He Thong Cong Thuc Mon Nguyen Ly Thong Ke1

7/27/2019 He Thong Cong Thuc Mon Nguyen Ly Thong Ke1

1/21

Ti liu n tp HP Nguyn l thng k 2013

SV: Ng ThLinh HaKT 4D - QUI Page 1

N TP MN NGUYN L THNG KChng I: Cc khi nim vthut ng.

Phn tvi cc khong cch tbng nhau.

max mini i

i

X Xh

n

Trong :

ih : Trskhong cch t.

maxiX : Lng bin ln nht ca tiu thc phn t.

miniX : Lng bin nhnht ca tiu thc phn t.

n : Stcn chia.Cc bc phn tthng k:

Bc 1: La chn tiu thc phn t.Bc 2: Xc nh stcn phn v khong cch t.

Bc 3: Phn phi cc n vvo tng t.Chng II: nh gi thng k.

STT Tn Cng thcnv

Ch thch

1

Stngi ng

thi( T GT ),

(TLH

T )

1

0

T G

yT

y

1

iTLH

i

yT

y

%,pv

1y : Mc ca hin tng knghin cu.

0y : Mc ca hin tng kgc.

T GT : Stng i ng thi nh gc (Tc PT

nh gc).

TLH

T : Stng i ng thi lin hon (Tc PT

lin hon).

2

Stngi khoch( KHT ),

( TKT )

0

KHKH

yT

y

1TK

KH

yT

y

Hqu:

DT KH TKT T T

DTTK

KH

TT

T

DTKH

TK

TT

T

%,pv

KHT : Stng i nhim vkhoch.

TKT : Stng i hon thnh khoch.

KHy : Mc ca hin tng kkhoch.

0y : Mc thc tca chtiu kgc so snh.

1y : Mc ca hin tng kbo co.

3

Stngi kt

cu ( KCT )

bq

KC

TT

yT

y

%,pv

bqy : Mc ca bphn.

TTy : Mc ca tng th.

4

Stngi cng( CDT )

C

mT

n

% m : Mc ca hin tng cn nh gi phbin.n : Mc ca hin tng no c lin quan.

5

Stng

i khnggian( KGT )

1

2

KG

xT

x

%,

pv1x : Mc ca hin tng khng gian thnht

cn phn tch.2x : Mc ca hin tng khng gian thhai

dng lm c sso snh.


2/21



6

Sbnhqun cnggin n

(X )

1

n

i

i

X

Xn

vti

X : Lng bin (i=1, 2, , n)

n : Sn vtrong tng th.

7

Sbnhqun cnggia quyn

(X )

1

1

n

i i

i

n

i

i

X f

Xf

vti

X : Lng bin (i=1, 2, , n)

i

f: Quyn s(Tn s)

i iX f : Gia quyn

8

Trsgia

( gX )

ax min

2m

g

X XX

(vi lng bin c khongcch t)

vtaxmX : Lng bin ln nht ca t.

minX : Lng bin nhnht ca t.

9

Sbnhqun

chung tcc s

bnh qunt(

tX )

1

1

k

i i

it k

i

i

X n

X

n

vti

X : Sbnh qun ti.

in : Quyn shoc sn vti.

K: Slng t.

10

Sbnhqun iuho giaquyn

(X )

1

1

n

i

i

ni

i i

M

XM

X

Khi c:

1 2 ... nM M M M

th:

1

1n

i i

nX

X

vti i iM X f : Gia quyn.

(Vn dng khi cha bit tn shay tn sn)

11

Sbnhqun nhngin n

(X )

1

1 2 3. . .....

n

ni

i

nn

X X

X X X X

vti

X : Lng bin (i=1, 2, 3,,n)

n : Sn v( Slng bin).

12

Sbnhqun nhngia quyn

(X )

1

21

1

1 2. .....

n

ii

i

n

i

ii

nff

i

i

ff f fn

n

X X

X X X

vti

X : Lng bin (i=1, 2, 3,,n)

if: Tn stngng.

13

Strung v(MEDIAN - eM)

+)Vi dy sclng bin khngc khong ccht:

*2 1, e qn k k N M x ( q l tgia) *2 ,

2

q p

e

x xn k k N M

( ,q p l tgia)

+)Vi dy slng bin ckhong cch t:

*Xc nh tcha eM : Cng dn tn sn khi no bng hoc vt qu2

if

th dng.

