgarit w=10x, s CSCN ca w phi bng s CSCN ca x

1.1.S o, ch s c ngha Php lu tha V d:

- S t nhin. o trc tip (trn my o pH) c pH =3.6[H+]= 10-pH=10-3,6=0.0002511 2,5x10-4 - S logarit. Cho bit pK = 4.75. K = 10-4.75 = 1.77827941 x 10-5 1.8x10-5


Chng 1: CC KHI NIM TRONG X L THNG K 1.2. Sai s v phn loi Sai s tuyt i Sai s tng i

1.2. Sai s v phn loi

L s sai khc ca mt gi tr nghin cu no so vi gi tr trung bnh (hoc gi tr thc). Sai khc ny c th m hay dng. C cng n v vi i lng o.

L t s ca sai s tuyt i i vi gi tr trung bnh (hoc gi tr thc). Khng c th nguyn.

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Chng 1: CC KHI NIM TRONG X L THNG K 1.2. Sai s v phn loi Sai s ngu nhinL sai s c ln khng th xc nh c, pht sinh do cc yu t ngu nhin tc ng. Khc phc: tuy khng th loi tr hon ton, nhng c th nh gi mc nh hng ca n thng qua mt s i lng tnh ton: Gi tr trung bnh lch chun


Sai s h thngKt qu khng i trong ton b cc ln o Nguyn nhn: Do dng c, thit b khng chnh xc, ha cht khng tinh khit iu kin mi trng trong qu trnh o Phng php o Ngi thc hin php o

1.2. Sai s v phn loi Sai s ngu nhin Sai s h thng

Kt qu

khng chc

sai

Loi tr

X l s liu

c



1.3. Cc i lng thng k c trng Cc i lng trung bnh Trung bnh cng Trung bnh nhn

1.3. Cc i lng thng k c trng Trung v (Median): l s ng gia tp s liu c sp xp t b n ln, chia dy s thnh 2 phn bng nhau v s s liu. V d: c dy s liu X1, X2, X3 , .... XN Nu N l s l th Med = X(N+1)/2 Nu N l s chn Med = (XN/2 + XN/2+1)/2Tn sut: trong N s liu, gi tr Xi xut hin ni ln: pi = ni/N S tri: l s c tn sut ln nht trong tp s liu kt qu

Trung bnh bnh phng

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1.3. Cc i lng thng k c trng Phng saiTrung bnh ca tng bnh phng sai khc gia cc gi tr ca tp s liu so vi gi tr trung bnh. c trng cho phn tn ca tp s liu.

1.3. Cc i lng thng k c trng

lch chunC cng th nguyn v cng ngha vi phng sai.

lch chun tng i = H s bin thin (Cv)

sai chun:

Khong bin thin:


Chng 1: CC KHI NIM TRONG X L THNG K 1.4. Phn bit ng, chnh xc

1.4. Phn bit ng, chnh xc

ng: Mc gn ca gi tr o so vi gi trthc ca i lng o. chnh xc: l mc st gn gia cc kt qu th ring l Xi so vi gi tr trung bnh. chnh xc cao, ng km chnh xc km, ng cao chnh xc cao, ng cao chnh xc km, ng km

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Chng 2: CC HM PHN B 1.1. Hm Gauss

1.5. lp li

im un

im un

lp li: ch mc ph hp gia cc kt quo cng 1 mu hay mc lp li ca cc kt qu thu c .Cng thc

Hm Gauss chun ho

Chng 2: CC HM PHN B 1.1. Hm Gauss

Chng 2: CC HM PHN B 1.1. So snh hm Gauss v Gauss chun ho

Hm Gauss chun ho c = 0 v =1

c im K vng Phng sai ca X

Hm Gauss ; c th nguyn ca X , c th nguyn

Hm Gauss chun ho ; khng th nguyn , khng th nguyn

nhn ca ng Gauss tng khi gim

Dng hnhS ln thc nghim

i xng hnh chung, c 2 i xng hnh chung, c 2 im un ti Z = 1 im un ti X =N l , (thng chp nhn N >30)

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Chng 2: CC HM PHN B1.1. Hm Gauss Xc sut 2 pha Xc sut 2 pha i xng Xc sut 1 pha

Chng 2: CC HM PHN B 1.2. Hm phn b Student

Cng thc hm

ng dng: - Tnh s ln o cn thit t c khong tin cy yu cu - Loi b s o lch th bo (Psym =0.9974)

