Module Twelve: Designs and Analysis for Factorial Treatments In most of experimental studies, through the brain storming and cause-effect diagram, we often

Module Twelve: Designs and Analysis for Factorial Treatments

In most of experimental studies, through the brain storming and cause-effect diagram, we often find there are more than one factor that may have significant impact to the response variables.

For example, in our concrete strength study, three factors may be of equal importance for the compressive strength:

(1) Type of sand (2) the amount of water and (3) the amount of cement

Our purpose is to determine the combination of sand size, amount of water and the amount of cement that will result the strongest compressive strength, as well as to determine the uncertainty due to each factor and uncertainty due to interaction of two or more factors.

In many engineering literatures, it is suggested to fix two factors and change the level of one factor. This approach is not very efficient nor is able to show the interaction effect between two factors.

Experimental design techniques will help us to identify the main effect of each factor, the interaction effects, and provide information to study the uncertainty due to individual factor as well as due the combined interaction effects.

Factorial Experiments with Two Factors

We first consider the two-factor factorial design study. Consider the concrete compressive strength study:

Purpose of the Study:

To determine the effects of sand type and the cement/sand ratio (weight ratio) and their combined interaction effect. A concrete specimen is a standard 50 mm cube specimen.

Treatment Design:

Two factors to be studies are : Sand type and amount of cement in terms of the cement/sand ratio in weight. Two commonly used sand types are: Small and Large grain. Three different sand/cement ratios in weight are 2.50, 2.75 and 3.00.

Experimental Design:

A two-factor full factorial design is planned for the experiment. A total of 2x3 treatment combinations. For each treatment, six specimens, 50mm cube each, will be formed. The standard mixture of water will be applied. The compressive strength will be tested 28 days after the specimen are formed.

Statistical Models for the Two-Factor Factorial Design – Fixed Effect Model

An appropriate statistical model to describe this design is:

ij.

A two-factor axb full factorial Design cab be described as a cell-mean model:

is the mean of each treatment combination. It is estimated by y

When is futher decomposed into main an

ijk ij ijk

ij

ij

y e

d interaction effects, the effects model is

, i = 1,2,...a; j =1,2,.., b; k =1,2,..,r

is the overall strength mean.

's are the effect of the ith level of Treatment A.

is

ijk i j ij ijk

i

j

y e

ijk ijk

the effect of the jth level of Treatment B.

is the interaction effect of ith level of A and jth level of B.

e is the random error with e ~ (0, )

ij

N

Relationship between cell mean model and effect model:

. . . .

. . .

The cell-mean model:

The effect model:

They are related in the foloowing way:

( ) ( ) ( ( ) ( ) )

( ) ( ) (

ijk ij ijk

ijk i j ij ijk

ij i j ij i j

i j ij i

y e

y e

. )

j

i j ij

Level B1 Level B2 Mean

Level A1

Level A2

Mean

What is an effect? Simple effect? main effect? Interaction effect?

Factor B

Level B1 Level B2 Mean

Factor A Level A1 20 ( 40 ( 30 (

Level A2 50 ( 14 ( 32 (

Mean 35 ( 27 ( 31 (

The effect of a factor is a change in the response caused by a change in the level of that factor. An effect can be expressed as a contrast. Three effects of interest are:

Simple Effect of a factor: is a contrast between levels of one factor at a level of another factor. In this example, 20 – 40 = - is a simple effect of factor B between levels B1 and B2 at Level A1 of factor A.

Can you find the other three simple effects for the above example?

Main effect of a factor: is a contrast between levels of one factor averaged over all levels of another factor.

The main effect of Factor A : (20+40)/2 – (50+14)/2 = -2

The main effect of factor B :

The Interaction effect between two factors: is the difference between simple effects of one factor at different levels of the other factor.

Consider Level B1: The change of A from Level A1 to Level A2 at Factor B = B1 is: 50-20 = 30, call is C1, which is the simple effect of factor A at B1 of factor B.

Consider Level B2: The change of A from Level A1 to Level A2 for Factor B = B2 is: 14-40 = -26, call it C2, which is the simple effect of factor A at B2 of factor B.

The interaction effect is the difference between C2 and C1 = -26-30 = -56

The changes for Factor from A1 to A2 are different between two levels of Factor B. This says that A and B are interacted. For this example, When B = B1, there is a huge increase in A from A1 to A2 of 30. However, when B = B2, there is a huge decrease in A from A1 to A2 of –26.

In real world applications, this happens often. When fertilizer A is given to a field, the production increases from low dosage to high dosage. Similar situation for B. However, when A and B both are applied at the same time, the production may be decreased. This is the interaction effect of fertilizer A and B.

When individual A and B work independently, each one has his/her progress. When both work as a team, the accomplishment can be much more than the sum of two independent workers, or possibly much less. This is interaction effect. The following figures demonstrates a several possible patterns of interaction between A and A factors, when both have two levels.

The following figures demonstrate some possible patterns of interaction between A and B for a 2x2 factorial design (a=2, b=2)

1 2 A

B=1

B=2

1 2 A

B=1

B=2

1 2 A

B=1

B=2

1 2 A

B=1

B=2

1 2 A

B=1

B=2

1 2 A

B=1

B=2

The relationship between observed data and the model

Factor B

Factor A 1 2 b Mean

1Cell Mean

Model Term

y111,y112,..,y11r y121,y122,..,y12r y1b1,y1b2,..,y1br

a ya11,ya12,..,ya1r ya21,ya22,..,ya2r yab1,yab2,..,yabr

Mean

Model Term

11.y 12.y1 .by 1..y

..ay.aby2.ay1.ay

.1.y.2.y . .by

...y

1 2 12 1 1 11 1 1b b 1

1

i.. ...

.j. ...

The main effect of ith level of factor A: is estimated by y

The main effect of jth level of factor B: is estimated by y

The interaction effect of ith level of factor A and jth l

i

i

y

y

ij. ... i..- ... .j. ... ij. i..- .j. ...

evel of B:

: is estimated by

(y -y ) - (y ) (y ) y -y y

(Consider a two level case for deomnstration)

ij

y y y

Analysis of Two-factors

When response, yijk’s are observed, we need a method to estimate treatment effects:

What is the main effect of factor A, factor B?

What is the interaction effect?

Is any if these effects significant?

If a effect is significant, where are the differences from?

If there is a control, is any other level of the factor significantly different from the control level?

Do the responses show any interesting patterns in relation to the levels of a factor?

And so on?

We asked similar questions for one-factor analysis before. Many of the techniques applied there will be applied here as well.

Case Study: A lab testing of life time of a battery is planned. Three life time is thought affected by two factors:Plate Material for the battery and and Environmental temperature.

