1 1 slide slides by john loucks st. edward’s university

1 1 Slide Slide

Slides by

JOHNLOUCKSSt. Edward’sUniversity

2 2 Slide Slide

Chapter 13, Part AChapter 13, Part A Experimental Design and Analysis of Experimental Design and Analysis of

Variance Variance Introduction to Experimental DesignIntroduction to Experimental Design and Analysis of Variance and Analysis of Variance

Analysis of Variance Analysis of Variance and the Completely Randomized Designand the Completely Randomized Design Multiple Comparison ProceduresMultiple Comparison Procedures

3 3 Slide Slide

Statistical studies can be classified as being Statistical studies can be classified as being either experimental or observational.either experimental or observational.

In an In an experimental studyexperimental study, one or more factors are , one or more factors are controlled so that data can be obtained about how controlled so that data can be obtained about how the factors influence the variables of interest.the factors influence the variables of interest.

In an In an observational studyobservational study, no attempt is made , no attempt is made to control the factors.to control the factors.

Cause-and-effect relationshipsCause-and-effect relationships are easier to establish in are easier to establish in experimental studies than in observational studies.experimental studies than in observational studies.

An Introduction to Experimental DesignAn Introduction to Experimental Designand Analysis of Varianceand Analysis of Variance

Analysis of variance (ANOVA) can be used to Analysis of variance (ANOVA) can be used to analyze the data obtained from experimental or analyze the data obtained from experimental or observational studies.observational studies.

4 4 Slide Slide


In this chapter three types of experimental In this chapter three types of experimental designs are introduced.designs are introduced.

• a completely randomized designa completely randomized design

• a randomized block designa randomized block design

• a factorial experimenta factorial experiment

5 5 Slide Slide


A A factorfactor （因子）（因子） is a variable that the is a variable that the experimenter has selected for investigation.experimenter has selected for investigation.

A A treatmenttreatment （处理或因子水平）（处理或因子水平） is a level of a factor.is a level of a factor.

Experimental unitsExperimental units are the objects of interest are the objects of interest in the experiment.in the experiment.

A A completely randomized designcompletely randomized design is an is an experimental design in which the treatments are experimental design in which the treatments are randomly assigned to the experimental units.randomly assigned to the experimental units.

6 6 Slide Slide

Analysis of Variance: A Conceptual Analysis of Variance: A Conceptual OverviewOverview

Analysis of VarianceAnalysis of Variance (ANOVA) can be used to test (ANOVA) can be used to test for the equality of three or more population means.for the equality of three or more population means. Analysis of VarianceAnalysis of Variance (ANOVA) can be used to test (ANOVA) can be used to test for the equality of three or more population means.for the equality of three or more population means.

Data obtained from observational or experimentalData obtained from observational or experimental studies can be used for the analysis.studies can be used for the analysis. Data obtained from observational or experimentalData obtained from observational or experimental studies can be used for the analysis.studies can be used for the analysis.

We want to use the sample results to test theWe want to use the sample results to test the following hypotheses:following hypotheses: We want to use the sample results to test theWe want to use the sample results to test the following hypotheses:following hypotheses:

HH00: : 11==22==33==. . . . . . = = kk

HHaa: Not all population means are equal: Not all population means are equal

7 7 Slide Slide

HH00: : 11==22==33==. . . . . . = = kk


If If HH00 is rejected, we cannot conclude that is rejected, we cannot conclude that allall population means are different.population means are different. If If HH00 is rejected, we cannot conclude that is rejected, we cannot conclude that allall population means are different.population means are different.

Rejecting Rejecting HH00 means that at least two population means that at least two population means have different values.means have different values. Rejecting Rejecting HH00 means that at least two population means that at least two population means have different values.means have different values.


8 8 Slide Slide

For each population, the response (dependent)For each population, the response (dependent)（反应变量）（反应变量） variable is normally distributed.variable is normally distributed. For each population, the response (dependent)For each population, the response (dependent)（反应变量）（反应变量） variable is normally distributed.variable is normally distributed.

The variance of the response variable, denoted The variance of the response variable, denoted 22,, is the same for all of the populations.is the same for all of the populations. The variance of the response variable, denoted The variance of the response variable, denoted 22,, is the same for all of the populations.is the same for all of the populations.

The observations must be independent.The observations must be independent. The observations must be independent.The observations must be independent.

