chapter 13 statistics in practice

111SlideSlideSlide

Chapter 13Chapter 13

STATISTICSSTATISTICS in in PRACTICEPRACTICE� Burke Marketing Services,

Inc., is one of the most

experienced market research

firms in the industry.

� In one study, a firm retained

Burke to evaluate potential new versions of a

children’s dry cereal.

� Analysis of variance was the statistical method used to study the data obtained from the taste tests.

� The experimental design employed by Burke and the subsequent analysis of variance were helpful in making a product design recommendation.

222SlideSlideSlide

Chapter 13 Analysis of Variance and Chapter 13 Analysis of Variance and

Experimental DesignExperimental Design

� Introduction to Analysis of Variance

�Analysis of Variance: Testing for the Equality of k Population Means

�Multiple Comparison Procedures

� Factorial Experiments

�An Introduction to Experimental Design

�Completely Randomized Designs

�Randomized Block Design

333SlideSlideSlide

IntroductionIntroduction to Analysis of Varianceto Analysis of Variance

Analysis of Variance (ANOVA) can be used to test

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 experimental

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 the

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: : µµ11 == µµ22 == µµ33 == . . . . . . = = µµkk

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

444SlideSlideSlide

Introduction to Analysis of VarianceIntroduction to Analysis of Variance

HH00: : µµ11 == µµ22 == µµ33 == . . . . . . = = µµkk

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

IfH0 is rejected, we cannot conclude that all population

means are different.

IfIf HH00 is rejected, we cannot conclude that all populationis rejected, we cannot conclude that all population

means are different.means are different.

RejectingH0 means that at least two population means

have different values.

RejectingRejecting HH00 means that at least two population meansmeans that at least two population means

have different values.have different values.

555SlideSlideSlide

� Sampling Distribution of Given H0 is Truexx


µµ 1x1x 3x3x2x2x

Sample means are close togetherSample means are close togetherbecause there is onlybecause there is only

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

22x

n

σσ =

22x

n

σσ =

666SlideSlideSlide


� Sampling Distribution of Given H0 is Falsexx

µ3µ3 1x1x 2x2x3x3x µ1µ1 µ2µ2

Sample means come fromSample means come fromdifferent sampling distributionsdifferent sampling distributions

and are not as close togetherand are not as close togetherwhen when HH00 is false.is false.

777SlideSlideSlide

For each population, the response variable is

normally distributed.

For each population, the response variable isFor each population, the response variable is

normally distributed.normally distributed.

Assumptions for Analysis of VarianceAssumptions for Analysis of Variance

The variance of the response variable, denoted σ 2,is the same for all of the populations.

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

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

888SlideSlideSlide

Analysis of Variance:Analysis of Variance:

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

� Between-Treatments Estimate of Population

Variance

� Within-Treatments Estimate of Population

Variance

� Comparing the Variance Estimates: The F Test

� ANOVA Table

999SlideSlideSlide



� Analysis of variance can be used to test for the

equality of k population means.

� The hypotheses tested is

H0:

Ha: Not all population means are equal

where mean of the jth population.

kµµµ === L21

jµ

101010SlideSlideSlide



� Sample data

= value of observation i for treatment j

= number of observations for treatment j

= sample mean for treatment j

= sample variance for treatment j

= sample standard deviation for treatment j

ijx

jn

jx2

js

js


� Statisitcs

� The sample mean for treatment j

� The sample variance for treatment j



j

n

i

ij

jn

x

x

j

∑=

=1

1

)(1

2

2

−

−

=

∑=

j

n

i

jij

jn

xx

s

j




� The overall sample mean

where nT = n1 + n2 +. . . + nk

� If the size of each sample is n, nT = kn then

T

k

j

n

i

ij

n

x

x

j

∑∑= =

=1 1

kn

x

k

nx

kn

x

x

k

j

ij

k

j

n

i

ij

k

j

n

i

ij

jj

∑∑∑∑∑== == =

===11 11 1

/




� Between-Treatments Estimate of Population

Variance

� The sum of squares due to treatments (SSTR)