Treatment Design: A plate material used for the battery, and the temperature of the environment. There are three types of plate materials common for battery. Three temperature levels that are common in real environment are chosen for the experiment.

Experimental Design: A two-factor full factorial experiment is planned for the study. Nine treatment combinations will be tested. For each treatment combination, six batteries will be tested.

This is a two-factor 3x3 full factorial design. The factors are fixed effects, since the levels of plate material and temperature are about the only choices for the study, although one can argue that temperature may not be fixed. Our interest is to compare the life time of the treatment, not about the variation of life time among different temperatures.

The life time data are (Data source: Montgomery, 1991):Row Matype Temp Life

1 1 15 74

2 1 15 130

3 1 15 155

4 1 15 180

5 2 15 126

6 2 15 150

7 2 15 159

8 2 15 188

9 3 15 110

10 3 15 138

11 3 15 160

12 3 15 168

13 1 70 34

14 1 70 40

15 1 70 75

16 1 70 80

17 2 70 106

18 2 70 115

19 2 70 122

20 2 70 126

21 3 70 120

22 3 70 139

23 3 70 150

24 3 70 174

25 1 125 20

26 1 125 58

27 1 125 70

28 1 125 82

29 2 125 25

30 2 125 45

31 2 125 58

32 2 125 70

33 3 125 60

34 3 125 82

35 3 125 96

36 3 125 104

Yijk is the lefe time. Factor A is Plate of Material (three levels) and Factor B is Temperature Three levels).

Each response can be decomposed in terms of treatment effects that estimate the corresponding terms of the effect model. And Sum of Squares of all responses can then be partitioned accordingly:

. .

... .. ... . . ... . ... .. ... . . ... .

... .. ... . . ... . .. . . ... .

( )

( ) ( ) [ [ ( ) ( )]] ( )

( ) ( ) ( ) ( )

ˆˆ ˆ + +

ijk ij ijk ij

i j ij i j ijk ij

i j ij i j ijk ij

i j

y y y y

y y y y y y y y y y y y y

y y y y y y y y y y y

ijkˆ + ( ) + eij

... . ... .

.. ...

The overall deviation can be partitioned in terms of factor effects:

Overall deviation = Treatment effect + Experimental error

( ) ( ) ( )

= ( )

ijk ij ijk ij

i

y y y y y y

y y

. . ... . .. . . ... . ( ) ( ) ( )

=Factor A Effect + Factor B effect + Interaction Effect + Experimental error

j ij i j ijk ijy y y y y y y y

2...

2 2.. ... . . ...

2 2. .. . . ... .

The overall Sum of Squares can be partitioned according to the effects of factors in the model:

( )

( ) ( )

( ) ( )

ijk

i j

ij i j ijk ij

y y

br y y ar y y

r y y y y y y

SSTO SS

The corresponding degrees of freedoom are:

(abr-1) = (a-1) + (b-1) + (a-1)(b-1) + ab(r-1)

A SSB SSAB SSE

This is the basis of the ANOVA table for two-factor models when replications are equal. Each sum of square component can be further decomposed based on the research interest. This is accomplished by setting up proper contrasts. The techniques we discussed for one-factor analysis can be extended here.

A typical procedure to conduct analysis for two-factor experiment

1. Conduct descriptive summary using both graphical and numerical techniques for detecting unusual observations and for demonstrating some interesting patterns that will be useful during the analysis.

2. Conduct the preliminary ANOVA analysis based on the raw data and residual analysis to check for the adequacy of assumptions, especially the constant variance and normality. Graphical methods are particularly useful here.

3. If transformation is needed, perform transformation, and conduct ANOVA analysis along with effect plots. If the result using the transformed data is very similar to that using raw data, use the raw data for the analysis.

4. Determine if further analysis is needed:

1. If interaction is significant (also closely examine the interaction plot to learn the interaction pattern), an analysis of simple effects of factor A (or B) at each level of factor B (or A) is recommended.

2. If main effect of a factor is significant, one should decide what further comparisons should be useful: Pairwise comparison, contrasts, trend analysis, comparison with control , and so on. (consult the one-way analysis for more details).

In the following, we will discuss the analysis of the Battery Life Time Testing Data (Data Source: Montgomery, 1991).

We start with descriptive and graphical summaries. Recall the Case Study:

Case Study: A lab testing of life time of a battery is planned. Three life time is thought affected by two factors:Plate Material for the battery and and Environmental temperature.

Treatment Design: A plate material used for the battery, and the temperature of the environment. There are three types of plate materials common for battery. Three temperature levels that are common in real environment are chosen for the experiment.

Experimental Design: A two-factor full factorial experiment is planned for the study. Nine treatment combinations will be tested. For each treatment combination, six batteries will be tested.

A total of 3x3x6 life time data are recorded.

20 60 100 140 180

95% Confidence Interval for Mu

40 50 60 70 80 90 100 110 120

95% Confidence Interval for Median

Variable: Life

A-Squared:P-Value:

MeanStDevVarianceSkewnessKurtosisN

Minimum1st QuartileMedian3rd QuartileMaximum

52.295

34.420

44.736

0.5240.144

83.166748.58892360.88

0.8658476.70E-02

12

20.000 44.500 74.500118.000180.000

114.039

82.498

117.370

Matype: 1

Anderson-Darling Normality Test


95% Confidence Interval for Sigma


Descriptive Statistics

20 60 100 140 180


60 70 80 90 100 110 120 130 140 150


Variable: Life

A-Squared:P-Value:



76.337

34.744

61.157

0.2610.641

107.500 49.0462405.55-2.1E-01-7.1E-01

12

25.000 61.000118.500144.000188.000

138.663

83.275

143.685

Matype: 2






20 60 100 140 180


100 110 120 130 140 150 160


Variable: Life

A-Squared:P-Value:



102.359

25.336

98.105

0.1780.896

125.083 35.7661279.17-3.2E-01-8.2E-01

12

60.000 98.000129.000157.500174.000

147.808

60.726

157.369

Matype: 3






1 2 3

0

100

200

Matype

Life

Dotplots of Life by Material Type

20 60 100 140 180


130 140 150 160 170


Variable: Life

A-Squared:P-Value:



124.696

22.452

127.052

0.2670.620

144.833 31.6941004.52-9.0E-01

0.99770512

74.000127.000152.500166.000188.000

164.971

53.813

165.895

Temp: 15






20 60 100 140 180


72 82 92 102 112 122 132 142


Variable: Life

A-Squared:P-Value:



79.826

30.018

76.316

0.2950.536

106.750 42.3751795.66-4.3E-01-3.9E-01

12

34.000 76.250117.500135.750174.000

133.674

71.948

135.579

Temp: 70






20 60 100 140 180


45 55 65 75 85


Variable: Life

A-Squared:P-Value:



47.855

18.186

48.421

0.2240.772

64.166725.6722659.061-3.0E-01-3.9E-01

12

20.000 48.250 65.000 82.000104.000

80.478

43.588

82.000

Temp: 125






15

70

12

5

0

100

200

Temp

Life

Dotplots of Life by Temperature

Variable Material N Mean Median TrMean StDev

Life 1 12 83.2 74.5 79.8 48.6

2 12 107.5 118.5 107.7 49.0

3 12 125.1 129.0 126.7 35.8

Variable Material SE Mean Minimum Maximum Q1 Q3

Life 1 14.0 20.0 180.0 44.5 118.0

2 14.2 25.0 188.0 61.0 144.0

3 10.3 60.0 174.0 98.0 157.5

Variable Temperature N Mean Median TrMean StDev

Life 15 12 144.83 152.50 147.60 31.69

70 12 106.8 117.5 107.3 42.4

125 12 64.17 65.00 64.60 25.67

Variable Temperature SE Mean Minimum Maximum Q1 Q3

Life 15 9.15 74.00 188.00 127.00 166.00

70 12.2 34.0 174.0 76.3 135.8

125 7.41 20.00 104.00 48.25 82.00

Bonferroni 95% CI for standard deviations

Lower Sigma Upper N Factor Levels

20.9231 45.3532 357.644 4 1 15

10.8871 23.5991 186.096 4 1 70

12.3875 26.8514 211.743 4 1 125

11.8182 25.6174 202.012 4 2 15

4.0460 8.7702 69.160 4 2 70

8.8860 19.2614 151.890 4 2 125

11.9829 25.9743 204.827 4 3 15

10.4006 22.5444 177.779 4 3 70

8.8939 19.2787 152.026 4 3 125

0 100 200 300 400

95% Confidence Intervals for Sigmas

Bartlett's Test

Test Statistic: 6.864

P-Value : 0.551

Levene's Test


P-Value : 0.504

Factor Levels

1

1

1

2

2

2

3

3

3

15

70

125

15

70

125

15

70

125

Test for Equal Variances for Life

Can you recall how the Bonferroni’s CI interval for is conducted?

How about Levene’s Test?

The corresponding ANOVA Table for Two-Factor Model

Source Df SS MS F P-value EMS

Treatment A a-1 SSA MSA=SSA/(a-1) MSA/MSE

Treatment B b-1 SSB MSB=SSB/(b-1) MSB/MSE

AB Interaction (a-1)(b-1)

SSAB MSAB=SSAB/

[(a-1)(b-1)]

MSAB/MSE

Error ab(r-1) SSE MSE=SSE/[ab(r-1)] 2

Total abr-1 SSTO

2

2

( )

( 1)( 1)

iji j

r

a b

2 2 ( 1)jj

ar b

2 2 ( 1)ij

br a

Sum of Squares in the ANOVA table is the decomposition of

SSTO into four components with the additive property:

SSTO = SSA + SSB + SSAB + SSE

And d.f. also satisfies the additive property:

DF(Total) = DF(A) + DF(B) + DF(AB) + DF(E)

The F-statistics test the following hypothesis:

0 a

0 a

0

tests: H : ' all equal to zero Vs. H : Not all ' equal to zero

tests: H : ' all equal to zero Vs. H : Not all ' equal to zero

tests: H : ' al

A i i

B j j

AB ij

MSAF s s

MSEMSB

F s sMSEMSAB

F sMSE

al equal to zero Vs. H : Not all ' equal to zeroij s

Similar to what we saw in the One-way ANOVA case, these tests are determined based on the EMS (Expected Mean Squares). They are provided in the Minitab output.

In analyzing the ANOVA results for two-factor models, we need to examine the Interaction effect before analyzing main effects. Since if there exists interaction effect, sometimes the results of main effects may be misleading.

Factor Type Levels Values

Matype fixed 3 1 2 3

Temp fixed 3 15 70 125

Analysis of Variance for Life, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P

Matype 2 10633.2 10633.2 5316.6 7.98 0.002

Temp 2 39083.2 39083.2 19541.6 29.34 0.000

Matype*Temp 4 9437.7 9437.7 2359.4 3.54 0.019

Error 27 17980.7 17980.7 666.0

Total 35 77134.8

Unusual Observations for Life

Obs Life Fit SE Fit Residual St Resid

1 74.000 134.750 12.903 -60.750 -2.72R

4 180.000 134.750 12.903 45.250 2.02R

Expected Mean Squares, using Adjusted SS

Source Expected Mean Square for Each Term

1 Matype (4) + Q[1, 3]

2 Temp (4) + Q[2, 3]

3 Matype*Temp (4) + Q[3]

4 Error (4)

Error Terms for Tests, using Adjusted SS

Source Error DF Error MS Synthesis of Error MS

1 Matype 27.00 666.0 (4)

2 Temp 27.00 666.0 (4)

3 Matype*Temp 27.00 666.0 (4)

Variance Components, using Adjusted SS

Source Estimated Value

Error 666.0

Q(1,3) is a Quadratic function of and (ij

Other terms are derived based on similar approaches.

Three F-tests all use the Source (4) as the error term.

The only variance component

Matype Temp

60

80

100

120

140

Life

Main Effect Plot for the Battery Life Study

50

100

150

50

100

150

Matype

Temp

1

2

3

15

70

125

1

2

3

15

70

125

Interaction Plot between Material Type and TemperatureAll three types of material show much longer life time at low temperature, at much shorter life time at high temperature environment. However, the performance varies greatly at middle temperature. Material Three seems to perform well at middle temperature as well.

The trend of the life time is clearly going shorter when temperature increases. However, the trends are different for different types of material. Material three maintains its high life time until very high temperatures. Material 1 and 2 sharply decrease from low temperature to middle temperature.

when temperature increases, Material 2 is affected by temperature the most.

0 50 100

-3

-2

-1

0

1

2

Temp

Sta

ndar

dize

d R

esid

ual

Residuals Versus Temp(response is Life)

1 2 3

-3

-2

-1

0

1

2

Matype

Sta

ndar

dize

d R

esid

ual

Residuals Versus Material Type(response is Life)

50 100 150

-3

-2

-1

0

1

2

Fitted Value

Sta

ndar

dize

d R

esid

ual

Residuals Versus the Fitted Values(response is Life)

Average: -0.0107593StDev: 1.06160N: 36

Anderson-Darling Normality TestA-Squared: 0.341P-Value: 0.476

-3 -2 -1 0 1 2

.001

.01

.05

.20

.50

.80

.95

.99

.999

Pro

babi

lity

TRES1

Normal Probability Plot

There is a slight evidence that there is larger variation at higher life time. When temperature is low, life time is higher, and also a somewhat larger variation among life time.