Assumptions for Analysis of VarianceAssumptions for Analysis of Variance


9 9 Slide Slide

Sampling Distribution of Given Sampling Distribution of Given HH00 is True is Truexx

1x1x 3x3x2x2x

Sample means are close togetherSample means are close together because there is onlybecause there is only

one sampling distributionone sampling distribution when when HH00 is true. is true.

22x n

2

2x n


10 10 Slide Slide

Sampling Distribution of Given Sampling Distribution of Given HH00 is False is Falsexx

33 1x1x 2x2x3x3x 11 22

Sample means come fromSample means come fromdifferent sampling distributionsdifferent sampling distributionsand are not as close togetherand are not as close together

when when HH00 is false. is false.


11 11 Slide Slide

Analysis of VarianceAnalysis of Variance

Between-Treatments Estimate of Population VarianceBetween-Treatments Estimate of Population Variance

Within-Treatments Estimate of Population VarianceWithin-Treatments Estimate of Population Variance

Comparing the Variance Estimates: The Comparing the Variance Estimates: The F F Test Test

ANOVA TableANOVA Table

12 12 Slide Slide

2

1

( )

MSTR1

k

j jj

n x x

k

2

1

( )

MSTR1

k

j jj

n x x

k

Between-Treatments EstimateBetween-Treatments Estimateof Population Variance of Population Variance 22

Denominator is theDenominator is thedegrees of freedomdegrees of freedom

associated with associated with SSTRSSTR

Numerator is calledNumerator is calledthe the sum of squares sum of squares

duedueto treatmentsto treatments (SSTR)(SSTR)

（处理平方和）（处理平方和）

The estimate of The estimate of 22 based on the variation of the based on the variation of the sample means is called the sample means is called the mean square due tomean square due to treatmentstreatments （处理均方）（处理均方） and is denoted by and is denoted by MSTRMSTR..

13 13 Slide Slide

The estimate of The estimate of 22 based on the variation of the based on the variation of the sample observations within each sample is called the sample observations within each sample is called the mean square errormean square error （误差均值）（误差均值） and is denoted by and is denoted by MSEMSE..

Within-Treatments EstimateWithin-Treatments Estimateof Population Variance of Population Variance 22

Denominator is Denominator is thethe

degrees of degrees of freedomfreedom

associated with associated with SSESSE

Numerator is Numerator is calledcalled

the the sum of sum of squaressquares

due to errordue to error (SSE)(SSE)（误差平方和）（误差平方和）

MSE

( )n s

n k

j jj

k

T

1 2

1MSE

( )n s

n k

j jj

k

T

1 2

1

14 14 Slide Slide

Comparing the Variance Estimates: The Comparing the Variance Estimates: The FF TestTest

If the null hypothesis is true and the ANOVAIf the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution ofassumptions are valid, the sampling distribution of MSTR/MSE is an MSTR/MSE is an FF distribution with MSTR d.f. distribution with MSTR d.f. equal to equal to kk - 1 and MSE d.f. equal to - 1 and MSE d.f. equal to nnTT - - kk..

If the means of the If the means of the kk populations are not equal, the populations are not equal, the value of MSTR/MSE will be inflated because MSTRvalue of MSTR/MSE will be inflated because MSTR overestimates overestimates 22.. Hence, we will reject Hence, we will reject HH00 if the resulting value of if the resulting value of MSTR/MSE appears to be too large to have beenMSTR/MSE appears to be too large to have been selected at random from the appropriate selected at random from the appropriate FF distribution.distribution.

15 15 Slide Slide

Sampling Distribution of MSTR/MSESampling Distribution of MSTR/MSE

Do Not Reject H0Do Not Reject H0

Reject H0Reject H0

MSTR/MSEMSTR/MSE

Critical ValueCritical ValueFF

Sampling DistributionSampling Distributionof MSTR/MSEof MSTR/MSE

Comparing the Variance Estimates: The Comparing the Variance Estimates: The FF TestTest

16 16 Slide Slide

MSTRSSTR

-

k 1MSTR

SSTR-

k 1

MSESSE

-

n kT

MSESSE

-

n kT

MSTRMSE

MSTRMSE

Source ofSource ofVariationVariation

Sum ofSum ofSquaresSquares

Degrees ofDegrees ofFreedomFreedom

MeanMeanSquareSquare FF

TreatmentsTreatments

ErrorError

TotalTotal

kk - 1 - 1

nnTT - 1 - 1

SSTRSSTR

SSESSE

SSTSST

nnT T - - kk

SST is SST is partitionedpartitioned

into SSTR and into SSTR and SSE.SSE.

