Bài báo cáo XSTK

8/11/2019 Bi bo co XSTK

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Cu 1.Trnh by li v d3.4 trang 207 v v d4.2 trang 216 Sch BTXSTK 2012 (N..HUY).V d 3.4:Hiu sut phn trm (%) ca mt phn ng ha hc c nghin

cu theo 3 yu t pH (A), nhit (B) v cht xc tc (C) c trnh bytrong bng sau:

Hy nh gi v nh hng ca cc yu t trn hiu sut phn ng ?

BI LM

1/ C s l thuyt:

Nhn xt:y l bi ton Phn tch phng sai ba yu t:- S phn tch ny c dng nh gi v s nh hng ca 3 yu ttrn cc gi tr quan st G (yu t A:i=1,2..r, yu t B: j=1,2..r, yu t C:k=1,2..r)

- M hnh: khi nghin cu nh hng ca 2 yu t, mi yu t c n mcth ngi ta dng m hnh hnh vung latin n n. V d:

B C D A

C D A B

D A B CA B C D

- M hnh vung latin 3 yu t c trnh by nh sau:Yu t C (T..k : vd T..1= Y111+Y421+Y331+Y241)

Yu tA

Yu t BB1 B2 B3 B4

A1 C1 9 C2 14 C3 16 C4 12A2 C2 12 C3 15 C4 12 C1 10A3 C3 13 C4 14 C1 11 C2 14A4 C4 10 C1 11 C2 13 C3 13


2/29

YutA

Yu t B

B1 B2 B3 B4 Ti

A1 C1 Y111 C2 Y122 C3 Y133 C4 Y144 T1...

A2 C2 Y212 C3 Y223 C4 Y234 C1 Y241 T2...A3 C3 Y313 C4 Y324 C1 Y334 C2 Y342 T3

A4 C4 Y414 C1 Y421 C2 Y412 C3 Y443 T4

T.j. T.1. T.2. T.3. T.4.

Bng ANOVA

Trc nghim:*Gi thit:

H0: 1 =2= k Cc gi tr trung bnh bng nhau.H1: i j C t nht 2 gi tr trung bnh khc nhau.*Gi tr thng k: v *Bin lun:

Nu (chp nhn H0(yu t A) )Nu (chp nhn H0(yu t B) )Nu (chp nhn H0(yu t C) )

Ngun sais

Bc tdo

Tng s bnh phgBnh phng

trung bnhGi tr

thng k

Yu t A(hng)

r-1 SSR=

r

i

i

rTT

r12

2

...

2

.. MSR=1r

SSR FR=

MSE

MSR

Yu t B(ct)

r-1 SSC=

r

i

j

rTT

r12

2

...

2

.. MSC=1r

SSC FC=

MSE

MSC

Yu t C r-1 SSF=

r

i

k

rTT

r12

2

...

2

.. MSF=1r

SSF F=

MSE

MSF

Sai s (r-1)(r-2) SSE=SST-(SSF+SSR+SSC) MSE=)2)(1( rr

SSE

Tng cng (r2-1) SST=r

TYijk 2

2

...2


3/29

2/ p dng Excel:

Thit lp bng tnh nh sau :

Tnh cc gi tr Ti..(tng theo hng t B n E)Chn B7 v nhp vo biu thc =SUM(B2:E2)Chn C7 v nhp vo biu thc =SUM(B3:E3)Chn D7 v nhp vo biu thc =SUM(B4:E4)Chn E7 v nhp vo biu thc =SUM(B5:E5)

Tnh cc gi tr T.j.(tng theo ct t hng th 2 n hng th 5)Chn B8 v nhp vo biu thc =SUM(B2:B5)Dng con tr ko k hiu in t B8 n E8

Tnh cc gi trT..kChn B9 v nhp biu thc =SUM(B2,C5,D4,E3)Chn C9 v nhp biu thc =SUM(B3,C2,D5,E4)Chn D9 v nhp biu thc =SUM(B4,C3,D2,E5)Chn E9 v nhp biu thc =SUM(B5,C4,D3,E2)

