Lecture ANOVA

7/24/2019 Lecture ANOVA

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DESIGN OF EXPERIMENTS

ANALYSIS OF VARIANCE


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Normal Distribution

f(y) =

b

a

dyyf )(P(a y b) =

1)(

dyyf

m

Gaussian Equation:

s

Measurement:yi= + ei


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Estimation of Parameters

1

1estimates the population mean

n

i

i

y yn

m

2 2 2

1

1( ) estimates the variance

1

n

i

i

S y y

n

s

The average (mean):

The Sample Variance (Standard Deviation)2:

Measurement:yi= + ei

m


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The Standard Normal Distribution

One way to simplify calculating probabilities is to use aNormal Deviate (z)

Subtract mfrom allyobservations and the newobservations will have a mean of 0

Divide these new observations by sand the new standarddeviation becomes 1.

s

m

y

z

Result:f(z) = a normal standarddistribution

2exp2

1)(

2z

zf


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Areas under the standard normal distribution curve

Y rea Y rea Y rea Y rea Y rea Y rea Y rea Y rea

-4,00 0,0000 -3,00 0,0013 -2,00 0,0228 -1,00 0,1587 0,00 0,5000 1,00 0,8413 2,00 0,9772 3,00 0,9987

-3,99 0,0000 -2,99 0,0014 -1,99 0,0233 -0,99 0,1611 0,01 0,5040 1,01 0,8438 2,01 0,9778 3,01 0,9987

-3,98 0,0000 -2,98 0,0014 -1,98 0,0239 -0,98 0,1635 0,02 0,5080 1,02 0,8461 2,02 0,9783 3,02 0,9987

-3,97 0,0000 -2,97 0,0015 -1,97 0,0244 -0,97 0,1660 0,03 0,5120 1,03 0,8485 2,03 0,9788 3,03 0,9988-3,96 0,0000 -2,96 0,0015 -1,96 0,0250 -0,96 0,1685 0,04 0,5160 1,04 0,8508 2,04 0,9793 3,04 0,9988

-3,95 0,0000 -2,95 0,0016 -1,95 0,0256 -0,95 0,1711 0,05 0,5199 1,05 0,8531 2,05 0,9798 3,05 0,9989

-3,94 0,0000 -2,94 0,0016 -1,94 0,0262 -0,94 0,1736 0,06 0,5239 1,06 0,8554 2,06 0,9803 3,06 0,9989

-3,93 0,0000 -2,93 0,0017 -1,93 0,0268 -0,93 0,1762 0,07 0,5279 1,07 0,8577 2,07 0,9808 3,07 0,9989

-3,92 0,0000 -2,92 0,0018 -1,92 0,0274 -0,92 0,1788 0,08 0,5319 1,08 0,8599 2,08 0,9812 3,08 0,9990

-3,91 0,0000 -2,91 0,0018 -1,91 0,0281 -0,91 0,1814 0,09 0,5359 1,09 0,8621 2,09 0,9817 3,09 0,9990

Y

mz= 0

sz= 1

Y

dzzf )(


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Portland Cement Formulation

Observation

(Sample),j

Formula 1

y1,j

Formula 2

y2,j1 16.85 17.50

2 16.40 17.63

3 17.21 18.25

4 16.35 18.00

5 16.52 17.86

6 17.04 17.75

7 16.96 18.22

8 17.15 17.90

9 16.59 17.96

10 16.57 18.15

Average: 16.76 17.92

1

2

1

1

1

16.76

0.100

0.316

10

y

S

S

n

Summary Statistics

Formulation 1

2

2

2

2

2

17.920.061

0.247

10

yS

S

n

Formulation 2

yij= i+ ij (i=1,2, j=1-10)


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14 15 16 17 18 19 20 21

0.316

0.247

17.9216.76

Frequency

Observation

12

1

1

1

16.760.100

0.316

10

yS

S

n

22

2

2

2

17.920.061

0.247

10

yS

S

n

Cement Formulation Data

Do these formulations differ?


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The Two-Sample t-Test

t0follows a

t-distribution

with n1+n2-2degreesof freedom

21

210

11

nnS

yyt

p

kdegrees

of freedom

2)1()1(

21

2

22

2

112

nnSnSnSp


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The Two-Sample t-Test

CementFormulation:t0= -9.13

m1= m2is consideredto be a rare event if

| t0| is 2.101.

The P-value (3.6810-8)is the risk ofwrongly rejectingthe null hypothesis ofequal means

2.5%2.5%

t distributionwith 18 DOF


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What If There Are More

Than Two Factor Levels?

The t-test does not directly apply

There are lots of practical situations wherethere are either more than two levels ofinterest, or there are several factors ofsimultaneous interest

The analysis of variance(ANOVA) is the

appropriate analysis engine for these typesof experiments

Used extensively today for industrial

experiments


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New synthetic fiber to make cloth for shirts Response variable: tensile strength

Cotton content vary between 15 and 35%

Each experiments replicated 5 times

What is the best weight % of cotton to use?

Cotton Fiber Example


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Does changingthecotton weight

percent change the

mean tensilestrength?

Is there an optimumlevel for cottoncontent?


