Pertemuan 06 Baru-Uji Hipotesis Satu Populasi.pptx

7/25/2019 Pertemuan 06 Baru-Uji Hipotesis Satu Populasi.pptx

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Metode Statistika II

Pertemuan 6 - Uji Hipotesis Rata-

rata

Oleh : Rudi Salam

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Uji Hipotesis Rata-rata

Known

LargeSamples

Unknown

Hypothesis

Tests for

SmallSamples

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Known

LargeSamples

Unknown

HypothesisTests for

SmallSamples

"he test statistic is:

#alculatin$ the "est Statistic

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Known

LargeSamples

Unknown

HypothesisTests for

SmallSamples



&ut issometimes

appro'imatedusin$ a (:

(continued)

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Known

LargeSamples

Unknown

HypothesisTests for

SmallSamples



*"he population must+e appro'imatel,

normal

(continued)

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Re/ie0: Steps in H,pothesis"estin$

. Speci1, the population /alue o1 interest

!. 2ormulate the appropriate null andalternati/e h,potheses

%. Speci1, the desired le/el o1 si$ni3cance

). 4etermine the rejection re$ion

. O+tain sample e/idence and computethe test statistic

6. Reach a decision and interpret theresult

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H,pothesis "estin$ 5'ample

Test the claim that the true mean# of TV sets in US homes is at

least 3.!ssume " $.%&

1. Specify the population value of interest The mean number of TVs in US homes

2. Formulate the appropriate null an alternative

hypotheses

!"# $ !%# & $ 'This is a lo(er tail test) $. Specify the esire level of si*nificance

Suppose that + .", is chosen for this test

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Reject H7 4o not rejectH7

). 4etermine the rejection re$ion

8 .7

'() '*.+,- 7

"his is a one-tailed test 0ith 8 .7.Since " is known9 the cuto /alue is a ( /alue:

Reject H7i1 ( ; (8 -.6) < other0ise do not

reject H7

H,pothesis "estin$ 5'ample(continued)

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. O+tain sample e/idence and computethe test statistic

Suppose a sample is taken 0ith the 1ollo0in$

results: n 8 779 ' 8 !.=) *8 7.= is assumedkno0n

"hen the test statistic is:

H,pothesis "estin$ 5'ample

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.


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Reject H7 4o not rejectH7

8 .7

'*.+,- $

6. Reach a decision and interpret the result

'.$

Since ( 8 -!.7 ; -.6)9 0e reject the nullh,pothesis that the mean num+er o1 "?s inUS homes is at least %


(

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Reject H7

8 .7

.%+%

,

4o not rejectH7 3

n alternate 0a, o1 constructin$ rejection re$ion:

.%,

Since ' 8 !.=) ;!.=6=)9 0e reject thenull h,pothesis


/

Ao0e'pressed in '9not (units

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2.--/1""

".-1./,$

n

zx 00 ===


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p-?alue pproach to "estin$

#on/ert Sample Statistic *e.$. to"est Statistic *Z or t statistic

O+tain the p-/alue1rom a ta+le orcomputer

#ompare the p-/alue0ith

I1 p-/alue ; 9 reject H7

I1 p-/alue

9 do not reject H7

'

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p-?alue pproach to "estin$

p-/alue: Pro+a+ilit, o1 o+tainin$ a

test statistic more e'treme * B or

than the o+ser/ed sample /alue$i/en H7is true

lso called o+ser/ed le/el o1

si$ni3cance

Smallest /alue o1 1or 0hich H7can

+e rejected

(continued)

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p'0alue.$%

8 .7

p-/alue e'ample

5'ample: Ho0 likel, is it to see a samplemean o1 !.=) *or somethin$ 1urther +elo0the mean i1 the true mean is 8 %.7C

.%+%,

3

.%,

/

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."22-2.")'z

1""".-

$."2.-/

z

$.")2.-/x'

=


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#ompare the p-/alue 0ith

I1 p-/alue ; 9 reject H7 I1 p-/alue 9 do not reject

H7

Here: p-/alue 8 .7!!=

8 .7Since .$% 1 .$-2 wereect the nullhypothesis

(continued)p-/alue e'ample

p'0alue

.$%

8 .7

.%+%,

3

.%, [email protected]


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5'ample: Upper "ail ( "est1or Mean *Dno0n

phone industr, mana$er thinks thatcustomer monthl, cell phone +ill ha/eincreased9 and no0 a/era$e o/er E!

per month. "he compan, 0ishes totest this claim. *ssume 8 7 iskno0n

H7: F B ! the a/era$e is not o/er E! permonth

H: F G ! the a/era$e is$reater than E! per

month

*i.e.9 sucient e/idence e'ists to support themana$ers claim

2orm h,pothesis test:

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5 l 2i d R j i


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Reject H74o not rejectH7

Suppose that 8 .7 is chosen 1or this test

2ind the rejection re$ion:

.*$

()*.