There is a slight evidence, material one seems to have larger life time variation.

Normality seems to be fine.

Least Squares Means for Life

Matype Mean SE Mean

1 83.17 7.450

2 107.50 7.450

3 125.08 7.450

Temp

15 144.83 7.450

70 106.75 7.450

125 64.17 7.450

Matype*Temp

1 15 134.75 12.903

1 70 57.25 12.903

1 125 57.50 12.903

2 15 155.75 12.903

2 70 117.25 12.903

2 125 49.50 12.903

3 15 144.00 12.903

3 70 145.75 12.903

3 125 85.50 12.903

These least square means are the same as the sample means when sample size is the same.

How to compute the SE Mean? Where do we use it?

This is the estimated standard error of Mean from sample data, which has the form:

for the Mean of the Level 1 of Material Type is

MSE

#obs used to compute the Mean

MSE 6667.45

br 3 4

SE

Can you compute the SE of 12.903 for each treatment combination?

1..y

32.y

What is next?

Now, we have conducted an ANOVA analysis, and checked that assumptions. Both normality and constant variance assumptions seem to be fine.

The interaction is significant. Main effects are also significant.

We have some observations about the patterns related to interaction effect.

The main effects indicate: Three materials produce battery with very different life time. The higher the temperature is , the lower the life time.

So, what is next?

There are two major tasks in data analysis: estimation and comparative testing. Measuring uncertainty is an estimation problem. While hypothesis testing is a comparative problem. What is next depending on the interest of the study.

In measuring uncertainty, we may be interested in determining the uncertainty of the experimental error, response mean due to a factor, response of treatment combination, or even uncertainty of mean difference between two factor levels.

IN comparative testing, we may be interested in pairwise comparison, contrasts comparison, trend analysis and so on. In the comparative testing, we need to estimate the uncertainty of the measurement we are comparing as well.

For the Battery Life Time study, we will conduct each of these to demonstrate how to measure uncertainty and how to conduct comparative analysis.

1. Since the interaction is significant, we would conduct a simple effect comparison. It may be more interested in comparing material type at each temperature.

011. 21. 31.

12. 22. 32.

This would be a multiple comparison among

(1) , . , the mean life time of three material type at temperature = 15

(2) , . , the mean life time of three material type at temperat

0

013. 23. 33.

ure = 70

(3) , . , the mean life time of three material type at temperature = 125

Each of these multiple comparisons involves three pair-wise comparisons:

For (1), the three comparisons are

We can apply Tukey’s pair-wise multiple comparison procedure for this purpose:

(Recall Tukey’s method: HSD(k,) = q(,k,df)

11 21 11 31 21 31 , ,

( , , )

11 12 11 12

Recall the Tukey's Method for pair-wise multiple comparisons:

( , )# of obs used to compute the estimate of

The best estimate of is y .

Similarly, the best estimate of

k dfij

MSEHSD k q

y

ij

ij

a ij

is y

100(1- )% confidence interval of is |y | ( , )

A test of two-sdied hypothesis to conclude : H : 0 , if |y | > ( , )

For this case, a 95% CI for the thr

ij ik ik

ij ik ik

ij ik ik

y

y HSD k

y HSD k

ij

(.05,3,27)

ee comparisons has

k = 3 (# of treatment levels) and df is df of MSE = ab(r-1)=27,

# of obs used to compute y is r = 4

666.0HSD(3,.05) (3.48)(12.9) 44.9

4q

Matype*Temp ( )

1 15 134.75 12.903

1 70 57.25 12.903

1 125 57.50 12.903

2 15 155.75 12.903

2 70 117.25 12.903

2 125 49.50 12.903

3 15 144.00 12.903

3 70 145.75 12.903

3 125 85.50 12.903

ijy

At Temperature = 15, three comparisons for Material Type:

11 21

11 31

21 31

| | | 134.75 155.75 | 11.0 < HSD(3,.05)=44.9, Not significant

| | | 134.75 144.00 | 9.25 44.9, Not significant

| | | 155.75 144.0 | 11.75 44.9, Not significant

y y

y y

y y

Hands-on Activity

1. Conduct a Tukey’s Multiple Comparison procedure to compare three types at Temperature = 700.

2. Conduct a Linear and Quadratic Trend analysis of responses in relation to Temperature for the Material Type 2.

(Hint: Orthogonal Polynomial Coefficients for three levels: are

Linear: -1, 0, 1 Quadratic: 1, -2, 1

3. Conduct a Tukey’s pairwise comparison for the Material Type.

4. If one is interested in quantifying the uncertainty of individual observation, yijk, what is s.d. of yijk?

5. If one is interested in quantifying the uncertainty of Mean response of each material type, what is it?

An interesting question for the Battery Life Time study is: Is the trend of Life time in relation to Temperature for Material Type 1 different from that for Material Type 2?

50

100

150

50

100

150

Matype

Temp

1

2

3

15

70

125

1

2

3

15

70

125

Interaction Plot between Material Type and Temperature The question is: Do these three linear lines have the similar slopes. That is,

If the rates of change of life time from Low temperature to High Temperature are similar or not.

The slower rate change means the life time is less sensitive to the temperature.

ANOVA Table with Sum of Squares Decomposition



Matype 2 10633.2 10633.2 5316.6 7.98 0.002

C1 1 SSC1

C2 1 SSC2

Temp 2 39083.2 39083.2 19541.6 29.34 0.000

Linear 1 SSL

Quadratic 1 SSQ

Matype*Temp 4 9437.7 9437.7 2359.4 3.54 0.019

C1*Linear 1 SS(C1*L)

C2*Linear 1 SS(C2*Q)

C1*Quadratic 1 SS(C1*Q)

C2*Quadratic 1 SS(C2*Q)

Error 27 17980.7 17980.7 666.0

Total 35 77134.8

This Sum of Squares partitioning technique is very useful , especially when we are interested in a specific part of the effects. The method is based on Contrasts.

SSA = SSC1+SSC2 and DF(A) = DF(C1) + DF(C2)

SSB = SSL + SSQ and DF(B) = DF(L) + DF (Q)

SSAB = SS(C1*L) + SS(C2*L) + SS(C1*Q) + SS(C2*Q)

The question :

Do the three linear response lines in relation to Temperature have the similar slopes?

Can be answered using the sum of square decomposition technique. This is simply to test if the combined two contrasts of of (C1*Linear) and(C2*Linear) is significant or not.

How do I know that?

Since C1*Linear + C2*Linear = Type*Linear(Temp). This information indeed reflects the linear pattern of temperature at different Material Types.