Chng 2: CC HM PHN B

Chng 2: CC HM PHN B 1.2. Hm phn b Student c im K vng Phng sai SX Dng hnh Hm Gauss f/(f-2) vi f > 2

1.2. Hm phn b Studentc im:

N :PP Gauss chunPP chun

K vng =0

F tng cng nhn

i xng hnh chung ti t = 0 , Hm Student chuyn thnh hm Gauss N chun ho (thc t N > 30) f = N -1 khi N < 30 Bc t do (f) f = N khi N > 30

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Chng 2: CC HM PHN B 1.2. Hm phn b Student ng dng:

Chng 2: CC HM PHN B

1.3. Hm phn b Fisher Hm c imS1 > S2

- Tnh s ln o cn thit t c khong tin cy yu cu - Loi b s o lch th bo (Psym = 0.997), so snh gi tr trung bnh thc nghim vi gi tr tht, so snh 2 gi tr trung bnh

ng cong v trn gc phn t th nht nn k vng > 0 Tnh i xng ph thuc vo f1 v f2. Theo minh ho dng ng cong trn hnh th f1 < f2.

Chng 2: CC HM PHN B 1.3. Hm phn b Fisher ng dng:

Chng 2: CC HM PHN B 1.4. Hm phn b Chi phng Hm

- Phn tch phng sai: Kim nh tnh ng nht thng k gia hai phng sai mu bt k khi cha bit phng sai tng th. - Phn tch hi quy, phn tch tng quan

f=5 f =2

f =10

Mc i xng tu thuc vo f

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26/04/2012

Chng 2: CC HM PHN B 1.4. Hm phn b Chi phng ng dng:

Chng 2: CC HM PHN B 1.5. Hm phn b Poisson Hm

- Kim nh mc suy gim cp chnh xc ca thit b o lng - Kim nh s ph hp vi phn b chun - Kim nh tnh ng nht thng k ca nhiu phng sai mu

Chng 2: CC HM PHN B 1.5. Hm phn b Poisson

Chng 3: X L KT QU THC NGHIM BNG THNG K

Kim tra s liu thc nghim ng dng trong qu trnh phng x:

- c lng khong tin cy- Loi b s o lch th bo

Biu din cc s o gin tip

nh gi kt qu phn tch theo thng k

S dng Excel tnh ton

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Chng 3: X L KT QU THC NGHIM BNG THNG K Nguyn tc thc hin mt chun thng k Cc loi sai lm: Khi chp nhn hay bc b gi thit thng k, ngi ta c th mc mt sai lm nht nh, nht l khi RTN xp x RLT l thuyt.Phn quyt thng k Chp nhn Ho Bc b H0 Nu gi thit H0 thc s l ng Khng mc sai lm Mc sai lm loi I sai Mc sai lm loi II Khng mc sai lm

Nguyn tc thc hin mt chun thng kTrnh t: + Chn phn b l thuyt: tu tnh hung thc nghim m chn phn b l thuyt thch hp, cn c bng thng k tm s phn v ca phn b ny. Chun 1 pha thng chn P = 0.95, chun 2 pha thng chn P = 0.025 v 0.975. + Tnh gi tr thc nghim RTN. Mt phn b l thuyt c th c nhiu cch tnh RTN. Chn cng thc tnh ph hp vi thc nghim. + Chp nhn hoc bc b gi thit thng k. C 2 loi gi thit thng k. Gi thit H0 cho rng RTN ph hp phn b l thuyt v gi thit H1 cho rng RTN khng ph hp phn b l thuyt. (Nu chp nhn H0 th mc nhin s bc b gi thit H1)

Trong xc sut thng k, ngi ta tm cch trnh sai lm loi I.

Chng 3: X L KT QU THC NGHIM BNG THNG KNguyn tc thc hin mt chun thng k Quy tc chn xc sut Khi bc b gi thit H0, ngi ta chn mc ngha 0.01, tc l chn s phn v R0.99 i vi chun 1 pha v chn khong phn v [R0.005; R0.995] vi chun 2 pha. Khi chp nhn gi thit H0, ngi ta chn mc ngha 0.05, tc l chn s phn v R0.95 i vi chun 1 pha v chn khong phn v [R0.025; R0.975] vi chun 2 pha. Khi RTN ri vo vng tranh chp: [R0.95; R0.99] vi chun 1 pha; [R0.025; R0.05] v [R0.975; R0.995] ca chun 2 pha, ngi ta phi tm thm thng tin b sung bng cch lm thm thc nghim hoc bng trc gic.


