Lecture 2.21 2015 Michael Stuart Design and Analysis of Experiments Lecture 2.2 1.Review Lecture 2.1 –Minute test –Why block? –Deleted residuals 2.Interaction

Lecture 2.2 1 2015 Michael Stuart

Design and Analysis of ExperimentsLecture 2.2

1. Review Lecture 2.1

– Minute test– Why block?– Deleted residuals

2. Interaction

3. Random Block Effects

4. Introduction to 2-level factorial designs

– a 22 experiment– introducing the Design Matrix

Certificate in StatisticsDesign and Analysis of Experiments


How Much

5432



How Fast

4321



Why block?• Blocking is useful when there are

known external factors (covariates)that affect variation between plots.

• Blocking reduces bias arising due toblock effects disproportionately affecting factor effectsdue to levels disproportionally allocated to blocks.

• Neighbouring plots are likely to bemore homogeneous than separated plots, so that

– blocking reduces variation in resultswhen treatments are compared within blocks

– (and increases precisionwhen results are combined across blocks).



Deleted residuals

Minitab does this automatically for all cases!

They are used to allow each case to be assessed using a criterion not affected by the case.

The residuals are not deleted,

it is the case that is deleted

while the corresponding "deleted residual is calculated

Simple linear regression illustrates:



Scatterplot



Scatterplot



Scatterplot



Deleted residual

Given an exceptional case,

deleted residual > residual using all the data

deleted s < s using all the data

deleted standardised residual

>> standardised residual using all the data

Using deleted residuals accentuates exceptional cases



Residuals






2. Interaction






Multi-factor designsreveal interaction

Pressure

Temperature

High

High

Low

Low65

75

70

60



Interaction defined

Factors interact when the effect of changing one factor depends on the level of the other.

Interaction displayed



Iron-deficiency anemia

Contributory factors:

– cooking pot type

• Aluminium (A), Clay (C) and Iron (I)

– food type

• Meat (M), Legumes (L) and Vegetables (V)

Pot Type Food Type

Meat Legumes Vegetables

Aluminium 1.77 2.36 1.96 2.14 2.40 2.17 2.41 2.34 1.03 1.53 1.07 1.30

Clay 2.27 1.28 2.48 2.68 2.41 2.43 2.57 2.48 1.55 0.79 1.68 1.82

Iron 5.27 5.17 4.06 4.22 3.69 3.43 3.84 3.72 2.45 2.99 2.80 2.92



Interaction

Pot Type Vegetable Type


Aluminium 2.06 2.33 1.23

Clay 2.18 2.47 1.46

Iron 4.68 3.67 2.79



Interaction


Legumes Vegetable Meat

Aluminium 2.33 1.23 2.06

Change effect 0.14 0.23 0.12

Clay 2.47 1.46 2.18


Iron 3.67 2.79 4.68


Interaction


Legumes Vegetable Meat

Aluminium 2.33 1.23 2.06


Clay 2.47 1.46 2.18


Iron 3.67 2.79 4.68


Two 2-level factors

Pressure

Temperature

High

High

Low

Low65

75

70

60 Pressure effect

Low T: 60 – 65 = –5High T: 75 – 70 = +5Diff: 5 – (–5) = 10

Temperature effect

Low P: 70 – 65 = 5High P: 75 – 60 = 15Diff: 15 – 5 = 10

Postgraduate Certificate in Statistics Design and Analysis of Experiments


Model for analysis

Iron content includes– a contribution for each food type

plus– a contribution for each pot type

plus– a contribution for each food type / pot type

combination

plus– a contribution due to chance variation



Model for analysis

Y = m + a + + b ab + e

where

m is the overall mean,

a is the food effect, above or below the mean, depending on which food type is used,

b is the pot effect, above or below the mean, depending on which pot type is involved

ab is the food/pot interaction effect, depending on which food type / pot type combination is used

