Statistics for Business and Economics Chapter 8 Design of Experiments and Analysis of Variance.

Statistics for Business and Economics

Chapter 8

Design of Experiments and Analysis of Variance

Learning Objectives

1. Describe Analysis of Variance (ANOVA)

2. Explain the Rationale of ANOVA

3. Compare Experimental Designs

4. Test the Equality of 2 or More Means• Completely Randomized Design• Factorial Design

Experiments

Experiment

• Investigator controls one or more independent variables

– Called treatment variables or factors– Contain two or more levels (subcategories)

• Observes effect on dependent variable – Response to levels of independent variable

• Experimental design: plan used to test hypotheses

Examples of Experiments

1. Thirty stores are randomly assigned 1 of 4 (levels) store displays (independent variable) to see the effect on sales (dependent variable).

2. Two hundred consumers are randomly assigned 1 of 3 (levels) brands of juice (independent variable) to study reaction (dependent variable).

Experimental Designs

Factorial

One-Way ANOVA

Completely Randomized

Two-Way ANOVA

Completely Randomized Design

Factorial

One-Way ANOVA

Two-Way ANOVA

• Experimental units (subjects) are assigned randomly to treatments

– Subjects are assumed homogeneous

• One factor or independent variable– Two or more treatment levels or

classifications

• Analyzed by one-way ANOVA

Factor (Training Method)

Factor levels(Treatments)

Level 1 Level 2 Level 3

Experimentalunits

Dependent 21 hrs. 17 hrs. 31 hrs.

variable 27 hrs. 25 hrs. 28 hrs.

(Response) 29 hrs. 20 hrs. 22 hrs.

Randomized Design Example

One-Way ANOVA F-Test

Factorial

One-Way ANOVA

Two-Way ANOVA

One-Way ANOVA F-Test

• Tests the equality of two or more (k) population means

• Variables– One nominal scaled independent variable

Two or more (k) treatment levels or classifications

– One interval or ratio scaled dependent variable

• Used to analyze completely randomized experimental designs

Conditions Required for a Valid ANOVA F-test:

1. Randomness and independence of errors• Independent random samples are drawn

2. Normality• Populations are approximately normally

distributed

3. Homogeneity of variance• Populations have equal variances

One-Way ANOVA F-Test Hypotheses

• H0: 1 = 2 = 3 = ... = k

— All population means are equal

— No treatment effect

• Ha: Not All i Are Equal— At least 2 pop. means

are different— Treatment effect—1 2 ... k is

1 = 2 = 3

1 2 3X

Why Variances?

• Same treatment variation

• Different random variation

Possible to conclude means are equal!

Pop 1 Pop 2 Pop 3

Pop 4 Pop 6Pop 5

Variances WITHIN differ

A Pop 1 Pop 2 Pop 3

Pop 4 Pop 6Pop 5

Variances AMONG differ

Different treatment variation

Same random variation

1. Compares two types of variation to test equality of means

2. Comparison basis is ratio of variances

3. If treatment variation is significantly greater than random variation then means are not equal

4. Variation measures are obtained by ‘partitioning’ total variation

One-Way ANOVA Basic Idea

One-Way ANOVA Partitions Total Variation

Total variationTotal variation

Variation due to treatment

Variation due to random samplingVariation due to

random sampling

• Sum of Squares Among• Sum of Squares Between• Sum of Squares Treatment• Among Groups Variation

• Sum of Squares Within• Sum of Squares Error• Within Groups Variation

Partitioning the Variation• Total variation can be split into two parts:

SST = Total Sum of Squares (Total variation)

SSA = Sum of Squares Among Groups (Treatment or Among-group variation)

SSW = Sum of Squares Within Groups (Random or Within-group variation)

