_bab 11 pengujian hipotesis dua populasi.ppt

8/15/2019 _Bab 11 Pengujian Hipotesis Dua Populasi.ppt

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INFERENSIA STATISTIKA

Bab 11Selang Kepercayaan dan

Pengujian Hipotesis Dua Populasi


2/50

Selang Kepercayaan dan

Pengujian Hipotesis:Inferensia untuk Dua Populasi

Inferensia Statistika

Terhadap Dua Populasi


3/50

Pendugaan Mean DuaPopulasi

Pendugaan Dua

Populasi

Mean

Populasi,

independent

samples

Sampel

Berpasangan/

Paired

samples

Proporsi

Populasi

Group 1 vsindependentGroup !

Same group"e#ore vs a#tertreatment

Proportion 1 vs Proportion !

$%amples:


4/50

n erens a untu e a uaMean

Mean Populasi,

Sampel

"e"as/independent

samples

&1 dan &! di'eta(ui

&1 dan &! un'no)n,n1 dan n! ≥ *+

&1 dan &! un'no)n,n1 atau n! *+

Goal: Mem"uat Selang

Kepercayaan untu' "eda

dua mean, -1 . -!

Penduga titi' untu'

"eda dua mean:

%1 . %!


5/50

ampe e asIndependent Samples

Mean Populasi,

independent

samples

&1 dand &! di'eta(ui



Sum"er data "er"eda 0ida' "er(u"ungan Saling "e"as

Sampel terpili( darisatu populasi tida'"erpengaru( ter(adapsampel terpili( daripopulasi lainnya

n#erensia dila'u'an ter(adap"eda ! mean sampel

Guna'an distri"usi 2 ataudistri"usi t3student, denganragam ga"ungan 4pooledvariance5


6/50

Mean Populasi,

independent

samples

&1 dan &! di'eta(ui



SK Beda Mean,Bila σ1 dan σ2 diketahui

6sumsi::

Sampel diam"il secara aca'

dan "e"as

Distri"usi populasi normal

atau u'uran 'edua sampel ≥ *+

Simpangan "a'u populasi

di'eta(ui


7/50

Mean Populasi,

independent

samples

&1 dan &! di'eta(ui



7dan t(e standard error dari

%1 . %! adala(

Bila &1 dan &! di'eta(ui dan 'edua

populasi "erdistri"usi normal atau

u'uran 'edua sampel minimal 8 *+

, guna'an statisti' uji 23value7

!

!

!

1

!

1

%% n

&

n

&&

!1+=−


S d


8/50

Population means,

independent

samples

&1 dan &! di'eta(ui



( )!

!

!

1

!

1/!!1

n

&

n

&2%% +±− α

Selang Kepercayaan

413951++ "agi -1 . -!:



9/50

Population means,

independent

samples

&1 dan &! di'eta(ui



Bila σ1 dan σ2 tidak diketahui, large

samples

6sumsi:

Sampel diam"il secara aca'

dan "e"as

sample si2es

are ≥ *+ Simpangan "a'u populasi tida'

di'eta(ui da, &1 8 &! 8& 4ragam (omogen5


10/50

Population means,

independent

samples

&1 dan &! di'eta(ui



Mem"uat Selang

Kepercayaan:

use sample standard

deviation s to estimate &

t(e test statistic is a 2 value


samples


11/50

Population means,

independent

samples

&1 dan &! di'eta(ui



( )!

!

!

1

!

1/!!1

n

s

n

s2%% +±− α

0(e con#idence interval #or

-1 . -! is:


samples

t h SK B d M


12/50

ontoh SK Beda Mean,Bila σ1 dan σ2 !idak Diketahui

Contoh ;jian suatu mata 'ulia( di"eri'an pada ,

sedang'an ma(asis)i memili'i mean =? dengan

simpangan "a'u ? 0entu'an selang 'epercayaaan @? "agi "eda mean ujianA

a)a":

n1 dan

n! cu'up "esar 4C*+5

8 >! 3 =? 8 ?