e represents chance variation



Estimating the model

Food Type

PotMeans

PotMain Effects

M L V

A 2.1 2.3 1.2 1.9 1.9 – 2.5 = – 0.6

PotType C 2.2 2.5 1.5 2.0 2.0 – 2.5 = – 0.5

I 4.7 3.7 2.8 3.7 3.7 – 2.5 = + 1.2

FoodMeans 3.0 2.8 1.8 2.5

FoodMain

Effects3.0 – 2.5= +0.5

2.8 – 2.5= +0.3

1.8 – 2.5= – 0.7



Estimating the model

Food Type

PotMeans

PotMain Effects

M L V

A 2.1 2.3 1.2 1.9 1.9 – 2.5 = – 0.6

PotType C 2.2 2.5 1.5 2.0 2.0 – 2.5 = – 0.5

I 4.7 3.7 2.8 3.7 3.7 – 2.5 = + 1.2

FoodMeans 3.0 2.8 1.8 2.5

FoodMain

Effects3.0 – 2.5= +0.5

2.8 – 2.5= +0.3

1.8 – 2.5= – 0.7



Interaction effects


M L V PotEffects

A 2.1 2.3 1.2 -0.7C 2.2 2.5 1.5 -0.5I 4.7 3.7 2.8 1.2

FoodEffects 0.4 0.3 -0.7 2.5

Interaction Effects

-0.2 0.2 0.1-0.3 0.2 0.10.5 -0.3 -0.2


Interaction effects


M L V PotEffects

A 2.1 2.3 1.2 -0.7C 2.2 2.5 1.5 -0.5I 4.7 3.7 2.8 1.2

FoodEffects 0.4 0.3 -0.7 2.5

Interaction Effects

-0.2 0.2 0.1-0.3 0.2 0.10.5 -0.3 -0.2


Estimating s

Calculate s from each cell, based on 4 – 1 = 3 df,

Estimate is average across all 9 cells, with 9 x 3 = 27 df

Pot Type Vegetable Type


Aluminium 1.77 2.36 1.96 2.14 2.40 2.17 2.41 2.34 1.03 1.53 1.07 1.30

Clay 2.27 1.28 2.48 2.68 2.41 2.43 2.57 2.48 1.55 0.79 1.68 1.82

Iron 5.27 5.17 4.06 4.22 3.69 3.43 3.84 3.72 2.45 2.99 2.80 2.92



Analysis of Variance

SS(Total)

= SS(Pot effects)

+ SS(Food effects)

+ SS(Interaction effects)

+ SS(Error)

Source DF SS MS F-Value P-ValuePot 2 24.9 12.4 92.3 0.000Food 2 9.3 4.6 34.5 0.000Pot*Food 4 2.6 0.66 4.9 0.004Error 27 3.6 0.13Total 35 40.5



Recall: Case StudyReducing yield loss in a chemical process

• Process: chemicals blended, filtered and dried

• Problem: yield loss at filtration stage

• Proposal: adjust initial blend to reduce yield loss

• Plan:

– prepare five different blends

– use each blend in successive process runs, in random order

– repeat at later times (blocks)



Results

Ref: BlendLoss.xls

Block Run Blend Loss, per cent

I 1 B 18.2 2 A 16.9 3 C 17.0 4 E 18.3 5 D 15.1 II 6 A 16.5 7 E 18.3 8 B 19.2 9 C 18.1 10 D 16.0 III 11 B 17.1 12 D 17.8 13 C 17.3 14 E 19.8 15 A 17.5



Initial data analysis

• Little variation between blocks

• More variation between blends

• Disturbing interaction pattern; see laterPostgraduate Certificate in Statistics Design and Analysis of Experiments


Analysis of Variance

Blend Loss analysis model included

Blend effects + Block effects + Chance variation,

– NO INTERACTION EFFECTS

Analysis of Variance for Loss

Source DF Seq SS Adj SS Adj MS F P

Blend 4 11.5560 11.5560 2.8890 3.31 0.071Block 2 1.6480 1.6480 0.8240 0.94 0.429Error 8 6.9920 6.9920 0.8740

Total 14 20.1960



Include interaction in model?