%7

4eect H$

Reject H7i1 ( G

.!=

5'ample: 2ind RejectionRe$ion

(continued)

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2indin$ #ritical ?alue - One"ail

Z .7 .7>

. .%>7.%=7 .%=%7

*..%>=7 .)7

.% .)).)6! .)z 7 *.%

.$%

Standard Aormal4istri+ution "a+le*Portion

Jhat is ( $i/en 87.7C

8 .7

5riticalValue

*.%

.6$

.%>>

.*$

.)7.7

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O+tain sample e/idence and compute thetest statistic

Suppose a sample is taken 0ith the1ollo0in$ results: n 8 6)9 ' 8 %. *87

0as assumed kno0n

"hen the test statistic is:

5'ample: "est Statistic(continued)

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Reject H7

.*$

4o not rejectH7 *.%

7

4eect H$

( .%%

#alculate the p-/alue and compare to

(continued)

p'0alue .*%6,

p -?alue Solution

7o not reect H$since p'0alue .*%6, 9

.*$

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.1-3/

.$1".,".--)'z

/1"

,2.",$.1z

,2."),$.1x'

===


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5'ample: "0o-"ail "est*Unkno0n

"he a/era$e cost o1 ahotel room in Ae0

Kork is said to +eE6= per ni$ht. random sample o1 !hotels resulted in '

8 E!.7 and

s 8 E.)7. "est atthe

8 7.7 le/el.

H$:

*+%

H!:

*+%

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5 l S l ti " " il


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8 $.$-

n

-

is unknown2so

use a tstatistic

5ritical Value:

t, ; .$+36

5'ample Solution: "0o-"ail

"est

7o not reect H$:not sucient e/idence

that true mean cost is dierent than E6=

Reject H7Reject H7

L!8.7!

-tL!4o not rejectH7 7

tL!

L!8.7!

'.$+36 .$+36*.,+

H$:

*+% H!:

*+%

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Uji Proporsi Satu Populasi

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Uji Hipotesis untuk Proporsi

Meli+atkan nilai kate$ori

4ua outcome ,an$ mun$kin

NSukses *memiliki karakteristik tertentu

Na$al *tidak memiliki karakteristik tertentuterse+ut

2raksi atau proporsi populasi dalamkate$ori NsuksesQ dinotasikan den$an p


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Proporsi

Proporsi sampel pada kate$ori suksesdin,atakan den$an p

Detika npdan n*-ppalin$ tidak +ernilai

9 pdapat diapproksimasi den$andistri+usi normal den$an rata-rata danstandar de/iasi

(continued)

number of successes in samplepsample size

xn= =

p p=

p

p(1 p)

n

=


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4istri+usisamplin$ dari padalah normal9

sehin$$astatistic ujin,aadalah ( :

Uji Hipotesis untuk Proporsi

np -an!8)67

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Z " 1 P i


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Z "est 1or Proportion:Solution

8 .7

n 8 779 p 8 .7

Reject H7at 8 .7

H$: p .

$% H!: p

.$%

5ritical Values: ;*.6+

Test Statistic:

7ecision:

5onclusion:

z$

4eect

4eect

.7!

.7!

*.6+

-!.)

"here is suciente/idence to rejectthe compan,s claim

o1 = response rate.

'*.6+

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2./4

,""

."-)."-'1

."-.",

n

p)p'1

ppz =

=

=


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4o not rejectH7 4eect H$4eect

H$ L!8 .7!

*.6+7

( '.,=

#alculate the p-/alue and compare to*2or a t0o sided test the p-/alue is al0a,s t0o sided

(continued)

p'0alue .

$*3+:

p -?alue Solution

4eect H$since p'0alue .$*3+ 1

.$-

( .,=

'*.6+

L!8 .7!

.$$+%

.$$+%

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"."1$2'.""-)

./3$2)2'.,

2./4)'x2./4)'z

==

=

+


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Soal

Se+uah uan$ lo$am dilantun !7 kali danmen$hasilkan muka. pakah dari sinicukup ada alasan untuk menolak hipotesis+ah0a uan$ tadi setan$kup dan me,akinitandin$ann,a +ah0a muka muncul kuran$dari 7C Se+utkan p-/alue n,a.


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#hapter Summar,

"erima Dasih

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Pertemuan 06 Baru-Uji Hipotesis Satu Populasi.pptx