Nguyn tc thc hin mt chun thng k Tn mt vi mc xc sut

P = 0.90: P = 0.95: P = 0.99: P = 0.999:

mc mc mc mc

sai khc du hiu sai khc ng k sai khc rt ng k sai khc ti cao

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Chng 3: X L KT QU THC NGHIM BNG THNG K Kim tra s liu thc nghimSai s th Sai s th: xut hin do vi phm nhng iu kin c bn ca php o kt qu sai lch hn so vi nhng ln o khc. Loi b sai s th

Chng 3: X L KT QU THC NGHIM BNG THNG K Kim tra s liu thc nghimLoi b sai s th trong tp s liu Xichun Dixon

chun GaussPhi bit trc

chun Student khng bit trc

Nguyn nhn: Do chn phng php phn tch khng n nh. Do ngi lm phn tch cu th, khng cn thn. Khc phc: lm cn trng, chnh xc, ng qui nh, o nhiu ln .

Cc chun thng k

khng bit trc

Sp xp dy s liu theo chiu tng dn X1, X2, ..., XNchn x* l X1 hoc XN |X X1| ?? |XN X| chn x* |X X1| ?? |XN X| chn x*

Chng 3: X L KT QU THC NGHIM BNG THNG K Kim tra s liu thc nghimLoi b sai s thchun Dixon

Chng 3: X L KT QU THC NGHIM BNG THNG K Kim tra s liu thc nghimV d: Loi b sai s th

chun Gauss

chun Student

So snh Qtn vi Qchun P(=0.95;N) Lu :

So snh z*tn vi zlt =3 Psym(=0.9974) Tnh XN-1 th loi b X*

So snh t*tn vi tP,f Psym(=0.997) Tnh XN-1 SX, N-1 th loi b X*

Kt qu phn tch hm lng Chloramphenicol trong 1 mu sa ti bng phng php cc ph xung vi phn, thu c cc kt qu cng ng in (mA) nh sau: 76.5; 79.8; 86.52; 86.3; 76.5; 90.00; 82.5; 87.05. Hy loi b s o lch th bo theo chun Student (chn Psym= 0.997), chun Dixon (chn P = 0.95)

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26/04/2012

Chng 3: X L KT QU THC NGHIM BNG THNG K Kim tra s liu thc nghimV d: Loi b sai s th

Chng 3: X L KT QU THC NGHIM BNG THNG K Biu din cc s o gin tip khong tin cy chun Gauss biu din s o chun Student

Loi b s o lch th bo trong dy s o lp li sau y: 0.380; 0.413; 0.400; 0.410; 0.411; 0.403; 0.401. a) Hy loi b s o lch th bo theo chun Student (chn Psym= 0.997), chun Dixon (chn P = 0.95). b) Hy loi b s o lch th bo theo chun Gauss a chu kz (chn Psym= 0.997) bit php o c = 0.010. ngha

a < X < b hay X nm trong khong (a, b)

Chng 3: X L KT QU THC NGHIM BNG THNG K V d: Biu din cc s o gin tip

Chng 3: X L KT QU THC NGHIM BNG THNG K nh gi kt qu phn tch theo thng k Mc ch: nh gi s sai bit ca cc tham s t c th nh gi, so snh cc phng php, cc quy trnh phn tch, cc thit b, kt qu cc PTN, hoc nh gi tay ngh cc k thut vin Cc chun so snh phng sai Cc chun so snh tr trung bnh Chun Student Chun Fisher Chun chi phng Chun Gauss Chun Duncan

Biu din kt qu phn tch hm lng crom trong nc thi bng phng php trc quang vi thuc th 1,5-diphenylcabazid, thu c dy s liu mt quang nh sau: 32.5; 33.7; 31.4; 34.1; 35.2; 34.5. a) Hy loi b s o lch th bo theo chun Student a chu k (chn Psym= 0.95 v 0.99). b) Tnh khong tin cy v biu n kt qu o mt quang. c) Xc nh bin gii tin cy ca tr trung bnh.