S1 8 > dan S! 8 ?

98 +,+ dan 9/! 8 +,+!, se(ingga 2 E9/! 8 !,+<

%1 . %!

t h SK B d M


13/50


adi selang 'epercayaan @? "agi µ1 3 µ! adala(:

F Dengan demi'ian, dengan ting'at 'epercayaan @? "edamean nilai ujian antara ma(asis)a dan ma(asis)i se"enarnya

terleta' di antara *,* dan >,


14/50

Population means,

independent

samples

&1 and &! 'no)n

&1 and &

! un'no)n,

n1 and n! ≥ *+

&1 and &! un'no)n,n1 or n! *+

6ssumptions:

populations are normally

distri"uted

t(e populations (ave eual

variances

samples are independent

SK Beda Mean,Bila σ1 dan σ2 tidak diketahui, Small Samples


15/50

Population means,

independent

samples

&1 dan &! di'eta(ui

&1 dan &

! un'no)n,

n1 adan n! ≥ *+


orming interval

estimates:

0(e population variancesare assumed eual, so use

t(e t)o sample standard

deviations and pool t(em to

estimate &

t(e test statistic is a t value

)it( 4n1 I n! . !5 degrees

o# #reedom



16/50

Population means,

independent

samples

&1 dan &! di'eta(ui

&1 dan &

! un'no)n,

n1 dan n! ≥ *+


0(e pooled standard

deviation is

( ) ( )

!nn

s1ns1ns

!1

!

!!

!

11p

−+

−+−=



17/50

Population means,

independent

samples

&1 dan &! di'eta(ui

&1 dan &

! un'no)n,

n1 dan n! ≥ *+


( )!1

p/!!1

n

1

n

1st%% +±− α


-1 . -! is:

J(ere tα/! (as 4n1 I n! . !5 d#,

and( ) ( )

!nn

s1ns1ns

!1

!

!!

!

11p −+

−+−=


ontoh SK Beda Mean


18/50


onto(: ;ji la"oratorium dila'u'an untu' menguji sta"ilitas dan

permea"ilitas pondasi "eton yang ter"uat dari "a(an campuran

aspal Dua jenis pondasi "eton masing3masing ter"uat dari "a(an

campuran yang mengandung * dan = aspal Data (asil

pengujian permea"ilitas pondasi 4inci per jam5 ter(adap empatspecimen atau cupli'an pondasi "eton disaji'an di "a)a( ini

0entu'an selang 'epercayaan @@ >+ 1+!+ @>+

Kandungan = 6spal: ><

ontoh SK Beda Mean


19/50


a)a":

n1 8 dan n! 8 4'eduanya *+5

8 411>@ I 7 I @>+5/ 8 1++=,!<

8 4>1=,=<

s1! 8 !+?*?,@!

s!! 8


20/50


98 +,+< dan 9/! 8 +,+! dan *>=F Karena selang 'epercayaan terse"ut menca'up +, ma'a tida'

dapat disimpul'an "a()a permea"ilitas 'edua jenis pondasi

"eton terse"ut "er"eda

( ) 21 p/221

n

1

n

1sxx +±− α t

4

1

4

1)1443,114(447,2)75,81725,1007( +±−

ontoh SK Beda Mean


21/50


onto(: Peneliti mela'u'an perco"aan untu' mengeta(ui 'ee#e'ti#an

suatu jenis o"at ter(adap jumla( cacing yang ada dalam perut dom"a

Setela( dom"a se"anya' 1 diin#e'si cacing, secara aca' separu(nya

di"eri o"at dan separu( lainnya se"agai 'ontrol Setela( ? minggu, perut

dom"a di"eda( dan jumla( cacing yang ada di(itung dan dicatat seperti

"eri'ut:

Dengan o"at: 1> * !>


22/50


a)a":

n1 8 = 4'ontrol5 dan n! 8 = 4di"eri o"at5 4'eduanya *+5

8 4+ I 7 I *@5/= 8 +

8 41> I 7 I 1*5/= 8 !>,


23/50


98 +,+< dan 9/! 8 +,+!