Analysis of Variance for Loss

Source DF Adj SS Adj MS F P

Blend 4 11.5560 2.8890 **

Block 2 1.6480 0.8240 **

Blend*Block 8 6.9920 0.8740 **

Error 0 * * *

Total 14 20.1960



ANOVA with no replication

Recall F-test logic:

– MS(Error) ≈ s2

– MS(Effect) ≈ s2 + effect contribution

– F = MS(Effect) / MS(Error) ≈ 1 if effect absent,

>>1 if effect present

No replication?

use MS(Interaction) as MS(Error)

If Block by Treatment interaction is absent,– OK

If Block by Treatment interaction is present,– conservative test



Fitted values show no interaction.


Recall : Estimating the model


Classwork 2.1.2Calculate fitted values

Diploma in StatisticsDesign and Analysis of Experiments

Block 1

–0.4 Block 2

0.1 Block 3

0.4

Blend A –0.6

16.5 17.0 17.3

Blend B 0.6

17.7 18.2 18.5

Blend C –0.1

17.0 17.5 17.8

Blend D –1.2

15.9 16.4 16.7

Blend E 1.3

18.4 18.9 19.2

ˆˆˆ

17.5 +


Classwork 2.1.2 (cont'd)

Make a Block profile plot

Block 1

–0.4 Block 2

0.1 Block 3

0.4

Blend A –0.6

16.5 17.0 17.3

Blend B 0.6

17.7 18.2 18.5

Blend C –0.1

17.0 17.5 17.8

Blend D –1.2

15.9 16.4 16.7

Blend E 1.3

18.4 18.9 19.2



Fitted values; NO INTERACTION

EDCBA

19.5

19.0

18.5

18.0

17.5

17.0

16.5

16.0

Blend

Fit

ted

Val

ue

s

1

2

3

Block

Line Plot of Fitted Values



Interaction?

Blend x Block interaction?

no general test without replication

EDCBA

20

19

18

17

16

15

Blend

Lo

ss, p

er

cen

tBlock 1

Block 2

Block 3

Block Profiles






2. Interaction






Part 3 Random block effects• Contribution of blend effect is predictable,

depends on the known makeup of each blend

• Contribution of block effect is not predictable, depends on current conditions at run time.

• Convention:

– Blend effect is fixed,

– Block effect is random

aA, aB, aC, aD, aE are fixed but unknown,

bI, bII, bIII are random numbers

• Assumption: b N( 0 , sB )



Random block effects

Recall F-test logic:

– MS(Error) ≈ s2

– MS(Effect) ≈ s2 + effect contribution

– F = MS(Effect) / MS(Error) ≈ 1 if effect absent,

>>1 if effect present

For Blend Effect, effect contribution =

For Block Effect, effect contribution =

No effect on logic of F-test

43

2

2B5



Minitab analysis

Analysis of Variance for Loss, using Adjusted SS for Tests

Source DF Adj SS Adj MS F-Value P-Value

Block 2 1.648 0.8240 0.94 0.429

Blend 4 11.556 2.8890 3.31 0.071

Error 8 6.992 0.8740

Total 14 20.196

Expected Mean Square

Source for Each Term

1 Block (3) + 5.0000 (1)

2 Blend (3) + Q[2]

3 Error (3)

Ref: DCM, p. 125, p. 133

43

22

2B

2 5

2






2. Interaction






Part 4Introduction to 2-level factorial designs

A 22 experiment

Project:

optimisation of a chemical process yield

Factors (with levels):

operating temperature (Low, High)

catalyst (C1, C2)

Design:

Process run at all four possible combinations of factor levels, in duplicate, in random order.



