Chuong 1 Chau2

GV: ThS. Nguyn Th Trm Chu

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XLSL V QHTN TRONG HA (30)

Bi ging

Gio trnh chnh: Nguyn Cnh, Qui hoch thc nghim (2003), Trng i Hc Bch Khoa TP.HCM.Ti liu tham kho:[1]. Nguyn Cnh- X l thng k v quy hoch trong Ha hc- HBK TPHCM, 2001.[2]. X.C. Acxadavova, V.Y. Capharob (1985), Ti u ha thc nghim trong ha hc v cng ngh ha hc (bn dch Nguyn Cnh, Nguyn nh Soa), i Hc Bch Khoa TP.HCM.[3]. Phn mm tin hc Stat graphic 15.2.


Ti liu

NI DUNG MN HC


Gii thiu

Tiu chun anh gia sinh vin* D lp - C mt trn lp nghe giang t 80% tng s thi gian tr ln.- Lm tiu lun 30%- Thuyt trnh cng thm 10% cho tiu lun- Thi gia hc phn: T lun 20%- Thi kt thc hc phn: T lun 50%* Thang im thi: Theo qui ch tn ch


Tiu chun

NI DUNG

1.1. Khi nim v mt s cc thng s ca i lng ngu nhin.

1.2. Xc nh cc thng s thc nghim.

1.3. Kim nh cc gi thit thng k.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

M U- Nghin cu cng ngh ho hc thc cht l thc hin cc th nghim ly s liu.- Khi th nghim phi chn mu, dng c, trang thit b ly cc s liu mt cch i din v chnh xc.- Khi c s liu phi tin hnh thanh lc x l cc s liu thu thp c.- Tnh ton v biu din kt qu nghin cu.

MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

Phng php c in:- Phng php thc nghim mt yu t.- Nghin cu chin lc ti u thc nghim.- Tm mt m hnh ton hc biu din hm mc tiu.- Chn c m hnh: Yu t no gi nguyn, yu t no thay i, mc tiu cn t ti u.- Phng php qui hoch ti u: Thay i ng thi nhiu yu t.- Phng php m hnh ha ton hc tnh ton cc qu trnh k thut, chn cng thc thc nghim, c lng cc tham s ca cng thc.


1.1. KHI NIM V MT S CC THNG S CA I LNG NGU NHIN* Khi nim: i lng ngu nhin (X) l tp hp tt c cc i lng m gi tr ca n mang li mt cch ngu nhin. => s xut hin l khng bit trc. - i lng ngu nhin X c gi l ri rc khi n nhn hu hn hoc v hn cc gi tr m c khc nhau.- i lng ngu nhin X c gi l lin tc nu n nhn gi tr bt k trong mt khong ca trc s.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

1.2. XC NH CC THNG S THC NGHIM1.2.1. Phn loi cc sai s o lng* lch gia cc gi tr thc v s o gi l sai s quan st: X = X -aTrong : a l gi tr thc ca mt vt. X l kt qu quan st c (gtr o c) X l lch gia a v X.* Sai s chia lm 3 loi: sai s th, sai s h thng v sai s ngu nhin.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

(1) Sai s th - L sai s phm phi do ph v nhng iu kin cn bn ca php o, dn n cc ln o c kt qu khc nhau nhiu.* Cch kh sai s th:- Kim tra cc iu kin c bn c b vi phm hay khng.- S dng mt phng php nh gi, loi b hoc gi li nhng kt qu khng bnh thng.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

(2) Sai s h thng- L sai s khng lm thay i trong mt lot php o, m thay i theo mt quy lut nht nh.* Nguyn nhn gy sai s: do khng iu chnh chnh xc dng c o, hoc mt i lng lun thay i theo mt quy lut no , nh nhit * khc phc: ngi ta t mt h s hiu chnh ng vi mi nguyn nhn.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

(3) Sai s ngu nhin- Sai s ngu nhin nh hng n chnh xc ca php o.- L sai s cn li sau khi kh sai s th v sai s h thng.- Sai s ngu nhin do nhiu yu t gy ra, hon ton ngu nhin khng bit trc c, nn khng loi tr c.- Khc phc n mc ti thiu bng cch: tng s ln thc nghim, lm th nghim cn thn hn, x l thng k kt qu thc nghimMT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

