hypothesis testing

5/28/2018 Hypothesis testing

1/31

1BBi 5i 5

Bi 5:

Kim nh gi thit

THNG K KINH DOANH


2/31

22BBi 5i 5

NHNG CH CHNH1. Kim nh gi thit l g?

Gi thit thng k v cc loi gi thit Qu trnh kim nh gi thit

Mc ngha v sai lm trong vic ra quyt

nh2. Cc bc kim nh gi thit

3. Kim nh gi thit v gi tr trung bnhLin h gia khong tin cy v kim nh gi thit

4. Kim nh gi thit v t l ca mt tng th

chung


3/31

33BBi 5i 5

GiGi thithitt ll mmtt giginhnh vv mmtt thamtham ss ccaattngng thth chungchung..

flfl

ThamTham ss ll trungtrung bbnhnhhohocc tt ll ccaa ttngng ththchungchung

flfl

ThamTham ss phphii cc xxccnhnh trtrcc khikhi phnphnttchch..

Ti cho rng im trungbnh ca c lp l 8.5!

1984-1994 T/Maker Co.


4/31

44BBi 5i 5

aararagiginhnh((bbngngss))kikimmnhnhVD:VD: SS llngng TVTV trungtrung bbnhnh ccaa cccc giagia nhnh MM tt

nhnhtt ll 33 ((HH00:: 3)3)

BBttuuvviigiginhnhrrngnggigithithittkhngkhnglunlun

lunlun NGNG..((TngTng ttnhnhkhkhii ninimm vv ttii chocho nn khikhi cc

bbngng chchngng chchngng minhminh ll cc ttii))

1.1 Cc loi gi thit:Gi thit khng,H0

Lun lun c du =

Gi thit khng c th hoc khng thbc b.


5/31

55BBi 5i 5

LL ssii llpp vvii gigi thithitt khngkhng

VD:VD: SS llngng TVTV trungtrung bbnhnh trongtrong cccc hhgiagia nhnhMM nhnh hnhn 3 (3 (HH11:: < 3)< 3)

KhngKhng baobao gigicc dduu ==

GiGi thithitt ii cc thth hohocc khngkhng thth chchppnhnhnn

1.1 Cc loi gi thit:Gi thit i, HH11


6/31

66BBi 5i 5

--tt gigi thithitt khngkhng (H0:(H0: 3)3)

--tt gigi thithitt ii ((HH11:: < 3)< 3)

CCcc gigi thithitt loloii trtrllnn nhaunhau vv baobao hhmm tttt cccccc trtrngng hhpp

ii khikhi iinn gigi thithitt ii trtrcc ss dd ddngng hnhn..Thu thThu thp thng tin mp thng tin mu u rrt ra kt ra kt lut lunn

vv gigi thithit ct cn kin kim m nh.nh.

1.2 Qu trnh kim nh gi thit


7/31

77BBi 5i 5

1.3. Qu trnh kim nh gi thit

Tng th chung

Gi nhtui trung bnh

ca tng th chungl 50

(Gi thit khng)

Trung bnhca tng th mul 20

Tng th muBC B

Gi thit khng

Khng,

khng tng t!

20=X c ?50=


8/31

88BBi 5i 5

Trung bnh catng th mu

= 50

Phn b ca tng th muChng ta s ly

mu v tnh ctrung bnh ca

tng th mu lgi tr ny...

... Nu ni rng y ltrung bnh ca tngth chung???

... V vy, chngta bc b gi

thit rng =50.

20H

0

Bc b H0


9/31

99BBi 5i 5

LL xxcc susutt bbcc bb HoHo khikhi nn ngngGGiillmiminnbbccbbccaaphnphnbbmmuu

LLaa chchnn gigi trtr (alpha)(alpha)CCccgigitrtriinnhhnhnh: 0.01, 0.05, 0.10: 0.01, 0.05, 0.10

cc chchnn trtrcc khikhi bbtt uu nghinnghin ccuu aa rara nhnhngng gigi trtr ttii hhnn chocho kikimm nhnh

1.3 Mc ngha


10/31

1010BBi 5i 5

Mc ngha v min bc b

H0: 3H1: < 3

0

0

0

H0: 3H1: > 3

H0: ==== 3H1: 3

/2

Gi trti hn

Min bc b


11/31

1111BBi 5i 5

LoLoii II

BBcc bb gigi thithitt khngkhng khikhi nn ngng

GyGy rara hhuu ququ nghimnghim trtrngng

XXcc susutt saisai ccaa loloii II ll GGii llmmccnghnghaa

LoLoii IIII KhngKhng bbcc bb gigi thithitt khngkhng khikhi nn saisai

XXcc susutt xxyy rara saisai sstt loloii IIII ll(Beta)(Beta)