SST = SSA + SSW

Total Variation

Group 1 Group 2 Group 3

Response, X

11 21 ijSS Total X X X X X X

Among Group Variation

Response, X

1 1 2 2 j jSSA n X X n X X n X X

With Group (Random) Variation

Response, X

11 1 21 2 ij jSSW X X X X X X

2 2 2 211 21 12 22

2 211 1 1 21 2 2

2 212 1 1 22 2 2

2 211 1 1 11 1 1

2 221 2 2 21 2

Consider the case i=2 and j=2

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) 2( )( )

( ) ( ) 2( )(

SST X X X X X X X X

X X X X X X X X

X X X X X X X

2 212 1 1 12 1 1

2 222 2 2 22 2 2

( ) ( ) 2( )( )

X X X X X X X X

2 211 1 1 11 1 1

2 221 1 1 21 1 1

2 212 2 2 12 2 2

The first two items can be rewritten as

( ) ( ) 2( )( )

Likewise, the last two items can be rewritten as

( ) ( ) 2( )( )

X X X X X X X X

22 22 2 2( ) 2( )( )X X X X X X

1 11 12 1

2 21 22 2

2 2 2 211 1 21 1 12 2 22 2

2 2 2 21 1 2 2

2 2 211 1 21 1 12 2

Because 2( )( 2 ) equals to zero

and 2( )( 2 ) equals to zero

, we have

( ) ( ) ( ) ( )

( ) ( ) ( ) (

X X X X X

SST X X X X X X X X

X X X X X X X X

X X X X X X

2 21 2

2( ) 2( )

X X X X

SSW SSA

Obtaining the Mean Squares

SSWMSW

SSAMSA

SSTMST

The Mean Squares are obtained by dividing the various sum of squares by their associated degrees of freedom

Mean Square Among(d.f. = c-1)

Mean Square Within(d.f. = n-c)

Mean Square Total(d.f. = n-1)

One-Way ANOVA Table

Source of Variation

Sum OfSquares

Degrees ofFreedom

Mean Square(Variance)

Among Groups

c - 1 MSA =

Within Groups

SSWn - c MSW =

Total SSTn – 1

c = number of groupsn = sum of the sample sizes from all groupsdf = degrees of freedom

FSTAT =

One-Way ANOVA F-Test Critical Value

If means are equal, F = MSA / MSW 1. Only reject large F!

Always One-Tail!

F(α; c – 1, n – c)

Reject H 0

Do NotReject H 0

One-Way ANOVA F-Test Example

As production manager, you want to see if three filling machines have different mean filling times. You assign 15 similarly trained and experienced workers, 5 per machine, to the machines. At the .05 level of significance, is there a difference in mean filling times?

Mach1Mach1 Mach2Mach2 Mach3Mach325.4025.40 23.4023.40 20.0020.0026.3126.31 21.8021.80 22.2022.2024.1024.10 23.5023.50 19.7519.7523.7423.74 22.7522.75 20.6020.6025.1025.10 21.6021.60 20.4020.40

One-Way ANOVA F-Test Solution

• H0:• Ha:• = • 1 = 2 = • Critical Value(s):

F0 3.89

1 = 2 = 3

Not All Equal

Summary TableSolution

From Computer

Treatment (Machines)

3 - 1 = 2 47.1640 23.5820 25.60

Error 15 - 3 = 12 11.0532 .9211

Total 15 - 1 = 14 58.2172

Source of Variation

Degreesof

Freedom

Sum of Squares

Mean Square

(Variance)F

One-Way ANOVA F-Test Solution

Test Statistic:

Decision:

Conclusion:

Reject at = .05

There is evidence population means are different

23 5820

921125.6

表 14.1 A 、 B 、 C 廠牌汽車耗油試驗

表 14.5 變異數分析表

變異來源平方和（SS）自由度（df）平均平方和（MS） F值

因子 17.652 2 8.83 55.19

隨機 1.868 12 0.16

總和 19.520 14

圖 14.6 汽車耗油的檢定

f (F )

拒絕域接受域

3.89 55.19 F

F 2,12

One-Way ANOVA F-Test Thinking Challenge

You’re a trainer for Microsoft Corp. Is there a difference in mean learning times of 12 people using 4 different training methods ( =.05)?