1.2.2. nh lut cng sai s- Cho Xi (i = 1 n): l i lng ngu nhin (x1, x2, xn)- ai (i = 1 n): l i lng khng ngu nhin Phng sai Sz2 c tnh:

- Gi thit: ,th: - Nu x1..xn l nhng quan st c lp, ta xc nh c:MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

1.2.3. Nhng c lng ca cc c trng s ca cc i lng ngu nhin(1) Gi tr trung bnh c tnh bng:- Trong : xi: s o i lng x th nghim i n: s lng mu o(2) Trung v l tr s ng gia mt chui: + Khi s mu l l c tnh: n = 2m 1

+ Khi s mu l chn c tnh: n = 2mMT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

(3) Phng sai c c lng bng phng sai mu:

(4) lch bnh phng trung bnh hay gi l lch chun:

=> Phng sai c trng cho chnh xc ca php o. MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

1.2.4. Phng sai ti hin (Sth2)- Phng sai ti hin xc nh sai s ti hin ca cc php o hng lot nhng th nghim.- C n th nghim song song, gi tr o c l y1, , yn

- Sai s ti hin:MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

Nu cn m mu, mi mu lm n th nghim cc phng sai l:S1Sn. Phng sai ti hin c tnh theo cng thc:

- Vi: fi l s bc t do ca th nghim song song th i, fi = ni - 1MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

1.2.5. Khong tin cy v xc sut tin cyGi: - gi tr trung bnh ca php o. - gi tr thc nghim thu c - cn tin cy ca php o, vi l xc sut tin cy c trng cho tin tng ca c lng, thng chn l: 0,90; 0,95; =>Khong tin cy c xc nh bng cng thc: - c tnh bng cng thc: Vi, tp,k tra bng Student (p=1- ; k=n-1)MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

VD: Cho dy s liu 82,50; 86,52; 90,00; 87,05; 86,301.Tnh gi tr trung bnh.2. Tnh gi tr trung v.3. Tnh phng sai.4. Tnh lch bnh phng trung bnh.5. Tnh khong tin cy ca php o.p n: - Gi tr tr trung bnh: = 86,474- Gi tr trung v (dy l): X0,5 = 86,52- Phng sai: = 7,147- lch bnh phng trung bnh: S = 2.673 Khong tin cy ca php o: 83,242 < < 89,797=> KL: Cc gi tr thc tm c 86,30 ; 86,52 ; 87,05 nm trong khong gi tr thc => khong gi tr tin cy.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

1.3. KIM NH GI THIT THNG K1.3.1. Kim nh s ng nht ca 2 phng saia. Kim nh s ng nht ca hai phng sai- C hai mu I v II s ln o tng ng l n1 v n2.- Mun kim nh: dng chun Fis so snh 2 phng sai ca 2 dy kt qu (S12 , S22), tin hnh cc bc:+ Gi thit S12 S22 do ng.nhn ngu nhin vi p=0,95.+ Tnh: ; iu kin: F>1, S12 > S22

+ So snh: Ft v Fp,k1,k2 (Vi: k1=n1-1, k2=n2-1)KL: + Nu Ft < Fp,k1,k2 th S12 S22 do nguyn nhn ngu nhin, 2 php o cng chnh xc. + Nu Ft > Fp,k1,k2 th S12 S22 do nguyn nhn khng ngu nhin, 2 php o khng cng chnh xc.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

VD. So snh kt qu xc nh hm lng S trong mu than nhn c kt qu sau:- Phng TN A: 1 = 1,01% , n1=5 ; S12= 1,71.10-4Phng TN B: 2 = 1,08% , n2=9 ; S22= 2,05.10-5p n: Ft = 8,34 > Fp,k1,k2 =3,8KL: 2 PTN cho kt qu khc nhau, khng cng chnh xc.BT: Phn tch mu dung dch bng hai phng php khc nhau, thu c kt qu nh sau:A : 17,77 17,79 17,83 17,70 17,69 17,75B : 17,78 17,74 17,80 17,76 17,67 17,70Hy so snh 2 kt qu phn tch trn.


b. Kim nh s ng nht ca nhiu phng sai* Gi s c m tp chnh, s lng ly t mu t m tp chnh bng nhau. kim nh s ng nht ca cc phng sai ta dng tiu chun Corchran.

