商务统计学课件-统计ch10-学生版

Business

Statistics:

ACourse6th

EditionChapter

10Two-Sample

TestsandOne-Way

ANOVA两样本检验和单边方差分析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:如何利用统计假设来比较两个不一样的：The

means

of

two

independent

populations

两个独立的总体的均值The

means

of

two

related

populations

两个相关的总体的均值The

proportions

of

two

independent

populations两个独立的总体的比例The

variances

of

two

independent

populations两个独立的总体的方差The

means

of

more

than

two

populations

两个以上总体的均值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独立样本总体均值检验Goal:Test

hypothesis

or

form

检验的假设或形式：a

confidence

interval

for

thedifference

between

two

populationmeans两总体均值差异的置信区间,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检验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检验Different

data

sourcesUnrelatedIndependent

两总体间不相关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两总体均值假设检验Lower-tail

test:左侧检验

H0:μ1

μ2H1:μ1

<μ2i.e.,H0:

μ1

–

μ2

0H1:

μ1

–

μ2

<

0Upper-tail

test:右侧检验

H0:μ1

≤μ2H1:μ1

>μ2i.e.,H0:

μ1

–

μ2

≤

0H1:

μ1

–

μ2

>

0Two-tail

test:双侧检验

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样品随机，独立抽取Populations

are

normallydistributed

or

both

samplesizes

are

at

least30人群呈正态分布或两个样本量均至少为30Population

variances

areunknown

but

assumed

equal总体方差未知，但假设他们相等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置信区间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东北游客游客Sample

Size2125Sample

mean3.272.53Sample

stddev1.301.16是否存在差异？DCOVA据说来自和东北的游客，在家界游玩的满意度存在差异，旅游学院老师希望通过问卷检验这一结论的真实性，收集了以下数据:Assuming

both

populations

areapproxima y

normal

with

equal

variances,

istherea

difference

in

mean

yield

(

=

0.05)?假设两总体呈正态分布，方差相同，请问两总体均值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样品随机，独立抽取Populations

arenormallydistributed

or

both

samplesizes

are

at

least

30人群呈正态分布或两个样本量均至少为30Population

variances

areunknownand cannot

beassumed

tobe

equal总体方差未知，不能假设他们相等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

大样本Copyright

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10-17Related

Populations

相关总体The

Paired

Difference

TestDCOVATests

Means

of

2

Related

PopulationsPaired

or

matched

samples

配对样本，期望与满意度Repeated

measures(before/after)重复检验，事前事后

Use

difference

between

paired

values:两个配对值之间的差异Eliminates

Variation

Among

Subjects消除两主体之间的变异Assumptions:Both

Populations

Are

Normally

Distributed

正态分布RelatedsamplesDi

=

X1i

-X2iRelated

PopulationsThe

Paired

Difference

TestThe

ith

paired

difference

is Di

,

whereRelatedsamplesDi

=

X1i

-

X2iThe

point

estimate

for

the

paireddifference

population

mean

μD

is匹配均值差异的点估计D:n

D

iD

i

1

nn

is

the

number

of

pairs

in

the

paired

sampleN是匹配样本的成对的数量n2(D

D)SD

i

i1

n

1The

sample

standarddeviation样本标准差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旅游有助于提高员工的服务质量，降量。我们收集了员工旅游前和旅游后一通常认为低顾客的个月的顾客量，数据如下，请问 旅游对员工服务质量的是否存在显著影响：Paired

Difference

Test匹配t检验: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两总体比例Copyright

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10-24difference

between

two

populationproportions,

两总体比例的差异DCOVAGoal:

test

a

hypothesis

or

form

检验的假设：a

confidence

interval

for

theπ1

–

π2The

point

estimate

for

thedifference

这个差异的点估计isPopulationproportionsp1

p2Two

Population

Proportions两总体比例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零假设：假设两总体比例相等The

pooled

estimate

for

theoverallproportion总的比例的混合估计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，购物的比例是否存在显著差异？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.

和女性的购物的比例存在显著差异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检验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检验Different

data

sourcesUnrelatedIndependent

两总体间不相关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两总体方差齐性检验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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HallWhere:DCOVAChap

10-342S22S1The

F

critical

value

is

found

from

the

F

tableF临界值可以从F表中查找P483There

are

two

degrees

of

freedom

required:

numeratorand

denominator

度由分子和分母决定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假设我们现在正在对来自不同地区的游客对家界旅游的满意度.以下数据是华北游客满意度和华南游客满意度的数据。华北游客华南游客Number2125Mean3.272.53Std

dev1.301.16华北游客和华南游客满意度的方差在

=0.05

的水平上是否存在差异呢？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的显著性水平下，没有充分的

表明，华北游客的满意度和华南游客的满意度方差存在差异FOne-Way ysis

Of

Variance(ANOVA)Setting单向方差分D析COVAWant

to

examine

differences

among

more

than

two

groups检验上个以上总体间差异The

groups

involvedare

classified

according

to

levels

of

afactor

of

interest(numerical

or

categorical)根据因素的层级进行的分组Different

levels

produce

different

groups不同的层级就是不同的组Think

of

each

group

as

asample

from

a

differentpopulation讲每个组假想成一个样本Observe

effects

on

the

dependentvariableAre

the

groups

thesame?他们是一样的吗？

When

there

is

only

1

factor

the

design

is

called

a

comple

yrandomizeddesign当只有一个因素的时候，称为完全随机设计Copyright

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10-40One-Way ysis

of

Variance单向方差分析Evaluate

the

difference

among

the

means

of

threeor

more

groups评估三个或以上的组之间的差异Examples:

不同出游方式的游客的购物花费AssumptionsPopulations

are

normally

distributed

正态分布Populations

have

equal

variances

方差齐（相等）Samples

are

randomly

and

independently

drawn样本随机独立抽取DCOVACopyright

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10-41All

population

means

are

equal

所有总体均值相等i.e.,

no

factor

effect

(no

variation

in

means

among

groups)各组不相互影响（各组均值方差不存在差异）At

least

one

population

mean

isdifferent

至少有一个总体均值是不一样的i.e.,there

is

a

factor

effect

存在因素影响Does

not

mean

that

all

population

means

are

different(some

pairs

may

be

the

same)并不意味着所有组的均值都不一样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离差的构成Total

variation

can

be

split

into

twoparts:SST

=Total

Sum

of

Squares总离差

(Total

variation)SSA

=Sum

of

Squares

Among

Groups组间离差

(Among-group

variation)SSW

=Sumof

Squares

Within

Groups组内离差

(Within-group

variation)SST

=

SSA

+

SSWDCOVAPartitioning

the

VariationTotal

Variation

=the

aggregate

variationof

the

individualdata

values

across

the

various

factor

levels(SST)不同因素层级间所有数据的总变异Among-Group

Variation

=

variation

among

the

factorsample

means(SSA)因素各层级均值的变异Within-Group

Variation

=

variation

that

exists

amongthe

data

values

withina

particular

factor

level(SSW)各因素层 所有数据值之间的变异/离差SST

=

SSA

+

SSW(continued)DCOVACopyright

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10-46Variation

Due

toFactor

(SSA)组间离差Copyright

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10-47Variation

Due

to

RandomError

(SSW)随机误差Total

Variation

(SST)Partition

of

Total

VariationDCOVA=+Total

Sum

of

Squares离差平方和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

总离差（总变异）(continued)Group

1Group

2Group

3Response,

XXCopyright

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10-49Among-Group

Variation

组间离差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

间均方=

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组内离差Where:SSW

=

Sum

of

squareswithin

groupsc