� The mean square due to treatments (MSTR)

1

)(1

2

−

−

=

∑=

k

xxn

MSTR

k

j

jj

∑=

−

k

j

jj xxn1

2)(




� Within-Treatments Estimate of Population Variance

� The sum of squares due to error (SSE)

� The mean square due to error (MSE)

∑=

−

k

j

jj sn1

2)1(

kn

sn

MSET

k

j

jj

−

−

=

∑=1

2)1(


BetweenBetween--Treatments EstimateTreatments Estimate

of Population Varianceof Population Variance

� A between-treatment estimate of σσσσ 2 is called the

mean square treatment and is denoted MSTR.

2

1

( )

MSTR1

k

j jj

n x x

k

=

−

=−

∑2

1

( )

MSTR1

k

j jj

n x x

k

=

−

=−

∑

Denominator representsDenominator represents

the the degrees of freedomdegrees of freedom

associated with SSTRassociated with SSTR

Numerator is theNumerator is the

sum of squaressum of squares

due to treatmentsdue to treatments

and is denoted SSTRand is denoted SSTR


� The estimate of σσσσ 2 based on the variation of the

sample observations within each sample is called

the mean square error and is denoted by MSE.

WithinWithin--Samples EstimateSamples Estimate

of Population Varianceof Population Variance

kn

sn

T

k

j

jj

−

−

=

∑=1

2)1(

MSEkn

sn

T

k

j

jj

−

−

=

∑=1

2)1(

MSE

Denominator representsDenominator represents

the the degrees of freedomdegrees of freedom

associated with SSEassociated with SSE

Numerator is theNumerator is the

sum of squaressum of squares

due to errordue to error

and is denoted SSEand is denoted SSE


Comparing the Variance Estimates:Comparing the Variance Estimates: TheThe FF TestTest

� If the null hypothesis is true and the ANOVA

assumptions are valid, the sampling distribution of

MSTR/MSE is an F distribution with MSTR d.f.

equal to k - 1 and MSE d.f. equal to nT - k.

� If the means of the k populations are not equal, the

value of MSTR/MSE will be inflated because MSTR

overestimates σσσσ 2.

� Hence, we will reject H0 if the resulting value of

MSTR/MSE appears to be too large to have been

selected at random from the appropriate F

distribution.


Test for the Equality ofTest for the Equality of kk Population MeansPopulation Means

FF = MSTR/MSE= MSTR/MSE

HH00: : µµ11 == µµ22 == µµ33 == . . . . . . = = µµkk

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

� Hypotheses

� Test Statistic


Test for the Equality ofTest for the Equality of k k Population MeansPopulation Means

� Rejection Rule

where the value of Fα α α α

is based on an

F distribution with k - 1 numerator d.f.and nT - k denominator d.f.

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

Critical Value Approach: RejectReject HH00 if if FF >> FFαα


Sampling Distribution of MSTR/MSESampling Distribution of MSTR/MSE

� Rejection Region

Do Not Reject H0Do Not Reject H0

Reject H0Reject H0

MSTR/MSEMSTR/MSE

Critical ValueCritical Value

FαFα

Sampling DistributionSampling Distribution

of MSTR/MSEof MSTR/MSE

αα


ANOVA TableANOVA Table

SST is partitionedSST is partitionedinto SSTR and SSE.into SSTR and SSE.

SSTSST’’s degrees of freedoms degrees of freedom

(d.f.) are partitioned into(d.f.) are partitioned into

SSTRSSTR’’ss d.f. and d.f. and SSESSE’’ss d.f.d.f.

TreatmentTreatment

ErrorError

TotalTotal

SSTRSSTR

SSESSE

SSTSST

kk –– 11

nnTT –– kk

nnTT -- 11

MSTRMSTR

MSEMSE

Source ofSource ofVariationVariation

Sum ofSum ofSquaresSquares

Degrees ofDegrees ofFreedomFreedom

MeanMeanSquaresSquares

MSTR/MSEMSTR/MSE

FF



SST divided by its degrees of freedom nT – 1 is the

overall sample variance that would be obtained if we

treated the entire set of observations as one data set.