Tnh gi trT..(tng cc phn t trong bng)Chn B1 v nhp biu thc =SUM(B2:E5)

*Tnh cc gi tr v - Cc gi tr v

Chn G7 v nhp biu thc =SUMSQ(B7:E7)Dng con tr ko k hiu in t G7 n G9

- Gi tr Chn G1 v nhp biu thc =POWER(B1,2)

- Gi tr Chn G11 v nhp biu thc =SUMSQ(B2:E5)


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*Tnh cc gi tr SS,SSC,SSF,SST v SSE- Cc gi tr SS,SSC v SSF

Chn I7 v nhp vo biu thc =G74-39601/POWER(4,2)Dng con tr ko k hiu in t I7 n I9

-

Gi tr SSTChn I11 v nhp biu thc =G11-G10/POWER(4,2)- Gi trSSE

Chn I1 v nhp biu thc =I11-SUM(I7:I9)

*Tnh cc gi tr SS,MSC,MSF v MSE- Cc gi tr MS,MSC v MSF

Chn 7 v nhp biu thc =I7(4-1)Dng con tr ko k hiu in t M7 n M9

-

Gi tr MSEChn 1 v nhp biu thc =I1((4-1)*(4-2))

*Tnh cc gi tr v FChn M7 v nhp vo biu thc =7.3958Dng con tr ko k hiu in t M7 n M9

Kt qu v bin lun:FR=3.10 chp nhn H0(pH)FC=11.95>F0.05(3.6) = 4.76 =>bc b H0(nhit )F =30.05 >F0.05(3.6) = 4.76 =>bc b H0(cht xc tc)

Vy ch c nhit v cht xc tc nh hng n hiu sut

V d 4.2: Ngi ta dng ba mc nhit gm 105, 120, 135oC kt hp viba khong thi gian l 15, 30, 60 pht thc hin mt phn ng tng hp.Cc hiu sut ca cc phn ng (%) c trnh by trong bng sau :

Thi gian (pht) Nhit (oC) Hiu sut (%)X1 X2 Y15 105 1,8730 105 2,0260 105 3,2815 120 3,0530 120 4,07

60 120 5,5415 135 5,03


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30 135 6,4560 135 7,26

Hy cho bit yu t nhit v thi gian/hoc yu t thi gian c linquan

tuyn tnh vi hiu sut ca phn ng tng hp?Nu c th vi iu kinnhit 115oC trong vng 50 pht, hiu sut phn ng s l bao nhiu?

BI LM

1/C s l thuyt:

Nhn xt:y l dng bi Hi quy tuyn tnh a tham s.

Trong phng trnh hiquy tuyn tnh a tham s, bin s ph thuc Y clin quan n k bin s c lp Xi(i =1,2,k) thay v ch c mt nh trong hi

quy tuyn tnh n gin.

Phng trnh tng qut:

= B0 + B1X1+ B2X2+ + BkXk

Phng trnh hi quy atham s c th c trnh by di dng ma trn:

1

N

= +

1 1 1

N N

k

k


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Bng ANOVA

Ngun sais

Bc tdo

Tng s bnhphng

Bnh phng trungbnh

Gi trthng k

Hi quy k SSR MSR=SSR/k F =

Sai sNk

1SSE

MSE = SSE/( Nk1)

Tng cng N1 SST=SSR + SSE

Gi tr thng k:

- Gi tr R-bnh phng:

Gi tr R2c hiu chnh (Adjusted R Square)

R2 =

=

( ) (R

3.81 l tt nht)Gi tr R2c hiu chnh (Adjusted R Square)

R2

ii =

( ) = R2

( )

(R2

ii s tr nn m hay khng xc nh nu R2hay N nh).

- lch chun:

S = ( )(S .3 l kh tt)Trc nghim thng k:

Tng t hi quy n gin, song cn ch :- Trong trc nghim t

H0: i = Cc h s hi quy khng c nghaH0: i C t nht vi h s hi quy c ngha

Bc t do ca gi tr t: = N k1

t =||

;

-

Trong trc nghim FH0: i = Phng trnh hi quy khng thch hp


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H0: i Phng trnh hi quy thch hp vi t nht vi Bi.Bc t do ca gi tr F: 1= 1, 2= Nk1.

