lecture 2 1 - school of computer science and statistics - …€¦ · · 2012-05-09lecture 2.1 1....

Design and Analysis of ExperimentsLecture 2.1

1. Review of Lecture 1.2

– Minute tests

– Randomized Blocks Design and Analysis,

with extensions

Lecture 2.1 1

© 2012 Michael Stuart

Diploma in Statistics

Design and Analysis of Experiments

2. Factorial design

– a 3 x 3 experiment

3. Randomized blocks revisited

– interaction?

– explaining ANOVA

Minute Test: How Much

Lecture 2.1 2




Minute Test: How Fast

Lecture 2.1 3




Randomized Blocks Design and Analysis

Review and Extensions

• Review

• Randomization procedure

• Deleted diagnostics

Lecture 2.1 4





• Effect of Blocking

• Adjusted sums of squares

Randomized block design

Where replication entails increased variation, replicate the full experiment in several blocks so that

• non-experimental variation within blocks is as small as possible,

Lecture 2.1 5




– comparison of experimental effects subject to minimal chance variation,

• variation between blocks may be substantial,

– comparison of experimental effects not affected

Illustrations of blocking variables

Agriculture:

fertility levels in a field or farm,

moisture levels in a field or farm,

genetic similarity in animals, litters as blocks,

Lecture 2.1 6


genetic similarity in animals, litters as blocks,

etc.




Clinical trials

age,

sex,

height, weight,

Lecture 2.1 7


height, weight,

social class,

medical history

etc.

(but analysis of covariance frequently used instead)Diploma in Statistics



Clinical trials

treatments applied to same individual at different times,

cross-over, carry-over, correlation,

Lecture 2.1 8


cross-over, carry-over, correlation,

body parts as blocks,

hands, feet, eyes, ears,

etc.




Industrial trials

similar machines as blocks,

time based blocks,

time of day, day of week, shift

Lecture 2.1 9


time of day, day of week, shift

etc.



Randomized blocksIllustration

• Process: chemicals blended, filtered and dried

• Problem: yield loss at filtration stage

• Proposal: adjust initial blend to reduce yield loss

• Plan:

Lecture 2.1 10




• Plan:

– prepare five different blends

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

– repeat at later times (blocks)

ResultsBlock 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

Lecture 2.1 11




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

Ref: BlendLoss.xls

Analysis of Variance

Source DF SS MS F P

Block 2 1.648 0.824 0.94 0.429

Blend 4 11.556 2.889 3.31 0.071

Error 8 6.992 0.874

Total 14 20.196

Lecture 2.1 12


S = 0.9349

F(Blends) is almost statistically significant, p = 0.07

F(Blocks) is not statistically significant, p = 0.4

→ Diagnostic analysis reveals exceptional case



Iterated analysis:delete Case 11

Source DF SS MS F P

Block 2 3.758 1.879 4.14 0.065

Blend 4 14.572 3.643 8.03 0.009

Error 7 3.176 0.454

Total 13 19.989

S = 0.674

Lecture 2.1 13


S = 0.674

F(Blends) is statistically significant, p = 0.01

F(Blocks) is almost significant, p = 0.065

Prediction standard deviation: S = 0.67



Randomized Block Design and Analysis


• Review



Lecture 2.1 14







Randomization procedure

1. enter numbers 1 to 5 in Column A of a spreadsheet, headed Run,

2. enter letters A-E in Column B, headed Method,

3. generate 5 random numbers into Column C, headed Random

Lecture 2.1 15




Random

4. sort Method by Random,

5. allocate Treatments as sorted to Runs in Block I,

6. repeat Steps 3 - 5 for Blocks II to V.

Go to Excel



• Review



Lecture 2.1 16







Source DF SS MS F P

Block 2 1.648 0.824 0.94 0.429

Blend 4 11.556 2.889 3.31 0.071

Error 8 6.992 0.874

Total 14 20.196

S = 0.9349

Minitab Analysis

Lecture 2.1 17


Unusual Observations for Loss, per cent

Obs Loss Fit SE Fit Residual St Resid

11 17.10 18.523 0.64 -1.43 -2.09 R

R denotes an observation with a large

standardized residual.



