Design and Analysis of Experiments Lecture 1.2

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 1.2 1

Design and Analysis of ExperimentsLecture 1.2

1. Review of Lecture 1.1

2. Application; a simple comparative experiment

3. Comparing several means

4. Randomised block design

5. Randomised block analysis


Lecture 1.2 2

Examinations Timetable

Please note that the definitive versions of the timetables are displayed on the College webpages at

www.tcd.ie/Examinations/Timetables/PDF/1291000.pdf


Lecture 1.2 3

Experiments

To find out what happens when you change something, it is necessary to change it.

Experiment and you'll see!

(BHH)


Lecture 1.2 4

Recall Example 2: walking babies

•How long does it take a baby to walk?

•Can this be affected by special training programs?

4 "training" programs:

1. special exercises

2. normal daily exercise

3. weekly check

4. end of study check

each of 24 babies allocated at random to groups of 6 in each program.


Lecture 1.2 5

Characteristics of an experiment

Experimental units:

entities on which observations are madee.g., babies

Experimental Factor:

controllable input variablee.g., Training

Factor Levels / Treatments:

values of the factore.g., training programmes

Response:

output variable measured on the unitse.g., walking age


Lecture 1.2 6

Exercise 1.2.1

Recall comparison of standard (old) and new processes for manufacture of electronic components, 50 components sampled per day, 6 days per week, for 8 weeks,

What were the

experimental units

factor

factor levels

response


Lecture 1.2 7

Two design principles

• Blocking

– identify homogeneous blocks of experimental units

– assess effects of experimental change within homogeneous blocks

– average effects across blocks

• Randomisation

– allocate experimental conditions to units at random

– minimise chances of "lurking variable" pattern coinciding with factor level allocation pattern


Lecture 1.2 8

Homework 1.1.1

Assess the statistical significance of the difference in defect rates, %, between the first period and second period for the new process.

First

Period Second Period

Difference

Both Processes 3.0 0.9 2.1

Old Process 3.3 0.8 2.5

New Process 2.7 1.0 1.7


Lecture 1.2 9

Illustration of a full factorial design,with 12 experimental runs

Pressure

Temperature

High

High

Low

Low321 YYY

121110 YYY987 YYY

654 YYY


Lecture 1.2 10

Exercise 1.2.2

Recall yield optimisation experiment. What were the

experimental units

factors

factor levels

response


Lecture 1.2 11

2 ApplicationA simple comparitive experiment

Wear of shoe solesmade of two materials, A and B,

worn on opposite feet by each of 10 boys,with randomallocation of materials to feet.


Lecture 1.2 12

Exercise 1.2.3

What were the

experimental units

factor

factor levels

response

blocks

randomisation procedure


Lecture 1.2 13

AnalysisWear of shoe soles

made of two materials, A and B,worn on opposite feet by each of 10 boys

10n386.0s41.0d d

Boy Material A Material B Difference

1 13.2 14.0 0.8 2 8.2 8.8 0.6 3 10.9 11.2 0.3 4 14.3 14.2 -0.1 5 10.7 11.8 1.1 6 6.6 6.4 -0.2 7 9.5 9.8 0.3 8 10.8 11.3 0.5 9 8.8 9.3 0.5

10 13.3 13.6 0.3


Lecture 1.2 14

Analysis

t test of differences

highly significant

4.3

10/386.0

041.0

n/s

0dt

d


Lecture 1.2 15

Selected critical values for the t-distribution .25 .10 .05 .02 .01 .002 .001

= 1 2.41 6.31 12.71 31.82 63.66 318.32 636.61 2 1.60 2.92 4.30 6.96 9.92 22.33 31.60 3 1.42 2.35 3.18 4.54 5.84 10.22 12.92 4 1.34 2.13 2.78 3.75 4.60 7.17 8.61 5 1.30 2.02 2.57 3.36 4.03 5.89 6.87 6 1.27 1.94 2.45 3.14 3.71 5.21 5.96 7 1.25 1.89 2.36 3.00 3.50 4.79 5.41 8 1.24 1.86 2.31 2.90 3.36 4.50 5.04 9 1.23 1.83 2.26 2.82 3.25 4.30 4.78 10 1.22 1.81 2.23 2.76 3.17 4.14 4.59 12 1.21 1.78 2.18 2.68 3.05 3.93 4.32 15 1.20 1.75 2.13 2.60 2.95 3.73 4.07 20 1.18 1.72 2.09 2.53 2.85 3.55 3.85 24 1.18 1.71 2.06 2.49 2.80 3.47 3.75 30 1.17 1.70 2.04 2.46 2.75 3.39 3.65 40 1.17 1.68 2.02 2.42 2.70 3.31 3.55 60 1.16 1.67 2.00 2.39 2.66 3.23 3.46 120 1.16 1.66 1.98 2.36 2.62 3.16 3.37 ∞ 1.15 1.64 1.96 2.33 2.58 3.09 3.29


Lecture 1.2 16

A simpler test

Sign test:

count the "+" signs; what are the chances of getting that number or more?