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Chun so snh phng sai Tnh hung: 1. nh gi lp li ca 2 quy trnh phn tch khc nhau bng cch so snh 2 phng sai mu S12 v S22 ca mi quy trnh thu c trong cng iu kin: + Cng 1 ch tiu phn tch trong cng 1 mu th trn ng nht; + c tin hnh bi cc k thut vin thnh tho.

Chun so snh phng sai Tnh hung: 2. nh gi tay ngh ca 2 k thut vin phn tch khc nhau bng cch so snh 2 phng sai mu S12 v S22 ca mi ngi thu c trong cng iu kin: + Cng ch tiu phn tch trong cng 1 mu th trn ng nht; + Cng 1 quy trnh phn tch, cng loi thit b phn tch, cng trong 1 PTN, cng tin hnh trong thi gian quy nh; + Mu th c chia thnh nhiu phn v cc KTV khng bit xut x cc mu.

Chng 3: X L KT QU THC NGHIM BNG THNG K Chun so snh phng sai

Chng 3: X L KT QU THC NGHIM BNG THNG K Chun so snh phng sai

Chun Fisher: so snh tng cp phng sai

V d:

Vi S12 > S22 Dng bng thng k Fisher tm FLT = F(P, f1, f2) So snh FTN v F LT Nu FTN < F LT Nu FTN > F LT chp nhn H0 chp nhn H0

Php chun HCl bng NaOH vi ch th bromothymol chm (ch th A) v methyl (ch th B) cho kt qu nh sau: SA = 0,00178M; fA = 25 v SB = 0,00099M; fB = 20. Vy dng ch th methyl c chnh xc hn ch th bromothymol chm?

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Chng 3: X L KT QU THC NGHIM BNG THNG K V d: Chun so snh phng sai

Chng 3: X L KT QU THC NGHIM BNG THNG K Chun so snh phng sai Chun chi phng

Php chun HCl bng NaOH vi ch th phenolphtalein (ch th A) v methyl (ch th B) cho kt qu nng mol (M) nh sau:Ln Ch th A Ch th B Ln Ch th A Ch th B Ln Ch th A Ch th B

1

2

3

4

5

6

7

8

9

0,137 0,122 10 0,132 0,125 19 0,133 0,130

0,136 0,125 11 0,134 0,123 20 0,137 0,131

0.130 0.130 12 0.129 0.133 21 0.135 0.134

0.129 0.121 13 0.125 0.129 22 0.126

0.132 0.120 14 0.127 0.122 23 0.137

0.134 0.137 15 0.139 0.133 24 0.138

0.130 0.124 16 0.129 0.125 25 0.125

0.128 0.118 17 0.127 0.113 26 0.122

0.133 0.119 18 0.131 0.117 27 0.135

Phn tch k mu th (tm c k phng sai mu), mi mu th c o lp li. Gi fj l s bc t do ca mu th j. Tng s ln o l N. S bc t do ti hin fth = N - k

Vy dng ch th phenolphtalein chnh xc hn ch th methyl ? Gi s khng c s o lch th bo.

chp nhn H0


Chng 3: X L KT QU THC NGHIM BNG THNG K Cc chun so snh tr trung bnh Chun Duncan Phn tch k mu th mi, mu th c o lp li v tm c k gi tr trung bnh. Gi fj l s bc t do ca mu th j. Tng s ln o l N. S bc t do ti hin fth = N k. S bc t do gia cc mu: fgia= k -1. Gi xij l gi tr ring l ca X trong th nghim lp li th i di nh hng ca yu t th j. Yu cu p dng chun Duncan: + Cc phng sai phi ng nht thng k (kim nh theo chun Fisher hoc chun chi phng. + C s sai bit gia cc gi tr trung bnh.

Cc chun so snh tr trung bnh Tnh hung: 1. Kim tra tnh ng nht ca mu. 2. Thm nh kt qu 2 phng php phn tch. 3. nh gi 2 mu c thuc cng 1 vt liu. 4. nh gi tc dng nh hng ca tng yu t j gy ra s khc bit h thng gia cc gi tr trung bnh.

