Business

Statistics:

ACourse6th

EditionChapter

10Two-Sample

TestsandOne-Way

ANOVA兩樣本檢驗和單邊方差分析Copyright

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10-1Copyright

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10-2Learning

ObjectivesIn

this

chapter,

you

learnHow

to

use

hypothesis

testing

for

comparing

thedifferencebetween:如何利用統計假設來比較兩個不一樣的：The

means

of

two

independent

populations

兩個獨立的總體的均值The

means

of

two

related

populations

兩個相關的總體的均值The

proportions

of

two

independent

populations兩個獨立的總體的比例The

variances

of

two

independent

populations兩個獨立的總體的方差The

means

of

more

than

two

populations

兩個以上總體的均值Two-Sample

TestsPopulationMeans,IndependentSamplesPopulationMeans,RelatedSamplesPopulationVariancesGroup

1

vs.Group

2Samegroupbefore

vs.

aftertreatmentVariance

1

vs.Variance

2Examples:PopulationProportionsProportion

1

vs.Proportion

2DCOVATwo-Sample

TestsCopyright

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10-3Difference

Between

Two

MeansDCOVAPopulation

means,independent

samples獨立樣本總體均值檢驗Goal:Test

hypothesis

or

form

檢驗的假設或形式：a

confidence

interval

for

thedifference

between

two

populationmeans兩總體均值差異的置信區間,μ1

–μ2The

point

estimate

for

thedifference

isX1

–

X2*σ1

and

σ2

unknown,assumed

equalσ1

and

σ2

unknown,not

assumed

equalCopyright

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10-4Difference

Between

Two

Means:Independent

SamplesPopulation

means,independentsamples*Use

Sp

to

estimate

unknownσ.Use

a

Pooled-Variance

ttest.混合方差t檢驗σ1

and

σ2

unknown,assumed

equalσ1

and

σ2

unknown,not

assumed

equalUse S1

and

S2

to

estimate

unknownσ1

and

σ2.Use

a

Separate-Variancet

test.不同方差t檢驗Different

data

sourcesUnrelatedIndependent

兩總體間不相關Sample

selected

from

oneCopyright

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10-5population

has

no

effect

on

thesample

selected

from

the

otherpopulationDCOVACopyright

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10-6Hypothesis

Tests

forTwo

Population

Means兩總體均值假設檢驗Lower-tail

test:左側檢驗

H0:μ1

μ2H1:μ1

<μ2i.e.,H0:

μ1

–

μ2

0H1:

μ1

–

μ2

<

0Upper-tail

test:右側檢驗

H0:μ1

≤μ2H1:μ1

>μ2i.e.,H0:

μ1

–

μ2

≤

0H1:

μ1

–

μ2

>

0Two-tail

test:雙側檢驗

H0:μ1

=μ2H1:

μ1

≠

μ2i.e.,H0:

μ1

–

μ2

=

0H1:

μ1

–

μ2

≠

0Two

Population

Means,

Independent

SamplesDCOVATwo

Population

Means,

Independent

SamplesLower-tail

test:H0:

μ1

–

μ2

0H1:

μ1

–

μ2

<

0Upper-tail

test:H0:

μ1

–

μ2

≤

0H1:

μ1

–

μ2

>

0Two-tail

test:H0:

μ1

–

μ2

=

0H1:

μ1

–

μ2

≠

0-ttReject

H0

if

tSTAT

<

-tReject

H0

if

tSTAT

>

t

/2

/2Copyright

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10-7-t/2

t/2Reject

H0

if

tSTAT

<

-t/2or

tSTAT

>

t/2Hypothesis

tests

for

μ1

–

μ2DCOVAPopulation

means,independentsamplesAssumptions:Samples

are

randomlyandindependently

drawn樣品隨機，獨立抽取Populations

are

normallydistributed

or

both

samplesizes

are

at

least30人群呈正態分布或兩個樣本量均至少為30Population

variances

areunknown

but

assumed

equal總體方差未知，但假設他們相等Hypothesis

tests

for

μ1

-

μ2

with

σ1and

σ2

unknown

and

assumed

equalDCOVA*σ1

and

σ2

unknown,assumed

equalσ1

and

σ2

unknown,not

assumed

equalCopyright

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10-8Population

means,independentsamples混合方差(continued)The

pooled

variance

is:Where

tSTAThas

d.f.