SST divided by its degrees of freedomSST divided by its degrees of freedom nnTT –– 1 1 is theis 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 formula

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= =

= − = +∑∑



ANOVA can be viewed as the process of partitioning

the total sum of squares and the degrees of freedom

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 appropriate degrees

of freedom provides the variance estimates and the F

value used to test the hypothesis of equal population

means.

Dividing the sum of squares by the appropriate degrees Dividing the sum of squares by the appropriate degrees

of freedom provides the variance estimates and the of freedom provides the variance estimates and the FF

value used to test the hypothesis of equal population value used to test the hypothesis of equal population

means.means.



� Example:

� National Computer Products, Inc. (NCP),

manufactures printers and fax machines at plants

located in Atlanta, Dallas, and Seattle.

� Object: To measure how much employees at these

plants know about total quality management.

� A random sample of six employees was selected from

each plant and given a quality awareness examination.



� Data

� Let

= mean examination score for population 1



1µ

2µ

3µ


TestTest for the Equality of for the Equality of kk Population MeansPopulation Means

� Hypotheses

H0: = =

Ha: Not all population means are equal

� In this example

1. dependent or response variable : examination score

2. independent variable or factor : plant location

3. levels of the factor or treatments : the values of a

factor selected for investigation, in the NCP

example the three treatments or three population

are Atlanta, Dallas, and Seattle.

1µ 2µ3µ


Test for the Equality of Test for the Equality of kk Population MeansPopulation Means

Three assumptions

1. For each population, the response variable is normally

distributed. The examination scores (response variable)

must be normally distributed at each plant.

2. The variance of the response variable, , is the same for

all of the populations. The variance of examination scores

must be the same for all three plants.

3. The observations must be independent. The examination

score for each employee must be independent of the

examination score for any other employee.

2σ



� ANOVA Table

� p-value = 0.003 < αααα = .05. We reject H0.


� Example: Reed Manufacturing


Janet Reed would like to know if

there is any significant difference in

the mean number of hours worked per

week for the department managers

at her three manufacturing plants

(in Buffalo, Pittsburgh, and Detroit).




A simple random sample of five

managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide.

Conduct an F test using αααα = .05.


11

22

33

44

55

4848

5454

5757

5454

6262

7373

6363

6666

6464

7474

5151

6363

6161

5454

5656

PlantPlant 11BuffaloBuffalo

Plant 2Plant 2PittsburghPittsburgh

PlantPlant 33DetroitDetroitObservationObservation

SampleSample MeanMean

Sample VarianceSample Variance

5555 6868 5757

26.026.0 26.526.5 24.524.5




H0: µµµµ 1 = µµµµ 2 = µµµµ 3

Ha: Not all the means are equal

where:

µµµµ 1 = mean number of hours worked per

week by the managers at Plant 1





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

� p -Value and Critical Value Approaches


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



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

MSTR = 490/(3 - 1) = 245

SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490

(Sample sizes are all equal.)

Mean Square Due to Treatments

= (55 + 68 + 57)/3 = 60x


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


MSE = 308/(15 - 3) = 25.667

SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308

Mean Square Due to Error

(continued)

F = MSTR/MSE = 245/25.667 = 9.55



TreatmentTreatment

ErrorError

TotalTotal

490490

308308

798798

22

1212

1414

245245

25.66725.667

Source ofSource of

VariationVariation

Sum ofSum of

SquaresSquaresDegrees ofDegrees of

FreedomFreedom

MeanMean

SquaresSquares

9.559.55

FF


� ANOVA Table



5. Determine whether to reject5. Determine whether to reject HH00..

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

The p-value < .05, so we reject H0.

With 2 numerator d.f. and 12 denominator

d.f.,the p-value is .01 for F = 6.93. Therefore, the

p-value is less than .01 for F = 9.55.

� p -Value Approach

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


5. Determine whether to reject5. Determine whether to reject HH00..

Because F = 9.55 > 3.89, we reject H0.