Diagnostic analysis

Lecture 2.1 18




Diagnostic plots

• The diagnostic plot, residuals vs fitted values

– checking for homogeneity of chance variation

Lecture 2.1 19




– checking for homogeneity of chance variation

• The Normal residual plot,

– checking the Normal model for chance variation

Deleted residuals

• Residual

– observed – fitted

• Standardised Residual

– divide by the usual estimate of σσσσ

Lecture 2.1 20




– divide by the usual estimate of σσσσ

• Standardised Deleted Residual

– residual calculated from data with suspect case deleted

– σσσσ estimated from data with suspect case deleted

Deleted residuals

For each potentially exceptional case:

– delete the case

– calculate the ANOVA from the rest

– use the deleted fitted model to calculate a

deleted fitted value

Lecture 2.1 21




deleted fitted value

– calculate deleted residual= obseved value – deleted fitted value

– calculate deleted estimate of s

– standardise

Minitab does this automatically for all cases!

Scatterplot, with exceptional case

Lecture 2.1 22




Corresponding residual plots,Standardized vs Deleted

Lecture 2.1 23




Randomized Blocks ExampleStandardized vs Deleted

Lecture 2.1 24




Randomized Blocks ExampleStandardized vs Deleted

Lecture 2.1 25






• Review



Lecture 2.1 26







Was the blocking effective?

Source DF SS MS F P

Block 2 1.648 0.824 0.94 0.429

Blend 4 11.556 2.889 3.31 0.071

Error 8 6.992 0.874

Total 14 20.196

S = 0.9349

Lecture 2.1 27


S = 0.9349

Source DF SS MS F P

Blend 4 11.556 2.889 3.34 0.055

Error 10 8.640 0.864

Total 14 20.196

S = 0.9295



Was the blocking effective?Summary

• F(Blocks) not statistically significant

– Blocks MS smaller than Error MS

• When blocks deleted from analysis

Lecture 2.1 28





– residual standard deviation almost unchanged

and

– F(Blends) almost unchanged

• Blocking appeared to be ineffective.

Effect of Blocking, revised analysis

Analysis of Variance, with blocks

Source DF SS MS F P

Blend 4 14.572 3.643 8.03 0.009

Block 2 3.758 1.879 4.14 0.065

Error 7 3.176 0.454

Total 13 19.989

Lecture 2.1 29




Total 13 19.989

Analysis of Variance, blocks excluded

Source DF SS MS F P

Blend 4 13.055 3.264 4.24 0.034

Error 9 6.933 0.770

Total 13 19.989

Effect of Blocking, revised analysisSummary

• F(Blocks) almost significant

– Blocks MS more than 4 times Error MS


– Blocks DF and Error DF combined

– Blocks SS and Error SS combined

Lecture 2.1 30




– Blocks SS and Error SS combined

– combined Error MS much larger than original

– variation between blocks is substantial

• F(Blends) is more significantwhen variation between blocks is allowed for– blocking was effective.



• Review



Lecture 2.1 31







Blend Loss Analysis:Case 11 deleted

General Linear Model: Loss versus Blend, Block

Analysis of Variance for Loss

Source DF Seq SS Adj SS Adj MS F P

Blend 4 13.0552 14.5723 3.6431 8.03 0.009

Block 2 3.7577 3.7577 1.8788 4.14 0.065

Lecture 2.1 32




Block 2 3.7577 3.7577 1.8788 4.14 0.065

Error 7 3.1757 3.1757 0.4537

Total 13 19.9886

S = 0.673548

Adjusted sums of squares

• Sequential sums of squares are calculated allowing for the presence of earlier effects in the model.

• Adjusted sums of squares are calculated allowing for the presence of all other effects in

Lecture 2.1 33




allowing for the presence of all other effects in the model

• No difference with balanced data,– separate sets of effects are uncorrelated,– values of one set not affected by values of

another

• Use Adjusted SS, ignore Sequential SS

Lecture 2.1


– Minute tests

– Randomized Block Design and Analysis,

with extensions

Lecture 2.1 34




2. Factorial design



– interaction?


Part 2Factorial Design

Iron-deficiency anemia

• the most common form of malnutrition in developing countries

• contributory factors:

Lecture 2.1 35


– cooking pot type

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

– food type

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



Study design and results

• 4 samples of each food type were cooked in each pot type,

• iron content in each sample measured in milligrams of iron per 100 grams of cooked food.