*Gi trgn ng ca strung vc xc nh theo cng thc:


3/21



min

12

e

e e

e

i M

e M M

M

f SM X h

f

Trong :

mineMX : Gii hn di ca tcha trung v.

eM

h : Trskhong cch tca tcha trung v.

eMf :Tn sca tcha trung v.

1eMS : Tng tn sca tng trc tcha trung v.

1

n

i

i

f

: Tng tn s.

14

SMt (MODE - oM)

+)Vi dy slng bin khngc khong ccht:

axmo fM X

(Mt l lng bin ln nht trong dy lng bin)

+)Vi dy lngbin c khongcch t:

-)TH c khong cch tu nhau:

*Tcha mt l tc tn sln nht maxf

T .

*Gi trgn ng ca mt c tnh theo cng thc:

min1

1 1

o o

o o

o o o o

M M

o M M

M M M M

f fM X h

f f f f

Trong :

minoM

X :Gii hn di ca tcha Mt.

oMh : Trskhong cch tca tcha Mt.

oMf :Tn sca tcha Mt.

1oMf : Tn sca tng sau tcha Mt.

1oMf : Tn sca tng trc tcha Mt.

-)TH khong cch tkhng u nhau:

*Tcha Mt l tc mt phn phi l ln nht PPmax

MT .

i

iPP

i

fM

h

Trong :

iPPM : Mt phn phi ca ti.i

f: Tn sca ti.

ih : Trskhong cch tca ti.

*Gi trgn ng ca Mt c tnh:

min1

1 1

M Moo

o o

M M M Mo o o o

PP PP

o M M

PP PP PP PP

M MM X h

M M M M

Trong :

minoM

X :Gii hn di ca tcha Mt.

oM

h : Trskhong cch tca tcha Mt.

MoPPM : Mt phn phi tcha Mt.


4/21



1MoPPM : Mt phn phi ttrn tcha Mt.

1MoPPM : Mt phn phi tdi tcha Mt.

15

Khongbin thin

(khong

chnh lch)(R )

ax minmR X X vt axmX : Lng bin ln nht.

minX : Lng bin nhnht.

16

lchtuyt i

bnh qun

( e )

+)TH khng c quyn s:

1

n

i

i

X X

en

+)TH c quyn s:

1

1

.n

i i

i

n

i

i

X X f

e

f

vti

X : Lng bin.

X : Trung bnh cng ca cc lng bin.

if: Tn s.

17Phng sai

( 2 )

+)TH khng c quyn s:

2

2 1

n

i

i

X X

n

+)TH c quyn s:

2

2 1

1

n

i i

i

n

i

i

X X f

f

iX : Lng bin.

X : Trung bnh cng ca cc lng bin.

if: Tn s.

18 lchchun ( )

2

19Hsbin

thin

100e

eV

X

100VX

20

Cc tham sbiu thhnh dng ca tham s:

Cch 1: So

snh 3 chtiu ctrng.

+)Nu ng cong phn phi i xng th:

e oX M M

+)Nu ng cong phn phi lch phi th:

e oX M M

+)Nu ng cong phn phi lch tri th:

e oX M M

Cch 2: Tnhhkhng ixng.

oA

X MK

*Khi AK >0 l phn phi lch phi.

*Khi AK


5/21



Tng thchung vtng thmu

Chtiu Tng thchung Tng thmuQuy m N n

Sbnh qun X Tltheo mt tiu thc p f

Phng sai 2 2 2X 22 2

o X X

Phng sai ca tng thmu:

22

2 i i i i

o

i i

X n X n

n n

Cch chn ngu nhin slng n vca tng thmu ( n ):

+)Cch 1: Chn hon li (chn lp):

Slng mu khng hnh thnh: nQ N (mu)

+)Cch 2: Chn khng hon li (chn khng lp):

Slng mu c hnh thnh l:

!

! !

NQ

N n n

(mu)

-Vi 2, ,p l bnh qun, tl, phng sai ca tng thchung.

-Vi 2, ,i i iX f l bnh qun, tl, phng sai ca mu thi (i=1,2,3,,q)

Chn hon li(Chn nhiu ln) Chn khng hon li(Chn 1 ln)

Phng sai cc bnhqun mu

22

X n

22 1

X

n

n N

Phng sai cc tlmu

2 1f

p p

n

2 1 1fp p n

n N

-Kvng cc bnh qun mu: iE x

-Kvng cc tlmu: iE f p

Sai schn mu:

Cch chnSuy rng

Chn hon li(Chn nhiu ln)

Chn khng hon li(Chn 1 ln)

Bnh qun

Tng th2

X n

Tng th

2

1X

n

n N

Mu2

1

o

X n

Mu

2

11

o

X

n

n N

Tl Tng th 1

p

p p

n

Tng th

11

p

p p n

n N


6/21



Mu 1

1f

f f

n

Mu

11

1f

f f n

n N

, pX : Cc sai sbnh qun chn mu khi c lng sbnh qun v tl.