=

(n1

+

n2

–

2)1

22221

1

2p(n

1)

(n

1)n

1S

n

1

SS2*

?

The

test

statistic

is:σ1

and

σ2

unknown,assumed

equalσ1

and

σ2

unknown,not

assumed

equalHypothesis

tests

for

μ1

-

μ2

withσ1and

σ2

unknown

and

assumed

equal2Copyright

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10-911

2STAT1 2

p

n

nS2

1

1

X

X

μ

μ

t

Population

means,independentsamples

2p

n1

n2

1

1/2X1

X2

tS置信區間The

confidence

intervalforμ1

–μ2

is:Where

tα/2

has

d.f.

=

n1

+n2

–

2*Confidence

interval

for

μ1

-

μ2

with

σ1and

σ2

unknown

and

assumed

equalDCOVAσ1

and

σ2

unknown,assumed

equalσ1

and

σ2

unknown,not

assumed

equalCopyright

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10-10Pooled-Variance

t

Test

Example東北游客游客Sample

Size2125Sample

mean3.272.53Sample

stddev1.301.16是否存在差異？DCOVA據說來自和東北的游客，在家界游玩的滿意度存在差異，旅游學院老師希望通過問卷檢驗這一結論的真實性，收集了以下數據:Assuming

both

populations

areapproxima y

normal

with

equal

variances,

istherea

difference

in

mean

yield

(

=

0.05)?假設兩總體呈正態分布，方差相同，請問兩總體均值Copyright

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10-11Pooled-Variancet

Test

Example:Calculating

the

Test

Statistic(continued)1.50212122211

2(n

1)

(n

1) (21

-

1)

(25

1)n

1S

n

1S

21

11.302

25

11.162

S

2P3.27

2.53

02.040t

X1

X2

μ1

μ2

Copyright

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10-12STAT1.5021

1

21

25

1

S

n1

n2

2p1

1

The

test

statistic

is:H0:

μ1

-

μ2

=

0 i.e.

(μ1

=

μ2)H1:

μ1

-

μ2

≠

0 i.e.

(μ1

≠μ2)DCOVAPooled-Variancet

Test

Example:Hypothesis

Test

SolutionH0:

μ1

-

μ2

=

0 i.e.

(μ1

=

μ2)H1:

μ1

-

μ2

≠0 i.e.

(μ1

≠μ2)

=

0.05df

=

21

+

25

-

2

=

44Critical

Values:

t

=

±

2.0154Test

Statistic:Decision:Reject

H0

at

=

0.05Conclusion:There

is

evidence

of

adifference

in

means.t02.0154-2.0154.025Reject

H0.0252.0402.040Copyright

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10-13DCOVAReject

H03.27

2.53STATt

21

25

1.5021

1

1

Pooled-Variancet

Test

Example:Confidence

Interval

for

μ1

-

μ2Copyright

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10-14DCOVASince

we

rejected

H0

can

webe

95%

confident

that

μDB

>μGD?95%

Confidence

Interval

for

μDB

-μGDSince

0

is

less

than

the

entire

interval,

wecan

be

95%confident

that

μNYSED

>

μGDSp

n1

n2

2

1 1

0.74

2.0154

0.3628

(0.009,

1.471)X1

X2

t/2Population

means,independentsamplesHypothesis

tests

for

μ1

-

μ2

withσ1and

σ2

unknown,

not

assumed

equal*σ1

and

σ2

unknown,assumed

equalσ1

and

σ2

unknown,not

assumed

equalDCOVACopyright

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10-15Assumptions:Samples

arerandomly

andindependently

drawn樣品隨機，獨立抽取Populations

arenormallydistributed

or

both

samplesizes

are

at

least

30人群呈正態分布或兩個樣本量均至少為30Population

variances

areunknownand cannot

beassumed

tobe

equal總體方差未知，不能假設他們相等Population

means,independentsamples*σ1

and

σ2

unknown,assumed

equalσ1

and

σ2

unknown,Hypothesis

tests

for

μ1

-

μ2

with

σ1

andnot

assumed

equalCopyright

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10-16σ2

unknown

and

not

assumed

equal(continued)DCOVAThe

formulae

for

this

test

arenot

covered

in

this

book.See

reference

8

from

thischapter

for

moredetails.This

test

utilizes

twoseparatesample

variances

to

estimatethe

degrees

of

freedom

for

thet

test

Or,

if

not

Normal,use

large

samples

大樣本Copyright

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10-17Related

Populations

相關總體The

Paired

Difference

TestDCOVATests

Means

of

2

Related

PopulationsPaired

or

matched

samples

配對樣本，期望與滿意度Repeated

measures(before/after)重復檢驗，事前事后

Use

difference

between

paired

values:兩個配對值之間的差異Eliminates

Variation

Among

Subjects消除兩主體之間的變異Assumptions:Both

Populations

Are

Normally

Distributed

正態分布RelatedsamplesDi

=

X1i

-X2iRelated

PopulationsThe

Paired

Difference

TestThe

ith

paired

difference

is Di

,

whereRelatedsamplesDi

=

X1i

-

X2iThe

point

estimate

for

the

paireddifference

population

mean

μD

is匹配均值差異的點估計D:n

D

iD

i

1

nn

is

the

number

of

pairs

in

the

paired

sampleN是匹配樣本的成對的數量n2(D

D)SD

i

i1

n

1The

sample

standarddeviation樣本標準差is

SDCopyright

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10-18DCOVA(continued)The

test

statistic

for

μD

is:Copyright

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10-19PairedsamplestSTAT

D

μ

DSDnWhere

tSTAT

has n

-

1

d.f.The

Paired

Difference

Test:Finding

tSTATDCOVALower-tail

test:H0:

μD

0H1:

μD

<

0Upper-tail

test:H0:

μD

≤

0H1:

μD

>

0Two-tail

test:H0:

μD

=

0H1:

μD

≠

0PairedSamplesThe

Paired

Difference

Test:Possible

Hypotheses-ttReject

H0

if

tSTAT

<

-tReject

H0

if

tSTAT

>

t

/2/2-t/2

t/2Reject

H0

if

tSTAT

<

-tor

tSTAT

>

tWhere

tSTAT

has n

-

1

d.f.Copyright

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10-20DCOVAThe

confidence

interval

for

μD

isPairedsamplesSD

Copyright

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10-21n

2i(D

D)

i1

n

1SDn

/

2D

twhereThe

Paired

DifferenceConfidence

IntervalDCOVA旅游有助于提高員工的服務質量，降量。我們收集了員工旅游前和旅游后一通常認為低顧客的個月的顧客量，數據如下，請問 旅游對員工服務質量的是否存在顯著影響：Paired

Difference

Test匹配t檢驗:ExampleSalesNumber

of

Complaints:Before

(1)

After

(2)(2)

-

(1)Difference,

DiC.B.T.F.M.H.R.K.M.O.62030446200-

2-14-

10-

4-21D

=

Din

5.672(D

D)

in

1DS

=

-4.2DCOVACopyright

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10-22Has

the

employees’

travel

experience

made

a

difference

inthe

number

of

complaints

(at

the

0.01

level)?1.66SD/

n

5.67/

5

D

μD

4.2

0

tSTATH0:

μD

=0H1: μD

0/2-

4.604

4.604Decision:

Do

not

rejectH0(tstat

is

not

in

the

reject

region)Conclusion:

There

isinsufficient

evidence

thereis

significant

change

in

thenumber

ofcomplaints.

=

.01 D

=

-

4.2t0.005

=

±

4.604d.f.