M1 M2 M3 M410 11 13 18

9 16 8 235 9 9 25

Use the following table.

Summary Table Solution*

Treatment(Methods)

4 - 1 = 3 348 116 11.6

Error 12 - 4 = 8 80 10

Total 12 - 1 = 11 428

Source of Variation

Degreesof

Freedom

Sum of Squares

Mean Square

(Variance)F

One-Way ANOVA F-Test Solution*

0 4.07

1 = 2 = 3 = 4

Not All Equal

One-Way ANOVA F-Test Solution*

Test Statistic:

Decision:

Conclusion:

Reject at = .05

1011 6.

ANOVA Assumptions• Randomness and Independence

– Select random samples from the c groups (or randomly assign the levels)

• Normality– The sample values for each group are from a normal

population

• Homogeneity of Variance– All populations sampled from have the same variance

– Can be tested with Levene’s Test

Randomized Block Design

• Reduces sampling variability (MSE)

• Matched sets of experimental units (blocks)

• One experimental unit from each block is randomly assigned to each treatment

Randomized Block Design Total Variation Partitioning

Variation Due to Random Sampling

Total Variation

Variation Due to Blocks

Variation Due to Treatment

Conditions Required for a Valid ANOVA F-test:

Randomized Block Design

1. The blocks are randomly selected, and all treatments are applied (in random order) to each block

2. The distributions of observations corresponding to all block-treatment combinations are approximately normal

3. All block-treatment distributions have equal variances

Partitioning the Variation

• Total variation can now be split into three parts:

SST = Total variationSSA = Among-Group variationSSBL = Among-Block variationSSE = Random variation

SST = SSA + SSBL + SSE

Sum of Squares for Blocks

Where:

c = number of groups

r = number of blocks

Xi. = mean of all values in block i

X = grand mean (mean of all data values)

2i. )XX(cSSBL

Partitioning the Variation• Total variation can now be split into three parts:

SST and SSA are computed as they were in One-Way ANOVA

SSE = SST – (SSA + SSBL)

Mean Squares

SSAgroups among square MeanMSA

SSBMSBL Mean square blocking

SSEMSE Mean square error

( 1)( 1)r c

Randomized Block ANOVA Table

Source of Variation

dfSS MS

Among Groups

SSA MSA

Error n-r-c+1SSE MSE

Total n - 1SST

c - 1 MSA

c = number of populations n=rc = total number of observationsr = number of blocks df = degrees of freedom

Among Blocks SSB r - 1 MSB

• Main Factor test: df1 = c – 1

df2 = (r – 1)(c – 1)

c..3.2.10 μμμμ:H

equal are means population all Not:H1

FSTAT =

Reject H0 if FSTAT > Fα

Testing For Factor Effect

Test For Block Effect

• Blocking test: df1 = r – 1

df2 = (r – 1)(c – 1)

r.3.2.1.0 ...:H μμμμ

equal are means block all Not:H1

FSTAT =

Reject H0 if FSTAT > Fα

Randomized Block Design Example

A production manager wants to see if three assembly methods have different mean assembly times (in minutes). Five employees were selected at random and assigned to use each assembly method. At the .05 level of significance, is there a difference in mean assembly times?EmployeeEmployee Method 1Method 1 Method 2Method 2 Method 3Method 3

11 5.45.4 3.63.6 4.04.022 4.14.1 3.83.8 2.92.933 6.16.1 5.65.6 4.34.344 3.63.6 2.32.3 2.62.655 5.35.3 4.74.7 3.43.4

Random Block Design F-Test Solution*

• H0:• Ha:• =• 1 = 2 = • Critical Value(s):

F0 4.46

1 = ., 2 = ., 3

Not all equal.05

Summary Table Solution*

Treatment(Methods)

3 - 1 = 2 5.43 2.71 12.9

Error 15 - 3 - 5 + 1 = 8 = 2*4

1.68 .21

Total 15 - 1 = 14 17.8

Source of Variation

Degreesof

Freedom

Sum of Squares

Mean Square

(Variance)F

Block(Employee)

5 - 1 = 4 10.69 2.67 12.7

Test Statistic:

Decision:

Conclusion:

Reject at = .05

.2112.9

• H0:• Ha:• =• 1 = 2 = • Critical Value(s):

F0 4.46

1,. = 2 ,. = 3 ,.