Bi ton c bn ca iu tra chn mu.

*Cng thc tng qut:

. 2

. 2

X X

p p

P X z z

P f p z z

(*)

Vi.

..

X X

p f

zz

z

(**)

-i lng .X Xz v .p fz c gi l phm vi sai schn mu, zc gi l hstin cy.

-Nu 30n th ,i iX fc xem nh phn phi chun vi 2

2

0

1

2

tz

z e dt

. Cc gi trca hm c

cho bng tnh sn (Bng phn phi chun ho N(0:1)).

-Nu 30n thi

X c xem nh phn phi theo quy lut Student.

Dng T n Tm tt Cch gii

Biton 1

Suy rng tiliu iu trachn mu.

Bi ton tm&p bit

...P a

Tgithit tnh (*) v (**) :+)Suy rng bnh qun:

X XX X

+)Suy rng tl:

p pf p f

Bi

ton 2

Tm xc sut(tin cy)khi suy rngti liu iutra chnmu.

Bi ton tm

... ?P bit

&X p

Tcng thc (**) ta c:

? ... 2 ?

X

X

p

f

z

z P z

z

Trong :

X : Phm vi sai schn mu bnh qun.

p : Phm vi sai schn mu tl.

z: Hstin cy.

Biton 3

Tnh slng n vtng thmu( n )

Bi ton tm n

bit ...P av

Theo githit ta suy ra:


7/21



2

2

2

2

... 2 ? ?

?

X X

p p

o

P z z z

n

f

Vi n c tnh nh sau:Suy rng Chn hon li Chn khng hon li

Bnhqun

2 2

2

X

zn

2 2

2 2 2

X

Nzn

N z

Tl 2

2

1

p

z p pn

2

2 2

1

1p

Nz p pn

N z p p

Nu hin tng nghin cu c phn phi chun th c thc tnh lch tiu chun theo khong bin thin:

Ta c: ax min 3 3mR X X

Theo quy tc 3 , nu 2,x N th hu ht cc gi trX sai lch vi khng qu 3 ln

ax min

6

mX X

Cc phng php chn mu ngu nhin

Chn mu ngu nhin n gin

Chn lp Chn khng lp Ch gii

22

X X n

2 1

X

n

N

n

2

o : Phng sai ca tng thmu iu tra.

2 : Phng sai ca tng thchung.n : Sn vca tng thmu.N : Sn vca tng thchung.2

1

o

X n

2 1

1

o

X

n

N

n

Chn mu hthng (chn my mc)

Nd

n

Trong :d: Khong cch chn.N : Slng n vtng thchung.n : Slng n vtng thmu.

Chn mu phn loi (phn loi)

Cng thc tnh sai sbnh qun chn mu:

Cch 1: Chia u slng cc n vca tng thmu cho st( in )

Chn lp Chn khng lp


8/21



Suy rngbnh qun

2 21 i iX

i

N

N n

2 211i i i

X

i i

N n

N n N

Suy rngtl

211 i i if

i

p p N

N n

2111i i i i

f

i i

p p N n

N n N

Trong : i : Sai sbnh qun chn mu ca ti.

iN : Sn vtrong ti.

Cch 2: Chia slng n vca tng thmu theo tlslng n vca tng ttrong tng thchung.Slng n vtng thmu c chia cho stth i l:

ii

Nn n

N

Chn lp Chn khng lp ViSuy rng

bnh qun

2

X n

2

1X

n

n N

2

2 i i

i

N

N

Suy rngtl

1f

p p

n

11f

p p n

n N

11

i i

i

p p Np p

N

Cch 3: Chia slng n vca tng thmu theo tlslng n vca tng ttrong tng thchungv lch chun ca tng ttrong tng thchung. Slng n vtng thmu c chia cho stth il:

i ii

i i

Nn nN

Chn lp Chn khng lp

Suy rngbnh qun

2 21 i i

X

N

N n

2 21

1i i

X

N n

N n N

Suy rngtl

11 i i if

p p N

N n

111

i i i

f

p p N n

N n N

Chn mu ckhi (mu chm)

Cng thc Ch gii

Suyrngbnhqun

2

1

X

X

R r

r R

+) Nu sn vcc khi khng bng nhau:

2

2.i i i

X

i

x x n

n

+)Nu sn vcc khi bng nhau:

2

2 i

X

x x

r

2

X : Phng sai gia cc sbnh qun khi

c chn.

ix : Sbnh qun ca mi khi c chn

(i=1,2,,r).

x : Sbnh qun ca cc khi c chn.