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Chng 3: X L KT QU THC NGHIM BNG THNG K Cc chun so snh tr trung bnh Chun Duncan kim tra s sai bit gia cc gi tr trung bnh Tnh gi tr trung bnh ca mi mu th j v ca k mu th

Chng 3: X L KT QU THC NGHIM BNG THNG K p dng chun Duncan nh s bc r = 1, 2, 3, ..., k

trong r < r Bc tng i gia 2 gi tr trung bnh: R = r r +1; fth = N k, tra bng phn v Q tm Q = Q(0.95, R, f th) v Q(0.99, R, f th) Tnh Q TN Ftn>F0,95, fgia, fth C s khc bit gia cc gi tr trung bnh

Chng 3: X L KT QU THC NGHIM BNG THNG K p dng chun Duncan Qtn < Q0,95, R, f th Qtn > Q0,95, R, f th Qtn > Q0,99, R, f th

Chng 3: X L KT QU THC NGHIM BNG THNG K V d: p dng chun Duncan so snh cc tr trung bnh

Hy so snh nh hng ca cc halogenua alkyl: CH3I (1); C3H7I(2); C4H9I(3); C2H5Br(4); C3H7Br(5) n hiu sut ca phn ng polimeho theo c ch gc t do, da vo bng s liu o hiu sut sau:Cht 1 2 3 4 5 6 7 8 (1) 79.8 86.3 86.5 92.3 76.5 87.1 82.5 90.0 (2) 87.3 69.6 81.1 78.0 83.7 64.8 67.3 75.5 (3) 42.5 64.3 79.0 61.0 31.3 72.9 58.7 52.5 (4) 76.0 83.5 72.8 89.0 76.5 87.5 74.5 93.2 (5) 70.7 64.7 38.5 77.0 91.5 68.0 38.1 80.0

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=

Chng 3: X L KT QU THC NGHIM BNG THNG K Cc chun so snh tr trung bnh Chun Gauss Gi thuyt H0: hai gi tr trung bnh sai khc khng ng k

Chng 3: X L KT QU THC NGHIM BNG THNG K Cc chun so snh tr trung bnh Chun Student Gi thuyt H0: hai gi tr trung bnh sai khc khng ng k

nA v nB l s ln o lp li ca A v B, l lch chun ca quy trnh ztnz0,99 = 2.58

NA v NB l s ln o lp li ca A v B,

So snh 2 phng sai mu SA2 v SB2 theo chun Fisher v chn gii php: chp nhn H0Bc b H0 + Nu FTN > FLT: khng th so snh 2 tr trung bnh + Nu FTN < FLT: tin hnh kim nh theo chun Student

Chng 3: X L KT QU THC NGHIM BNG THNG K Cc chun so snh tr trung bnh Chun Student

Chng 3: X L KT QU THC NGHIM BNG THNG K V d: Mt hn hp KNO3; KCl nghin mn. xc nh hm lng kali trong mu ngi ta ly ra 5 mu ngu nhin, mi mu c phn tch lp li 4 ln, thu c kt qu sau: STT 1 2 3 Mu PTN 1 2 3 4 5 Hm lng K (%) 12.42 ; 12.28 ; 12.33 ; 12.36 12.27 ; 12.24 ; 12.19 ; 12.19 12.41 ; 12.48 ; 12.51 ; 12.39 12.42 ; 12.43 ; 12.47 ; 12.40 12.19 ; 12.28 ; 12.20 ; 12.32

f = nA + nB -2 tTN < t0,95,f tTN > t0,99,f chp nhn H0 bc b H0

4 5

Dng chun Gauss v chun Student xc nh cc gi tr trung bnh ca 5 mu PTN trn c khc nhau ng k hay khng.

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Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUYH s tng quan Spearrman Php phn tch tng quan H s tng quan Kendall Php phn tch hi quy tuyn tnh mt bin

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch tng quan 1. Tnh hung thc nghim: yu cu thit lp mi quan h gia hai i lng X v Y. Trong X v Y l hai s o c th o c trong thc nghim vi chnh xc cn thit ( lch chun mu tng i RSD)

Phng trnh hi quy

lch chun d Sresidue, lch chun ca cc h s hi quy (S(a), S(b))

Tnh cc h s hi quy (a, b)

Phn bit LOD, LOQ v nhy S dng Exel tnh ton

Php phn tch tng quan

X v Y c tn ti mi lin kt?

