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Statistical Analysis. Professor Lynne Stokes Department of Statistical Science Lecture 6 Solving Normal Equations and Estimating Estimable Model Parameters. Regression Models. Model. Residuals. Least Squares. Sum of Squared Residuals. Solution: Solve the Normal Equations. - PowerPoint PPT Presentation
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1 Statistical Analysis Professor Lynne Stokes Department of Statistical Science Lecture 6 Solving Normal Equations and Estimating Estimable Model Parameters
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Page 1: Statistical Analysis

1

Statistical Analysis

Professor Lynne StokesDepartment of Statistical Science

Lecture 6Solving Normal Equations andEstimating Estimable Model

Parameters

Page 2: Statistical Analysis

2

Regression Models

Residuals

Least Squares

Solution: Solve the Normal Equations

ˆXyr

)X-(y)X-(y)( Minimize toˆ Choose

yXˆXX

ModeleXy

Sumof SquaredResiduals

Page 3: Statistical Analysis

3

Regression Solution

Under usual assumptions, the least squares estimator is Unique Unbiased Minimum Variance Consistent Known sampling distribution Universally used

yXXXˆ 1

Page 4: Statistical Analysis

4

Analysis of Completely Randomized Designs

Fixed Factor EffectsFactor levels specifically chosenInferences desired only on the factor levels

included in the experimentSystematic, repeatable changes in the mean response

Page 5: Statistical Analysis

5

Flow Rate Experiment

Filter Flow RatesA 0.233 0.197 0.259 0.244B 0.259 0.258 0.343 0.305C 0.183 0.284 0.264 0.258D 0.233 0.328 0.267 0.269 MGH Fig 6.1

Fixedor

Random?

Page 6: Statistical Analysis

6

Flow Rate Experiment

A B C D

0.30

0.25

0.20

AverageFlowRate

Conclusion ?

Filter Type

0.35 Filter EffectsA -0.028B 0.030C -0.014D 0.013

Page 7: Statistical Analysis

7

Statistical Model for Single-Factor, Fixed Effects Experiments

Response OverallMean

(Constant)

MainEffect

for Level

i

Error

Modelyij = + i + eij i = 1, ..., a; j = 1, ..., ri

i: Effect of Level i = change in the mean response

Page 8: Statistical Analysis

8

Statistical Model for Single-Factor, Fixed Effects Experiments

Cell Means Model

yij = i + eij i = 1, ..., a; j = 1, ..., ri

yˆ ii

Effects Modelyij = + i + eij i = 1, ..., a; j = 1, ..., ri

Fixed Effects ModelsConnection: i = + i

yy~ , yˆ ii

Page 9: Statistical Analysis

9

Solving the Normal EquationsSingle-Factor, Balanced Experimentyij = + i + eij i = 1, ..., a j = 1, ..., r n = ar

Matrix Formulation

y = X + e

y = (y11 y12 ... y1r ... ya1 ya2 ... yar)’

]1I : )11[(

1001............01010011

=X raar

rrrr

rrrr

rrrr

1

...

a

Page 10: Statistical Analysis

10

Solving the Normal Equations

Residuals

Least Squares

Solution: Solve the Normal Equations

~Xyr

)X-(y)X-(y)( Minimize to~ Choose

yX~XX

Page 11: Statistical Analysis

11

Solving the Normal Equations

X X = X y~

Normal Equations

n r r rr rr r

r r

yyy

ya a

.........