From the ANOVA point of view, we are trying to test part of the interaction effect.

When we observe the data information, the question we ask here is the same as the following:

Linear Trend of Temperature

At Material Type 1

At Material Type 2

At Material Type 3

1 11 13ˆ 134.75 57.5 77.25L y y

2 21 23ˆ 155.75 49.5 96.25L y y

3 31 33ˆ 144.00 85 59.0L y y

Are L1, L2 and L3 significantly different?

One can apply pairwise comparison technique such as Tukey’s method to make three comparisons. We have learned how to do this.

Or one can use the Sum of Squares partition technique to test if Material Type*Linear(Temp) is significant or not. We will discuss this techniques by

hand and by Minitab.

Determine the Sum of Square for Linear(Temp)*Material Type

Material Type has three levels. This means we can partition the Material Type into two orthogonal contrasts. Two meaningful orthogonal contrasts for comparing Material Type could be:

(1) C1: A contrast for comparing Type 1 with Type 3.

(2) C2: A contrast for comparing Average of Type 1 and Type 3) with Type 2.

How to set up contrasts for these two comparisons?

How do I know these two comparisons are orthogonal?

Material Type

C1: Type 1 Vs Type 3 -1 0 1

C2: (Type1+type3)/2 Vs Type 2 1 -2 1

Sum of Squares for each contrast is

The multiple, br, is the number of observations used to compute

2

..

2

i i

i

k ySSC br

k

..iy

NOTE: One can partition the two df of Material Type using different set of orthogonal contrasts as wish.

Temperature

Linear -1 0 1

Quadratic 1 -2 1

Sum of Squares for each contrast is

The multiple, ar, is the number of observations used to compute

2

. .

2

j j

j

k ySSC ar

k

. .jy

Similarly, the trend of temperature is a contrast:

How about the interaction between

C1(Type)*Linear(Temp) and C2(Type)*Linear(Temp)

If we can obtain the Sum of Squares for each of the Interaction term, Adding these two together, it is the Sum of Squares for the

(Material Type)*Linear(Temp)

Determine Sum of Square of C1*Linear and C2*Linear for the Life Time Study

It is important to understand that an interaction term such as C1*Linear is still a CONTRAST. If we know how to set a proper contrast for C1*Linear, we can determine the corresponding SS.

How to set up a proper contrast for C1*Linear?

X (multiplication) Linear Contrast of Temperature

C1 of Material Type -1 0 1

-1 1 0 -1

0 0 0 0

1 -1 0 1

As we know a contrast is just a weighted sum of the mean responses.1. 2. 3.

.1 .2 .3

For Material Type, the mean responses are : , ,

For Temperature, the mean responses are : , ,

For Interaction, the mean responses are: the 9 ' .

The coefficients for settin up C1*Line

ij s

ar contrast is the multiplications of the coefficients

of C1 and Linear. They are given in the aboce table.

The contrast for C1*Linear is

Mean Responses

kij , coefficients 1 0 -1 0 0 0 -1 0 1

The corresponding estimate from sample is .ij iji j

k y

2

.

2

ij.

The corresponding Sum of Squares is

SS(C1*L) = (r)

The multiple (r) is the number of observations used to compute y

The Standard Error of (C1*L) can also be estimat

ij iji j

iji j

k y

k

2

ed by:

SE(C1*L) = iji j

MSEk

r

Determine SS(C1*L) for the Battery Life CaseMatype*Temp ( )

1 15 134.75 12.903

1 70 57.25 12.903

1 125 57.50 12.903

2 15 155.75 12.903

2 70 117.25 12.903

2 125 49.50 12.903

3 15 144.00 12.903

3 70 145.75 12.903

3 125 85.50 12.903

.ijyThe estimate of the contrast C1*Linear is

(134.75 - 57.5 - 144.0 + 85.5) = 18.75

2

.

2

2

SS(C1*L) = (r)

(18.75)(4) 351.56

(1 1 1 1)

ij iji j

iji j

k y

k

The SE(C1*L) is

2 666SE(C1*L) = (1 1 1 1) 25.806

4iji j

MSEk

r

Hands-On Activity

1. Set up the contrast for C2*Linear.

2. Estimate the C2*Linear contrast, compute the corresponding Sum of Squares, and SE of the estimate.

3. Add the SS of C1*L and C2*L together and conduct an F-test to test if the Material Type*Linear(Temp) significant or not, and make an appropriate conclude of this F-test.

Use Minitab to to conduct a general linear model analysis

You must specify the model terms in the Model box. This is an abbreviated form of the statistical model that you may see in textbooks. Because you enter the response variable(s) in Responses, in Model you enter only the variables or products of variables that correspond to terms in the statistical model. Minitab uses a simplified version of a statistical model as it appears in many textbooks. Here are some examples of statistical models and the terms to enter in Model. A, B, and C represent factors.

Case Statistical model Terms in model

Factors A, B crossed yijk = m + ai + bj + abij + ek(ij) A B A*B

Factors A, B, C crossed yijkl = m + ai + bj + ck + abij + acik + bcjk + abcijk + el(ijk) A B C A*B A*C B*C A*B*C

3 factors nested

(B within A, C within A and B)

yijkl = m + ai + bj(i) + ck(ij) + el(ijk) A B(A) C(AB)

Crossed and nested (B nested within A, both crossed with C)

yijkl = m + ai + bj(i) + ck + acik + bcjk(i) + el(ijk) A B(A) C A*C B*C

Models with covariates

You can specify variables to be covariates in GLM. You must specify the covariates in Covariates, but you can enter the covariates in Model, though this is not necessary unless you cross or nest the covariates (see table below).

GLM allows terms containing covariates crossed with each other and with factors, and covariates nested within factors. Here are some examples of these models, where A is a factor.

Case Covariates Terms in model

test homogeneity of slopes (covariate crossed with factor) X A X A*X

same as previous X A | X

quadratic in covariate (covariate crossed with itself) X A X X*X

full quadratic in two covariates (covariates crossed) X Z A X Z X*X Z*Z X*Z

separate slopes for each level of A (covariate nested within a factor)

X A X(A)

Rules for Expression Models

1. * indicates an interaction term. For example, A*B is the interaction of the factors A and B.

2. ( ) indicate nesting. When B is nested within A, type B(A). When C is nested within both A and B, type C(A B). Terms in parentheses are always factors in the model and are listed with blanks between them.

3. Abbreviate a model using a | or ! to indicate crossed factors and a - to remove terms.

4. Terms in parentheses are always factors in the model and are listed with blanks between them. Thus, D*F (A B E) is correct but D*F (A*B E) and D (A*B*C) are not.