9/21



Suyrng t

l

11

r r

f

f f R r

r R

+)Nu sn vcc khi khng bng nhau:

i i

r

i

f nf

n

+)Nu sn vcc khi bng nhau:

i

r

ff

r

rf : Tlbnh qun ca cc khi c chn.

Vi 1,2,...,i r l tlca mi khi cchn.

Chng IV: KIM NH GITHUYT

Githuyt thng k.

Cp githuyt:

0

1 0

H : Gi thuyt gc

H : Gi thuyt i ca H

TH Cng thc Hnh minh ho Ni dungKimnh 2pha

0 0

1 0

:

:

H

H

Bc bgithuyt 0H khi tham sc trng

ca mu cao hn hoc thp hn so vi githit vtng thchung. Min bc bnm pha tri v pha phi ca ng phn phi.

Kimnhphatri

0 0

1 0

:

:

H

H

Bc bgithuyt 0H khi tham sc trng

ca mu nhhn mt cch ng kso vigi trca githuyt 0H . Min bc bnm

ph tri ca ng phn phi.Kim

nhphaphi

0 0

1 0

:

:

H

H

Bc bgithuyt0

H khi tham sc trng

ca mu ln hn mt cch ng kso vi gitrgithuyt 0H . Min bc bnm pha

phi ca ng phn phi.Sai lm v mc ngha trongkim nh.

Khi la chn gia 2 githuyt 0 1&H H ta c thmc phi 2 loi sai lm sau:

Sai lm loi 1: Bc bgithuyt 0H khi githuyt ny ng. Sai lm loi 2: Tha nhn gia thuyt 0H khi n sai. (nguy him hn v sai li tha nhn l ng).

Kt lunThc t

Chp nhn 0H Chp nhn 1H

0H ng Kt lun ng Sai lm loi 1

0H sai Sai lm loi 2 Kt lun ng

: Mc ngha (Xc sut mc sai lm loi 1).

: Mc ngha (Xc sut mc sai lm loi 2).

1 : Lc lng ca kim nh (Xc sut bc b 0H khi 0H sai)

Thng thng, c ly bng 0,01 (1%), 0,02 (2%), 0,05 (5%), Tmc ngha kim nh c thxc

nh min bc bgithuyt 0H v min tha nhn.

Mi n tha nhn

Mi n tha nhn

Mi n tha nhn


10/21



CC BC TIN HNH KIM NH GITHUYT THNG K:

1. Xy dng cp githuyt: 0H v 1H .2. Xc nh mc ngha (Xc sut mc sai lm loi 1).3. Chn tiu chun kim nh.4. Tnh gi trtiu chun kim nh (z)

Kt lun:

Nu Z min bc b: Bc b 0H , chp nhn 1H . Nu Z min bc b: cha c sbc b 0H (chp nhn githuyt 0H ).

KIM NH V SO SNH STRUNG BNH

*)Kim nh githuyt vgi trtrung bnh ca 1 tng thchung.

1

Phng sai ca tng thchung bit (

2

bit)Tiu chun kim nh:

0

X nZ

i lng (Z) phn phi theo quy lut chun ho N(0,1), c cc TH sau:Kim nhpha phi

0 0

1 0

:

:

H

H

Nu 0,5Z Z Bc b 0H , chp

nhn 1H

Kim nhpha tri

0 0

1 0

:

:

H

H

Nu 0,5 Z Z Bc b 0H , chp

nhn 1H

Kim nhhai pha

0 0

1 0

:

:

H

H

Nu 0,5 2Z Z Bc b 0H , chp

nhn 1H

2

Phng sai ca tng thchung cha bit ( 2 cha bit), mu ln ( 30)n Tiu thc kim nh:

00

X nZ

Vi 20 01

n

n

0 : lch tiu chun mu

iu chnh.

