Bi 1: bng sau y cho ta phn b thu nhp ca 2 nhm tui: nhm t 40 50 tui v nhm t 50 60 tui trong s cc cng nhn lnh ngh Thy in nm 1930

Thu nhp Nhm tui 0-1 1-2 2-3 3-4 4-6 >=6

40-50 71 430 1072 1609 1178 158 50-60 54 324 894 1202 903 112

C s khc nhau v phn b mc thu nhp gia 2 nhm tui ny trong s cc cng nhn lnh ngh hay khng? Mc ngha = 5%.

Bi gii Loi bi: kim nh tnh c lp, s dng phn mm MS Excel.

1. Nhp bng d liu thc t v tnh cc tng ni,mj:

nhm tui 0-1 1-2 2-3 3-4 4-6 >=6 ni

40-50 71 430 1072 1609 1178 158 4518 50-60 54 324 894 1202 903 112 3489

THC T

mj 125 754 1966 2811 2081 270 n = 8007

ni = sum(hng) mj = sum(ct)

2. Tnh d liu k vng ij theo cng thc: ij = ni * mj / n, ta c bng sau:

nhm tui 0-1 1-2 2-3 3-4 4-6 >=6

40-50 70.53 425 1109 1586 1174 152 K

VNG 50-60 54.47 329 857 1225 907 118

3. Tnh 2 = 20.05(6-1)(2-1) = 20.05(5) = CHIINV(0.05,5) = 11.07 4. Tnh = CHITEST(bng_thc_t,bng_k_vng) =

=CHITEST(C2:H3,C7:H8)

5. Tnh 20 = 20.05() = CHIINV(C11,5)= 4.267

6. Kt lun: v 20 < 2 nn phn b thu nhp gia hai nhm tui ny trong s cc cng nhn lnh ngh l nh nhau.

Bi 2: tin hnh phn tch phng sai i vi cc s liu sau:

mu I mu II mu III mu IV 22 27 20 18 19 25 18 16 13 22 21 24 19 27 21 19 23 19 16 22 15 23 17 22

16 21 20 24 18 28 18 20 23 17 20 25 19 27 18

Bi gii Loi bi: phn tch phng sai, s dng phn mm MS Excel. Khi nim thng k: php phn tch phng sai c dng trong cc trc nghim so snh cc gi tr trung bnh ca hai hay nhiu mu c ly t cc phn s. y c th c xem nh phn m rng ca trc nghim t hay z ( so snh hai gi tr trung bnh)

Mc ch ca s phn tch phng sai mt yu t l nh gi s nh hng ca mt yu t (nhn to hay t nhin) no trn cc gi tr quan st. V bi ton ch c kt qu sau khi thng k, nn ta c th t li 1 bi ton da trn cc s liu thng k ny d dng hnh dung:

Gi s c 4 ngi nng dn, trng cng 1 loi cy nh nhau, by gi chng ta s tm hiu xem s tri cy thu c ca loi cy ny, khi c trng bi 4 ngi nng dn kia c khc nhau hay khng. ngha ca vic phn tch ny cho ta kt qu, nng sut ca loi cy ny c ph thuc vo ngi trng n hay khng ?

y l bi ton phn tch phng sai mt nhn t:

Gi s nhn t A c k mc X1, X2 , , Xk vi Xj c phn phi chun N(a,s2) c mu iu tra:

Vi mc ngha a , hy kim nh gi thit : H0 : a1 = a2 = = ak

H1 : Tn ti j1 #j 2 sao cho aj1aj2 t:

1. SST : tng bnh phng cc lch:

SST = ( )21 1

njk

j ixij x

= =

.

2. SSA: tng bnh phng lch ring ca cc nhm so vi x 3. SSA = SST - SSE (SSE : tng bnh phng do sai s) 4. MSA: trung bnh, bnh phng ca nhn t

MSA = 1

SSAk

5. MSE: trung bnh bnh phng ca sai s:

MSE = SSEn k

Nu H0 ng th F = MSAMSE

c phn phi theo Fisher bc t do k-1; n-k

Bng ANOVA:

Cc bc tin hnh : 1. Nhp d liu theo bng sau:

2. chn menu tools nh trong hnh:

3. Chn Anova: single factor

4. Nhp d liu nh trong hnh

Sau khi nhp cc thng s, bng s liu c gi ra nh sau:

5. Kt qu v bin lun: F = 10,68 > F crit = F0.05 = 2.87 Bc b gi thit H0 . Vy s tri cy thu c khc nhau khi c trng bi 4 ngi nng dn

Bi 3) Tui v huyt p ca 20 bnh nhn tr em ( di 14 tui ), chn ngu nhin c cho trong bng sau y :

X 14 1 9 7 9 12 1 3 9 1 14 1 9 7 9 12 1 3 9 1

Y 100 83 112 152 104 90 92 85 120 130 110 73 132 122 134 98 82 65 140 110

Trong X l tui cn Y l huyt p.tnh t s tng quan, h s tng quan v h s xc nh ca Y i vi X. vi mc ngha =5%, c kt lun g v mi tng quan gia X v Y ( phi tuyn hay tuyn tnh) ? tm ng hi quy tuyn tnh ca Y i vi X. Tnh sai s tiu chun ca ng hi quy.

Bi gii Loi bi: tng quan v hi quy, s dng phn mm MS Excel.

I )C s l thuyt : 1) Phn tch tng quan tuyn tnh:

Gi s X v Y l hai LNN. Chng ta bit rng X v Y gi l c lp nu vic LNN ny nhn mt gi tr no cng khng nh hng g n phn b xc sut ca LNN kia.Tuy nhin trong nhiu tnh hung thc t, X v Y khng c lp vi nhau. iu ny thng gp khi X v Y l hai php o no tin hnh trn cng mt c th. V vy o mc ph thuc gia hai LNN X v Y, ngi ta a ra khi nim v h s tng quan. H s tng quan l thuyt ca X v Y, k hiu l , v c cng thc:

Trong l gi tr trung bnh v lch chun ca X v l gi tr trung bnh v lch chun ca Y. nm trong khong [-1,1] . Khi = 0 th khng c tng quan tuyt tnh gia X v Y. (X,Y) c phn b chun th = 0 khi v ch khi X v Y c lp. Khi | | cng gn 1 th s ph thuc tuyn tnh gia X v Y cng mnh. Nu | | = 1, th Y l mt hm tuyn tnh ca X.

Do thng rt kh tm v mun bit chng ta cn bit phn b ca tp hp chnh bao gm tt c cc gi tr ca cp (X,Y). V th chng ta c bi ton c lng v kim nh h s tng quan cn c trn mt mu quan st (x1,y1), (x2,y2),,(xn,yn) cc gi tr ca (X,Y). V c lng cho c thay th bng i lng r (r c gi l h s tng quan).

r =

tnh ton thun li r c th c vit di dng sau: r =

r cng nm trong [-1,1], nu thu c gi tr r nm ngoi on [-1,1] c ngha l ta tnh ton sai. Chng ta c bi ton kim nh :

Ho : = 0 ( X, Y khng tng quan) Vi i gi thit : H1 : 0 Nu (X,Y) c phn b chun hai chiu th di gi thit Ho, LNN

T = c phn b Student vi n-2 bc t do. V vy test thng k thch hp cho bi ton kim nh thng k cho bi ny l :

T = Ta s bc b Ho, nu |T| > c, c l phn v mc ca phn b Student vi bc n-2 bc t do. 2)phn tch hi quy: Cho h cc bin ngu nhin (X,Y).Gi s theo kt qu ta nhn c n im ( x1 ; y1 ),( x2 ; y2 )( xn ; yn ) (trong cc im ny c th trng nhau).Cn tm h s tng quan ca h cc bin ngu nhin ny. Ch y ti lut s ln,th vi n ln trong cc cng thc tnh

2x,

2y v

C xy ta c th thay cc k vng M(X) v M(Y) bng trung bnh cng cc gi tr ca cc bin ngu nhin tng ng.Ta c cc ng thc xp x sau y:

M(X) x =n

1

=

n

iix

1; M(Y) y =

n

1

=

n

iiy

1;

=

n

iix xxn 1

222 1 ;