5. Also, one set of parentheses cannot be used inside another set. Thus, C (A B) is correct but C (A B (A)) is not.

6. An interaction term between a nested factor and the factor it is nested within is invalid.

Examples of what to type in the Model text box

Two factors crossed: A B A*B (or enter A|B for a full factorial model.)

Three factors crossed: A B C A*B A*C B*C A*B*C (or enter A|B|C for a full factorial model).

Three factors nested: A B(A) C(A B)

Crossed and nested (B nested within A, and both crossed with C): A B(A) C A*C B*C(A)

When a term contains both crossing and nesting, put the * (or crossed factor) first, as in C*B(A), not B(A)*C

Use Minitab to conduct Sum of Squares partitions – the Battery Life Case

The following is the Minitab command that is used to produce the result.

This is created by using the Pull-down Menu and enabling the commands.

GO TO Editor and choose ‘enable Commands’ will provide you the actual Minitab program in the output.

MTB > GLM 'Life' = Matype Temp Temp*temp Matype*temp Matype*temp*temp;

SUBC> Covariates 'Temp';

SUBC> Brief 1 .

This model enables us to conduct the sum of square partitions as we discussed here.

Steps for running Minitab procedure: Generalized Linear Model:

1. Go to Stat, choose ANOVA, then select ‘General Linear Model’.

2. In the dialog box, enter Response variable. In the Model box, enter

Matype Temp Temp*temp Matype*temp Matype*temp*temp

Choose ‘Covariate’ and enter ‘Temp’ as covariate.

3. For other selections, please consult the One-Way analysis.

General Linear Model: Life versus Matype


Matype fixed 3 1 2 3



Matype 2 10633.2 1178.7 589.3 0.88 0.424

Temp 1 39042.7 1406.3 1406.3 2.11 0.158

Temp*Temp 1 40.5 40.5 40.5 0.06 0.807

Matype*Temp 2 2315.1 7066.9 3533.4 5.31 0.011

Matype*Temp*Temp 2 7122.6 7122.6 3561.3 5.35 0.011

Error 27 17980.8 17980.8 666.0

Total 35 77134.8

The ANOVA results produced by Minitab using the model statement:

MTB > GLM 'Life' = Matype Temp Temp*temp Matype*temp Matype*temp*temp;

SUBC> Covariates 'Temp';

SUBC> Brief 1 .

Temp : Linear, Temp*Temp: Quadratic, Matype*Temp : Type*Linear(Temp),

Matype*Temp*Temp: Type*Quadratic(Temp)

Hands-on Project

The yield of a chemical process is suspected to be affected by the pressure and temperature of the process. Each factor has three choices in the chemical process. Pressure: 200, 215 and 230. Temperature: Low, medium and high. A factorial experiment with two replications is performed. The yield data are collected:

Row Pressure Temp Yield

1 200 Low 90.4

2 200 Low 90.2

3 200 Medium 90.1

4 200 Medium 90.5

5 200 High 90.3

6 200 High 90.8

7 215 Low 89.8

8 215 Low 89.6

9 215 Medium 90.1

Row Pressure Temp Yield

10 215 Medium 90.2

11 215 High 90.8

12 215 High 90.7

13 230 Low 90.6

14 230 Low 90.8

15 230 Medium 90.2

16 230 Medium 90.5

17 230 High 90.3

18 230 High 90.0

Conduct a proper analysis and make some recommendations based on the findings.

Two-Factor Design – Random effect modelThrough cause-effect diagram and team discussion, it was suspected that the surface finish of a metal part is influenced by the feed rate and the depth of cut. The variability is of a particular concern, since uneven metal finish will result leaking of the finish products using this metal. Three feed rates and four depths of cuts are randomly chosen. A factorial experiment with three replications is performed. The surface roughness is measured and recorded. The lower the roughness, the better the surface.

Row Depth Feed Rough

1 0.12 0.20 74

2 0.12 0.20 68

3 0.12 0.20 60

4 0.12 0.25 90

5 0.12 0.25 85

6 0.12 0.25 89

7 0.12 0.30 99

8 0.12 0.30 106

9 0.12 0.30 103

10 0.14 0.20 79

11 0.14 0.20 67

12 0.14 0.20 75


13 0.14 0.25 99

14 0.14 0.25 102

15 0.14 0.25 94

16 0.14 0.30 107

17 0.14 0.30 106

18 0.14 0.30 98

19 0.16 0.20 80

20 0.16 0.20 84

21 0.16 0.20 86

22 0.16 0.25 91

23 0.16 0.25 95

24 0.16 0.25 98


25 0.16 0.30 99

26 0.16 0.30 105

27 0.16 0.30 100

28 0.18 0.20 99

29 0.18 0.20 102

30 0.18 0.20 103

31 0.18 0.25 108

32 0.18 0.25 111

33 0.18 0.25 99

34 0.18 0.30 111

35 0.18 0.30 110

36 0.18 0.30 106

The levels of two factors are randomly chosen, and the variability among levels are the main concern. This is clearly a random effect model. The variance components are the main interest. An appropriate model is:

Statistical Model for tow random effect factor Factor Experiment

ijk The observation , y can be expressed as:

( ) , i=1,2,...,a; j = 1,2,..., b; k = 1,2,...,r.

where is the unknow grand mean,

~ (0, ), which are the ranom effects of Factor

ijk i j ij ijk

i

y e

N

ijk

A.

~ (0, ), which are the random effects of Factor B

~ (0, ), which are the random effects of A and B interaction

e ~ (0, ), which is the random error due to replications.

And these components

j

ij

N

N

N

2 2 2 2

2 2 2 2ijk

are independent.

Note we have four variance components: , , , .

The observation y ~ ( , )N

The main goal is to estimate the four variance components to understand the source of uncertainty.

Based on the model, the ANOVA and the corresponding EMS, which will be used to estimate the variance components , are given by:Source Df SS MS F P-value EMS

Treatment A a-1 SSA MSA=SSA/(a-1) MSA/MSE

Treatment B b-1 SSB MSB=SSB/(b-1) MSB/MSE

AB Interaction (a-1)(b-1)

SSAB MSAB=SSAB/

[(a-1)(b-1)]

MSAB/MSE

Error ab(r-1) SSE MSE=SSE/[ab(r-1)] 2

Total abr-1 SSTO

2 2r

2 2 2r ar

2 2 2r br

The F-statistics test the following hypotheses:

2 20 a

2 20 a

2 20 a

tests: H : = 0 Vs. H : > 0

tests: H : = 0 Vs. H : > 0

tests: H : = 0 Vs. H : > 0

A

B

AB

MSAF

MSEMSB

FMSEMSAB

FMSE

Based on the EMS, the variance components are estimated by:2

2

2

2

2 2 2 2 2

ˆ

ˆ

ˆ

ˆ

ˆ ˆ ˆ ˆ ˆy

MSE

MSA MSAB

brMSB MSAB

arMSAB MSE

r

2

2 2( / 2, ( 1)) (1 / 2, ( 1))

100(1- )% confidence interval for is

SSE SSELower Bound = Upper Bound =

ab r ab r

IN many applications, the variance components are presented in terms of percent of variance component, or the s.d. components , which as known as measurement uncertainties are presented, instead of variance.