� Critical Value Approach

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

Reject H0 if F > 3.89


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

Based on an F distribution with 2 numerator

d.f. and 12 denominator d.f., F.05 = 3.89.



� Summary


Multiple Comparison ProceduresMultiple Comparison Procedures

� Suppose that analysis of variance has provided

statistical evidence to reject the null hypothesis of

equal population means.

� Fisher’s least significant difference (LSD)

procedure can be used to determine where the

differences occur.


FisherFisher’’s LSD Procedures LSD Procedure

1 1MSE( )

i j

i j

x xt

n n

−=

+1 1MSE( )

i j

i j

x xt

n n

−=

+

� Test Statistic

� Hypotheses

µ µ−0 : i jH µ µ−0 : i jH

µ µ≠: a i jH µ µ≠: a i jH



where the value of ta/2 is based on a

t distribution with nT - k degrees of freedom.

� Rejection Rule

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

p-value Approach:

Critical Value Approach:

Reject Reject HH00 ifif tt < < --ttaa/2 /2 oror tt > > ttaa/2 /2


� Test Statistic


Based on the Test StatisticBased on the Test Statistic xxii -- xxjj

/21 1LSD MSE( )

i jt n nα

= +/21 1LSD MSE( )

i jt n nα

= +where

−i jx x−i jx x

Reject Reject HH00 if > LSDif > LSD−i jx x−i jx x

�Hypotheses

�Rejection Rule

µ µ−0 : i jH µ µ−0 : i jH

µ µ≠: a i jH µ µ≠: a i jH





Recall that Janet Reed wants to know

if there is any significant difference in

the mean number of hours worked per

week for the department managers

at her three manufacturing plants.

Analysis of variance has provided

statistical evidence to reject the null

hypothesis of equal population means.

Fisher’s least significant difference (LSD) procedure

can be used to determine where the differences

occur.


For αααα = .05 and nT - k = 15 – 3 = 12

degrees of freedom, t.025 = 2.179

98.6)5

15

1(667.25179.2LSD =+=

/21 1LSD MSE( )

i jt n nα

= +/21 1LSD MSE( )

i jt n nα

= +

MSE value wasMSE value was

computed earliercomputed earlier




� LSD for Plants 1 and 2



• Conclusion

• Test Statistic

−1 2x x−1 2x x = |55 - 68| = 13

Reject H0 if > 6.98−1 2x x−1 2x x

• Rejection Rule

µ µ−0 1 2: H µ µ−0 1 2: H

µ µ≠1 2: aH µ µ≠1 2: aH

• Hypotheses (A)

The mean number of hours worked at Plant 1 is

not equal to the mean number worked at Plant 2.





• Conclusion

• Test Statistic

−1 3x x−1 3x x = |55 −−−− 57| = 2

Reject H0 if > 6.98−1 3x x−1 3x x

• Rejection Rule

µ µ−0 1 3: H µ µ−0 1 3: H

µ µ≠1 3: aH µ µ≠1 3: aH

• Hypotheses (B)

There is no significant difference between the mean

number of hours worked at Plant 1 and the mean

number of hours worked at Plant 3.





• Conclusion

• Test Statistic

−2 3x x−2 3x x = |68 - 57| = 11

Reject H0 if > 6.98−2 3x x−2 3x x

• Rejection Rule

µ µ−0 2 3: H µ µ−0 2 3: H

µ µ≠2 3: aH µ µ≠2 3: aH

• Hypotheses (C)

The mean number of hours worked at Plant 2 is

not equal to the mean number worked at Plant 3.


� The experimentwise Type I error rate gets larger

for problems with more populations (larger k).

Type I Error RatesType I Error Rates

ααEWEW = 1 = 1 –– (1 (1 –– αα))((k k –– 1)!1)!

� The comparisonwise Type I error rate αααα indicates

the level of significance associated with a single

pairwise comparison.

� The experimentwise Type I error rate ααααEW is the

probability of making a Type I error on at least

one of the (k – 1)! pairwise comparisons.


An Introduction to Experimental DesignAn Introduction to Experimental Design

� Statistical studies can be classified as being either

experimental or observational.