MINITAB

Boy Material A Material B Difference

1 13.2 14.0 0.8 2 8.2 8.8 0.6 3 10.9 11.2 0.3 4 14.3 14.2 -0.1 5 10.7 11.8 1.1 6 6.6 6.4 -0.2 7 9.5 9.8 0.3 8 10.8 11.3 0.5 9 8.8 9.3 0.5

10 13.3 13.6 0.3


Lecture 1.2 17

Was the blocking effective?

10987654321

16

14

12

10

8

6

4

2

0

Boy

Dat

a

Material AMaterial BDifference

Variable

Profile Plots of Material A, Material B, Difference


Lecture 1.2 18

Was the blocking effective?

Ignore blocking and use a two sample t-test of the effectiveness of the new material.

Boy Material A Material B

1 13.2 14.0 2 8.2 8.8 3 10.9 11.2 4 14.3 14.2 5 10.7 11.8 6 6.6 6.4 7 9.5 9.8 8 10.8 11.3 9 8.8 9.3

10 13.3 13.6

Mean 10.6 11.0

Standard Deviation 2.45 2.52


Lecture 1.2 19

Effect of pairing

Paired T for Material B - Material A

N Mean StDev SE MeanMaterial B 10 11.0400 2.5185 0.7964Material A 10 10.6300 2.4513 0.7752Difference 10 0.410000 0.387155 0.122429

95% CI for mean difference: (0.133046, 0.686954)T-Test of mean difference = 0 (vs not = 0): T-Value = 3.35 P-Value =0.009

Two-sample T for Material B vs Material A

N Mean StDev SE MeanMaterial B 10 11.04 2.52 0.80Material A 10 10.63 2.45 0.78

Difference = mu (Material B) - mu (Material A)Estimate for difference: 0.41000095% CI for difference: (-1.924924, 2.744924)T-Test of difference = 0 (vs not =): T-Value = 0.37 P-Value = 0.716


Lecture 1.2 20

Effect of pairing

Paired t:

df = 9, t9,.05 = 2.26

Two-sample t:

df = 17, t17,.05 = 2.11

B

2B

A

2A

AB

n

s

n

s

XXt

37.0

n

s0d

td

10/39.0

41.0 35.3


Lecture 1.2 21

Diagnostic analysis

Dif

fere

nc

e

Score

2.0

1.5

1.0

0.5

0.0

-0.5

-1.0

210-1-2

Mean

0.622

0.41

StDev 0.3872

N 10

AD 0.261

P-Value

Probability Plot of DifferenceNormal - 95% CI


Lecture 1.2 22

Homework 1.2.1

A maintenance manager tested a new method for repairing machines by recording the time since the previous repair prior to using the new method and the time to next failure after using the new method for each of 10 machines, with the following results:

Formally test the effect of changing to the new method.Criticise the design.Was the blocking effective?

Machine 1 2 3 4 5 6 7 8 9 10

Current 155 222 346 287 115 389 183 451 140 252

New 211 345 419 274 244 420 319 505 396 222


Lecture 1.2 23

3 Comparing several means(Observational study)

Burst strength (kPa) of 10 samplesof each of four filter membrane types

Exercise 1.2.4Make dotplots of the breaking strengths

Membrane A Membrane B Membrane C Membrane D

95.5 90.5 86.3 89.5 103.2 98.1 84.0 93.4 93.1 97.8 86.2 87.5 89.3 97.0 80.2 89.4 90.4 98.0 83.7 87.9 92.1 95.2 93.4 86.2 93.1 95.3 77.1 89.9 91.9 97.1 86.8 89.5 95.3 90.5 83.7 90.0 84.5 101.3 84.9 95.6


Lecture 1.2 24

Comparing several means

Burst strength (kPa) of 10 samplesof each of four filter membrane types

Variable Membrane N Mean StDev Minimum Maximum RangeStrength A 10 93 4.8 85 103 19 B 10 96 3.4 91 101 11 C 10 85 4.3 77 93 16 D 10 90 2.8 86 96 9