2/ p dng Excel:

Nhp bng d liu vo bng tnh:D liu bt buc phi c nhp theo ct

S dng egression:VoData->Data Analysis.Chn mcRegression.Chn O.

a/Trong hp thoi Regression, ln lt n nh cc chi tit:

Phm vi ca bin s Y (Input Y Range): $C$1:$C$10 Phm vi ca bin s X (Input X Range): $A$1:$A$10 Nhn d liu (Labels)

Mc tin cy (Confidence Level): chn mc 95%

Ta u ra (Output Range): $A$14


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V mt s ty chn khc nh ng hi quy (Line Fit Plots), biu thcsai s (Residuals Plots)


9/29

Phngtrnh hi quy: )= 2.7367 +0.04454X1 (R2=0.2139, S=1.8112); N=9; k=1

t0= t Stat(Intercept)= 2.129 < t0,05(7)=2.365 (hay P value=0.0708>=0.05)

Chp nhn gi thit H0.

t1= t Stat(X1) = 1.3802=0.05)

Chp nhn gi thit H0.

F=1.9049=0.05)

Chp nhn gi thit H0.

Vy phng trnh hi quy ny khng thch hp.


10/29

Kt lun:Yu t thi gian khng c lin quan tuyn tnh vi hiu sut phnng tng hp.

b/Trong hp thoiRegression

, ln lt n nh cc chi tit:

Phm vi ca bin s Y (Input Y Range): $C$1:$C$10 Phm vi ca bin s X (Input X Range): $B$1:$B$10 Nhn d liu (Labels) Mc tin cy (Confidence Level): chn mc 95% Ta u ra (Output Range): $A$45 V mt s ty chn khc nh ng hi quy (Line Fit Plots), biu thc

sai s (Residuals Plots)


11/29

Phng trnh hi quy: )= -11.1411 +0.12856X2 (R2=0.7638; S=0.9929);N=9; k=1

t0= t Stat(Intercept) =3.4178 > t0,05(7)=2.365 (hay P value=0.0112 t0,05(7)=2.365 (hay P value=.21 < =.5)

Bc b gi thit H0.

F=22.6309 > F0,05(1.7)=5.59 (hay FS=0.0021


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Kt lun:Yu t nhit c lin quan tuyn tnh vi hiu sut phn ng tnghp.

c/Trong hp thoi Regression, ln lt n nh cc chi tit:

Phm vi ca bin s Y (Input Y Range): $C$1:$C$10

Pham vi ca bin s X (Input X Range): $A$1:$B$10

Nhn d liu (Labels)

Mc tin cy (Confidence Level): chn mc 95%

Ta u ra (Output Range): $A$76 V mt s ty chn khc nh ng hi quy (Line Fit Plots), biu thc

sai s (Residuals Plots)

Phng trnh hi quy: ) = -12.7 + 0.0445X1+ 0.1286X2 (R2=0.9777; S=0.3297); N=9; k=2


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14/29

Kt lun:Hiu sut phn ng tng hp c lin quan tuyn tnh vi c hai yu tl nhit v thi gian.

D on hiu sut phn ng ti t =115oC v thi gian l 50phtTi B94, nhp = B91 + B92*5 + B93*115

t qu d on hiu sut phn ng l 4.3187316

Cu 2: Bng sau y cho ta mt mu gm 11 quan st (xi, yi) ttp hpchnh cc gi trca cp LNN (X, Y):

X 0,9 1,22 1,32 0,77 1,3 1,2

Y -0,3 0,1 0,7 -0,28 -0,25 0,02

X 1,32 0,95 1,45 1,3 1,2

Y 0,37 -0,70 0,55 0,35 0,32

a) Tm ng hi quy ca Y i vi X.b) Tnh sai s tiu chun ca ng hi quy.c) Tnh t s F kim nh s ng n ca gi thit: C hi quy tuyn

tnh ca Y theo X.