Not all equal.05

Test Statistic:

Decision:

Conclusion:

Reject at = .05

There is evidence block means are different

.2112.7

Factorial Experiments

Factorial

One-Way ANOVA

Two-Way ANOVA

Factorial Design

• Experimental units (subjects) are assigned randomly to treatments

– Subjects are assumed homogeneous

• Two or more factors or independent variables

– Each has two or more treatments (levels)

• Analyzed by two-way ANOVA

Two-Way ANOVA Data Table

Level i Factor

Level j Factor

Observation k

Factor Factor BA 1 2 ... b

1 X111 X121 ... X1b1

X112 X122 ... X1b2

2 X211 X221 ... X2b1

X212 X222 ... X2b2

: : : : :

a Xa11 Xa21 ... Xab1

Xa12 Xa22 ... Xab2

Treatment

Factorial Design Example

Factor 2 (Training Method)FactorLevels

Level 1 Level 2 Level 3

Level 1 15 hr. 10 hr. 22 hr.Factor 1(Motivation)

(High)11 hr. 12 hr. 17 hr.

Level 2 27 hr. 15 hr. 31 hr.(Low)

29 hr. 17 hr. 49 hr.Treatment

Advantages of Factorial Designs

• Saves time and effort– e.g., Could use separate completely

randomized designs for each variable

• Controls confounding effects by putting other variables into model

• Can explore interaction between variables

Graphs of Interaction

Effects of motivation (high or low) and training method (A, B, C) on mean learning time

Interaction No Interaction

AverageResponse

Two-Way ANOVA

Factorial

One-Way ANOVA

Two-Way ANOVA

• Tests the equality of two or more population means when several independent variables are used

• Same results as separate one-way ANOVA on each variable

– No interaction can be tested

• Used to analyze factorial designs

Interaction

• Occurs when effects of one factor vary according to levels of other factor

• When significant, interpretation of main effects (A and B) is complicated

• Can be detected

– In data table, pattern of cell means in one row differs from another row

– In graph of cell means, lines cross

Graphs of Interaction

Effects of motivation (high or low) and training method (A, B, C) on mean learning time

Interaction No Interaction

AverageResponse

Two-Way ANOVA Total Variation Partitioning

Variation Due to Random Sampling

Variation Due to Interaction

SS(AB)

Total Variation

Variation Due to Treatment A

Variation Due to Treatment B

SSA SSB

Conditions Required for Valid F-Tests in Factorial

Experiments

1. Normality• Populations are approximately normally

distributed

2. Homogeneity of variance• Populations have equal variances

3. Independence of errors• Independent random samples are drawn

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquare

A(Row)

r - 1 SS(A) MS(A) MS(A)MSE

B(Column)

c - 1 SS(B) MS(B) MS(B)MSE

AB(Interaction)

(r - 1)(c - 1) SS(AB) MS(AB) MS(AB)MSE

Error n - rc SSE MSE

Total n - 1 SS(Total)

Two-Way ANOVA Summary Table

Same as other designs

Two-Way ANOVA Hypotheses

• Test for Main Effect of Factor AH0: No difference among mean levels of factor A

Ha: At least two factor A mean levels differ

• Test StatisticF = MS(A) / MSE

• Degrees of Freedom1 = (r – 1) 2 = n – rc

• Test for Main Effect of Factor BH0: No difference among mean levels of factor B