24/50

"#i $ipotesis bagi Beda DuaMean

Pengujian (ipotesis tentang -1 . -! Guna'an situasi yang serupa dengan SK:

Simpangan "a'u di'eta(ui atau'a( tida'

di'eta(ui

;'uran sampel atau Sample si2es n ≥ *+ atau'a( n *+


25/50

"#i $ipotesis bagi -1 . -!

Mean populasi, sampel "e"as

&1 dan &! di'eta(ui



Guna'an statisti' uji !

Guna'an s untu' menduga &,

de'ati dengan statisti' uji !

Guna'an s untu' menduga&, guna'an degan statisti'

uji t dan simpangan "a'u

ga"ungan

po es s ag - -


26/50

Mean populasi,

sampel "e"as

&1 dan &! di'eta(ui

&1

dan &!

un'no)n,n1 dan n! ≥ *+


( ) ( )

!

!

!

1

!

1

!1!1

n&

n&

--%%2

+

−−−=

Statisti' uji "agi

-1 . -! adala(:

po es s ag -1 . -! "ila σ1 dan σ2 diketahui

Daerah Kritis/Daerah Penolakan


27/50

Dua mean populasi, sampel "e"as

Lo)er tail test:

H+: -1 . -! 8 +

H 6: -1 . -! +

;pper tail test:

H+: -1 . -! 8 +

H 6: -1 . -! C +

0)o3tailed test:

H+: -1 . -! 8 +

H 6: -1 . -! +

/! /!

32α 32α/!2α 2α/!0ola' H+ i# 2 32α 0ola' H+ i# 2 C 2α 0ola' H+ i# 2 32α/!

or 2 C 2α/!

Daerah Kritis/Daerah Penolakan"#i $ipotesis bagi %1 & %2


28/50


29/50

%a&ab

Step 1 State the null and alternate h'potheses(

4'ey)ord: Olonger t(an5

H+: QS 8 Q;H1: QS C Q;

Step ) Sele#t the le*el of si$nifi#an#e(

0(e +1 signi#icance level is stated in t(e pro"lem

"#i $ipotesis bagi -1 . -! "ila σ

1

dan σ2

diketahui


30/50

Step + Deter"ine the appropriate test statisti#( Because "ot( population standard deviations are 'no)n, )e can use z-distribution as t(e test statistic

Step , For"ulate a de#ision rule(

Neject H+ i# R C Rα

R C !**

Step - Co"pute the *alue of z and "ake a de#ision

0(e computed value o# *1* is larger t(an t(e critical value o# !** ur decision is to reject t(e null (ypot(esis 0(e

di##erence o# !+ minutes "et)een t(e mean c(ec'out time using t(e standard met(od is too large to (ave occurred "y

c(ance Je conclude t(e ;3Scan met(od is #aster13.3064.0

2.0

100

30.0

50

40.0

3.55.5

22

22

==

+

−=

+

−=

u

u

s

s

u s

nn

X X z

σ σ

"#i $ipotesis bagi -1 . -! "ila σ

1

dan σ2

diketahui

"#i $ipotesis bagi "il σ dan σ


31/50

Mean populasi,

sampel "e"as

&1 dan &! di'eta(ui

&1

dan &!



Statisti' uji "agi

-1 . -! adala(:

( ) ( )

!

!

!

1

!

1

!1!1

ns

ns

--%%2

+

−−−=

"#i $ipotesis bagi -1 . -! "ila σ1 dan σ2

tidak diketahui, ukuran sampel besar

"#i $ipotesis bagi "ila σ dan σ


32/50

Mean populasi,

sampel "e"as

&1 dan &! di'eta(ui

&1

dan &!