Mc lin kt l cng (mnh m) hay mm (yu)

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch tng quan

r: h s tng quanr = + 1: X v Y lin kt cng, ng bin. r = - 1: X v Y lin kt cng, nghch bin. r = 0: X v Y khng c lin kt no. 0.7 r 1: tng quan mnh 0.3 r < 0.7: tng quan yu Khi xy dng ng chun, yu cu: 0.9975 r 1

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Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch tng quan Gi s c tp hp mu {xi, yi}N H s tng quan r (Pearson)

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch tng quan nh gi mi quan h gia 2 hay nhiu bin thng qua h s tng quan. - Tp s liu tun theo phn phi chun. - Gi tr cc bin l c lp nhau. - Loi b gi tr bt thng trc khi tnh h s tng quan. H s tng quan Spearrman H s tng quan Kendall So snh nhiu hn 2 cp s liu Cp ph hp: (Xi Xj).(Yi-Yj)>0 Cp khng ph hp: (Xi Xj).(Yi-Yj)>0

d: s sai khc gia 2 th hng ca 2 tham s X, Y

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch tng quanZ A B C D E F G H K L X 30 34 32 47 20 24 27 25 22 16 Y 25 38 30 40 7 10 22 35 28 12

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch tng quanZ A B C D E F G H K L X 30 34 32 47 20 24 27 25 22 16 Y 25 38 30 40 7 10 22 35 28 12 X2 Y2 XxY t.hng X t.hng Y d t.hng d2 t.hng

Kho st yu t X v yu t Y c s tng quan tc ng ln i tng Z

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Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch hi quy Nhim v: tm ra biu thc l thuyt gia X v Y ph hp tt nht vi kt qu thc nghim

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch hi quy tuyn tnh mt bin 1. Phng trnh hi quy: Y = aX + b

iu kin v chnh xc: X l bin c lp, Y l bin ph thuc. RSDX Flt,f1=1,f2=N-2 chng t c s ph hp gia PTHQ vi thc nghim h s a, b lm trn theo U(a); U(b) R 100: PTHQ v thc nghim ph hp hon ton

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Php phn tch hi quy tuyn tnh mt bin 7) nh gi khong bt n chun

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY

nhy: kh nng pht hin s thay i tn hiu khi thay i mt lng nh nht v nng cht phn tch Phn bit LOD, LOQ v nhy LOD: l nng nh nht ca cht phn tch m h thng phn tch cn cho tn hiu phn tch LOQ: l nng thp nht ca cht phn tch m h thng phn tch nh lng c vi tn hiu phn tch

N: s im ca PTHQ m: s ln o tn hiu phn tch/s lng cn...

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Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY LOD, LOQ LOD: l i lng lin quan n khong bt n ca php o, khng th v khng cn xc nh chnh xc LOD m ch c lng

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Cng thc tnh LOD, LOQ da vo ng chun Cch thc nghim tm LOD, LOQ: + o mu trng vi s ln o 21 ln, t tm lch chun (Sblank) ca mu nn. + Dng ng chun y = a + bx

LOD ph thuc thit b, iu kin vn hnh thit b, phng php phn tch, tay ngh ca phn tch vin LOD = 3 Sblank / b LOD thit b (IDL) LOD phng php (MDL) LOQ = 10 Sblank / b

Cng thc tnh LOD, LOQ da vo ng chunXc nh LOD, LOQ khi xc nh NH4+ bng phng php trc quang, ngi ta tin hnh o mt quang 21 ln dung dch nn, cho kt qu nh sau:

Chng 4: PHP PHN TCH TNG QUAN V PHN TCH HI QUY Tnh LOD, LOQ da vo ng chun

Ln o A Ln o A Ln o A

1 8 15

2 9 16

3 10 17

4 11 18

5 12 19

6 13 20

7 14 21

0.0057 0.0045 0.0063 0.0058 0.0037 0.0053 0.0058 0.0059 0.0041 0.0064 0.0059 0.0035 0.0033 0.0028 0.0027 0.0035 0.0033 0.0028 0.0047 0.0023 0.0068

Sblank khi o mt quang 21 ln dung dch nn: 0.0014 ng chun: C = 0.0013 + 17.457 A

v o mt quang ca cc dung dch c nng tng dn, cho kt qu nh sau:

C(ppm) A

0.1

0.2

0.3

0.4

0.6

1

2

LOD = 3 * 0.0014/ 17.457 = 0.00024 LOQ = 10 * 0.0014/17.457 = 0.00080

0.0056 0.0104 0.0160 0.0227 0.0347 0.0610 0.1126

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Chng 5: NH LUT LAN TRUYN SAI S nh ngha khong bt n ca s o S o u vo u ra