1.4 Nhng sai lm trong ra quyt nh


12/31

1212BBi 5i 5

1.4.1 Nhng kt qu c th cH0: ng

Hi ng xt x Kim nh gi thit

Tnh hung thc t

Tuyn n V ti C ti Quyt

nh H0 ng H0 sai

V ti ng SaiKhng

bc b

H0

1 - Sai

loi II ( )

C ti Sai ng Bc bH0

Sailoi I( )

Lc lngkim nh

(1 - )

Tnh hung thc t


13/31

1313BBi 5i 5

Gim sc xut sai loi ny s

lm tng sc xut tng loi kia

& c mi lin h ngc chiu


14/31

1414BBi 5i 5

GiGi trtr ngng ccaa thamtham ss ttngng thth chungchungTngTngkhikhichnhchnh llchchgigiaathamthamsskikimm nhnh vvgigitrtrngnggigimm

MMcc nghnghaa TngTngkhikhi gigimm llchch chuchunn ccaa ttngng thth chungchung

TngTngkhikhi tngtng CCmmuunn

TngTngkhikhinngigimm

1.4.2. Nhng nhn t nh hng

ti sai lm loi II,

n


15/31

1515BBi 5i 5

1.1. tt gigi thithittHH00 HH00 :: 33

2.2. tt gigi thithittHH11 HH11 :: < 3 < 3 < 3 < 3 < 3 < 3 < 3 < 3

3.3. ChChnn = .05= .05

4.4. ChChnnnn n = 100n = 100

5.5. ChChnn tiutiu chuchunn kikimm nhnh::kikimmnhnhZZ

((hohoccgigitrtrp)p)

VD: Kim nh vi gi nh s lng TV trung bnh cc h gia nh M t nht l 3.


16/31

1616BBi 5i 5

6.6. XXcc nhnh gigi trtr ttii hhnn Z =Z = --1.6451.645

7. Thu7. Thu ththpp ddliliuu iiuutratra 100100hhgiagianhnh

8.8. TTnhnh totonn kikimm nhnh ththngng kk == --22

9. Ra9. Ra quyquytt nhnh ththngng kk BBccbbgigithithitt khngkhng

10.10. KKtt lulunn SSllngng TVTVtrungtrungbbnhnhcccchhgiagia nhnhMM

ngng llnhnhhnhn3.3.

VD: Kim nh vi gi nh s lng TV trung bnh

cc h gia nh M t nht l 3.


17/31

1717BBi 5i 5

3.1. Kim nh Z khi bit ThayThay ii ththngng kk mmuu (VD: X) sang(VD: X) sang bibinn ZZ

chuchunn hhaa

SoSo ssnhnh vvii gigi trtr ZZ ttii hhnnNNuuththngngkkkikimmnhnhZZtrongtrongmiminnttiihhnn,,

bbccbbH0;H0;ngngcc lliiththkhngkhngbbccbbH0H0

Tiu chun kim nhn

XXZ

X

X

====

====


18/31

1818BBi 5i 5

GiGi nhnh::

** TTngng thth chungchung phnphn bb chuchunn** NNuu khngkhng chuchunn,, ssddngng mmuu llnn

** GiGi thithitt khngkhng chch cc dduu hohocc KiKimm nhnh ZZ::

NNu cha biu cha bit phng sait phng sai, m, mu lu ln thn th ddng ng

llch tiu chuch tiu chun mn mu s thay chou s thay cho

n

xx

z x

x

====

====

a. Kim nh Z mt pha


19/31

1919BBi 5i 5

H0: 0000H1: < 0

H0: 0H1: > 0

Z0

Bc bH0

Z0

Bc bH0

Min bc b


20/31

2020BBi 5i 5

MMtt hhpp ngng cccc trungtrung bbnhnhcc chchaa hnhn 368368 grams?grams?LLyy ngnguu nhinnhin mmtt mmuu

ggmm 2525 hhpp vv chocho kktt ququX = 372.5X = 372.5.. CngCng tyty nhnh rr ll 1515 grams.grams. KiKimm nhnhttii mmcc nghnghaa ==0.050.05..

368 gm.

V d

H0: 368H1: > 368

_


21/31

2121BBi 5i 5

Bng xc sut chun ha

Tm gi tr ti hn

Z .04 .06

1.6 .5495 .5505 .5515

1.7 .5591 .5599 .5608

1.8 .5671 .5678 .5686

.5738 .5750

Z0

Z = 1

1.645

.50-.05

.45

.05

1.9 .5744

Z ch ra iu g khi = 0.05?