Dimana tα/! memili'i derajat "e"as 4n1

I n! . !5 dan simpangan "a'u

ga"ungan:

( ) ( )!nn

s1ns1n

s!1

!

!!

!

11

p −+

−+−

=

( ) ( )

21

p

2121

n

1

n

1s

μμxxt

+

−−−=

Statisti' uji "agi

-1 . -! adala(:


tidak diketahui, ukuran sampel ke'il

"#i $ipotesis bagi - - "ila σ dan σ


33/50

Contoh)ens La)n are, nc, manu#actures and assem"les la)nmo)ers t(at are

s(ipped to dealers t(roug(out t(e ;nited States and anada 0)o di##erent

procedures (ave "een proposed #or mounting t(e engine on t(e #rame o# t(e

la)nmo)er 0(e uestion is: Is there a differen#e in the "ean ti"e to

"ount the en$ines on the fra"es of the la&n"o&ers.

0o evaluate t(e t)o met(ods, it )as decided to conduct a time and motion

study 6 sample o# #ive employees )as timed using t(e Jelles met(od and

si% using t(e 6t'ins met(od 0(e results, in minutes, are s(o)n "elo):

s t(ere a di##erence in t(e mean mounting timesT ;se t(e 1+ signi#icancelevel





34/50

Step 1 State the null and alternate h'potheses(

4Key)ord: Os t(ere a difference5

H+: Q1 8 Q!

H1: Q1 Q!

Step ) State the le*el of si$nifi#an#e(

0(e +1+ signi#icance level is stated in t(e pro"lem

Step + Find the appropriate test statisti#(

Because t(e population standard deviations are not

'no)n "ut are assumed to "e eual, )e use t(e

pooled t 3test

Step , State the de#ision rule(

Neject H+ i# t C tα/!,n1In!3! or t 3 tα/!, n1In!3!

t C t+**

Step - Co"pute the *alue of t and "ake a de#ision

0(e decision is not to re/e#t the null h'pothesis,

"ecause 0(22) falls in the re$ion bet&een 01(3++ and

1(3++

Je conclude t(at t(ere is no di##erence in t(e mean times

to mount t(e engine on t(e #rame using t(e t)o met(ods

3+??!





35/50

UouVre a #inancial analyst #or a "ro'erage #irm s t(ere a

di##erence in dividend yield "et)een stoc's listed on t(e

WUS$ X W6SD6YT Uou collect t(e #ollo)ing data:

N4SE NASDA5

Nu"ber )1 )-

Sa"ple "ean +()6 )(-+

Sa"ple std de* 1(+ 1(12

6ssuming eual variances, is

t(ere a di##erence in average

yield 4 7 (-5T





36/50

( ) ( ) ( ) ( )1!!


37/50

8 91 0 9) 7 i(e( :917 9);

8A 91 0 9) < i(e( :91 < 9);

7 (-

df 7 )1 = )- 0 ) 7 ,,Nilai kritis t 7 > )(1-,

Statistik u/i De#ision

Con#lusion !olak $( pada α )(*(+

!here is eiden'e ofa di-eren'e inmeans*

t + !+1


38/50

0(e test statistic #or d isPaired

samples

1n

5d4d

s

n

1i

!

i

d−

−

=

∑=

n

s -dt dd−=

J(ere tα/! (as n 3 1 d#

and sd is:

n is t(e

num"ero# pairs in

t(e

paired

sample

$.pothesis !esting forPaired Samples

$ th i ! ti f


39/50

Lo)er tail test:

H+: -d 8 +H 6: -d +

;pper tail test:

H+: -d 8 +H 6: -d C +

0)o3tailed test:

H+: -d 8 +H 6: -d +

Paired Samples

$.pothesis !esting forPaired Samples

/! /!