SST’s degrees of SST’s degrees of freedomfreedom

(d.f.) are partitioned (d.f.) are partitioned intointo

SSTR’s d.f. and SSE’s SSTR’s d.f. and SSE’s d.f.d.f.

ANOVA TableANOVA Table （方差分析表）（方差分析表）

pp--ValueValue

17 17 Slide Slide


SST divided by its degrees of freedom SST divided by its degrees of freedom nnTT – 1 is the – 1 is the overall sample variance that would be obtained if weoverall sample variance that would be obtained if we treated the entire set of observations as one data set.treated the entire set of observations as one data set.

SST divided by its degrees of freedom SST divided by its degrees of freedom nnTT – 1 is the – 1 is the overall sample variance that would be obtained if weoverall sample variance that would be obtained if we treated the entire set of observations as one data set.treated the entire set of observations as one data set.

With the entire data set as one sample, the formulaWith the entire data set as one sample, the formula for computing the total sum of squares, SST, is:for computing the total sum of squares, SST, is: With the entire data set as one sample, the formulaWith the entire data set as one sample, the formula for computing the total sum of squares, SST, is:for computing the total sum of squares, SST, is:

2

1 1

SST ( ) SSTR SSEjnk

ijj i

x x

2

1 1

SST ( ) SSTR SSEjnk

ijj i

x x

18 18 Slide Slide


ANOVA can be viewed as the process of partitioningANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedomthe total sum of squares and the degrees of freedom into their corresponding sources: treatments and error.into their corresponding sources: treatments and error.

ANOVA can be viewed as the process of partitioningANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedomthe total sum of squares and the degrees of freedom into their corresponding sources: treatments and error.into their corresponding sources: treatments and error.

Dividing the sum of squares by the appropriateDividing the sum of squares by the appropriate degrees of freedom provides the variance estimatesdegrees of freedom provides the variance estimates and the and the FF value used to test the hypothesis of equal value used to test the hypothesis of equal population means.population means.

Dividing the sum of squares by the appropriateDividing the sum of squares by the appropriate degrees of freedom provides the variance estimatesdegrees of freedom provides the variance estimates and the and the FF value used to test the hypothesis of equal value used to test the hypothesis of equal population means.population means.

19 19 Slide Slide

Test for the Equality of Test for the Equality of kk Population Population MeansMeans

FF = MSTR/MSE = MSTR/MSE

HH00: : 11==22==33==. . . . . . = = kk


HypotheseHypothesess

Test Test StatisticStatistic

20 20 Slide Slide

Test for the Equality of Test for the Equality of kk Population Population MeansMeans

Rejection Rejection RuleRule

where the value of where the value of FF is based on an is based on anFF distribution with distribution with kk - 1 numerator d.f. - 1 numerator d.f.and and nnTT - - kk denominator d.f. denominator d.f.

Reject Reject HH00 if if pp-value -value << pp-value Approach:-value Approach:

Critical Value Approach:Critical Value Approach: Reject Reject HH00 if if FF >> FF

21 21 Slide Slide

AutoShine, Inc. is considering marketing a AutoShine, Inc. is considering marketing a long-long-

lasting car wax. Three different waxes (Type 1, lasting car wax. Three different waxes (Type 1, Type 2,Type 2,

and Type 3) have been developed.and Type 3) have been developed.

Example: AutoShine, Inc.Example: AutoShine, Inc.

In order to test the durabilityIn order to test the durability

of these waxes, 5 new cars wereof these waxes, 5 new cars were

waxed with Type 1, 5 with Typewaxed with Type 1, 5 with Type

2, and 5 with Type 3. Each car was then2, and 5 with Type 3. Each car was then

repeatedly run through an automatic carwash repeatedly run through an automatic carwash until theuntil the

wax coating showed signs of deterioration.wax coating showed signs of deterioration.

Testing for the Equality of Testing for the Equality of kk Population Population Means:Means:

A Completely Randomized Experimental A Completely Randomized Experimental DesignDesign

22 22 Slide Slide

The number of times each car went through The number of times each car went through thethe

carwash before its wax deteriorated is shown carwash before its wax deteriorated is shown on theon the

next slide. AutoShine, Inc. must decide which next slide. AutoShine, Inc. must decide which waxwax

to market. Are the three waxesto market. Are the three waxes

equally effective?equally effective?

Example: AutoShine, Inc.Example: AutoShine, Inc.