Lecture 2.1 36




Pot Type Vegetable 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

Classwork 2.1.0

• What were the– experimental units– experimental runs– observational units– response– factors

Lecture 2.1 37


– factors– levels– treatments– unit structure– treatment structure– design– replication



Initial Data Analysis

Pot Type Vegetable Type

Meat Legumes Vegetables

Aluminium 2.06 2.33 1.23

Clay 2.18 2.47 1.46

Iron 4.68 3.67 2.79

Lecture 2.1 38




Model for analysis

Iron content includes

– a contribution for each food type

plus

– a contribution for each pot type

plus

Lecture 2.1 39


plus

– a contribution for each food type / pot type combination

plus

– a contribution due to chance variation.

Minitab: Pot Food Pot * FoodDiploma in Statistics


Analysis of Variance

Analysis of Variance for Iron

Source DF SS MS F P

Pot 2 24.8940 12.4470 92.26 0.000

Food 2 9.2969 4.6484 34.46 0.000

Pot*Food 4 2.6404 0.6601 4.89 0.004

Lecture 2.1 40


Pot*Food 4 2.6404 0.6601 4.89 0.004

Error 27 3.6425 0.1349

Total 35 40.4738

S = 0.367297



Summary

Cooking in iron pots adds substantially to the average iron content of all cooked foods.

However, it adds considerably more to the iron content of meat,

Lecture 2.1 41


– around 2.5 to 2.6 mgs per 100gms on average,

than to that of legumes or vegetables,

– around 1.2 to 1.5 mgs per 100gms on average

The iron content is very similar using aluminiumand clay for all three food types.



Interaction

Lecture 2.1 42




Interaction

Lecture 2.1 43




Diagnostic plots

Lecture 2.1 44




Diagnostic plots

Lecture 2.1 45


Slight suggestion of skewness, butconclusions are sufficiently strong to ignore this



Lecture 2.1


– Homework 1.2.1

– Randomized Block Design and Analysis,

with extensions

Lecture 2.1 46




2. Factorial design



– interaction?


Part 3Randomized blocks revisited

Blend Loss analysis included

Blend effects + Block effects + Chance variation,

– NO INTERACTION EFFECTS

Analysis of Variance for Loss,%, using Adjusted

Lecture 2.1 47


Analysis of Variance for Loss,%, using Adjusted

SS for Tests


Blend 4 11.5560 11.5560 2.8890 3.31 0.071

Block 2 1.6480 1.6480 0.8240 0.94 0.429

Error 8 6.9920 6.9920 0.8740

Total 14 20.1960



Include interaction in model?

Analysis of Variance for Loss, using Adjusted SS

for Tests


Blend 4 11.5560 11.5560 2.8890 **

Block 2 1.6480 1.6480 0.8240 **

Lecture 2.1 48


Block 2 1.6480 1.6480 0.8240 **

Blend*Block 8 6.9920 6.9920 0.8740 **

Error 0 * * *

Total 14 20.1960



Classwork 2.1.2

Calculate fitted values:

Overall Mean + Blend Deviation + Block deviation

Block 1

–0.4 Block 2

0.1 Block 3

0.4

Blend A –0.6

Blend B

Lecture 2.1 49




17.5 +

Blend B 0.6

Blend C –0.1

Blend D –1.2

Blend E 1.3

Classwork 2.1.1 (cont'd)

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 17.0 17.5 17.8

Lecture 2.1 50




Make a Block profile plot

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

19.5

19.0

18.5

18.0

17.5

Valu

es

1

2

3

Block

Line Plot of Fitted Values

Lecture 2.1 51




EDCBA

17.5

17.0

16.5

16.0

Blend

Fit

ted

Actual plot: Interaction?

20

19

18

17

16

Lo

ss, p

er

ce

nt

Block 1

Block 2

Block 3

Block Profiles

Lecture 2.1 52




EDCBA

16

15

Blend

Blend effects are similar for Blocks 1 and 2 but quite different for Block 3.

Caution: no general test without replication.

Interaction?

20

19

18

pe

r ce

nt

Block 1

Block 2

Block 3

Block Profiles

Lecture 2.1 53




Blend x Block interaction?

EDCBA

17

16

15

Blend

Lo

ss, p

Interaction?

20

19

18

17

Lo

ss, p

er

ce

nt

A

B

C

D

E

Blend

Blend profiles

Lecture 2.1 54




321

16

15

Block

L

Homework 2.1.1 Adjust some of the points in the graph to reduce the interaction. Relate to previous graph.