3tα 3tα/!tα tα/!Neject H+ i# t 3tα Neject H+ i# t C tα Neject H+ i# t 3tα/2

or t C tα/2 J(ere t (as n 3 1 d#

(continued)


40/50

6ssume you send your salespeople to a Ocustomerservice training )or's(op s t(e training e##ectiveT

Uou collect t(e #ollo)ing data:

Paired Samples 0ample

Nu"ber of Co"plaints :); 0 :1;Salesperson Before :1; After :); Differen#e? d

i

C(B( 2 , 0 )

T(F( ) 2 01,

@(8( + ) 0 1R(K(

@(( , 0 ,

0)1

d = Σ din


41/50

Has t(e training made a di##erence in t(e num"er o#

complaints 4at t(e ++1 level5T

- !d 8

1??


42/50

!o Population Proportions

Goal: orm a con#idence interval #or

or test a (ypot(esis a"out t(e

di##erence "et)een t)o population

proportions, p1 . p!

0(e point estimate #or

t(e di##erence is p1 . p!

Population

proportions

6ssumptions:

n1p1 ≥ < , n1413p15 ≥ <

n!p! ≥ < , n!413p!5 ≥


43/50

on3den'e Interal for !o Population Proportions

Population

proportions

( ) !!!

1

11/!!1

n

5p41p

n

5p41p

2pp

−+

−±− α


p1 . p! is:

$ th i ! t f


44/50

$.pothesis !ests for !o Population Proportions

Population proportions

Lo)er tail test:

H+: p1 8 p!H 6: p1 p!

ie,

H+: p1 . p! ≥ +

H 6: p1 . p! +

;pper tail test:

H+: p1 8 p!H 6: p1 C p!

ie,

H+: p1 . p! Z +

H 6: p1 . p! C +

0)o3tailed test:

H+: p1 8 p!H 6: p1 p!

ie,

H+: p1 . p! 8 +

H 6: p1 . p! +


45/50


Population

proportions

!1

!1

!1

!!11

nn

%%

nn

pnpn

p +

+

=+

+

=

0(e pooled estimate #or t(eoverall proportion is:

)(ere %1 and %! are t(e num"ers #rom

samples 1 and ! )it( t(e c(aracteristic o# interest

Since )e "egin "y assuming t(e null

(ypot(esis is true, )e assume p1 8 p!

and pool t(e t)o p estimates


46/50


Population

proportions

( ) ( )

+−

−−−=

!1

!1!1

n1

n15p14p

pppp2

0(e test statistic #or

p1 . p! is:

$.pothesis !ests for


47/50

$.pothesis !ests for !o Population Proportions

Population proportions

Lo)er tail test:

H+: p1 . p! 8 +

H 6: p1 . p! +

;pper tail test:

H+: p1 . p! 8 +

H 6: p1 . p! C +

0)o3tailed test:

H+: p1 . p! 8 +

H 6: p1 . p! +

/! /!

32α 32α/!2α 2α/!Neject H+ i# 2 32α Neject H+ i# 2 C 2α Neject H+ i# 2 32α/2

or 2 C 2α/2

l


48/50

0ample !o population Proportions

s t(ere a signi#icant di##erence "et)een t(e

proportion o# men and t(e proportion o#

)omen )(o )ill vote Ues on Proposition 6T

n a random sample, *? o# =! men and *1 o#


49/50

0(e (ypot(esis test is:

H+: p1 . p! 8 + 4t(e t)o proportions are eual5

H 6: p1 . p! + 4t(ere is a signi#icant di##erence "et)een proportions5

0(e sample proportions are:

Men: p1 8 *?/=! 8


50/50

0(e test statistic #or p1 . p! is:

0ample !o population Proportions

+!<

31@? 1@?

+!<

31*1

De#ision Do not re/e#t H

Con#lusion There is not

si$nifi#ant e*iden#e of a

differen#e in proportions

&ho &ill *ote 'es bet&een

( )

( ) ( )1*1

_bab 11 pengujian hipotesis dua populasi.ppt

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