Chng 5: NH LUT LAN TRUYN SAI S Cc loi sai s ca s o Sai s ngu nhinL sai s gy ra bi nhng

Sai s h thngGy ra bi mt s nguyn nhn rt xc nh lm cho s o c tr s khc bit so vi gi tr thc

Tm phn b phn tn (KBO chun) S bc t do Tnh khong bt n chun v s bc t do tng ng ca s o u vo Tnh khong bt n chun m rng v s bc t do ca s o u ra u (Xi) u (Yi)

nguyn nhn khng th xc nh c, pht sinh do cc yu t ngu nhin tc ng lm cho kt qu khc bit trong nhng ln o in ht nhau. S khc bit cng thu hp th php o cng chnh xc

Biu th chnh xc

Biu th ng

Chng 5: NH LUT LAN TRUYN SAI S Lan truyn sai s Cho biu thc sau: Y = f (X1, X2, X3 ....), vi X1, X2, X3 ... l cc s o u vo, Y l s o u ra. Sai s ca cc s o X1, X2, X3 ... lan truyn sang sai s ca s o u ra Y.

Chng 5: NH LUT LAN TRUYN SAI S Lan truyn sai s h thng

Nh nghin cu thng nh gi sai s ca kt qu thu c qua php tnh vi hai hoc mt s kt qu o m mi kt qu ny u mang theo sai s ca mnh. Phng php cng cc sai s ring bit ph thuc vo cc php tnh s hc c thc hin vi cc i lng c bao hm sai s v i lng c tnh ton ra. Ngoi ra nh hng ca sai s h thng v sai s ngu nhin n i lng tnh ton c cng khc nhau.

Php cng sai s h thng Phng php nh gi sai s h thng i vi tng hoc hiu khc vi phng php nh gi chng i vi php nhn hoc chia. Sai s ca tng hoc hiu Xt biu thc: y=a+bc y a, b v c l nhng gi tr ca ba i lng o c. Nu a, b v c l nhng sai s h thng tuyt i gn lin vi nhng php o ca nhng i lng y. Sai s ca kt qu l: y = a + b c

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Chng 5: NH LUT LAN TRUYN SAI S

Chng 5: NH LUT LAN TRUYN SAI S Lan truyn sai s ngu nhin

Lan truyn sai s h thng

Khi cng hoc tr, sai s tuyt i ca tng hoc hiu c xc nh bng sai s tuyt i ca cc s hng. V d: Cho: a = 0.50 (+ 0.02) b = 4.10 ( 0.03) c = 1.97 ( 0.05) (S trong ngoc l sai s h thng tuyt i) Thc hin php tnh sau: y = a + b - c Kt qu: y = 2.63 Sai s ca tng: y = 0.02 + ( 0.03) ( 0.05) = + 0.04 Biu din kt qu: y = 2.63 (+0.04)

Sai s ca tch hoc thng Xt biu thc: y = a.b/c y a, b v c l nhng gi tr ca ba i lng o c. Nu a, b v c l nhng sai s h thng tuyt i gn lin vi nhng php o ca nhng i lng y. Sai s ca kt qu l: Sai s tng i ca php nhn hoc chia c xc nh bng sai s tng i ca cc thnh phn tham gia vo kt qu c tnh ton.

Chng 5: NH LUT LAN TRUYN SAI S Lan truyn sai s h thng


V d: Hy tnh kt qu ca php tnh sau (trong ngoc l nhng gi tr sai s h thng tuyt i):

y = 0.104

Cng sai s ngu nhin lch tiu chun tuyt i l phng php thun li nht nh gi sai s ngu nhin ca nhng kt qu thc nghim. Khc vi sai s h thng khng nn ghi du cho lch tiu chun bi v n c th dng hoc m vi xc sut nh nhau, do lch tiu chun ca kt qu tnh c nm trong mt s vng.