Bi lm

Nhn xt:y l bi ton phn tch hi quy tuyn tnh

C s l thuyt:

x = B0+ BX

B0= - B

B =

X -bin s ph thuc (dependent reponse variable)

Y bin s c lp (independent predictor variable)

B0v B cc h s hi quy (regression coefficients)


15/29

Bng ANOVA

NgunBc t do

(DF)

Tngbnh

phng(SS)

Trung bnhbnh

phng(MS)

T s F

Hi quy 1 SSR MSR M SR

FMSE

Sai s n2 SSE MSE

Tngcng

n1 SST

Gi tr thng k:Gi tr R-bnh phng (R-square):

R=(100R

2: %ca bin i trn Y c gii thch bi X)

lch chun (Standard Error):

S= (S phn tn ca d liu cng t th gi tr ca S cng gn zero)

Trc nghim thng k:

i vi mt phng trnh hi quy, x = B0+ BX , ngha thng k ca cch s Bi(B0hay B) c nh gi bng trc nghim t (phn phi Student) trongkhi tnh cht thch hp ca phng trnh x = f(X) c nh gi bng trcnghim F(phi b Fischer)

Trc nghim t-Gi thit:

o H0: i= H s hi quy khng c nghao H0: i H s hi quy c ngha

-Gi tr thng k

t=| |

;

Phn b Student = N -2


16/29

-Bin lun: Nu t < t(N-2) => Chp nhn gi thit H0

Trc nghim F:

-Gi thit:o

H0: i= Phng trnh hi quy khng thch hpo H0: i Phng trnh hi quy thch hp

-Gi tr thng k:

F =

Phn b Fischer v1= 1, v2= N2

-t lun:

Nu F < F(1,N -2) => chp nhn gi thit H0

p dng excel

Bc 1: Nhp bng s liu

Bc 2: vo Data /Data analysis, chn Regression.


17/29

Bc 3: Nhp cc s liu vo bng sau:

Nhn O, ta c kt qu:

a) ng hi quy ca Y i vi X l: Y=1.5479X 1.7395b) Sai s tiu chun ca ng hi quy: .2896c) Ta thy F = 12.6367 > c = 5.12

(Tra bng phn b Fisher vi bc t do n1= 1, n2= 9 mc .5)

=> C hi quy tuyn tnh ca Y theo X


18/29

Cu 3:Bng sau y cho ta phn bthu nhp ca hai nhm tui: Nhm t4050 tui v nhm t5060 tui trong scc cng nhn lnh nghThy in nm 1930.

Nhm

tuiThu nhp

01 12 23 34 46 6

4050 71 430 1072 1609 1178 158

5060 54 324 894 1202 903 112

C s khc nhau v phn b thu nhp gia hai nhm tui ny trong scc cng nhn lnh ngh hay khng? Mc ngha = 2%.

Bi lmI. C s l thuyt:

a.Dng bi ton:Kim nh tnh c lpb.Khi nim thng k:i vi mt th nghim c hai kt qu (binomial experiment) - th d, i

vi mt thuc c k n: c hay khng bn thng so snh hai ts vi nhau (thc nghim vi l thuyt hay thc nghim vi thc nghim).Song i vi mt th nghim c nhiu kt qu (multinomial experiment) thd, bc s nh gi tnh trng ca cc bnh nhn c iu tr bi thuc trongmt khong thi gian bn cn so snh nhiu t s. Trc nghim khi bnh

phng (2) cho php bn so snh khng nhng hai m cn nhiu t s (hay t

l hoc xc sut) mt cch tin li. 2l phn phi v xcsut, khng c tnhi xng v ch c gi tr . Gi s bn c mt cng trnh nghin cu vi Nth nghim c lp, mi th nghim c k kt qu v mi kt qu mang mttrong cc xc sut thc nghim l Pi (i = 1, 2, k). Nu gi Pi,0 l cc gi trl thuyt tng ng vi Pi th cc tn s l thuyt s l Ei= NPi,0. iu kin p dng trc nghim 2mt cch thnh cng l cc tn s l thuyt Ei5.

c. Gi thuyt:H0: P1= P1,0; P2= P2,0; ; Pk= Pk,0 Cc cp Piv Pi,0ging nhau.