Kt lun: (nh phn trn)

3

Phng sai ca tng thchung cha bit (

2

cha bit), mu nh ( 30)n Tiu thc kim nh:

0

X nt

S

Vi S(hay 0 ): lch

chun mu iu chnh.

i lng ( t) sphn phi theo quy lut Student vi 1n bc tdo, ccc TH sau:

Kim nhpha phi

0 0

1 0

:

:

H

H

Nu , 1 nt t Bc b 0H , chp

nhn 1H

Kim nhpha tri

0 0

1 0

:

:

H

H


nhn 1H

Kim nhhai pha

0 0

1 0

:

:

H

H

Nu 2, 1 nt t Bc b 0H , chp

nhn 1H *)Kim nh 2 gi trtr ung bnh ca 2 tng thchung (2 mu c lp)


11/21



1 bit phng sai ca 2 tng thchung ( 2 21 2

, bit)

Tiu thc kim nh:

1 2

2 2

1 2

1 2

X XZ

n n

2

1 1 1

2

2 2 2

,

,

X N

X N

, c cc TH sau:

Kim nh

pha phi

0 0

1 0

:

:

H

H

Nu 0,5Z Z Bc b 0H ,

chp nhn 1H

Kim nhpha tri

0 0

1 0

:

:

H

H

Nu 0,5 Z Z Bc b 0H ,

chp nhn 1H

Kim nhhai pha

0 0

1 0

:

:

H

H

Nu

0,5 2Z Z Bc b 0H ,

chp nhn 1H

2 bit phng sai ca 2 tng thchung, mu ln ( 2 21 2

, bit; 1 230, 30 n n )

Tiu thc kim nh:

1 2

1 2

2 2

0 0

1 2

X X

Z

n n

Kt lun: (nh phn trn)

3 Cha bit phng sai ca 2 tng thchung, mu nh( 2 21 2, cha bit; 1 230, 30 n n )

Tiu thc kim nh:

1 2

2 2

1 2

X Xt

s s

n n

1 2

2

1 2

1 1

X X

sn n

;

Vi 2s l gi trchung ca 2

phng sai mu1 2

2 2

0 0, :

1 2

2 2

1 0 2 02

1 2

1 1

2

n ns

n n

C2 tng thchung c phn phi chun th tc phn phi Student

vi 1 2 2 n n bc tdo, c cc TH sau:Kim nhpha phi

0 0

1 0

:

:

H

H

Nu

1 2, 2

n nt t

Bc b 0H ,

chp nhn 1H

Kim nh

pha tri0 0

1 0

:

:

H

H

Nu

1 2, 2 n nt t

Bc b 0H ,

chp nhn 1H

Kim nhhai pha

0 0

1 0

:

:

H

H

Nu

1 22, 2

n nt t

Bc b 0H ,

chp nhn 1H

*)Kim nh 2 gi trtr ung bnh ca 2 tng thchung ( 2 mu phthuc)

So snh 1 2, Xt lch trung bnh d .

Tiu chun kim nh: 0

0

d

d nt

Trong :22

2

0 0

.

1 1 1

d d

i

i

dd

n

dn n d

n n n


12/21



2

1 1 1

2

2 2 2

,

,

X N

X N

, c cc TH sau:

Kim nhpha phi

0 0

1 0

:

:

H

H

Nu

, 1 nt t Bc b 0H , chp

nhn 1H Kim nhpha tri

0 0

1 0

:

:

H

H


nhn 1H

Kim nhhai pha

0 0

1 0

:

:

H

H

Nu 2, 1

n

t t

Bc b 0H , chp

nhn 1H

*)Kim nh tl(p)

1 Kim nh githuyt vtlca tng thchung

K p dng: n ln 0 0. 5 1 5 n p n p Tiu thc kim nh:

0

0 01

f p nZ

p p

Vi xn

fn

Ta c cc TH sau:Kim nhpha phi

0 0

1 0

:

:

H

H

Nu 0,5Z Z Bc b 0H , chp

nhn 1H

Kim nhpha tri

0 0

1 0

:

:

H

H

Nu 0,5 Z Z Bc b 0H , chp

nhn 1H

Kim nhhai pha

0 0

1 0

:

:

H

H

Nu

0,5 2Z Z Bc b 0H , chp

nhn 1H

2 Kim nh 2 tlca 2 tng thchung

K p dng: Khi 1 2,n n ln 1 1 1 1 2 2 2 2; 1 ; ; 1 5 n f n f n f n f Zphn phi chun (0,1)N

Tiu thc kim nh:

1 2

1 2

1 11

f fZ

f fn n

Vi:

1 21 1 2 2

1 2 1 2

x x

n nn f n f f

n n n n

xk

k

k

nf

n

Zvn c phn phi xp xchun N(0,1). Vi mc ngha cho trc, tavn kt lun nh trn.