Lecture 1.2 25

One-way ANOVA: Strength versus Membrane

Source DF SS MS F PMembrane 3 709.2 236.4 15.54 0.000Error 36 547.8 15.2Total 39 1257.0

S = 3.901 R-Sq = 56.42% R-Sq(adj) = 52.79%



Lecture 1.2 26

One-way ANOVA: Strength versus Membrane

Source DF SS MS F PMembrane 3 709.2 236.4 15.54 0.000Error 36 547.8 15.2Total 39 1257.0

S = 3.901 R-Sq = 56.42% R-Sq(adj) = 52.79%

Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ------+---------+---------+---------+---A 10 92.840 4.831 (----*----)B 10 96.080 3.391 (----*----)C 10 84.630 4.287 (----*----)D 10 89.890 2.764 (----*----) ------+---------+---------+---------+--- 85.0 90.0 95.0 100.0



Lecture 1.2 27


Tukey 95% Simultaneous Confidence IntervalsAll Pairwise Comparisons among Levels of Membrane

Membrane = A subtracted from:Membrane Lower Center Upper ------+---------+---------+---------+-B -1.46 3.24 7.94 (---*----)C -12.91 -8.21 -3.51 (----*---)D -7.65 -2.95 1.75 (----*----) ------+---------+---------+---------+- -10 0 10 20Membrane = B subtracted from:Membrane Lower Center Upper ------+---------+---------+---------+---C -16.15 -11.45 -6.75 (----*---)D -10.89 -6.19 -1.49 (----*----) ------+---------+---------+---------+--- -10 0 10 20Membrane = C subtracted from:Membrane Lower Center Upper ------+---------+---------+---------+---D 0.560 5.260 9.960 (---*----) ------+---------+---------+---------+--- -10 0 10 20


Lecture 1.2 28

Diagnostic analysis

95.092.590.087.585.0

10

5

0

-5

-10

Fitted Value

Res

idu

al

10

5

0

-5

-10

210-1-2

Res

idu

al

Score

N 40

AD 0.736

P-Value 0.051

Versus Fits(response is Strength)

Normal Probability Plot(response is Strength)


Lecture 1.2 29

4 Randomised block design

Example 1: treating crops with fertiliser to improve yield.

Four fertilisers being tested:

divide a single field into four plots (treatment units) to form one block,

assign treatments at random to the four plots,

repeat with several other fields to form several blocks,

choose blocks in varying locations, for generalising.


Lecture 1.2 30

Randomised block design

Example 2: treating large sheets of rubber to improve abrasion resistance.

Four treatments being tested:

cut a single piece into four treatment units to form one block,

assign treatments at random to the four units,

repeat with several other pieces to form several blocks.


Lecture 1.2 31


Piece 1

B

Piece 2

A

Piece 3

C

Piece 4

D

Block 1


Lecture 1.2 32


Piece 1

A

Piece 2

D

Piece 3

C

Piece 4

B

Block 2


Lecture 1.2 33


Block 3

Piece 1

B

Piece 2

A

Piece 3

D

Piece 4

C


Lecture 1.2 34


Block 1

Block 2

Block 3


Lecture 1.2 35


Example 3: assessing new laboratory test methods.

Four methods being assessed:

assess each method in a single laboratory,

randomise the time order in which the methods are run,

repeat in several other laboratories to form several blocks.


Lecture 1.2 36


Example 4: assessing process changes.

Five versions of the process being assessed:

assess each version on five successive days,

randomise the time order in which the versions are used,

repeat over several weeks to form several blocks.


Lecture 1.2 37

5 Randomised block designIllustration

Manufacture of an organic chemical using a filtration process

• Three step process:

– input chemical blended from different stocks

– chemical reaction results in an intermediate liquid product

– liquid filtered to recover end product.


Lecture 1.2 38

Randomised block designIllustration

• 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)


Lecture 1.2 39

Results


Lecture 1.2 40

General Linear Model ANOVA

General Linear Model: Loss, per cent versus Blend, Block

Analysis of Variance for Loss,%, using Adjusted SS for Tests

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

S = 0.934880 R-Sq = 65.38% R-Sq(adj) = 39.41%

Unusual Observations for Loss, per cent

Loss, perObs cent Fit SE Fit Residual St Resid 12 17.1000 18.5267 0.6386 -1.4267 -2.09 R


