06 Analysis of Variance

8/13/2019 06 Analysis of Variance

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Materials Engineering 14Analysis of Variance


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Recall:

What are the twotypes of

experiments?

Variable Screening

Optimization Experiments


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Variable Screening


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Optimization


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Complete Randomized Design

(CRD)

Randomized Block Design

(RBD)

vs


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Response Variable

dependent variable

Factors

qualitative or quantitative

Factor Levels

values of the factor utilized

Treatments

factor-level combinations

Experimental Unit

sample for which a data can be obtained

Some Terms


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a design for which random samples ofexperimental units are independentlyselected for each treatment

objective: compare the treatment means

H0: 1=2=...=a H1: at least two of the a treatment means differ

The Completely Randomized Design

(CRD)


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A manufacturer of paper used formaking grocery bags is interested inimproving the tensile strength of the

product. Product engineeringthinks that the tensile strengthis a function of the hardwoodconcentration in the pulp andthat the range of hardwoodconcentrations of practical interest isbetween 5% and 20%.

Example


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A team of engineers responsible for the studydecides to investigate 4 levels of hardwoodconcentration: 5%, 10%, 15% and 20%. They

decided to make up 6 test specimens at eachconcentration level, using a pilot plant. All 24specimens are tested on a laboratory tensiletester, in random order.

Example


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Typical Data for a Single-Factor

Experiment

TreatmentObservations

Totals Averages

1 y11 y12 ... ... ... y1n y1. y1.

2 y21 y22 ... ... ... y2n y2. y2.

a ya1 ya2 ... ... ... yan ya. y2.

y.. y..


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HardwoodConc. (%)

ObservationsTotals Averages

1 2 3 4 5 6

5 7 8 15 11 9 10

10 12 17 13 18 19 15

15 14 18 19 17 16 18

20 19 25 22 23 18 20

Table of Results


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H0: 1=2=3=4 Different hardwood concentrations do not affect

the mean tensile strength of the paper

H1: at least two means are not equal

Different hardwood concentrations affect the

mean tensile strength of the paper

Step 1: Formulating Hypotheses


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The usual values of are 0.25, 0.10, 0.05, and0.01

For the example, we will be using an of 0.01(99% confidence)

Use F-tables f(0.01,v1=n-1,v2=N-a)

Step 2: Deciding on the confidence

interval


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Computing for Sum of Squares


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measures the total variability in the data

Sum of Squares Total (SST)

a

i

n

j

ijTotalN

yySS

1 1

2

..2


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HardwoodConc. (%)


1 2 3 4 5 6

5 7 8 15 11 9 10 60 10.00

10 12 17 13 18 19 15 94 15.67

15 14 18 19 17 16 18 102 17.00

20 19 25 22 23 18 20 127 21.17

383 15.96

Computing for Total Sum of Squares

96.512

24

)383()20(...)8()7(

2222

TSS


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measures the variations between treatmentmeans

Sum of Squares for Treatments

(SSTreatments)

n

i

iTreatments

N

y

n

ySS

1

2

..

2


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HardwoodConc. (%)


1 2 3 4 5 6

5 7 8 15 11 9 10 60 10.00

10 12 17 13 18 19 15 94 15.6715 14 18 19 17 16 18 102 17.00

20 19 25 22 23 18 20 127 21.17

383 15.96

Computing for Treatment Sum of

Squares

79.382

24

)383(

6

)127()102()94()60( 22222

TreatmentsSS


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measures the sampling variability within thetreatments

accounts for the sampling error

Sum of Squares for Error (SSE)

TreatmentsTE SSSSSS


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HardwoodConc. (%)


1 2 3 4 5 6

5 7 8 15 11 9 10 60 10.00

10 12 17 13 18 19 15 94 15.6715 14 18 19 17 16 18 102 17.00

20 19 25 22 23 18 20 127 21.17

383 15.96

Computing for Error Sum of Squares

17.130

)79.382()96.512(

TreatmentsTE SSSSSS


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Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares

Computedf

Treatments SSTreatments

a-1 MST

MST/MS

E

Error SSE N-a MSE

Total SST (a-1)+(N-a)

Step 4: Summarizing of Results

aN

SSMS

aSSMS

ErrorError

TreatmentsTreatments

1

Perform Analysis of Variance (ANOVA)