Chng V: Dy sthi gian

SBNH QUN CNG THEO THI GIAN

1

Dy sthi k

1

n

i

i

y

yn

Trong :( 1,2,..., )iy i n :Mc thi.

n : Sthi k.

2 Dy sthi im TH khong cch thi gian bng nhau


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12 1...

2 2

1

nn

yyy y

yn

Trong :( 1,2,..., )iy i n :Mc thi.

n : Sthi k.

TH khong cch thi gian khng bng nhau1

1

.n

i i

in

i

i

y t

y

t

Trong :( 1,2,..., )

i

y i n :Mc thi.

n : Sthi k.

it : khong thi gian c mc iy .

LNG TNG (GIM) TUYT I

1

Lng tng (gim) tuyt i lin hon

1i i iy y iy : Mc tuyt i thi gian

( 2,3,..., )i i n .

1iy : Mc tuyt i thi gian 1i

+) 1i iy y th 0i : Phn nh quy m

hin tng tng.+) 1i iy y th 0i : Phn nh quy m

hin tng gim.

2Lng tng (gim) tuyt i nh gc

1i iy y i

y : Mc tuyt i thi gian ( 2,3,..., )i i n .

1y : Mc tuyt i thi ku.

3

Lng tng (gim) tuyt i bnh qun

2 1

1 1 1

n

i

i n ny y

n n n

Chtiu ny c ngha khi cc lng tng (gim) tuyt i lin honxp xnhau, ngha l trong sut thi knghin cu, hin tng tng(gim) vi mt lng tng i u.

4

Mi lin hgia lng tng (gim) tuyt i lin hon v lng tng (gim) tuyt i nh gc.

1

n

i n

i

TC PHT TRIN

1

Tc pht trin lin hon

1

ii

i

yt

y i


1iy : Mc tuyt i thi gian 1i

2

Tc pht trin nh gc

1

ii

yT

y i


1y : Mc tuyt i thi ku.

3

Tc pht trin bnh qun

1 11 12 3

2 1

. .....n

nn nn ni n i n

i

yt t t t t T

y

Chtiu ny c ngha khi cc tc pht trin linhon xp xnhau, ngha l trong sut thi knghin cu,tc pht trin vi mt lng tng i u.

4

Mi lin hgia tc pht trin lin hon v tc pht trin nh gc.

2

n

i n

i

t T

5

Thng ca 2 tc pht trin nh gc lin nhau

1

ii

i

Tt

T

TC TNG (GIM) TNG I


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1 Tc tng (gim) tng i lin hon1

1 1

1i i ii ii i

y ya t

y y

100i ia t

2 Tc tng (gim) tng i nh gc1

1 1

1i ii iy y

A Ty y

100i iA T

3 Tc tng (gim) tng i bnh qun 1i ia t (ln) 100i ia t (%)

STNG (GIM) TUYT I NG VI 1% TC TNG (GIM) LIN HON

1

1

(%) 100.100

i i ii

ii

i

yg

a

y

STNG (GIM) TUYT I NG VI 1% TC TNG (GIM) NH GC

1 1

1

.ons

.100 100.100

i ii

i i

y yg c t

y

PHNG PHP SBNH QUN DI NG (SBNH QUN TRT)

-Nu tnh sbnh qun trt cho nhm 3 mc , sc:

1

1 2 32 2

2 3 43 3

2 11 1

( ) :

( ) :3

( ) :

3...

( ) :3

( ) :

n n nn n

n

y

y y yy y

y y yy y

y y yy y

y

-Nu tnh sbnh qun trt cho nhm 4 mc , sc:

1

2

1 2 3 43 3

2 3 4 54 4

4 3 2 12 2

3 2 11 1

( ) :

( ) :

( ) :4

( ) :4

...

( ) :4

( ) :4

( ) :

n n n nn n

n n n nn n

n

y

y

y y y yy y

y y y yy y

y y y yy y

y y y yy y

y

Tng tvi cc TH khc.


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PHNG PHP BIN NG THI V

Chsthi v:

0

i

TVi

y

I y

Trong :

TViI : Chsthi vca thi gian i .

iy : Strung bnh cc mc ca cc thi gian cng tn i .