H1: t nht c mt cp Piv Pi,0khc nhau.

Gi tr thng k:2

ij ij2

1 ij

( )k

i

O E

E

Oi: cc tn s thc nghim (observed frequency);

Ei: cc tn s l thuyt (expected frequency).

Bin lun:

Nu 2 2 Bc b gi thuyt H0(DF = k-1)


19/29

Trong chng trnh MS-EXCEL c hm CHITEST c th tnh:

- Gi tr 2 theo biu thc:2

ij ij2

1 1 ij

( )[

r c

j i

O E

E

Oij: tn s thc nghim ca thuc hng i v ct j;Eij: tn s l thuyt ca thuc hng i vi ct j;r: s hng;c: s ct.

- Xc sut P(X >2) vi bc t do DF = (r-1)(c-1); trong , r l s hngv c l s ct trong bng ngu nhin (contingency table).

Nu P(X >2) > Chp nhn gi thuyt H0v ngc li.

d. Gii thut:- Tnh cc tng s- Tng hng (row totals)- Tng ct (column totals)-Tng cng (grand total)- Tnh cc tn s l thuyt- Tn s l thuyt = tng hng x tng ct tng cng

II. p dng Excel:o H0 : Phn b thu nhp gia 2 nhm tui trong s cc cng nhn lnh

ngh l nh nhau

Bc 1: Nhp bng s liu:

Bc 2:Tnh cc tng hng, ct : chn (B3:H5) n biu tng (AutoSum).


20/29

Chn B7 nhp biu thc = B$5*$H3$H$5Chn (B9:G1): n F2+Ctrl+Enter

Bc 3: Tnh gi tr, s dng hm CHITEST v hm CHIINVChn B12: Chn hm =CHITEST(B3:G4,B9:G1)Chn B13: Chn hm =CHIINV(.2,5)Chn B14: Chn hm =CHIINV(B12,5)

Bin Lun:P(X >

2) =.511582 > =.2Chp nhn gi thuyt H0.Hoc 20= 4.2675<

2a = 11.0705 Chp nhn gi thuyt H0.

t lun: Nn phnb thu nhp gia 2 nhm tui trong s cc cng nhn

lnh ngh l nh nhau

Cu 4. Mt nhm gm 105 nh doanh nghip M c phn loi cn c theothu nhp hng nm v tui ca h. Kt qu thu c nh sau:

Tui

Thu nhp

Di100 000 $

T 100 000$ - 399 599

$

Trn400 000

$

Di 40 6 9 5T 40 n

5418 19 8

Trn 54 11 12 17

Vi mc ngha 1%, kim nh gi thit cho rng tui v mc thunhp khng c quan h vi nhau.

Bi lm

I. C s l thuyt:a.Dng bi ton:Kim nh tnh c lp


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b.Khi nim thng k:i vi mt th nghim c hai kt qu (binomial experiment) - th d, i

vi mt thuc c k n: c hay khng bn thng so snh hai ts vi nhau (thc nghim vi l thuyt hay thc nghim vi thc nghim).

Song i vi mt th nghim c nhiu kt qu (multinomial experiment) thd, bc s nh gi tnh trng ca cc bnh nhn c iu tr bi thuc trongmt khong thi gian bn cn so snh nhiu t s. Trc nghim khi bnh

phng (2) cho php bn so snh khng nhng hai m cn nhiu t s (hay tl hoc xc sut) mt cch tin li. 2l phn phi v xc sut, khng c tnhi xng v ch c gi tr . Gi s bn c mt cng trnh nghin cu vi Nth nghim c lp, mi th nghim c kkt qu v mi kt qu mang mttrong cc xc sut thc nghim l Pi (i = 1, 2, k). Nu gi Pi,0 l cc gi trl thuyt tng ng vi Pi th cc tn s l thuyt s l Ei= NPi,0. iu kin

p dng trc nghim 2

mt cch thnh cng l cc tn s l thuyt Ei 5.c. Gi thuyt:H0: P1= P1,0; P2= P2,0; ; Pk= Pk,0 Cc cp Piv Pi,0ging nhau.H1: t nht c mt cp Piv Pi,0khc nhau.