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Step 4: Summarizing of Results

aN

SSMS

a

SS

MS

ErrorError

Treatments

Treatments

1

Source ofVariation

Sum ofSquares

dfMean

SquaresComputed

f

Hardwoodconcentration

382.79 3 127.60 19.6

Error 130.17 20 6.51

Total 512.96 23


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H0 is true if the ratio f=MST/MSE is a value ofthe random variable F having the F-distributionwith n-1 and (N-a) degrees of freedom

The null hypothesis is rejected at the value ofsignificance when

f > f[a-1, N-a)]

Step 5: Decision


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Example 2

In many integrated circuit manufacturing steps,wafers are completely coated with a layer ofmaterial such as silicon dioxide or a metal. The

unwanted material is then selectively removed byetching, whose rate is dependent on the radio-frequency (RF) power setting.An engineer isinterested in investigating the relationship

between the RF power setting and the etchrate. Four test levels of RF power: 160, 180, 200,and 220 W (with 5 wafers at each level of RF)were prepared.


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Example 2

RF Power(W)

Observed Etch Rate (/min)Totals Averages

1 2 3 4 5

160 575 542 530 539 570 2,756 551.2

180 565 593 590 579 610 2,937 587.4

200 600 651 610 637 629 3,127 625.4

220 725 700 715 685 710 3,535 707.0

12,355 617.75

Use

=0.05


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Example 2ANOVA for the experiment

Source of VariationSum of

Squaresdf

MeanSquare

F0 Ftable

RF Power

Error

Total


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Example 2ANOVA for the experiment

Source of VariationSum of

Squaresdf

MeanSquare

F0 Ftable

RF Power 66,870.55 3 22,290.18 66.80 3.24

Error 5,339.20 16 333.70

Total 72,209.75 19


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The Analysis of Variance


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Description

more number of levels of a factor is tested

examples:

the effect of curing temperature on the

compressive strength of concrete blocks

the effect of dosage size of a certain drug on itscurative properties


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Statistically-based experiments will

lead to better and improved processes

development of new processes

Control Variables are used to describe the

said processes. Examples of which are time,temperature, feed rate, amount of material,concentration, etc.


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The Pros improved process yield

reduced variability andcloser conformance to

specifications

reduced design anddevelopment time

reduced cost


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In Engineering Design

evaluation and comparison of configurationsand materials

selection of design parameters thatwill makethe product work well under varying fieldconditions

determination of key product design parametersthat affect product performance


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Steps involved in an experiment

Conjecture

Experiment

Analysis

Conclusion


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Exercise

An engineer is interested in how the meanabsorption of moisture in concrete varies among5 different concrete aggregates. The samples are

exposed to moisture for 48 hours. It is decidedthat 6 samples are to be tested for eachaggregate, requiring a total of 30 samples to betested.


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Summary of Results

Aggregate 1 2 3 4 5

Run 1 551 595 639 417 563

Run 2 457 580 615 449 631

Run 3 450 508 511 517 522

Run 4 731 583 573 438 613

Run 5 499 633 648 415 656

Run 6 632 517 677 555 679

Absorption of Moisture in Concrete Aggregates


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RBD


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blocks, or matched sets of experimental units,are formed

each block consists of p experimental units; eachblock (b) should contain units which are assimilar as possible

an experimental unit from each block is assignedto each treatment

Randomized Block Design


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Source ofVariation

SS df MS f

Treatment SST a-1 MST MST/MSE

Block SSB b-1 MSBError SSE N-a-b+1 MSE

Total SS(Total) N-1

General ANOVA Table for RBD


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Formulas

BlocksTreatmentsTotalError

n

iiTotal

b

i

BBlocks

T

a

i

Treatments

SSSSSSSS

xxSS

xxpSS

xxbSS

i

i

1

2

1

2

2

1

)(

)(

)(


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Suppose the USGA wants to compare the meandistances associated with four brands of golf ballswhen struck by a driver, and will employ human

golfers. Assuming that 10 balls of each brand are tobe utilized in the experiment

Suppose that an RBD is used, utilizing arandom sample of 10 golfers with each golfer

using a driver to hit four balls, one of eachbrand, in random sequence

Use one way ANOVA to see if there is a difference amongthe mean distance for the brands of golf balls