� In an experimental study, one or more factors are

controlled so that data can be obtained about how

the factors influence the variables of interest.� In an observational study, no attempt is made to

control the factors.

� Cause-and-effect relationships are easier to

establish in experimental studies than in

observational studies.


An Introduction to Experimental DesignAn Introduction to Experimental Design

� A factor is a variable that the experimenter has

selected for investigation.

� A treatment is a level of a factor.

� Experimental units are the objects of interest in the

experiment.

� A completely randomized design is an

experimental design in which the treatments are

randomly assigned to the experimental units.

� If the experimental units are heterogeneous,

blocking can be used to form homogeneous groups,

resulting in a randomized block design.


The between-samples estimate of σ σ σ σ 2 is referred

to as the mean square due to treatments (MSTR).

2

1

( )

MSTR1

k

j jj

n x x

k

=

−

=−

∑2

1

( )

MSTR1

k

j jj

n x x

k

=

−

=−

∑

� Between-Treatments Estimate of Population Variance

Completely Randomized DesignCompletely Randomized Design

denominator is thedenominator is the

degrees of freedomdegrees of freedom

associated with SSTRassociated with SSTR

numerator is callednumerator is called

the the sum of squares duesum of squares due

to treatmentsto treatments (SSTR)(SSTR)


The second estimate of σ σ σ σ 2, the within-samples estimate, is referred to as the mean square due to error (MSE).

�Within-Treatments Estimate of Population Variance


denominator is thedenominator is thedegrees of freedomdegrees of freedomassociated with SSEassociated with SSE

numerator is callednumerator is calledthe the sum of squaressum of squaresdue to errordue to error (SSE)(SSE)

MSE =

−∑

−

=

( )n s

n k

j jj

k

T

1 2

1MSE =

−∑

−

=

( )n s

n k

j jj

k

T

1 2

1


MSTRSSTR

-=k 1

MSTRSSTR

-=k 1

MSESSE

-=n kT

MSESSE

-=n kT

MSTR

MSE

MSTR

MSE

�ANOVA Table





MeanMeanSquaresSquares FF

TreatmentsTreatments

ErrorError

TotalTotal

kk -- 11

nnTT -- 11

SSTRSSTR

SSESSE

SSTSST

nnTT -- kk


AutoShine, Inc. is considering marketing a long-

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

and Type 3) have been developed.


� Example: AutoShine, Inc.

In order to test the durability

of these waxes, 5 new cars were

waxed with Type 1, 5 with Type

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

repeatedly run through an automatic carwash until the

wax coating showed signs of deterioration.



The number of times each car went through the

carwash is shown on the next slide. AutoShine,

Inc. must decide which wax to market. Are the

three waxes equally effective?

� Example: AutoShine, Inc.


11

22

33

44

55

2727

3030

2929

2828

3131

3333

2828

3131

3030

3030

2929

2828

3030

3232

3131

Sample MeanSample Mean

Sample VarianceSample Variance

ObservationObservationWaxWaxType 1Type 1

WaxWaxType 2Type 2

WaxWaxType 3Type 3

2.52.5 3.33.3 2.52.5

29.029.0 30.430.4 30.030.0



�Hypotheses


where:

µµµµ1 = mean number of washes for Type 1 wax



H0: µµµµ1 = µµµµ2 = µµµµ3

Ha: Not all the means are equal


Because the sample sizes are all equal:


MSE = 33.2/(15 - 3) = 2.77

MSTR = 5.2/(3 - 1) = 2.6

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

SSTR = 5(29–29.8)2 + 5(30.4–29.8)2 + 5(30–29.8)2 = 5.2

�Mean Square Error

�Mean Square Between Treatments

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


�Rejection Rule


where F.05 = 3.89 is based on an F distribution

with 2 numerator degrees of freedom and 12

denominator degrees of freedom

p-Value Approach: Reject H0 if p-value < .05

Critical Value Approach: Reject H0 if F > 3.89


� Test Statistic


There is insufficient evidence to conclude that

the mean number of washes for the three wax

types are not all the same.