- m s lng mu tp chnh bng nhau G1-p(f,m)- so snh gi tr tnh c vi bng nu:G < G1-p chp nhnG > G1-p bc bMT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

* Nu s lng mu ly t cc tp khng bng nhau ta dng tiu chun Bartlet

+ Trong :

- Sau so snh vi gi tr cc bng=> Nu (f) chp nhn, ngc li th bc b.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

1.3.2. So snh hai gi tr trung bnh t 2 dy TN c lp (2 PTN, 2 ngi TN):dng chun Student* Phng TNA: x1, x2, x3,., xn1 (n1 ln TN); , S12 * Phng TNB: x1, x2, x3,., xn2 (n2 ln TN); , S22=> so snh 2 gi tr , tin hnh nh sau:- Gi thit do nguyn nhn ngu nhin vi p=0,95.- Phng sai mu S2 c tnh bng cng thc:


- Tnh chun tTN theo cng thc:+ TH1: Nu 2 phng sai tng thch (S12 S22 do ng.nhn ngu nhin), chun tTN c tnh bng cng thc:

- TH2: Nu 2 phng sai k0 tng thch (S12 S22 do ng.nhn k0 ngu nhin), dng tiu chun tTN gn ng:


- So snh gi tr tTN v tp,k (f), Vi k = n1 + n2 2 =>KL: +Nu | tTN | tp,k (f), tc l tTN < tp,k < tTN th do ng.nhn ngu nhin,2 PTN, 2 ngi TN cho kt qu nh nhau. + Nu tp,k > tTN v tTN > -tp,k th do nguyn nhn khng ngu nhin, 2 PTN, 2 ngi TN cho kt qu khc nhau.


VD. Xc nh nng dd chun HCl theo 2 cht gc cho kt qu TN nh sau:(1) Chun HCl theo Na2CO3 (mol/l): 0,1250 0,1248 0,1252 0,1254(2) Chun HCl theo Na2B4O7.10H2O (mol/l): 0,1254 0,1258 0,1253 0,1255Hy so snh kt qu ca 2 php chun ny.p n: = 0,1251 ; = 0,1255 S2 = 5,67.10-8 ; tTN = 2,38 ; tp,k = t(0,95;6) = 2,45KL: -tp,k < tTN < tp,k => cho kt qu ging nhau.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

BI TPBT1. Kt qu chun ha dd HCl theo 2 cht gc sau:(1) Theo Na2CO3 (mol/l): 0,1050 0,1047 0,1052 0,1051 0,1066(2) Theo Na2B4O7.10H2O (mol/l):0,1053 0,1056 0,1052 0,1054 0,1052 0,1082a, Dng phn b Fisher so snh 2 phng sai ca 2 dy kt qu. b, Hy so snh gi tr trung bnh ca 2 dy kt qu trn.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

BT2. nh gi chnh xc ca php xc nh Bari trong BaCl2 vi cc kt qu th nghim nh sau: % Bari : 56,24 56,18 56,09BT3. Khi xc nh grafit trong gang xm, thu c kt qu thc nghim sau: %grafit : 2,86 2,89 2,90 2,91 2,89nh gi chnh xc ca php xc nh trn.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

BT4. Ngi ta pht hin thy mt t tc trong tay nn nhn ca mt v n mng. Vic phn tch hm lng Zn trong tc bng phng php hp th phn t tay nn nhn vi tc ngi phc v b nghi vn c kt qu nh sau:(1) Tc ngi phc v, %Zn: 250 ; 265 ; 258 ; 268 ppm.(2)Tc trong tay nn nhn,%Zn:234;245;249;242;237 ppm C th khng nh ngi phc v nm trong din nghi vn khng?Gi : So snh 2 gi tr trung bnh, nu do nguyn nhn ngu nhin => KL cho 2 kt qu nh nhau, ngi phc v nm trong din nghi vn. V ngc liMT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

BT5. Khi xc nh hm lng st c trong mi tn qung ta thu c kt qu sau: (%st): 36,28 22,68 49,23 34,78Hy nh gi chnh xc ca kt qu.BT6. nh gi chnh xc kt qu nh lng oxi trong mt mu hp kim ca Titan vi cc s liu thc nghim sau:%X: 0,116 0,118 0,124 0,118 0,145p n: c gi tr 0,145 khng nm trong khong gi tr thc => loi b.MT S THNG S CA I LNG NGU NHIN Chng 1: (6tit)

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