=

n

-

1

=

4Test

Statistic:Paired

Difference

Test:SolutionReject

Reject/2-

1.66DCOVACopyright

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10-23Two

Population

Proportions兩總體比例Copyright

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10-24difference

between

two

populationproportions,

兩總體比例的差異DCOVAGoal:

test

a

hypothesis

or

form

檢驗的假設：a

confidence

interval

for

theπ1

–

π2The

point

estimate

for

thedifference

這個差異的點估計isPopulationproportionsp1

p2Two

Population

Proportions兩總體比例Populationproportionsp

X1

X2n1

n2Copyright

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10-25In

the

null

hypothesis

we

assume

thenull

hypothesis

is

true,

so

we

assume

π1=

π2

and

pool

the

two

sampleestimates零假設：假設兩總體比例相等The

pooled

estimate

for

theoverallproportion總的比例的混合估計is:where

X1

and

X2

are

the

number

of

items

ofinterest

in

samples

1

and

2DCOVATwo

Population

ProportionsPopulationproportions

n1

n2

p

(1

p)1

1

p

p

π

π

12

1

2ZSTATThe

test

statistic

forπ1

–

π2is

a

Z

statistic:(continued)2Copyright

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10-262111n

n2p

X1

X2n,

p

X

2n,

p

X1whereDCOVACopyright

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10-27Hypothesis

Tests

forTwo

Population

ProportionsPopulation

proportionsLower-tail

test:H0:

π1

π2H1:

π1

<

π2i.e.,H0:

π1

–

π2

0H1:

π1

–

π2

<

0Upper-tail

test:H0:

π1

≤

π2H1:

π1

>

π2i.e.,H0:

π1

–

π2

≤0H1:

π1

–

π2

>

0Two-tail

test:H0:

π1

=

π2H1:

π1

≠

π2i.e.,H0:

π1

–

π2

=

0H1:

π1

–

π2

≠

0DCOVAHypothesis

Tests

forTwo

Population

ProportionsPopulation

proportionsLower-tail

test:H0:

π1

–

π2

0H1:

π1

–

π2

<

0Upper-tail

test:H0:

π1

–

π2

≤

0H1:

π1

–

π2

>

0Two-tail

test:H0:

π1

–

π2

=

0H1:

π1

–

π2

≠

0-zzReject

H0

if

ZSTAT

<

-ZReject

H0

if

ZSTAT

>

Z

/2

/2Copyright

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10-28-z/2

z/2Reject

H0

if

ZSTAT

<

-Zor

ZSTAT

>

Z(continued)DCOVAHypothesis

Test

Example:Two

population

Proportions游客和女性游客，購物的比例是否存在顯著差異？In

a

random

sample,

36

of

72

menand35

of

50

women

indicated

they

wouldpurchase.Test

at

the

.05

levelofsignificanceDCOVACopyright

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10-29The

hypothesis

:H0:

π1

–

π2

=

0H1:

π1

–

π2

≠

0(the

two

proportions

are

equal)(there

is

a

significant

difference

between

proportions)The

sample

proportions

are:Men:Women:p1

=

36/72

=

0.50p2

=

35/50

=

0.70

36

35

71

0.58272

50

122p

X1

X2n1

n2The

pooled

estimate

for

the

overall

proportion

is:Hypothesis

Test

Example:Two

population

Proportions(continued)DCOVACopyright

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10-30The

test

statistic

for

π1

–

π2

is:(continued).025961.96.025-1.-2.20Conclusion:

There

isevidence

of

a

difference

inproportions

who

willpurchase

between

men

andwomen.

和女性的購物的比例存在顯著差異zSTAT

72

50

.582(

1

.582

)

1

1

n1

n2

.50

.70

0

2.20

Decision:

Reject

H0

1 1

p(

1

p)

p

p

π

π

1

2

1

2Reject

H0Reject

H0Hypothesis

Test

Example:Two

population

Proportions

DCOVACritical

Values

=

±1.96For

=

.05Copyright

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10-31Confidence

IntervalforTwo

Population

ProportionsDCOVAPopulationproportions1

2Copyright

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10-32n

np1

(1

p1

)

p2

(1

p2

)p

p

Z1

2/2The

confidence

interval

forπ1

–

π2

is:Difference

Between

Two

Means:Independent

SamplesPopulation

means,independentsamples*Use

Sp

to

estimate

unknownσ.Use

a

Pooled-Variance

ttest.混合方差t檢驗σ1

and

σ2

unknown,assumed

equalσ1

and

σ2

unknown,not

assumed

equalUse S1

and

S2

to

estimate

unknownσ1

and

σ2.Use

a

Separate-Variancet

test.不同方差t檢驗Different

data

sourcesUnrelatedIndependent

兩總體間不相關Sample

selected

from

oneCopyright

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10-33population

has

no

effect

on

thesample

selected

from

the

otherpopulationDCOVATesting

for

the

Ratio

Of

TwoPopulation

Variances兩總體方差齊性檢驗Tests

for

TwoPopulationVariancesF test

statistic1

2H0:

σ

2

=

σ

2H1:

σ12

≠

σ221

2H0:

σ

2

≤

σ

212H1:

σ

2

>

σ

2*HypothesesFSTAT1n1

=

sample

size

of

sample

1S

2

=

Variance

of

sample

1(the

larger

sample

variance)S

2

=

Variance

of

sample

2

(the

smaller

sample

variance)2n2

=

sample

size

of

sample

2n1

–1

=

numerator

degrees

offreedomn2

–

1

=

denominator

degrees

of

freedomCopyright

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10-342S22S1The

F

critical

value

is

found

from

the

F

tableF臨界值可以從F表中查找P483There

are

two

degrees

of

freedom

required:

numeratorand

denominator

度由分子和分母決定The

larger

sample

variance

isalways

thenumerator大的是分子WhenIn

the

F

table,numerator

degrees

of

freedom

determine

the

column分子是列denominator

degrees

of

freedom

determine

the

row分母是行The

F

Distribution1

1df =

n –

1

;df2

=n2

–12S

2S

2FSTAT

1DCOVACopyright

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10-35Finding

the

Rejection

Region12H1:

σ

2

≠

σ

21

2

1

2H0:

σ

2

=

σ

2

H0:

σ

2

≤

σ

212H1:

σ

2

>

σ

2F0FαReject

H0Do

notreject

H0Reject

H0

if

FSTAT

>

FαF0/2Reject

H0Do

notreject

H0

Fα/2Reject

H0

if

FSTAT

>

Fα/2Copyright

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10-36DCOVAF

Test:

An

ExampleDCOVA假設我們現在正在對來自不同地區的游客對家界旅游的滿意度.以下數據是華北游客滿意度和華南游客滿意度的數據。華北游客華南游客Number2125Mean3.272.53Std

dev1.301.16華北游客和華南游客滿意度的方差在

=0.05

的水平上是否存在差異呢？Copyright

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10-37F

Test:

Example

SolutionCopyright

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10-38(there

is

no

difference

between

variances)(there

is

a

difference

between

variances)DCOVAForm

the

hypothesis

test:Find

the

F

critical

value

for

=

0.05:Numerator

d.f.

=

n1

–

1

=

21

–1

=

20Denominator

d.f.

=

n2

–

1

=

25

–1

=

24Fα/2

=

F.025,

20,

24

=

2.33H

:

σ

2

σ

21

1

2H

:

σ

2

σ

20

1

2Chap

10-39The

test

statistic

is:1.2561.1622S

2S

21.302FSTAT

1

/2

=

.025Reject

H0F0.025=2.33Do

notreject

H0H0:

σ

2

=

σ

21

2H1:

σ12

≠

σ22F

Test:

ExampleSolution0FSTAT

=

1.256

is

not

in

the

rejectionregion,

so

we

do

not

reject

H0DCOVA(continued)Conclusion:

There

is

insufficient

evidence

of

adifference

in

variances

at

=

.05在0.05的顯著性水平下，沒有充分的

表明，華北游客的滿意度和華南游客的滿意度方差存在差異FOne-Way ysis

Of

Variance(ANOVA)Setting單向方差分D析COVAWant

to

examine

differences

among

more

than

two

groups檢驗上個以上總體間差異The

groups

involvedare

classified

according

to

levels

of

afactor

of

interest(numerical

or

categorical)根據因素的層級進行的分組Different

levels

produce

different

groups不同的層級就是不同的組Think

of

each

group

as

asample

from

a

differentpopulation講每個組假想成一個樣本Observe

effects

on

the

dependentvariableAre

the

groups

thesame?他們是一樣的嗎？

When

there

is

only

1

factor

the

design

is

called

a

comple

yrandomizeddesign當只有一個因素的時候，稱為完全隨機設計Copyright

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10-40One-Way ysis

of

Variance單向方差分析Evaluate

the

difference

among

the

means

of

threeor

more

groups評估三個或以上的組之間的差異Examples:

不同出游方式的游客的購物花費AssumptionsPopulations

are

normally

distributed

正態分布Populations

have

equal

variances

方差齊（相等）Samples

are

randomly

and

independently

drawn樣本隨機獨立抽取DCOVACopyright

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10-41All

population

means

are

equal

所有總體均值相等i.e.,

no

factor

effect

(no

variation

in

means

among

groups)各組不相互影響（各組均值方差不存在差異）At

least

one

population

mean

isdifferent

至少有一個總體均值是不一樣的i.e.,there

is

a

factor

effect

存在因素影響Does

not

mean

that

all

population

means

are

different(some

pairs

may

be

the

same)并不意味著所有組的均值都不一樣H0

:

μ1

μ2

μ3

μcCopyright

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10-42Hypotheses

of

One-Way

ANOVADCOVAH1

:

Not

all

of

the

population

means

are

the

sameOne-Way

ANOVAThe

Null

Hypothesis

is

TrueAll

Means

are

thesame:(No

Factor

Effect)H0

:

μ1

μ2

μ3

μcH1

:

Not

all

μj

are

the

sameCopyright

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10-43μ1

μ

2

μ3DCOVAOne-Way

ANOVAμ1

μ2

μ3H0

:

μ1

μ2

μ3

μcH1

:

Not

all

μj

are

the

sameThe

Null

Hypothesisis

NOT

trueAt

least

one

of

the

means

is

different(Factor

Effect

is

present)orμ1

μ2

μ3Copyright

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10-44DCOVA(continued)Copyright

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10-45Partitioning

the

Variation離差的構成Total

variation

can

be

split

into

twoparts:SST

=Total

Sum

of

Squares總離差

(Total

variation)SSA

=Sum

of

Squares

Among

Groups組間離差

(Among-group

variation)SSW

=Sumof

Squares

Within

Groups組內離差

(Within-group

variation)SST

=

SSA

+

SSWDCOVAPartitioning

the

VariationTotal

Variation

=the

aggregate

variationof

the

individualdata

values

across

the

various

factor

levels(SST)不同因素層級間所有數據的總變異Among-Group

Variation

=

variation

among

the

factorsample

means(SSA)因素各層級均值的變異Within-Group

Variation

=

variation

that

exists

amongthe

data

values

withina

particular

factor

level(SSW)各因素層 所有數據值之間的變異/離差SST

=

SSA

+

SSW(continued)DCOVACopyright

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10-46Variation

Due

toFactor

(SSA)組間離差Copyright

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10-47Variation

Due

to

RandomError

(SSW)隨機誤差Total

Variation

(SST)Partition

of

Total

VariationDCOVA=+Total

Sum

of

Squares離差平方和cnjSST

(

X

ij

X

)2j

1

i

1Where:SST

=

Total

sum

of

squaresc

=

number

of

groups

or

levelsnj

=

number

of

observations

in

group

jXij

=

ith

observation

from

group

jX

=

grand

mean

(mean

of

all

datavalues)SSTCopyright

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10-48=

SSA

+

SSWDCOVATotal

Variation

總離差（總變異）(continued)Group

1Group

2Group

3Response,

XXCopyright

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10-49Among-Group

Variation

組間離差Where:SSA

=

Sum

of

squares

among

groupsc

=

number

ofgroupsnj

=

sample

size

from

group

jXj

=

sample

mean

from

group

jX

=

grand

mean

(meanof

alldata

values)2jc

n

(

X

X

)SSA

jj

1SST

=

SSA

+

SSWCopyright

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10-50Among-Group

VariationVariation

Due

toDifferences

Among

Groupsi

j2jc

jn

(

X

X

)SSA

j

1MSA

SSAc

1Mean

Square

Among

間均方=

SSA/degrees

of

freedom(continued)DCOVACopyright

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10-51Among-Group

VariationGroup

1Group

2Group

31XX2DCOVA(continued)SSA

n

(

X

1

X

)2

n

(

X

2

X

)2

n

(

X

c

X

)21

2

cResponse,

XX

3

XCopyright

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10-52Within-Group

Variation組內離差Where:SSW

=

Sum

of

squareswithin

groupsc