btl xstk nhom 6
DESCRIPTION
BtL XSTK nhom 6TRANSCRIPT
-
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.