Gi tr thng k:2

ij ij2

1 ij

( )k

i

O E

E

Oi: cc tn s thc nghim (observed frequency);

Ei: cc tn s l thuyt (expected frequency).

Bin lun:

Nu 2 2

Bc b gi thuyt H0(DF = k-1)

Trong chng trnh MS-EXCEL c hm CHITEST c th tnh:

- Gi tr 2 theo biu thc:2

ij ij2

1 1 ij

( )[

r c

j i

O E

E

Oij: tn s thc nghim ca thuc hng i v ct j;Eij: tn s l thuyt ca thuc hng i vi ct j;r: s hng;c: s ct.

- Xc sut P(X >2) vi bc t do DF = (r-1)(c-1); trong , r l s hngv c l s ct trong bng ngu nhin (contingency table).

Nu P(X >2) > Chp nhn gi thuyt H0v ngc li.


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d. Gii thut:- Tnh cc tng s- Tng hng (row totals)- Tng ct (column totals)

-Tng cng (grand total)- Tnh cc tn s l thuyt- Tn s l thuyt = tng hng x tng ct tng cng

II. p dng Excel:

Bc 1: Nhp bng s liu:

Bc 2: Dng hm AutoSum tnh tngTnh tng hng, tng ct: chn (B6:E7) n biu tng (AutoSum).

Chn B1 nhp biu thc =B$7*$E4$E$7.Chn vng (B1:D12) n F2+Ctrl+Enter


23/29

Bc 3: Dng hm CHITESTTnh gi tr P:

Chn B14 v chn hm CHITEST. Cc thng sActual_range: B4:D6

Expected_range: B10:D12Chn O, ta c bng:

P(X >2) =.14375984 > =.1Chp nhn gi thuyt H0.

Kt lun: Tui v mc thu nhp khng c mi quan h vi nhau

Cu 5: Vi mc ngha 1%, Hy phn tch tnh hnh kinh doanh ca mts ngnh ngh qun 4 qun ni thnh trn c s s liu v doanh thu camt s ca hng nh sau:

Ngnh nghkinh doanh

Khu vc kinh doanh

Q1 Q2 Q3 Q4in lnhVt liu xydngDch v tinhc

2.5:2.7:2.0:3.00.6:10.4

1.2:1.0:9.8:1.8

3.1:3.5:2.715.0

2.0:2.2:1.8

2.0:2.49.5:9.3:9.11.2:1.3:1.2

5.0:5.419.5:17.5

5.0:4.8:5.2

Bi lm


24/29

Dng ton: Phn tch phng sai 2 yu t (c lp)

I.C s l thuyt:Trn thc t mt bin lng chu tc ng khng ch mt nhn t m c

th hai (hay nhiu nhn t). Chng hn nng sut cy trng chu nh hng canhn t ging v ca nhn t t. t qu hc tp ca mt sinh vin chu nhhng khng nhng bi nhn t ging vin m cn bi nhn t s s ca lphc

Phng php phn tch phng sai hai nhn t nhm pht hin nh hngca mi nhn t cng nh tc ng qua li ca hai nhn t n bin lngang xt

Gi s chng ta quan tm ti nhn t A v B. Nhn t A c xem xt cc mc A1, A2,Arv nhn t B c xem xt cc mc B1, B2,Bc

+ Tng bnh phng chung, k hiu l SST, c tnh theo cng thc sau:

jknc r

2

i jk

k 1 j 1 i 1

SST x x

r

TYij

c

j

r

i

2

..2

11

+ Tng bnh phng cho nhn t A, k hiu l SSFA c tnh theo cng

thc:

c

2B ok ok

k 1

SSF n x x

+ Tng bnh phng cho nhn t B, k hiu l SSFB c tnh theo cngthc:

c

2B ok ok

k 1

SSF n x x

+Tng bnh phng do sai s, k hiu l SSE, c tnh theo cng thc

jknc r

2i jk jk

k 1 j 1 i 1

SSF x x

+ Tng bnh phng do tng tc (Sum of Squares for Interaction) k hiu lSSI, c tnh theo cng thc