Standard Order

Temperature Catalyst

1 Low 1 2 High 1 3 Low 2 4 High 2 5 Low 1 6 High 1 7 Low 2 8 High 2

Design set up



Go to Excel

Standard Order

Temperature Catalyst Run

Order 1 Low 1 6 2 High 1 8 3 Low 2 1 4 High 2 4 5 Low 1 3 6 High 1 7 7 Low 2 2 8 High 2 5

Randomisation



Design set up:Run order

Standard Order

Temperature Catalyst Run

Order 3 Low 2 1 7 Low 2 2 5 Low 1 3 4 High 2 4 8 High 2 5 1 Low 1 6 6 High 1 7 2 High 1 8

NB: Reset factor levels each time



Classwork 2.2.3

What were the

experimental units

factors

factor levels

treatments

response

blocks

allocation procedure



Results (run order)

Standard Order

Run Order

Temperature Catalyst Yield

3 1 Low 2 52

7 2 Low 2 45

5 3 Low 1 54

4 4 High 2 83

8 5 High 2 80

1 6 Low 1 60

6 7 High 1 68

2 8 High 1 72



Results (standard order)

Standard Order

Run Order

Temperature Catalyst Yield

1 6 Low 1 60

2 8 High 1 72

3 1 Low 2 52

4 4 High 2 83

5 3 Low 1 54

6 7 High 1 68

7 2 Low 2 45

8 5 High 2 80



Analysis (Minitab)

• Main effects and Interaction plots

• ANOVA results

– with diagnostics

• Calculation of t-statistics



HighLow

75

70

65

60

55

50

21

Temperature

Mea

n

Catalyst

21

85

80

75

70

65

60

55

50

CatalystM

ean

Low

High

Temperature

Main Effects Plot for YieldData Means

Interaction Plot for YieldData Means

Main Effects and Interactions



Minitab DOE command;

Estimated Effects and Coefficients for Yield

Term Effect Coef SE Coef T PConstant 64.25 1.31 49.01 0.000Temperature 23.0 11.50 1.31 8.77 0.001Catalyst 1.5 0.75 1.31 0.57 0.598Temperature*Catalyst 10.0 5.00 1.31 3.81 0.019

S = 3.70810

Effect = Coef x 2

SE(Effect) = SE(Coef) x 2

Analyze Factorial Design subcommand



Minitab DOEAnalyze Factorial Design

Estimated Effects and Coefficients for Yield (coded units)


S = 3.70810 R-Sq = 95.83% R-Sq(adj) = 92.69%

Analysis of Variance for Yield (coded units)

Source DF Seq SS Adj SS Adj MS F PMain Effects 2 1062.50 1062.50 531.25 38.64 0.0022-Way Interactions 1 200.00 200.00 200.00 14.55 0.019Residual Error 4 55.00 55.00 13.75 Pure Error 4 55.00 55.00 13.75Total 7 1317.50






S = 3.70810 R-Sq = 95.83% R-Sq(adj) = 92.69%








S = 3.70810 R-Sq = 95.83% R-Sq(adj) = 92.69%





ANOVA results

ANOVA superfluous for 2k experiments

"There is nothing to justify this complexity other than a misplaced belief in the universal value of an ANOVA table".

BHH, Section 5.10, p.188

“The standard form of the ‘analysis of variance’ • • • does not seem to me to be useful for 2n data.

Daniel (1976), Section 7.1, p.128



Diagnostic Plots

80706050

2

1

0

-1

-2

Fitted Value

Del

eted

Res

idu

al

3

2

1

0

-1

-2

-3

210-1-2D

elet

ed R

esid

ual

Score

N 8

AD 0.261

P-Value 0.600

Versus Fits(response is Yield)

Normal Probability Plot(response is Yield)



Direct CalculationTemp Cat

Rep 1

Rep 2

Rep1 – Rep2 2is

½(Rep1–Rep2)² Degrees of Freedom

Low 1 60 54 High 1 72 68 Low 2 52 45 High 2 83 80

Sum

Mean = s²

s

LowY HighY LowHigh YY

)YY(SE LowHigh

)YY(SE

YYt

LowHigh

LowHigh



Classwork 2.2.2

Calculate a confidence interval for the Temperature effect.