�Conclusion

F = MSTR/MSE = 2.6/2.77 = .939

The p-value is greater than .10, where F = 2.81.

(Excel provides a p-value of .42.)

Therefore, we cannot reject H0.







ErrorError

TotalTotal

22

1414

5.25.2

33.233.2

38.438.4

1212


2.602.60

2.772.77

.939.939

�ANOVA Table


• For a randomized block design the sum of

squares total (SST) is partitioned into three

groups: sum of squares due to treatments, sum

of squares due to blocks, and sum of squares due

to error.

�ANOVA Procedure

Randomized Block DesignRandomized Block Design

SST = SSTR + SSBL + SSESST = SSTR + SSBL + SSE

• The total degrees of freedom, nT - 1, are partitioned

such that k - 1 degrees of freedom go to treatments, b - 1 go to blocks, and (k - 1)(b - 1) go to the error

term.


MSTRSSTR

-=k 1

MSTRSSTR

-=k 1

MSTR

MSE

MSTR

MSE






ErrorError

TotalTotal

kk -- 11

nnTT -- 11

SSTRSSTR

SSESSE

SSTSST


�ANOVA Table

BlocksBlocks SSBLSSBL bb -- 11

((k k –– 1)(1)(bb –– 1)1)

=SSBL

MSBL-1b

=SSBL

MSBL-1b

MSESSE

=− −( )( )k b1 1

MSESSE

=− −( )( )k b1 1



� Example: Crescent Oil Co.

Crescent Oil has developed three

new blends of gasoline and must

decide which blend or blends to

produce and distribute. A study

of the miles per gallon ratings of the

three blends is being conducted to determine if the

mean ratings are the same for the three blends.



� Example: Crescent Oil Co.

Five automobiles have been

tested using each of the three

gasoline blends and the miles

per gallon ratings are shown on

the next slide.



29.8 29.8 28.828.8 28.428.4TreatmentTreatmentMeansMeans

11

22

33

44

55

3131

3030

2929

3333

2626

3030

2929

2929

3131

2525

3030

2929

2828

2929

2626

30.33330.333

29.33329.333

28.66728.667

31.00031.000

25.66725.667

Type of Gasoline (Treatment)Type of Gasoline (Treatment)BlockBlockMeansMeansBlend XBlend X Blend YBlend Y Blend ZBlend Z

AutomobileAutomobile(Block)(Block)


�Mean Square Due to Error


MSE = 5.47/[(3 - 1)(5 - 1)] = .68

SSE = 62 - 5.2 - 51.33 = 5.47

MSBL = 51.33/(5 - 1) = 12.8

SSBL = 3[(30.333 - 29)2 + . . . + (25.667 - 29)2] = 51.33

MSTR = 5.2/(3 - 1) = 2.6

SSTR = 5[(29.8 - 29)2 + (28.8 - 29)2 + (28.4 - 29)2] = 5.2

The overall sample mean is 29. Thus,�Mean Square Due to Treatments

�Mean Square Due to Blocks







ErrorError

TotalTotal

22

1414

5.205.20

5.475.47

62.0062.00

88

2.602.60

.68.68

3.823.82

�ANOVA Table


BlocksBlocks 51.3351.33 12.8012.8044


�Rejection Rule


For αααα = .05, F.05 = 4.46

(2 d.f. numerator and 8 d.f. denominator)

p-Value Approach: Reject H0 if p-value < .05

Critical Value Approach: Reject H0 if F > 4.46


�Conclusion


There is insufficient evidence to conclude that

the miles per gallon ratings differ for the three

gasoline blends.

The p-value is greater than .05 (where F =

4.46) and less than .10 (where F = 3.11). (Excel

provides a p-value of .07). Therefore, we cannot

reject H0.

F = MSTR/MSE = 2.6/.68 = 3.82� Test Statistic


Factorial ExperimentsFactorial Experiments

� In some experiments we want to draw conclusions about more than one variable or factor.

� Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required.

� For example, for a levels of factor A and b levels of factor B, the experiment will involve collecting data on ab treatment combinations.