What to do with no replication?

Recall F-test logic:

– MS(Error) ≈ σσσσ2

– MS(Effect) ≈ σσσσ2 + effect contribution

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

>>1 if effect present

No replication?

Lecture 2.1 55




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

Explaining ANOVA

ANOVA depends on a decompostion of "Total variation" into components:

Total Variation = Blend effect + Block effect

+ chance variation;

Lecture 2.1 56


+ chance variation;

∑∑∑∑ ++++−−−−−−−−++++

∑∑∑∑ −−−−++++∑∑∑∑ −−−−====∑∑∑∑ −−−−

••••••••••••••••

====

••••••••••••

====

••••••••••••••••••••

j,i

2jiij

k

1j

2j

k

1i

2i

j,i

2ij

)YYYY(

)YY(k)YY(n)YY(

.



Decomposition of results

Overall Deviations Blend Deviations Block Deviations Residuals

YYrc −−−− = YYr −−−− + YYc −−−− + YYYY crrc ++++−−−−−−−−

Blocks

I II III Mean A 16.9 16.5 17.5 17.0 B 18.2 19.2 17.1 18.2 Blends C 17.0 18.1 17.3 17.5

D 15.1 16.0 17.8 16.3 E 18.3 18.3 19.8 18.8

Mean 17.1 17.6 17.9 17.5

Lecture 2.1 57


I II III I II III I II III I II III

A -0.6 -1.0 0.0 -0.6 -0.6 -0.6 -0.4 0.1 0.4 0.4 -0.5 0.2

B 0.7 1.7 -0.4 0.6 0.6 0.6 -0.4 0.1 0.4 0.5 1.0 -1.4

C -0.5 0.6 -0.2 = -0.1 -0.1 -0.1 + -0.4 0.1 0.4 + 0.0 0.6 -0.5

D -2.4 -1.5 0.3 -1.2 -1.2 -1.2 -0.4 0.1 0.4 -0.8 -0.4 1.1

E 0.8 0.8 2.3 1.3 1.3 1.3 -0.4 0.1 0.4 -0.1 -0.6 0.6

SSTO = 20.20 SS(Blends) = 11.56 SS(Blocks) = 1.65 SSE = 6.99

dfTO = 14 df(Blends) = 4 df(Blocks) = 2 dfE = 8



Overall Deviations Blend Deviations Block Deviations Residuals

YYrc −−−− = YYr −−−− + YYc −−−− + )YY(

)YY(

r

rc

−−−−−−−−

−−−−

Alternative form of Residuals Blocks

I II III Mean A 16.9 16.5 17.5 17.0 B 18.2 19.2 17.1 18.2 Blends C 17.0 18.1 17.3 17.5

D 15.1 16.0 17.8 16.3 E 18.3 18.3 19.8 18.8

Mean 17.1 17.6 17.9 17.5

Lecture 2.1 58


rc r c

)YY( c

r

−−−−−−−−

I II III I II III I II III I II III

A -0.6 -1.0 0.0 -0.6 -0.6 -0.6 -0.4 0.1 0.4 0.4 -0.5 0.2

B 0.7 1.7 -0.4 0.6 0.6 0.6 -0.4 0.1 0.4 0.5 1.0 -1.4

C -0.5 0.6 -0.2 = -0.1 -0.1 -0.1 + -0.4 0.1 0.4 + 0.0 0.6 -0.5

D -2.4 -1.5 0.3 -1.2 -1.2 -1.2 -0.4 0.1 0.4 -0.8 -0.4 1.1

E 0.8 0.8 2.3 1.3 1.3 1.3 -0.4 0.1 0.4 -0.1 -0.6 0.6

SSTO = 20.20 SS(Blends) = 11.56 SS(Blocks) = 1.65 SSE = 6.99

dfTO = 14 df(Blends) = 4 df(Blocks) = 2 dfE = 8



Exercise 2.1.1

• See Exercise 2_1_1.doc in the Exercises section of the module web page.

Lecture 2.1 59




Reading

EM §7.3.2, §7.4.2

DCM §§4-1, 5-1 to 5-3

DV §6.2, 6.4

Lecture 2.1 60




lecture 2 1 - school of computer science and statistics - …€¦ · · 2012-05-09lecture 2.1 1....

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