Lecture 1.2 41

5% critical values for the F distribution

1 1 2 3 4 5 6 7 8 10 12 24 ∞

2

1 161 200 216 225 230 234 237 239 242 244 249 254 2 18.5 19.0 19.2 19.2 19.3 19.3 19.4 19.4 19.4 19.4 19.5 19.5 3 10.1 9.6 9.3 9.1 9.0 8.9 8.9 8.8 8.8 8.7 8.6 8.5 4 7.7 6.9 6.6 6.4 6.3 6.2 6.1 6.0 6.0 5.9 5.8 5.6 5 6.6 5.8 5.4 5.2 5.1 5.0 4.9 4.8 4.7 4.7 4.5 4.4 6 6.0 5.1 4.8 4.5 4.4 4.3 4.2 4.1 4.1 4.0 3.8 3.7 7 5.6 4.7 4.3 4.1 4.0 3.9 3.8 3.7 3.6 3.6 3.4 3.2 8 5.3 4.5 4.1 3.8 3.7 3.6 3.5 3.4 3.3 3.3 3.1 2.9 9 5.1 4.3 3.9 3.6 3.5 3.4 3.3 3.2 3.1 3.1 2.9 2.7 10 5.0 4.1 3.7 3.5 3.3 3.2 3.1 3.1 3.0 2.9 2.7 2.5 12 4.7 3.9 3.5 3.3 3.1 3.0 2.9 2.8 2.8 2.7 2.5 2.3 15 4.5 3.7 3.3 3.1 2.9 2.8 2.7 2.6 2.5 2.5 2.3 2.1 20 4.4 3.5 3.1 2.9 2.7 2.6 2.5 2.4 2.3 2.3 2.1 1.8 30 4.2 3.3 2.9 2.7 2.5 2.4 2.3 2.3 2.2 2.1 1.9 1.6 40 4.1 3.2 2.8 2.6 2.4 2.3 2.2 2.2 2.1 2.0 1.8 1.5

120 3.9 3.1 2.7 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.6 1.3 ∞ 3.8 3.0 2.6 2.4 2.2 2.1 2.0 1.9 1.8 1.8 1.5 1.0


Lecture 1.2 42

Conclusions (prelim.)

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

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

Prediction standard deviation: S = 0.93


Lecture 1.2 43

Deleted diagnostics

19181716

2

1

0

-1

-2

-3

Fitted Value

Del

eted

Res

idu

al

3

2

1

0

-1

-2

-3

210-1-2

Del

eted

Res

idu

al

Score

N 15

AD 0.245

P-Value 0.712

Versus Fits(response is Loss)

Normal Probability Plot(response is Loss)


Lecture 1.2 44

Iterated analysis:delete Case 12

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 0.009Block 2 3.7577 3.7577 1.8788 4.14 0.065Error 7 3.1757 3.1757 0.4537

Total 13 19.9886

S = 0.673548


Lecture 1.2 45

Deleted diagnostics

2019181716

2

1

0

-1

-2

-3

Fitted Value

Del

eted

Res

idu

al

3

2

1

0

-1

-2

-3

210-1-2

Del

eted

Res

idu

al

Score

N 14

AD 0.189

P-Value 0.881

Versus Fits(response is Loss)

Normal Probability Plot(response is Loss)


Lecture 1.2 46

Conclusions (prelim.)

F(Blends) is almost statistically significant, p = 0.01

F(Blocks) is not statistically significant, p = 0.65

Prediction standard deviation: S = 0.67


Lecture 1.2 47

Explaining ANOVA

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

Total Variation = Blend effect + Block effect

+ chance variation;

j,i

2jiij

k

1j

2j

k

1i

2i

j,i

2ij

)YYYY(

)YY(k)YY(n)YY(

.


Lecture 1.2 48

Decomposition of results

Overall Deviations Blend Deviations Block Deviations Residuals

YYrc = YYr + YYc + YYYY crrc

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

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 1.2 49





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



Blocks


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 1.2 50





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



Blocks


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 1.2 51





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



Blocks


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 1.2 52





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



Blocks


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 1.2 53





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



Blocks


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 1.2 54

Caution!

321

20

19

18

17

16

15

Block

Lo

ss, p

er

cen

t

A

B

C

D

E

Blend

Blend profiles

Blend x Block interaction?


Lecture 1.2 55

Caution!

Blend x Block interaction?

EDCBA

20

19

18

17

16

15

Blend

Lo

ss, p

er

cen

t

Block 1

Block 2

Block 3

Block Profiles


Lecture 1.2 56

Reading

EM §4.4, 7.2

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Design and Analysis of Experiments Lecture 1.2

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