The analysis of the Surface Roughness dataGeneral Linear Model: Rough versus Depth, Feed


Depth random 4 0.12 0.14 0.16 0.18

Feed random 3 0.20 0.25 0.30

Analysis of Variance for Rough, using Adjusted SS for Tests


Depth 3 1801.56 1801.56 600.52 4.60 0.053

Feed 2 3230.72 3230.72 1615.36 12.37 0.007

Depth*Feed 6 783.28 783.28 130.55 6.84 0.000

Error 24 458.00 458.00 19.08

Total 35 6273.56

Unusual Observations for Rough

Obs Rough Fit SE Fit Residual St Resid

3 60.000 67.333 2.522 -7.333 -2.06R

The ANOVA results indicate all three uncertainty components are statistically significant when compared to the random error. Different Feeding Rates, different Depth all introduces huge variation. In particular, the significance of Depth*Feed interaction component indicates there is a huge inconsistence of roughness due to different feeding rates for different level of depths.



1 Depth (4) + 3.0000(3) + 9.0000(1)

2 Feed (4) + 3.0000(3) + 12.0000(2)

3 Depth*Feed (4) + 3.0000(3)

4 Error (4)



1 Depth 6.00 130.55 (3)

2 Feed 6.00 130.55 (3)

3 Depth*Feed 24.00 19.08 (4)


Source Estimated Value % Variance

Depth 52.22 22.5%

Feed 123.73 53.3%

Depth*Feed 37.15 16.0%

Error 19.08 8.2%

More than 50% of the variability is due to feeding Rate. A further analysis would needed to determine the causes of the uncertainty due to Feed Rate. The Depth contributes 22.5% of the variability. The Interaction is about 16%.

These components are all statistically significant.

This EMS provides information for making proper F-tests.

The overall uncertainty of a measurement of roughness is

ˆ 52.22 123.73 37.15 19.08 232.18 15.24y

Test for Equal Variances

Bonferroni 95% CI for s.d. of Residuals

Lower Sigma Upper N Factor Levels

2.82680 7.02377 153.803 3 0.12 0.20

1.06481 2.64575 57.935 3 0.12 0.25

1.41340 3.51188 76.901 3 0.12 0.30

2.45908 6.11010 133.796 3 0.14 0.20

1.62653 4.04145 88.498 3 0.14 0.25

1.98529 4.93288 108.018 3 0.14 0.30

1.22954 3.05505 66.898 3 0.16 0.20

1.41340 3.51188 76.901 3 0.16 0.25

1.29373 3.21455 70.391 3 0.16 0.30

0.83779 2.08167 45.583 3 0.18 0.20

2.51337 6.24500 136.750 3 0.18 0.25

1.06481 2.64575 57.935 3 0.18 0.30

0 50 100 150

95% Confidence Intervals for Sigmas

Bartlett's Test


P-Value : 0.903

Levene's Test


P-Value : 0.955

Factor Levels

0.12

0.12

0.12

0.14

0.14

0.14

0.16

0.16

0.16

0.18

0.18

0.18

0.20

0.25

0.30

0.20

0.25

0.30

0.20

0.25

0.30

0.20

0.25

0.30

Test for Equal Variances for Residuals

Average: -0.0000000StDev: 3.61742N: 36

Anderson-Darling Normality TestA-Squared: 0.658P-Value: 0.079

-5 0 5

.001

.01

.05

.20

.50

.80

.95

.99

.999

Pro

babi

lity

RESI1

Normal Probability Plot

70 80 90 100 110

-5

0

5

Fitted Value

Res

idua

l

Residuals Versus the Fitted Values(response is Rough)

Both Normality and Constant Variance seem to be fine. No transformation will be needed.

Hands-on Project of Random Effect Model: Two Factor Experiment (Data Source: Kuehl 2000)

Spectrophotometer is used in medical clinical laboratories. The consistency of measurements from day to day among machines is very critical. An uncertainty study is conducted to evaluate the variability of measurements among machines operate over several days, and to study if the machine uncertainty is within an acceptable standards for applications.

Treatment Design: A factorial design is planned with treatments are Four randomly chosen machines, which will be tested on four randomly selected days.

Experimental Design: For each day, eight replicate serum samples will be tested. Two are randomly assigned to each machine for testing. The same well-trained technician prepares the serum samples and operates the machine throughout the experiment. The measurement is the Triglyseride levels(mg/dl) in serum samples.

Conduct an appropriate analysis and make suggestions for improvement

Row Day Machine Trig

1 1 1 142.3

2 1 1 144.0

3 1 2 148.6

4 1 2 146.9

5 1 3 142.9

6 1 3 147.4

7 1 4 133.8

8 1 4 133.2

9 2 1 134.9

10 2 1 146.3

11 2 2 145.2

12 2 2 146.3

13 2 3 125.9

14 2 3 127.6

15 2 4 108.9

16 2 4 107.5

Row Day Machine Trig

17 3 1 148.6

18 3 1 156.5

19 3 2 148.6

20 3 2 153.1

21 3 3 135.5

22 3 3 138.9

23 3 4 132.1

24 3 4 149.7

25 4 1 152.0

26 4 1 151.4

27 4 2 149.7

28 4 2 152.0

29 4 3 142.9

30 4 3 142.3

31 4 4 141.7

32 4 4 141.2

Mixed-Models with Nested and Crossed Factors Designs

An extension from two factors to three or more are straightforward if the factors are all fixed or all random. However, in many laboratory testing studies, the factors may be mixed, that is some are fixed and some are random. This occurs often when both nested and crossed factors are in the experiment. The following case study demonstrate the analysis of such an experiment.

Factor A is nested in B means: Physically, factor A is within B. For example, Day is within Week. Subsamples are within the sample that is split into these subsamples. The level of factor A is not identical across all levels of another factor B.

Factor A (a levels) and B (b levels) are crossed: the treatment combination is axb. And the experiment units are assigned to each combination. Then A and B are crossed. Each level of every factor occurs with all levels of th eother factors, and the interaction amomng factors can be quantified.

Case Study: Spectrophotometer is used in medical clinical laboratories. The consistency of measurements over multiple runs and from day to day is very critical. An uncertainty study is conducted to evaluate the variability of measurements among machines operate several runs per day over several days, and to study if the machine uncertainty is within an acceptable standards for applications.