Ha: At least two factor B mean levels differ

• Test StatisticF = MS(B) / MSE

• Degrees of Freedom1 = (c – 1) 2 = n – rc

• Test for Factor InteractionH0: The factors do not interact

Ha: The factors do interact

• Test StatisticF = MS(AB) / MSE

• Degrees of Freedom1 = (r – 1)(c – 1) 2 = n – rc

• Test for Treatment MeansH0: The ab treatment means are equal

Ha: At least two of the treatment means differ

• Test StatisticF = MST / MSE

• Degrees of Freedom1 = rc – 1 2 = n – rc

Two-Way ANOVA Equations

( )r c n

ijki j k

SST X X

..i )XX(ncSSA

.j. )XX(nrSSB

Total Variation:

Factor A Variation:

Factor B Variation:

.j.i..ij. )XXXX(nSSAB

2.ijijk )XX(SSE

Interaction Variation:

Sum of Squares Error:

(continued)

where:

Mean Grandnrc

r) ..., 2, 1, (i A factor of level i of Meannc

r = number of levels of factor A

c = number of levels of factor B

n’ = number of replications in each cell

(continued)

where:

c) ..., 2, 1, (j B factor of level j of Meannr

ij cell of Meann

ijk.ij

r = number of levels of factor A

c = number of levels of factor B

n’ = number of replications in each cell

(continued)

Factorial Design Example

Human Resources wants to determine if training time is different based on motivation level and training method. Conduct the appropriate ANOVA tests. Use α = .05 for each test.

Training MethodFactorLevels

Self–paced Classroom Computer

15 hr. 10 hr. 22 hr.

Motivation

High11 hr. 12 hr. 17 hr.

27 hr. 15 hr. 31 hr.Low

29 hr. 17 hr. 49 hr.

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquare

A(Row)

1 546.75 546.75

B(Column)

2 531.5 265.75

AB(Interaction)

2 123.5 61.76

Error 6 188.5 31.42

Total 11 SS(Total)

Main Factor A F-Test Solution

• H0:

• Ha:• =• 1 = 2 = • Critical Value(s):

F0 5.99

No difference between motivation levels

Motivation levels differ

Main Factor A F-Test Solution

Test Statistic:

Decision:

Conclusion:

( )17.4

Reject at = .05

There is evidence motivation levels differ

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquare

A(Row)

1 546.75 546.75

B(Column)

2 531.5 265.75

AB(Interaction)

2 123.5 61.76

Error 6 188.5 31.42

Total 11 SS(Total)

Main Factor B F-Test Solution

• H0:

• Ha:• =• 1 = 2 = • Critical Value(s):

F0 5.14

No difference between training methods

Training methods differ

Main Factor B F-Test Solution

Test Statistic:

Decision:

Conclusion:

( )8.46

Reject at = .05

There is evidence training methods differ

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquare

A(Row)

1 546.75 546.75

B(Column)

2 531.5 265.75

AB(Interaction)

2 123.5 61.76

Error 6 188.5 31.42

Total 11 SS(Total)

Interaction F-Test Solution

F0 5.14

The factors do not interact

The factors interact

Interaction F-Test Solution

Test Statistic:

Decision:

Conclusion:

( )1.97

MS ABF

Do not reject at = .05

There is no evidence the factors interact

Treatment Means F-Test Solution

• H0:

• Ha:• = • 1 = 2 = • Critical Value(s):

F0 4.39

The 6 treatmentmeans are equal

At least 2 differ

.05 5 6

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquare

Model 5 546.75 240.35

Error 6 188.5 31.42

CorrectedTotal

11 735.25

Treatment Means F-Test Solution

Test Statistic:

Decision:

Conclusion:

7.65MST

Reject at = .05

Conclusion

1. Described Analysis of Variance (ANOVA)

2. Explained the Rationale of ANOVA

3. Compared Experimental Designs

4. Tested the Equality of 2 or More Means• Completely Randomized Design• Factorial Design

Statistics for Business and Economics Chapter 8 Design of Experiments and Analysis of Variance.

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