Factor . . . Car waxFactor . . . Car waxTreatments . . . Type I, Type 2, Type 3Treatments . . . Type I, Type 2, Type 3Experimental units . . . CarsExperimental units . . . Cars

Response variable . . . Number of washesResponse variable . . . Number of washes

23 23 Slide Slide

1122334455

27273030292928283131

33332828313130303030

29292828303032323131

Sample MeanSample MeanSample VarianceSample Variance

ObservationObservationWaxWax

Type 1Type 1WaxWax

Type 2Type 2WaxWax

Type 3Type 3

2.52.5 3.3 3.3 2.5 2.529.0 30.429.0 30.4 30.0 30.0



24 24 Slide Slide

HypothesesHypotheses

where: where:

1 1 = mean number of washes using Type 1 wax= mean number of washes using Type 1 wax



HH00: : 11==22==33

HHaa: Not all the means are equal: Not all the means are equal



25 25 Slide Slide

Because the sample sizes are all Because the sample sizes are all equal:equal:

MSE = 33.2/(15 - 3) = 2.77MSE = 33.2/(15 - 3) = 2.77

MSTR = 5.2/(3 - 1) = 2.6MSTR = 5.2/(3 - 1) = 2.6

SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2

SSTR = 5(29–29.8)SSTR = 5(29–29.8)22 + 5(30.4–29.8) + 5(30.4–29.8)22 + 5(30–29.8) + 5(30–29.8)22 = 5.2 = 5.2

Mean Square ErrorMean Square Error

Mean Square Between TreatmentsMean Square Between Treatments

1 2 3( )/ 3x x x x 1 2 3( )/ 3x x x x = (29 + 30.4 + 30)/3 = 29.8= (29 + 30.4 + 30)/3 = 29.8



26 26 Slide Slide

Rejection RuleRejection Rule

where where FF.05.05 = 3.89 is based on an = 3.89 is based on an FF distribution distributionwith 2 numerator degrees of freedom and 12with 2 numerator degrees of freedom and 12denominator degrees of freedomdenominator degrees of freedom

pp-Value Approach: Reject -Value Approach: Reject HH00 if if pp-value -value << .05 .05

Critical Value Approach: Reject Critical Value Approach: Reject HH00 if if FF >> 3.89 3.89



27 27 Slide Slide

Test StatisticTest Statistic

There is insufficient evidence to conclude thatThere is insufficient evidence to conclude thatthe mean number of washes for the three waxthe mean number of washes for the three waxtypes are not all the same.types are not all the same.

ConclusionConclusion

F F = MSTR/MSE = 2.60/2.77 = .939 = MSTR/MSE = 2.60/2.77 = .939

The The pp-value is greater than .10, where -value is greater than .10, where FF = 2.81. = 2.81. (Excel provides a (Excel provides a pp-value of .42.)-value of .42.) Therefore, we cannot reject Therefore, we cannot reject HH00..



28 28 Slide Slide




MeanMeanSquaresSquares FF

TreatmentsTreatments

ErrorError

TotalTotal

22

1414

5.25.2

33.233.2

38.438.4

1212

2.602.60

2.772.77

.939.939




pp-Value-Value

.42.42

29 29 Slide Slide

Example: Reed ManufacturingExample: Reed Manufacturing

Janet Reed would like to know ifJanet Reed would like to know ifthere is any significant difference inthere is any significant difference inthe mean number of hours worked per the mean number of hours worked per week for the department managersweek for the department managersat her three manufacturing plantsat her three manufacturing plants(in Buffalo, Pittsburgh, and Detroit). (in Buffalo, Pittsburgh, and Detroit). An An FF test will be conducted using test will be conducted using = .05. = .05.


An Observational StudyAn Observational Study

30 30 Slide Slide


A simple random sample of fiveA simple random sample of fivemanagers from each of the three plantsmanagers from each of the three plantswas taken and the number of hourswas taken and the number of hoursworked by each manager in theworked by each manager in theprevious week is shown on the nextprevious week is shown on the nextslide.slide.