Example:


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Golfer(block)

Brand A Brand B Brand C Brand D Totals

1 202.4 203.2 223.7 203.6 832.9

2 242.0 248.7 259.8 240.7 991.2

3 220.4 227.3 240.0 207.4 895.14 230.0 243.1 247.7 226.9 947.7

5 191.6 211.4 218.7 200.1 821.8

6 247.7 253.0 268.1 244.0 1012.8

7 214.8 214.8 233.9 195.8 859.3

8 245.4 243.6 257.8 227.9 974.7

9 224.0 231.5 238.2 215.7 909.4

10 252.2 255.2 265.4 245.2 1018

Totals 2270.5 2331.8 2453.3 2207.3

Table of Results


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Computing for the Sum of Squares Total

2.15919

40

)9262()2.245(...)2.203()4.202(

2222

TotalSS

Golfer (block) Brand A Brand B Brand C Brand D Totals

1 202.4 203.2 223.7 203.6 832.9

2 242.0 248.7 259.8 240.7 991.2

3 220.4 227.3 240.0 207.4 895.1

4 230.0 243.1 247.7 226.9 947.7

5 191.6 211.4 218.7 200.1 821.8

6 247.7 253.0 268.1 244.0 1012.8

7 214.8 214.8 233.9 195.8 859.3

8 245.4 243.6 257.8 227.9 974.7

9 224.0 231.5 238.2 215.7 909.4

10 252.2 255.2 265.4 245.2 1018

Totals 2270.5 2331.8 2453.3 2207.3


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Computing for the Sum of Squares

Treatments

7.3298

40

)9262(

10

)3.2207...()5.2270( 222

TreatmentsSS


1 202.4 203.2 223.7 203.6 832.9

2 242.0 248.7 259.8 240.7 991.2

3 220.4 227.3 240.0 207.4 895.1

4 230.0 243.1 247.7 226.9 947.7

5 191.6 211.4 218.7 200.1 821.8

6 247.7 253.0 268.1 244.0 1012.8

7 214.8 214.8 233.9 195.8 859.3

8 245.4 243.6 257.8 227.9 974.7

9 224.0 231.5 238.2 215.7 909.4

10 252.2 255.2 265.4 245.2 1018

Totals 2270.5 2331.8 2453.3 2207.3


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Computing for the Sum of Squares Blocks

9.12073

40

)9262(

4

)1018(...)2.991()9.832( 2222

BlocksSS


1 202.4 203.2 223.7 203.6 832.9

2 242.0 248.7 259.8 240.7 991.2

3 220.4 227.3 240.0 207.4 895.1

4 230.0 243.1 247.7 226.9 947.7

5 191.6 211.4 218.7 200.1 821.8

6 247.7 253.0 268.1 244.0 1012.8

7 214.8 214.8 233.9 195.8 859.3

8 245.4 243.6 257.8 227.9 974.7

9 224.0 231.5 238.2 215.7 909.4

10 252.2 255.2 265.4 245.2 1018

Totals 2270.5 2331.8 2453.3 2207.3


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Computing for the Sum of Squares Error

6.546

)9.12073()7.3298()2.15919(

BlocksTreatmentsTError SSSSSSSS


1 202.4 203.2 223.7 203.6 832.9

2 242.0 248.7 259.8 240.7 991.2

3 220.4 227.3 240.0 207.4 895.1

4 230.0 243.1 247.7 226.9 947.7

5 191.6 211.4 218.7 200.1 821.8

6 247.7 253.0 268.1 244.0 1012.8

7 214.8 214.8 233.9 195.8 859.3

8 245.4 243.6 257.8 227.9 974.7

9 224.0 231.5 238.2 215.7 909.4

10 252.2 255.2 265.4 245.2 1018

Totals 2270.5 2331.8 2453.3 2207.3


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ANOVA TableSource of

Variationdf SS MS f

Treatment 3 3298.7 1099.6 54.31

Block 9 12073.9 1341.5 66.26

Error 27 546.6 20.2Total 39 15919.2

2.20

27

6.546

5.13419

9.12073

6.1099

3

7.3298

Error

Blocks

Treatments

MS

MS

MS

26.662.20

1.1345

31.542.20

6.1099

Blocks

Treatments

F

F


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06 Analysis of Variance