0y : Strung bnh ca tt ccc mc trong dy s.

CC PHNG PHP DON THNG K NGN HN (di 3 nm)

1 Don da vo lng tng (gim) tuyt i bnh qunM hnh don:

.n L ny y L

Vi 11

ny y

n

Trong :

ny : Mc cui cng trong dy sthi gian

: Lng tng (gim) tuyt i bnh qun.L : Thi gian don (tm xa don).

2 Don da vo tc pht trin bnh qunM hnh don:

( )Ln L ny y t

Vi 11

nn

yt

y

Trong :

ny : Mc cui cng trong dy sthi gian

: Lng tng (gim) tuyt i bnh qun.L : Thi gian don (tm xa don).

t: Tc pht trin bnh qun.3 Don da vo hm xu th

-)T Ptr ng thng: xy a bx

-)Vn dng trong dy sthi gian ta c Ptr: ty a bt

-)Xc nh ,a b

CCH 1: p dng phng php bnh phng nhnht gii Hpt:

2

y na b t

ty a t b t

CCH 2:

2

.

t

a y bt

ty t yb

M hnh don: n Ly a b t L

PHN TCH CC THNH PHN CA DY STHI GIAN

+)Kt hp cng: t t t t y f s z .

+)Kt hp nhn: . .t t t t y f s z .

Trong :+) Xu th:

tf.

+) Thi v: ts .

+)Ngu nhin: tz .

+)Hm xu thc dng: . if t t T

0 1tf a a t Vi 1,2,3,...t thtthi gian trong dy s.

1 2

12 1

2.. 1

S na T

m mm n n


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n : Snm.

m : Squ trong nm 4m

0 1

. 1

. 2

T m na a

m n

1 Phn tch cc thnh phn theo kt hp cngSdng cc cng thc sau:

0 1

1

2t j j

ms s y y a j

vi 1,2,3,4j

t t t t z y f s

iT T . iS t T

04

iyy

i

i

yy

n

2 Phn tch cc thnh phn theo kt hp nhnSdng cc cng thc sau:

.t js s H

j

mH

s

tj

t

ys y

tt

t

yy

s

.

tt

t t

yz

f s

Chng VI: Chs

PHNG PHP CHS

+)Phng php chscc th i :

Chsc thvcht lng

1

0

p

pi

p

Stuyt i: 1 0p p p

Trong :

1p : Gi chng ho kbo co.

0p : Gi chng ho kgc.

1q : Lng hng ho bn ra kbo co.

0q : Lng hng ho bn ra kgc.Chsc thv

slng

1

0

q

qi

q

Stuyt i: 1 0q q q

+) Phng php chschung I :

Chstng hp vgi


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1

0

.

.P

p qI

p q

Trong :

PI : Chstng hp gi c.

1p : Gi bn bnh qun 1vsp knghin cu.

0p : Gi bn bnh qun 1vsp kgc.

q : Lng hng ha tiu th.

1.Chstng hp gi cLaspeyres 1 0

0 0

.

.

L

p

p qI

p q

Vi: 1

0

p

pi

p 1 0.pp i p nn ta c:

1 0

0 0

. .

.

pL

p

i p qI

p q

2.Chstng hp gi cPassche1 1

0 1

.

.

P

P

p qI

p q

Vi: 1

0

p

pi

p 10

p

ppi

nn ta c:

1 1

0 1

.

.P

P

p

p qI

p q

i

Chstng hp vlng hng ha tiu th

1

0

.

.q

q pI

q p

Trong :

qI : Chstng hp lng hng tiu th.

1q v 0q : Lng hng ha tiu thknghin cu v kgc.

p : Gi bn bnh qun 1 vhh.1.Chstng hp lng hng hatiu thLaspeyres

1 0

0 0

.

.

L

q

q pI

q p

2.Chstng hp lng hng hatiu thPassche

1 1

0 1

.

.

p

q

q pI

q p

Vi: 1

2

q

qi

q 10

q

qq

i nn ta c:

1 1

1 1

.

.q

q

q p

I q p

i

*)Phng php chslin hp:

1

0

.

.X

X YI

X Y

Trong :

XI : Chschung ca chtiu X.

1X : Chtiu knghin cu.

0X : Chtiu kgc so snh.