Factor . . . Manufacturing plantFactor . . . Manufacturing plantTreatments . . . Buffalo, Pittsburgh, DetroitTreatments . . . Buffalo, Pittsburgh, DetroitExperimental units . . . ManagersExperimental units . . . Managers

Response variable . . . Number of hours workedResponse variable . . . Number of hours worked

31 31 Slide Slide

1122334455

48485454575754546262

73736363666664647474

51516363616154545656

Plant 1Plant 1BuffaloBuffalo

Plant 2Plant 2PittsburghPittsburgh

Plant 3Plant 3DetroitDetroitObservationObservation

Sample MeanSample MeanSample VarianceSample Variance

5555 68 68 57 5726.026.0 26.5 26.5 24.5 24.5



32 32 Slide Slide

HH00: : 11==22==33

HHaa: Not all the means are equal: Not all the means are equalwhere: where: 1 1 = mean number of hours worked per= mean number of hours worked per

week by the managers at Plant 1week by the managers at Plant 1 2 2 = mean number of hours worked per= mean number of hours worked per week by the managers at Plant 2week by the managers at Plant 23 3 = mean number of hours worked per= mean number of hours worked per week by the managers at Plant 3week by the managers at Plant 3

1. Develop the hypotheses.1. Develop the hypotheses.

pp -Value and Critical Value Approaches -Value and Critical Value Approaches



33 33 Slide Slide

2. Specify the level of significance.2. Specify the level of significance. = .05= .05


3. Compute the value of the test statistic.3. Compute the value of the test statistic.

MSTR = 490/(3 - 1) = 245MSTR = 490/(3 - 1) = 245SSTR = 5(55 - 60)SSTR = 5(55 - 60)22 + 5(68 - 60) + 5(68 - 60)22 + 5(57 - 60) + 5(57 - 60)22 = 490 = 490

= (55 + 68 + 57)/3 = 60= (55 + 68 + 57)/3 = 60xx(Sample sizes are all equal.)(Sample sizes are all equal.)

Mean Square Due to TreatmentsMean Square Due to Treatments



34 34 Slide Slide

3. Compute the value of the test statistic.3. Compute the value of the test statistic.

MSE = 308/(15 - 3) = 25.667MSE = 308/(15 - 3) = 25.667

SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308Mean Square Due to ErrorMean Square Due to Error

(con’t.)(con’t.)

FF = MSTR/MSE = 245/25.667 = 9.55 = MSTR/MSE = 245/25.667 = 9.55




35 35 Slide Slide

TreatmentTreatment

ErrorError

TotalTotal

490490

308308

798798

22

1212

1414

245245

25.66725.667




MeanMeanSquareSquare

9.559.55

FF




pp-Value-Value

.0033.0033

36 36 Slide Slide

5. Determine whether to reject 5. Determine whether to reject HH00..

We have sufficient evidence to conclude that We have sufficient evidence to conclude that the mean number of hours worked per week the mean number of hours worked per week by department managers is not the same at by department managers is not the same at all 3 plant.all 3 plant.

The The pp-value -value << .05, .05, so we reject so we reject HH00..

With 2 numerator d.f. and 12 With 2 numerator d.f. and 12 denominator d.f.,denominator d.f.,the the pp-value is .01 for -value is .01 for FF = 6.93. = 6.93. Therefore, theTherefore, thepp-value is less than .01 for -value is less than .01 for FF = 9.55. = 9.55.

pp –Value Approach –Value Approach

4. Compute the 4. Compute the pp –value. –value.



37 37 Slide Slide

5. Determine whether to reject 5. Determine whether to reject HH00..

Because Because FF = 9.55 = 9.55 >> 3.89, we reject 3.89, we reject HH00..

Critical Value ApproachCritical Value Approach

4. Determine the critical value and rejection rule.4. Determine the critical value and rejection rule.

Reject Reject HH00 if if FF >> 3.89 3.89

We have sufficient evidence to conclude that We have sufficient evidence to conclude that the mean number of hours worked per week the mean number of hours worked per week by department managers is not the same at by department managers is not the same at all 3 plant.all 3 plant.

Based on an Based on an FF distribution with 2 numerator distribution with 2 numeratord.f. and 12 denominator d.f., d.f. and 12 denominator d.f., FF.05.05 = 3.89. = 3.89.



38 38 Slide Slide

Multiple Comparison ProceduresMultiple Comparison Procedures （多重比较方法）（多重比较方法）

Suppose that analysis of variance has Suppose that analysis of variance has provided statistical evidence to reject the null provided statistical evidence to reject the null hypothesis of equal population means.hypothesis of equal population means.

Fisher’s least significant difference (LSD)Fisher’s least significant difference (LSD) （最小显著差异）（最小显著差异） procedure can be used to determine where the differences occur.procedure can be used to determine where the differences occur.