� The term factorial is used because the experimental conditions include all possible combinations of the factors.


• The ANOVA procedure for the two-factor

factorial experiment is similar to the completely

randomized experiment and the randomized block

experiment.

�ANOVA Procedure

SST = SSA + SSB + SSAB + SSESST = SSA + SSB + SSAB + SSE

• The total degrees of freedom, nT - 1, are partitioned

such that (a – 1) d.f go to Factor A, (b – 1) d.f go to

Factor B, (a – 1)(b – 1) d.f. go to Interaction, and

ab(r – 1) go to Error.

TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment

• We again partition the sum of squares total (SST)

into its sources.


=SSA

MSA-1a

=SSA

MSA-1a

MSA

MSE

MSA

MSE





Factor AFactor A

ErrorError

TotalTotal

aa -- 11

nnTT -- 11

SSASSA

SSESSE

SSTSST

Factor BFactor B SSBSSB bb -- 11

abab((rr –– 1)1) =−

SSEMSE

( 1)ab r=

−

SSEMSE

( 1)ab r


=SSB

MSB-1b

=SSB

MSB-1b

InteractionInteraction SSABSSAB ((a a –– 1)(1)(bb –– 1)1) =− −

SSABMSAB

( 1)( 1)a b=

− −

SSABMSAB

( 1)( 1)a b

MSB

MSE

MSB

MSE

MSAB

MSE

MSAB

MSE


� Step 3 Compute the sum of squares for factor B

2

1 1 1

SST = ( )a b r

ijki j k

x x= = =

−∑∑∑2

1 1 1

SST = ( )a b r

ijki j k

x x= = =

−∑∑∑

2

1

SSA = ( . )a

ii

br x x=

−∑2

1

SSA = ( . )a

ii

br x x=

−∑

2

1

SSB = ( . )b

jj

ar x x=

−∑2

1

SSB = ( . )b

jj

ar x x=

−∑


� Step 1 Compute the total sum of squares

� Step 2 Compute the sum of squares for factor A


� Step 4 Compute the sum of squares for interaction

2

1 1

SSAB = ( . . )a b

ij i ji j

r x x x x= =

− − +∑∑2

1 1

SSAB = ( . . )a b

ij i ji j

r x x x x= =

− − +∑∑


SSE = SST SSE = SST –– SSA SSA –– SSB SSB -- SSABSSAB

� Step 5 Compute the sum of squares due to error


A survey was conducted of hourly wages

for a sample of workers in two industries

at three locations in Ohio. Part of the

purpose of the survey was to

determine if differences exist

in both industry type and

location. The sample data are shown

on the next slide.

� Example: State of Ohio Wage Survey



� Example: State of Ohio Wage Survey


12.7012.5012.00II

12.1012.0012.50II

13.0012.6012.40II

12.2012.0012.10I

12.7011.2011.80I

12.9011.8012.10I

ColumbusClevelandCincinnatiIndustry


� Factors


• Each experimental condition is repeated 3 times

• Factor B: Location (3 levels)

• Factor A: Industry Type (2 levels)

�Replications






Factor AFactor A

ErrorError

TotalTotal

11

1717

.50.50

1.431.43

3.423.42

1212

.50.50

.12.12

4.194.19

�ANOVA Table

Factor BFactor B 1.121.12 .56.5622


InteractionInteraction .37.37 .19.1922

4.694.69

1.551.55


�Conclusions Using the p-Value Approach


(p-values were found using Excel)

Interaction is not significant.

•Interaction: p-value = .25 > αααα = .05

Mean wages differ by location.

•Locations: p-value = .03 < αααα = .05

Mean wages do not differ by industry type.

•Industries: p-value = .06 > αααα = .05


�Conclusions Using the Critical Value

Approach


Interaction is not significant.

•Interaction: F = 1.55 < Fαααα

= 3.89

Mean wages differ by location.

•Locations: F = 4.69 > Fαααα= 3.89

Mean wages do not differ by industry type.

•Industries: F = 4.19 < Fαααα

= 4.75

chapter 13 statistics in practice

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