All effects may be estimated and tested in this way.

Homework 2.2.1

Test the statistical significance of and calculate confidence intervals for the Catalyst effect and the Temperature by Catalyst interaction effect.



ApplicationFinding the optimum

More Minitab results

Least Squares Means for Yield

Mean SE MeanTemperature Low 52.75 1.854 High 75.75 1.854

Catalyst 1 63.50 1.854 2 65.00 1.854

Temperature*Catalyst Low 1 57.00 2.622 High 1 70.00 2.622 Low 2 48.50 2.622 High 2 81.50 2.622



Optimum operating conditions

• Highest yield achieved

– with Catalyst 2

– at High temperature.

• Estimated yield: 81.5%

• 95% confidence interval:

81.5 ± 2.78 × 2.622,

i.e., 81.5 ± 7.3,

i.e., ( 74.2 , 88.8 )Certificate in StatisticsDesign and Analysis of Experiments


Exercise 2.2.1

As part of a project to develop a GC method for analysing trace compounds in wine without the need for prior extraction of the compounds, a synthetic mixture of aroma compounds in ethanol-water was prepared.

The effects of two factors, Injection volume and Solvent flow rate, on GC measured peak areas given by the mixture were assessed using a 22 factorial design with 3 replicate measurements at each design point. The results are shown in the table that follows.

What conclusions can be drawn from these data? Display results numerically and graphically. Check model assumptions by using appropriate residual plots.



Measurements for GC study

Injection volume, L Solvent flow rate,

mL/min 100 200

13.1 126.5

400 15.3 118.5

17.7 122.1

48.8 134.5

200 42.1 135.4

39.2 128.6 .

(EM, Exercise 5.1, pp. 199-200)



Introducing the Design Matrix

Standard Order

Run Order

Temperature Catalyst

1 6 Low 1

2 8 High 1

3 1 Low 2

4 4 High 2

5 3 Low 1

6 7 High 1

7 2 Low 2

8 5 High 2



Design Matrix

Design Point

Run Order

Temperature A

Catalyst B

1 6 – –

2 8 + –

3 1 – +

4 4 + +

5 3 – –

6 7 + –

7 2 – +

8 5 + +



Design Matrix with Y’s

Design Point

Run Order

Temperature A

Catalyst B

Yield

1 6 – – Y1

2 8 + – Y2

3 1 – + Y3

4 4 + + Y4

5 3 – – Y5

6 7 + – Y6

7 2 – + Y7

8 5 + + Y8



Design Matrix with Data

Design Point

Run Order

Temperature A

Catalyst B

Yield

1 6 – – 60

2 8 + – 72

3 1 – + 52

4 4 + + 83

5 3 – – 54

6 7 + – 68

7 2 – + 45

8 5 + + 80



Augmented Design Matrix with Y’s

Design Point

Run Order

A B AB Yield

1 6 – – + Y1

2 8 + – – Y2

3 1 – + – Y3

4 4 + + + Y4

5 3 – – + Y5

6 7 + – – Y6

7 2 – + – Y7

8 5 + + + Y8



Augmented Design Matrix with Data

Calculate effects as Mean(+) – Mean(–)

Design Point

Run Order

A B AB Yield

1 6 – – + 60

2 8 + – – 72

3 1 – + – 52

4 4 + + + 83

5 3 – – + 54

6 7 + – – 68

7 2 – + – 49

8 5 + + + 80



Dual role of the design matrix

• Prior to the experiment, the rows designate the design points, the sets of conditions under which the process is to be run.

• After the experiment, the columns designate the contrasts, the combinations of design point means which measure the main effects of the factors.

• The extended design matrix facilitates the calculation of interaction effects



Reading

EM §5.3

DCM §6-2, §6-2


Documents

Lecture 2.21 2015 Michael Stuart Design and Analysis of Experiments Lecture 2.2 1.Review Lecture 2.1 –Minute test –Why block? –Deleted residuals 2.Interaction