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

...

~~~

...~

...

0 00 0

0 0

12

12

Check

Page 12: Statistical Analysis

12

Solving the Normal Equations

Normal Equations

Check

Linearly Dependent

a + 1 Parameters, a Linearly Independent Equations

aa

22

11

a21

y~r + ~r...

y ~r+ ~ry ~r+~ry~r + ... + ~r ~r+~n

Infinite Number of Solutions

Page 13: Statistical Analysis

13

Solving the Normal Equations

Normal Equations

One Solution

ii

a

ii

a

i iy

0 01 1

= yy

~

~~

aa

22

11

y~r + ~r...

y ~r+ ~ry ~r+~ry ~n

iii y~~ˆ

Page 14: Statistical Analysis

14

Solving the Normal Equations

Normal Equations

Another Solution

ii y~0~0

iii y~~ˆ

aa

22

11

a21

y~r ...

y ~r y ~r y~r ... ~r ~r

Page 15: Statistical Analysis

15

Solving the Normal Equations

~ ~~

a

i = 1, . . . , a - 1

0 yy ya

i i a

Another Solution

Normal Equations

iii y~~ˆ

)1a(1-a

22

11

1-a21

y ~r+ ~r...

y ~r+ ~ry ~r+~ry~r + ... + ~r ~r+~n

Page 16: Statistical Analysis

16

Solving the Normal Equations

Solutions are not estimates Estimable Functions

All solutions provide one unique estimator Estimators are unbiased

All solutions to the normal equationsproduce the same estimates of “estimable functions”

of the model means

Page 17: Statistical Analysis

17

Solving the Normal Equations

Two-Factor, Balanced Experiment

Matrix Formulationy = X + e

yijk = ij + eijk = + i + j + ()ij + eijk i = 1, ..., aj = 1, ..., bk = 1, ..., r

X = [ 1 : XA : XB : XAB ]

1a1b11ab

n = abr

Page 18: Statistical Analysis

18

Solving the Normal Equations

Two-Factor, Balanced Experiment

Matrix Formulationy = X + e

yijk = ij + eijk = + i + j + ()ij + eijk i = 1, ..., aj = 1, ..., bk = 1, ..., r

X = [ 1 : XA : XB : XAB ]

Numberof

Parameters1 + a + b + ab

rank( X ) < 1+a+b+ab

n = abr

1a1b11ab

Page 19: Statistical Analysis

19

Solving the Normal Equations

X X = X y~

Normal Equations

ab

1

1

ab

1

1

y...

y...

yy

...

~...~~

r...0...00..................0...0...brbrr...ar...brn

Check

Page 20: Statistical Analysis

20

Solving the Normal Equations

Matrix Linear Dependencies One Solution 1n None XA 1 : Columns of XA Sum to 1n a= 0

Eliminates a columnFrom XA

a – 1 “degrees of freedom”

Page 21: Statistical Analysis

21

Solving the Normal Equations

Matrix Linear Dependencies One Solution 1n None XA 1 : Columns Sum of XA to 1n a= 0 XB 1 : Columns Sum of XB to 1n b = 0

Eliminates a columnFrom XB

b – 1 “degrees of freedom”

Page 22: Statistical Analysis

22

Solving the Normal Equations

Matrix Linear Dependencies One Solution 1n None XA 1 : Columns sum to 1n a= 0 XB 1 : Columns sum to 1n b = 0 XAB 1 + (a - 1) + (b - 1) :

Sum over all columns = 1n ()ab = 0Eliminates a column

from XAB

Page 23: Statistical Analysis

23

Solving the Normal Equations

Matrix Linear Dependencies One Solution 1n None XA 1 : Columns Sum to 1n a= 0 XB 1 : Columns Sum to 1n b = 0 XAB 1 + (a - 1) + (b - 1) :

Sum over all columns = 1n ()ab = 0 Sums of columns over each i = 1,...,a-1 & each j = 1,...,b-1 ()ib = 0 equal one of the remaining i=1,...,a-1 columns of XA and XB ()aj = 0

j=1,...,b-1(a – 1)(b – 1) “degrees of freedom”

Page 24: Statistical Analysis

24

Solving the Normal Equations

Matrix Linear Dependencies One Solution XA 1 : Columns sum to 1n a= 0 XB 1 : Columns sum to 1n b = 0 XAB 1 + (a - 1) + (b - 1) :

Sum over all columns = 1n ()ab = 0 Sums of columns over each i = 1,...,a-1 & each j = 1,...,b-1 ()ib = 0 equal one of the remaining i=1,...,a-1 columns of XA and XB ()aj = 0

j=1,...,b-1Constraints : 1 + 1 + {1 + (a - 1) + (b - 1)} = a + b + 1Degrees of Freedom : (1 + a + b + ab) - (a + b + 1)

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

Page 25: Statistical Analysis

25

Solving the Normal Equations

~~ ~~~ ~~

~ ~ ~

yy

y

y

ab

i ib

a

j aj

b

ij i j

i = 1, . . . , a - 10

j = 1, . . . , b - 1

0

( ) i a ; j b

( ) 0 i = a or j = bij

ij

Check

Page 26: Statistical Analysis

26

Solving the Normal Equations

j i, yyyy)(

b , ... 1, = j yy~a , ... 1, = i yy~

y~

jiijij

jj

ii

Check

Another Solution

Page 27: Statistical Analysis

27

Flow Rate Experiment