=

n

iiy yyn 1

222 1 ;

yxn

n

iiixy yxC

=

1

1

T ta tm h s tng quan theo cng thc

yxxy

xy

Cr =

Nu 1nr xy >=3 th s lin h gia cc bin ngu nhin X v Y tin cy.Nu lin h gia X v Y c thit lp th xp x tuyn tnh y

x theo

x c cho bi cng thc hi quy tuyn tnh )( xx

x

yxyx ryy =

, hay baxyx

+=

Cn xp x tuyn tnh xy theo y c cho bi cng thc hi quy tuyn tnh

)( yyx

yxyy rxx =

hay dcxxy +=

Cn ch rng baxyx

+= v dcxxy += l cc ng thng khc

nhau.ng th nht nhn c do kt qu gii bi ton cc tiu ha tng bnh phng lch theo ng thng ng,cn ng th hai nhn c khi gii bi ton cc tiu ha tng bnh phng lch theo ng thng nm ngang. dng phng trnh hi quy tuyn tnh cn phi: 1)theo bng xut pht ca cc gi tr (X,Y) tnh ;,,,,, rc xyxyyxyx 2)kim nh gi thit tn ti s lin h gia X v Y; 3)lp cc phng trnh ca c hai ng hi quy v biu din th ca cc phng trnh .

II)Thut ton bng MS EXCEL: . Gi Thit:

Ho: X,Y khng tng quan vi nhau ( = 0) H1: X,Y tng quan vi nhau.

- Nhp d liu vo bng tnh :

A B

1 X Y 2 14 100 3 1 83 4 9 112 5 7 152 6 9 104 7 12 90 8 1 92 9 3 85 10 9 120 11 1 130 12 14 110 13 1 73 14 9 132 15 7 122 16 9 134 17 12 98 18 1 82 19 3 65 20 9 140 21 1 110

- S dng lnh data analysis. -Chn chng trnh correlation. -Nhp vng d liu : (A1,B21). -Check mc labels in first column. ENTER -MS EXCEL s xut cho ta bng sau:

X Y X 1

Y 0.344696 1

Vy h s tng quan r = 0.344696 Do ta c 13 cp quan st nn n = 20 c phn b Student vi 18 bc t do . Ta c T = = = 1.557 Vi bc t do l 18, =5%, ta tm c hng s c l 2.1009. Do T < c , vy ta chp nhn gi thit Ho ngha l X,Y khng tng quan vi nhau.

+ ng hi quy tuyn tnh ca Y i vi X: - S dng lnh data analysis. - Chn chng trnh Regression. Trong hp thoi ca Regression ln lt n nh cc chi tit:

Phm vi ca bin s Y (Input Y Range) $B$1:$B$21 Phm vi ca bin s X (Input X Range) $A$1:$A$21 Nhn d liu (Labels) Mc tin cy ( Confidence Level): 95% Ta u ra (Output Range)

V 1 s ty chn khc nh ng hi quy ( Line Fit Plots), biu thc sai s (ResidualsPlots)....

Bng kt qu:

Theo s liu tnh ton ta c : A=95,179 B=1,746

Vy Phng trnh hi quy X = f(X): X = 95,179 +1,746 X (R2= 0.119 ; Sai s tiu chun S = 22.877)

Bi 4) cho bng s liu v mu tc ca 422 ngi nh sau

Mu Tc Nam N en 56 32

Hung 37 66 Nu 84 90 Vng 19 38

Vi mc ngha = 1%, nhn nh xem liu c mi quan h gia mu tc v gii tnh hay khng?

Bi gii Loi bi: kim nh tnh c lp, s dng phn mm MS Excel.

Nhp bng d liu thc t v tnh cc tng ni,mj:

mu tc nam n ni en 56 32 88

hung 37 66 103 nu 84 90 174

THC T

vng 19 38 57 mi 196 226 n = 422

ni = sum(hng) mj = sum(ct)

Tnh d liu k vng ij theo cng thc: ij = ni * mj / n, ta c bng sau:

mu tc nam n en 40.9 47.1

hung 47.8 55.2 nu 80.8 93.2

K VNG

vng 26.5 30.5

Tnh 2 = 20.05(4-1)(2-1) = 20.01(3) = CHIINV(0.01,3) = 11.34 Tnh = CHITEST(bng_thc_t,bng_k_vng) = =CHITEST(C2:D5,C12:D15) = 0.0002

Tnh 20 = 20.05() = CHIINV(H2,3)= 19.22

Kt lun: v 20 > 2 nn c mi quan h gia mu tc v gii tnh.