Y: Chtiu ghp to ra chtiu lin hp.Quy tc: Trong phn tch thng k, mun nghin cu sbin ng ca chtiu cht lng th quyn sthng l chtiu slng c lin quan c cnh knghin cu. Mun nghin cu sbin ng cachtiu slng th quyn sthng l chie tiu cht lng c lin quan c cnh kgc.


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-)Chslin hp cht lng:

1 1

0 1

.

.p

p qI

p q

1 1 0 0. .pq p q p q 1 1 0 00 0 0 0

. .%

. .

pq

pq

p q p q

p q p q

-)Chslin hp vslng:

0 1

0 0

.

.q

p qI

p q

0 1 0 0. .pq p q p q 0 1 0 00 0 0 0

. .%

. .

pq

pq

p q p q

p q p q

*)Phng php chstheo chtiu bnh qun:

1

0

X

XI

X

Trong :

XI : Chtiu bnh qun ca chtiu X.

1X : Mc bnh qun ca chtiu nghin cu knghin cu.

0X :Mc bnh qun ca chtiu nghin cu kgc.

-)Chtiu gi bnh qun ( p ):

1 1 01

0 0 0 1

.

.pp q qp

Ip p q q

Vi:

1 1 1

1

1 1

.p q pp

q q

; 0 0 00

0 0

.p q pp

q q

1 0p p p

0

% p

pp

Vi knghin cu:

1 0 1.pq p p q

1 0 10 0 0 0

.%

. .

pq

pq

p p q

p q p q

-)Chtiu sn lng bnh qun ( q ):

Mc gi ging nhau

11

00

q

qqI

qq

;

Vi:

1

1

qq

n

; 00

qq

n

Mc gi khc nhau:

1

0q

qI

q

Vi:

1 1

1

1

.p qq

p

; 0 00

0

.p qq

p

Kgc

1 0 0pq q q p 1 0 00 0 0 0

%. .

pq

pq

q q p

p q p q

HTHNG CHS

*)Phng trnh kinh t:

D P Q D : Doanh thu.P: Gi bn.

Q : Sn lng.WQ N Q : Sn lng sn xut.

W : Nng sut lao ng.


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N : Snhn cng, lao ng (ngi).F L N F : Qutin lng.

L : n gi lng trn 1 cng nhn.N : Snhn cng, lao ng (ngi).

C z Q C: Chi ph sn xut.z: Gi thnh n vsn phm.

Q : Sn lng sn xut.* )Hthng chsca chspht trin:

Chspht trin = Chshon thnh khoch Chsnhim vkhoch

Chspht trin doanh thu:

1 1 1 1 1

0 0 1 0 0

. . .

. . .

KH

KH

p q p q p q

p q p q p q

* )Hthng chsca chspht trin:

Chsdoanh thu = Chsgi cChslng bn ra.

.pq p qI I I

1 1 1 1 0 1

0 0 0 1 0 0

. . .

. . .

p q p q p q

p q p q p q

+)Stuyt i:

1 1 0 0 1 0 1 0 1 0 0. . . . . .pq p q p q p q p q p q p q

+)Stng i:

1 1 0 0 1 1 0 1 0 1 0 0

0 0 0 0 0 0

. .% pq

p q p q p q p q p q p q

p q p q p q

Xt TH Qutin lngc 2 nhn tnh hng ta c:

F L NI I I

1 11

0 0 0

.

.F

L NFI

F L N

Phn tch 2 nhn tnh hng:

011 1

0 01 0

F

FF FI

F F F

Vi: 01 0 1.F L N


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1 0 1 1 0 1

0 0 0 1 0 0

. . .

. . .F

L N L N L NI

L N L N L N

+)Stuyt i:

1 0 1 01 01 0F F F F F F F

+)Stng i:

1 0 1 01 01 0

0 0 0

% F F F F F F

FF F F

Xt TH Sn lng sn xut ra chu 3 nhn tnh hng:

1 1 1011

0 0 001 0

W . WW

W . W WQ

N NI

N N

Vi: 1 111

W .W

N

N

0 1

01

1

W .W

N

N

0 0

0

0

W .W

N

N

+)Stuyt i:

1 1 0 0 1 01 1 01 0 1 1 0 1W . W . W W . W W . .WQ N N N N N N

+)Stng i:

0 1 1 01 0 1 1 0 01 1 0 00 0 0 0 0 1 1 0

W W . W W . .WW . W .%

W . W . W . W .

N N N NN NQ

N N N N


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He Thong Cong Thuc Mon Nguyen Ly Thong Ke1

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