39 39 Slide Slide

Fisher’s LSD ProcedureFisher’s LSD Procedure

1 1MSE( )

i j

i j

x xt

n n

1 1MSE( )

i j

i j

x xt

n n



0 : i jH 0 : i jH : a i jH : a i jH

40 40 Slide Slide

Fisher’s LSD ProcedureFisher’s LSD Procedure

where the value of where the value of ttaa/2 /2 is based on ais based on a

tt distribution with distribution with nnTT - - kk degrees of freedom. degrees of freedom.


Reject Reject HH00 if if pp-value -value <<

pp-value Approach:-value Approach:

Critical Value Approach:Critical Value Approach:

Reject Reject HH00 if if tt < - < -ttaa/2 /2 or or tt > > ttaa/2 /2

41 41 Slide Slide


Fisher’s LSD ProcedureFisher’s LSD ProcedureBased on the Test Statistic Based on the Test Statistic xxii - - xxjj

__ __

/ 21 1LSD MSE( )

i jt n n / 2

1 1LSD MSE( )i j

t n n wherewhere

i jx xi jx x

Reject Reject HH00 if > LSD if > LSDi jx xi jx x



0 : i jH 0 : i jH : a i jH : a i jH

42 42 Slide Slide



Recall that Janet Reed wants to knowRecall that Janet Reed wants to knowif there is any significant difference inif there is any significant difference inthe mean number of hours worked per the mean number of hours worked per week for the department managersweek for the department managersat her three manufacturing plants. at her three manufacturing plants.

Analysis of variance has providedAnalysis of variance has providedstatistical evidence to reject the nullstatistical evidence to reject the nullhypothesis of equal population means.hypothesis of equal population means.Fisher’s least significant difference (LSD) Fisher’s least significant difference (LSD) procedureprocedurecan be used to determine where the differences can be used to determine where the differences occur.occur.

43 43 Slide Slide

For For = .05 and = .05 and nnTT - - kk = 15 – 3 = 12 = 15 – 3 = 12

degrees of freedom, degrees of freedom, tt..025 025 = 2.179 = 2.179

LSD 2 179 25 667 15

15 6 98. . ( ) .LSD 2 179 25 667 1

51

5 6 98. . ( ) .

/ 21 1LSD MSE( )

i jt n n / 2

1 1LSD MSE( )i j

t n n

MSE value wasMSE value wascomputed earliercomputed earlier


44 44 Slide Slide

LSD for Plants 1 and 2LSD for Plants 1 and 2


• ConclusionConclusion

• Test StatisticTest Statistic1 2x x1 2x x = |55 = |55 68| = 13 68| = 13

Reject Reject HH00 if if > 6.98 > 6.981 2x x1 2x x

• Rejection RuleRejection Rule

0 1 2: H 0 1 2: H 1 2: aH 1 2: aH

• Hypotheses (A)Hypotheses (A)

The mean number of hours worked at Plant 1 isThe mean number of hours worked at Plant 1 isnot equalnot equal to the mean number worked at Plant 2. to the mean number worked at Plant 2.

45 45 Slide Slide







0 1 3: H 0 1 3: H 1 3: aH 1 3: aH

• Hypotheses (B)Hypotheses (B)

There is There is no significant differenceno significant difference between the mean between the mean number of hours worked at Plant 1 and number of hours worked at Plant 1 and the meanthe mean number of hours worked at Plant 3.number of hours worked at Plant 3.

46 46 Slide Slide







0 2 3: H 0 2 3: H 2 3: aH 2 3: aH

• Hypotheses (C)Hypotheses (C)

The mean number of hours worked at Plant 2 isThe mean number of hours worked at Plant 2 is not equalnot equal to the mean number worked at Plant 3. to the mean number worked at Plant 3.

47 47 Slide Slide

The experiment-wise Type I error rate gets larger for The experiment-wise Type I error rate gets larger for problems with more populations (larger problems with more populations (larger kk).).

Type I Error RatesType I Error Rates

EWEW = 1 – (1 – = 1 – (1 – ))((k k – 1)!– 1)!

The The comparison-wise Type I error ratecomparison-wise Type I error rate （比较性第一类（比较性第一类错误概率）错误概率） indicates the level of significance indicates the level of significance associated with a single pairwise comparison.associated with a single pairwise comparison.

The The experiment-wise Type I error rate(experiment-wise Type I error rate( 试验性第一类试验性第一类错误）错误） EWEW is the probability of making a Type I error is the probability of making a Type I error on at least one of the (on at least one of the (kk – 1)! pairwise comparisons. – 1)! pairwise comparisons.

48 48 Slide Slide

End of Chapter 13, Part AEnd of Chapter 13, Part A

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