25/29

C r

2

j k j o k o

k 1 j 1

SSI x x x x

+ Trung bnh bnh phng ca nhn t A, k hiu l MSFA, c tnh theo

cng thc

AA

SSFM SF

r 1

+Trung bnh bnh phng ca nhn t B, k hiu l MSFB, c tnh theocng thc

B

B

SSFMSF

c 1

+ Trungbnh bnh phng ca sai s, k hiu l MSE, c tnh bi

SSEM SE

n cr

+ Trung bnh bnh phng ca tng tc, k hiu l MSI, c tnh bi

( )( )

SSIM SI

c 1 r 1

(c

Ch rng:

(r 1) + (c 1) + (c 1) (r 1) + n rc = n 1 = bc t do tng cng

+ T s F cho nhn t A, k hiu bi FAc tnh nh sau,

AA

M SFFM SE

Tng t t s F cho nhn t B, FBc tnh bi

BB

M SFF

MSE

vt s F cho tng tc gia A v B, k hiu l FABc tnh bi:

AB

MSIF =

MSE


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Vi mc ngha cho ta k hiu f (u, v) l phn v mc ca phn bFisher vi bc t do (u, v).

Ta c quy tc quyt nh nh sau:

+ Nu FA> f (r 1, n cr) th ta bc b gi thit.:

AoH Cc mc A1,... Arc hiu qu trung bnh nh nhau

+ Nu FB> f (c1, ncr) th ta bc b gi thit

:BoH Cc mc B1, B2, ... Bcc hiu qu trung bnh nh nhau

+ Nu FAB> f ((r1)(c1), nrc)

Ta bc b gi thit:

:AB

oH C s tng tc gia A v B.

Trn thc hnh tnh ton chng ta thc hin nh sau:

Gi s Tjktng cc gi tr trong mu (j, k). hiu

,

,

c r

j o j k ok j k

k 1 j 1

c r

j o j k ok j k

k 1 j 1

T T T T

n n n n

jo ok i jk

jo ok

T T T x

n n n

2ijkA x

(3)

Ta c cc ng thc sau:2T

SST An

(4)

2r 2j o

Aj oj 1

T TSSF

n n

(5)

c 2 2ok

Bokk 1

T T

SSF n n (6)


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2c rjk

jkk 1 j 1

TSSE A

n

(7)

A BSSI SST SSF SSF SSE

(8)c bit nu tt c cc mu bng nhau njk= m vi mi j,k th:

,jo okn cm n r m

do

r2

jo2

j 1A

TT

SSFcm n

(5)

r 2ok 2

k 1B

TT

SSFr m n

(6)

2j k

k j

T

SSE Am

(7)Trc ht ta cn tnh cc i lng Tjk. Tip theo tnh cc gi tr

Tjo, njo, nok, Tok, n, T v A theo cc cng thc (1), (2), (3).

T tnh SST, SSFA, SSFB, SSE v SSI theo cc cng thc (4),(5), (6), (7) (hoc (5), (6), (7) nu njk= m).

II. p dng Excel:

Gi thuyt:o

H0:

Trung bnh ca cc ngnh ngh kinh doanh bng nhau Trung bnh ca cc khu vc kinh doanh bng nhau C s tng tc gia ngnh ngh v khu vc kinh doanh


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29/29

t lun:

FR= 2.8929 < F0.01=5.25 => chp nhn gi thuyt H0(ngnh ngh kinh

doanh) => Tnh hnh kinh doanh gia cc ngnh ngh l nh nhau FC= 0.8189 < F0.01= 4.38 => chp nhn gi thuyt H0(khu vc kinh

doanh) => Tnh hnh kinh doanh gia cc qun l nh nhau FI= 0.6498 < F0.01 = 3.35 => chp nhn gi thuyt H0(khu vc v ngnh

ngh kinh doanh) => Tnh hnh kinh doanh gia cc qun v ngnh nghtng tc vi nhau

HT

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