= .05

Gi tr ti hn= 1.645


22/31

2222BBi 5i 5

== 0.050.05

nn = 25= 25 GiGi trtr ttii hhnn: 1.645: 1.645

Tiu chun kim nh:

Quyt nh:

Kt lun:

Khng bc b vi = .05

Khng c bng chngchng t 1 hp ng cc trungbnh cha hn 368 grams.

Z0 1.645

.05

Bc b

Tnh ton ra kt lun

H0: 368H1:> 368

50.1====

====

n

XZ


23/31

2323BBi 5i 5

H0: = = = = 368H

1

: 368

MMtt hhpp ngng cccc trungtrung bbnhnh cc

chchaa ngng 368368 grams?grams? LLyy

ngnguu nhinnhin mmtt mmuu ggmm 2525hhpp vv chocho kktt ququ X = 372.5X = 372.5..CngCng tyty nhnh rr ll 1515 grams.grams.

KiKimm nhnh ttii mmcc nghnghaa ==0.050.05..

_

b. Kim nh Z hai phaV d

368 gm.


24/31

2424BBi 5i 5

== 0.050.05

nn = 25= 25 GiGi trtr ttii hhnn:: 1.961.96

Tiu chun kim nh:

Ra quyt nh:

Kt lun:

Khng bc b vi = .05

Khng c bng chngchng t 1 hp ng cc trung

bnh cha ng 368 grams.

V d

Z0 1.96

.025

Bc b

-1.96

.025

H0: = = = = 386H1: 386

50.1

25

153685.372

====

====

====

n

XZ


25/31

2525BBi 5i 5

Vi X = 372.5oz, = 15 v n = 25,Khong tin cy 95% l:

372.5 - (1.96) 15/ 25 n 372.5 + (1.96) 15/ 25Hay: 366.62 378.38

_c. Lin h vi khong tin cy

Vy, nu khong tin cy ny cha gi thit trung bnh

(368), chng ta khng bc b gi thit khng.


26/31

2626BBi 5i 5

vv

GiGi nhnh:: TTngng thth chungchung phnphn bb chuchunn

NNuu khngkhng chuchunn,, chch llchch nhnh vv llyy ccmmuullnn

vv QuQu trtrnhnh kikimm nhnh thamtham ssvv KiKimm nhnh tt

n

S

Xt

====

3.2. Kim nh t khi cha bit ,,,, mu nh(n


27/31

2727BBi 5i 5

V d: Kim nh 1 pha

Mt hp ng cc trung bnhc cha hn 368 grams? Lyngu nhin mt mu gm 36

h

p v cho kt qu

X = 372.5v S=15. Kim nh vi mc

ngha =0.01.368 gm.

H0: 368H1: > > > > 368Cha cho


28/31

2828BBi 5i 5

== 0.010.01

nn

= 36,= 36,

dfdf

= 35= 35

GiGi trtr tti hi hn:n:2.43772.4377

Tiu chun kim nh:

Ra quyt nh:

Kt lun:

Z0 2.4377

.01

Bc b

H0: 368H1: > > > > 368

Khng bc b vi = .01

Khng c bng chngchng minh mt hp ng cctrung bnh c cha hn 368gr

80.1

3615

3685.372====

====

====

nS

Xt

V d: Kim nh 1 pha


29/31

2929BBi 5i 5

LL 11 phnphn ss hohocc %% ccaa ttngng thth chungchung theotheo

tiutiu ththcc nhnh ttnhnh nnoo NNuu tiutiu ththcc nhnh ttnhnh cc haihai bibiuu hihinn vv phnphn

bb theotheo quyquy lulutt nhnh ththcc

BaoBaoggmmccbibiuuhihinncc vvkhngkhngcctheotheotiutiuththccnghinnghinccuu

TT ll ccaa ttngng thth mmuu ((ppss))

mucavnstng

hinbiucvns=

n

X=ps


30/31

3030BBi 5i 5

Vn : 1 cng ty marketing cn phi nhnc 4% li phc p sau khi gi thqua

ng bu in.

Tip cn: kim nh s i hi ny,iutra mt mu ngu nhin gm 500 n v, trong c 25 li phc p.

Gii php: Kim nh ti mc ngha =.05

Kim nh gi thit v t l

ca tng th chung - Kim nh ZV d


31/31

3131BBi 5i 5

= .05= .05

n = 500n = 500

Khng bc b vi = .05

H0: p ==== .04

H1: p .04

Tiu chun kim nh:

Ra quyt nh:

Kt lun:Chng ta khng c bng

chng bc b mc tiu ca

cng ty v t l phc p l 4%.

Z p - p

p (1 - p)

n

s =.04 -.05

.04 (1 - .04)

500

= -1.14

Gi tr ti hn: 1.96

Z0

Bc b Bc b

.025.025

V d

hypothesis testing

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