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In general, there will be alevelsof the factor, or a

treatments, and nreplicatesof the experiment, run inrandomorder, i.e.a completely randomized design

N= antotal runs Objectiveis to test for differences between the a means


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Models for the Data

Consider the normal (means) model:

yi j= i+ i j (mi: mean for the i:th treatment)

If we define:i= + t i

We will get an effects model:yi j= + t i+ i j

where t iis the i:th treatmenteffect


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The basic single-factor ANOVA model is

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

ij i iji ayj n

m t e

The name analysis of varianceis derived froma partitioning of the total variability in theresponse variable into its components parts

Analysis of Variance (ANOVA)

Treatments

Replications


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ANOVA1,2,...,

,1,2,...,

ij i ij

i ay

j nm t e

Definitions:

n

y

y

n

j

ij

i

1.

N

y

y

a

i

n

j

ij

1 1

..

Observation Mean: Overall Mean:

(within treatments) (all measurements)


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The Analysis of Variance

a measure of thetotal variability

Total Corrected Sum of Squares:

a

i

n

j

i jT yySS1 1

2

.. )(

The basic ANOVA partitioning is:

2

.

1 1

...

1 1

2

.. )]()[()( ii j

a

i

n

j

i

a

i

n

j

i jT yyyyyySS

TreatmentAverage


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2.

1 1

... )]()[( iij

a

i

n

j

i yyyy

))((2)()( .1 1...

2

.1 1

2

1 1... ii j

a

i

n

jiii j

a

i

n

j

a

i

n

jiT yyyyyyyySS

a

i

i yyn1

2

...)(

n

j

iij yy1

. 0)(


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E rrorsTreatmentsii j

a

i

n

j

a

i

iT SSSSyyyynSS

2

.

1 11

2

... )()(


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A large value of SSTreatments

reflects largedifferences in treatment means

A small value of SSTreatments likely indicates nodifferences in treatment means

Formal statistical hypotheses are:

T Treatments E SS SS SS

0 1 2

1

:

: At least one mean is different

aH

H

m m m


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While sums of squares cannot be directly compared totest the hypothesis of equal means, mean squarescan.

A mean square (MS) is a sum of squares divided by itsdegrees of freedom:

1

)(1

2

2

n

yy

S

n

i

i

Recall the equation for thesample variance:

1

)(

1

2

1

...

a

yyn

aSSMS

a

i

i

TreatmentsTreatments

aN

yy

aNSSMS

a

i

i

n

j

ij

Er ro rEr ro r

1

2

.

1

)(


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If the treatment means are equal, the treatment anderror mean squares will be (theoretically) equal.

1= 2= = a MSTreatment~ MSError

If treatment means differ, the treatment meansquare will be larger than the error mean square.

i k MSTreatment> MSError

If the error mean squares are larger than thetreatment mean squares, there is a PROBLEM


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Rejectthe null hypothesis(equal treatment means) if 0 , 1, ( 1)a a n

F F

It turns out that the ratio ofMSTreatmentsandMSErrorfollows the F distributionwith (a-1)and (N-a) degrees of freedom

Error

Treatments

MS

MS

F 0

So the test statisticfor the hypothesis of nodifference in mean is:


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Cotton Fiber Example

ANOVA

Source of

Variation SS df MS F P-value F crit

Between Groups 475.76 4 118.94 14.75682 9.13E-06 2.866081

Within Groups 161.2 20 8.06

Total 636.96 24

76.140

Error

Treatments

MS

MSF


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The F Distribution

F0 = 14.76 >> F0.01,4,20

F0.05,4,20= 2.87 (5% probability)

F0.01,4,20= 4.43 (1% probability)

With a-1 (5-1=4) and N-a (25-5=20)

Degrees of Freedom (F4,20)


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The Statistical EffectsModel1,2,...,

( ) 1,2,...,

1,2,...,ijk i j ij ijk

i a

y j b

k n

m t t e

ti= effect of treatment A

j= effect of treatment B

(t)ij= synergistic effect of

treatments A and B (interaction term)

MORE FACTORS?


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Extension of ANOVA to Factorials

2 2 2

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

1 1 1 1 1

2 2. .. . . ... .

1 1 1 1 1

( ) ( ) ( )

( ) ( )

a b n a b

ijk i j

i j k i j

a b a b n

ij i j ijk ij

i j i j k

y y bn y y an y y

n y y y y y y

Degrees of Freedom:

SST= abn-1

SSA= a-1

SSB= b-1

SSAB= (a-1)(b-1)

SSE= ab(n-1)

Can calculate mean

squares as before todetermine if the

variables have effect


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ANOVA Table Fixed Effects Case

Available computer softwares can

easily perform these calculations.


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Factorials with More Than

Two Factors

Basic procedure is similar to the two-factorcase; all a,b,c,,ktreatment combinationsare run in random order

ANOVA identity is also similar:

T A B AB AC

ABC AB K E

SS SS SS SS SS

SS SS SS

Computers are great, are they not?


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Download 30-day trial version fromweb site:

http://www.minitab.com/

1) Click on Products tab

2) MiniTab 17: Click: Learn More

3) Click: try Minitab 17,

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Lecture ANOVA