Biu din: y = 0.010 (+0.004)

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Cng sai s ngu nhin y=a+bc y a, b v c l nhng gi tr ca ba i lng o c. Nu Sa, Sb v Sc l nhng lch tiu chun tuyt i gn lin vi nhng php o ca nhng i lng y. lch tiu chun tuyt i, Sy, ca tng hoc hiu c xc nh:

V d: Cho: a = 0.50 ( 0.02) b = 4.10 ( 0.03) c = 1.97 ( 0.05) (S trong ngoc l lch tiu chun) Thc hin php tnh sau: y = a + b - c Kt qu: y = 2.63 Sai s ca tng:

Biu din kt qu: y = 2.63 ( 0.06)



Sai s ca tch hoc thng Xt biu thc: y = a.b/c y a, b v c l nhng gi tr ca ba i lng o c. Nu Sa, Sb v Sc l nhng lch tiu chun tuyt i gn lin vi nhng php o ca nhng i lng y. lch tiu chun tuyt i, Sy, ca tch hoc thng c xc nh:

V d: Cho: a = 0.55 ( 0.02) b = 4.10 ( 0.03) c = 1.97 ( 0.08) (S trong ngoc l lch tiu chun) Thc hin php tnh sau: y = a.b/c Kt qu: y = 1.14 Sai s kt qu: Sy = 0.055* 1.14 = 0.0627 Biu din kt qu: y = 1.10 ( 0.06)

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Lan truyn sai s ngu nhin

1. Nu thu c 3 thng tin khc nhau ca ui l uiA; uiB; uiC (gi chung l uiX) th cc gi tr ny c tnh theo 1 trong 2 cch sau: - S liu thc nghim t php o lp li: Nu bit trc , o lp li n ln th uiX = (Xi)/ Nu cha bit , th uiX = S(Xi)/ - S liu y t chng ch, cam kt ca nh sn xut: Nu chng ch cho bit Xi c dung sai l a th uiX = a/ Nu nh sn xut cam kt Xi c dung sai l a th uiX = a/1.96

p dng vo cc bi ton thc t

;



2. Tnh h s nhy C1, C2 C3..

5. Tnh s bc t do ca s o u vo

3. Tnh khong bt n chun ca cc s o u vo: tnh xong lm trn thnh s nguyn 4. Tnh khong bt n chun ca cc s o u ra: 6. Tnh s bc t do ca s o u ra

tnh xong lm trn thnh s nguyn

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.ui



7. Tnh khong bt n m rng U(Y) =ui.t0.95,(f =

hiu chun pipet 25mL, ngi ta cn lng nc cha trong pipet trn cn phn tch, thu c kt qu sau (gam): 25.0264 25.0195 25.0063 25.0052 24.9981 24.9972 24.9991 25.0067 25.0024 25.0288 Theo chng ch ca hng sn xut, th th tch 25mL c dung sai 0.03mL. Chp nhn nc c d20 = 1.000 g/mL Hy tnh khong bt n m rng ca pipet sau khi hiu chun v biu din kt qu.

8. Biu din kt qu:



Tnh phn t lng v xc nh khong bt n m rng ca phn t C2H4O2, bit: C = 12.011 0.001 H = 1.00795 0.00007 O = 15.9994 0.0003

Tnh khong bt n chun v khong bt n m rng trn php cn. Php cn trn mt cn phn tch c chng nhn sau: Cam kt ca hng sn xut, vi lng cn < 50 gam c dung sai l 0.1mg vi xc sut P = 0.95. Chng nhn hiu chun, vi lng cn < 50 gam c lch chun =0.07mg. Khi cn c kt qu sau:

Chai khng Chai ng C2H4O2

31.0234 g 36.1284 g

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Chng 5: NH LUT LAN TRUYN SAI STnh nng ca dung dch chun t pha (khng dng ng chun) v KB m rng nh sau: Cn 5.1050 gam cht chun gc KHP (kaihydrophotphat) (theo chng ch c tinh khit 99.9 0.1%) ri ho tan v nh mc trong bnh nh mc 250mL. Php cn trn cn phn tch c chng nhn sau: Cam kt ca hng sn xut, vi lng cn < 50 gam c dung sai l 0.1mg vi xc sut P = 0.95. Chng nhn hiu chun, vi lng cn < 50 gam c lch chun =0.07mg. Bnh nh mc c hiu chun bng cch o khi lng nc cha trong bnh trn cn k thut c kt qu nh sau:

Theo chng ch ca hng sn xut, dung sai ca bnh nh mc 250mL l 0.15mL. Theo h s gin n ca nc vi xc sut P = 0.995 20 30C l 2.1* 10-4.

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