Treatment Design: A factorial design is planned with treatments are three commonly used standard concentrations of glucose and three randomly chosen days, and within each combination of Concentration and Day, two runs of testing was performed, and two replicates of each run were tested.

Experimental Design: Four replicate serum samples are prepared for each of the three concentrations of the glucose standards each day. Two samples of each concentration are randomly assigned to each run of the day. Six samples (two samples for each concentration) are tested at a random order on each run.

This is a crossed and Nested design: Glucose concentration and Day are crossed. Runs are within Day, and replicates are within runs.

An appropriate Model to describe this design is:

( ) ( )

k(j)

( ) ( )

1, 2,... ; 1, 2,... ; 1, 2,..., ; 1, 2,..., .

is the fixed effect for facotr 'Concentration'.

is a random effect for factor 'Day'.

c is the random eff

ijkl i j k j ij ik j ijkl

i

j

y c c e

i a j b k c l r

( )

ect for the factor Run, which is nested within Day.

( ) is the random effect for the interaction between Concentration and Day.

( ) is the random effect for Concentraion by Run interaction nest

ij

ik jc

ijkl

ed within Day.

e is the random experimental error.

Five variance components can be identified from the model:

We will show how to use Minitab to estimate these components and conduct appropriate F-tests

2 2 2 2 2( ) ( ), , , ,c c

Analysis of Nested, Crossed Factorial Design – the Case Study of Glucose Concentrations


StGcon fixed 3 1 2 3

Day random 3 1 2 3

Run(Day) random 6 1 2 3 4 5 6

Analysis of Variance for YGcon, using Adjusted SS for Tests


StGcon 2 108263.6 108263.6 54131.8 1227.51 0.000

Day 2 24.9 24.9 12.4 0.12 0.889 x

StGcon*Day 4 176.4 176.4 44.1 1.47 0.321

Run(Day) 3 263.1 263.1 87.7 2.92 0.122

StGcon*Run(Day) 6 180.2 180.2 30.0 20.92 0.000

Error 18 25.9 25.9 1.4

Total 35 108934.1

x Not an exact F-test.

Unusual Observations for YGcon

Obs YGcon Fit SE Fit Residual St Resid

19 134.400 132.200 0.847 2.200 2.60R

20 130.000 132.200 0.847 -2.200 -2.60R

The Concentration*Run(day) interaction is significant-the inconsistency form run to run for different concentrations



1 StGcon (6) + 2.0000(5) + 4.0000(3) + Q[1]

2 Day (6) + 2.0000(5) + 6.0000(4) + 4.0000(3) + 12.0000(2)

3 StGcon*Day (6) + 2.0000(5) + 4.0000(3)

4 Run(Day) (6) + 2.0000(5) + 6.0000(4)

5 StGcon*Run(Day) (6) + 2.0000(5)

6 Error (6)



1 StGcon 4.00 44.1 (3)

2 Day 3.24 101.8 (3) + (4) - (5)

3 StGcon*Day 6.00 30.0 (5)

4 Run(Day) 6.00 30.0 (5)

5 StGcon*Run(Day) 18.00 1.4 (6)

The EMS provides information to help us to determine the appropriate F-test. In this case, the error term for testing Day is not straightforward. It is estimated by using components:

EMS(3)+EMS(4)-EMS(5)=(6)+2(5)+6(4)+4(3) as the denominator of the F-test for Day. The df must also be estimated


Source Estimated Value

Day -7.444

StGcon*Day 3.516

Run(Day) 9.611

StGcon*Run(Day) 14.300

Error 1.436

Least Squares Means for YGcon

StGcon Mean

1 42.15

2 136.56

3 172.11

Day

1 117.78

2 115.81

3 117.22

(Day)StGcon*Run

1 1 1 41.90

1 1 2 41.30

1 2 1 136.25

1 2 2 143.15

1 3 1 163.25

1 3 2 180.85

2 1 3 40.05

2 1 4 42.25

2 2 3 131.35

2 2 4 132.20

2 3 3 173.75

2 3 4 175.25

3 1 5 42.30

3 1 6 45.10

3 2 5 136.30

3 2 6 140.10

3 3 5 166.05

3 3 6 173.50

Negative component should treated as zero. The variance component due to is negligible

The largest uncertainty is the inconsistency from run to run for different concentrations within day. It may be due to the operation or the samples for each concentration from run to run

40 41 42 43 44 45

YGcon_1

Dotplot for YGcon_1

Day

1

2

3

130 135 140

YGcon_2

Dotplot for YGcon_2

Day

1

2

3

165 170 175 180

YGcon_3

Dotplot for YGcon_3

Day

1

2

3

The concentrations are similar from day to day. The interaction of Concentration by Day is insignificant.

40 41 42 43 44 45

YGcon_1

Dotplot for YGcon_1

Run

1

2

3

4

5

6

130 135 140

YGcon_2

Dotplot for YGcon_2

Run

1

2

3

4

5

6

165 170 175 180

YGcon_3

Dotplot for YGcon_3

Run

1

2

3

4

5

6

Run (1,2) – Day 1, Run(3,4)- Day 2 Run(5,6) – Day 3

Within each day, the average for run 1 and average of run2 differs greatly from concentration to concentration – THis is also shown in the ANOVA table: Concentration by Run(day) is significant.

In conclusion, the Day-to-Day operation seems to be consistent for different concentrations. This is important, since other wise, some days will result more accurate results than others days depending on the glucose concentration levels.

Within each day, the run-to-run results are highly dependent on the concentration level. This needs a closer examination to find out what cause this inconsistency. It could be the operation of the instrument, could be that the instrument is more sensitive to high concentration, or it could be the inconsistency of the preparation of concentrations.

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

SRES-Concentration 2

SR

ES

-co

n-1

Scatter Plot of Residuals Between Con 1 and Con 2

Residual Analysis for Outliers Between Two Concentrations

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

SRES_Con 3

SR

ES

_C

on

1

Scatter Plot of Residuals Between Con 1 and Con 3

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

SRES_Con 3

SR

ES

_C

on

2

Scatter Plot of Residuals Between Con2 and Con 3

-3 -2 -1 0 1 2 3

-1.0-0.50.00.51.0

SRES_2

SR

ES

_1

-2 -1 0 1 2

-3

-2

-1

0

1

2

3

SRES_3

SR

ES

_2

-2 -1 0 1 2

-1.0-0.50.00.51.0

SRES_3

SR

ES

_1

Scatter & Marginal Plots of Residuals : Two Concentrations

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Module Twelve: Designs and Analysis for Factorial Treatments In most of experimental studies, through the brain storming and cause-effect diagram, we often