Filter Flow RatesA 0.233 0.197 0.259 0.244B 0.259 0.258 0.343 0.305C 0.183 0.284 0.264 0.258D 0.233 0.328 0.267 0.269 MGH Fig 6.1

Fixedor

Random?

Page 28: Statistical Analysis

28

Quantifying Factor Effects

EffectChange in average response

due to changes in factor levels

1y 2y 3y ky y

1 2 3 k. . .

. . .

Factor Level

Average

Overall Average

Effect of Level t : ty y-

Page 29: Statistical Analysis

29

Quantifying Factor Effects

EffectChange in average response

due to changes in factor levels

1 2 3 k. . .Factor Level

Average

Overall Average

Effect of changing from Level s to Level t :

st

st

y -y =)y-y( - )yy(

1y 2y 3y ky y. . .

Page 30: Statistical Analysis

30

Quantifying Factor Effects

Main Effects for Factor Ay i y Change in average response due to

changes in the levels of Factor A

Main Effects for Factor By j y Change in average response due to

changes in the levels of Factor B

Interaction Effects for Factors A & B(y - y ) - (yij j i y )

Effect of Level i ofFactor A at Level jof Factor B

Effect of Level iof Factor A

Page 31: Statistical Analysis

31

Quantifying Factor Effects

Main Effects for Factor A

Main Effects for Factor B

Interaction Effects for Factors A & B

y i y

y j y

(y - y ) - (y

y - y - yij

ij i

j i

j

y

y

)

Change in average response due tochanges in the levels of Factor A

Change in average response due tochanges in the levels of Factor B

Change in average response duejoint changes in Factors A & Bin excess of changes in the maineffects

Page 32: Statistical Analysis

32

Two-Level Factors

Common to Use y - y2 1

Note: If r1 = r2 , y y1 = - (y y2 )

Effect of Level 1:Effect of Level 2:

y y1

y y2

Page 33: Statistical Analysis

33

Factors at Two Levels

Most common choice for designs involving many factors

Many efficient fractional factorial and screening designs available

Can use p two-level factors in place of factors whose number of levels is 2p

Page 34: Statistical Analysis

34

Calculating Two-Level Factor Effects: Pilot Plant Study

Main EffectDifference between the average responses at thetwo levels

M(Temp) = Average @ 180o - Average @ 160o

= 75.8 - 52.8 = 23.0

M(Conc) = Average @ 40% - Average @ 20%= 61.8 - 66.8 = -5.0

M(Catalyst) = Average @ C2 - Average @ C1= 65.0 - 63.5 = 1.5 BHH Section 10.3

MGH Section 5.3

Page 35: Statistical Analysis

35

Calculating Two-Level Factor Effects

Two-Factor Interaction EffectHalf the difference between the main effects of onefactor at each level of the second factor

M(Conc @ C2) = Average @ 40%&C2 - Average @ 20%&C2= 62.5 - 67.5 = -5.0

M(Conc @ C1) = Average @ 40%&C1 - Average @ 20%&C1= 61.0 - 66.0 = -5.0

I(Conc,Cat) = {M(Conc @ C2) - M(Conc @ C1)} / 2= 0

BHH Section 10.4MGH Section 5.3

Page 36: Statistical Analysis

36

Calculating Two-Level Factor EffectsTwo-Factor Interaction Effect

Half the difference between the main effects of onefactor at each level of the second factor

M(Temp @ C2) = Average @ 180o&C2 - Average @ 160o&C2= 81.5 - 48.5 = 33.0

M(Temp @ C1) = Average @ 180o&C1 - Average @ 160o&C1= 70.0 - 57.0 = 13.0

I(Temp,Cat) = {M(Temp @ C2) - M(Temp @ C1)} / 2= (33.0 - 13.0) / 2 = 10.0

Page 37: Statistical Analysis

37

Cell Means and Effects Model Estimability

Three-Factor Balanced Experiment

yijkl = ijk + eijkl i = 1 , ... , a ; j = 1 , ... , b ;k = 1, ... , c ; l = 1 , ... , r

ijk = + i + j + k + ()ij + ()ik + ()jk + ()ijk

Page 38: Statistical Analysis

38

Cell Means Models: Estimable Functions

ijk ijky

All cell means are estimable

Page 39: Statistical Analysis

39

Cell Means Models: Estimable Functions

ijk ijky

All cell means are estimable

All linear combinations of cell means are estimable

c

c yijk ijk

ijk ijk

(includes , , etc. ) i ij

Does not dependon parameter constraints

Page 40: Statistical Analysis

40

Cell Means Models: Estimable Functions

ijk ijky

All cell means are estimable

Some linear combinations of cell means are uninterpretable

1 23

1 2

Some linear combinations of cell means are essential

Page 41: Statistical Analysis

41

Cell Means and Effects Models

i

iiii

)()()()(

Imposing parameter constraintssimplifies the relationships;

makes the parameters more interpretable

Page 42: Statistical Analysis

42

Parameter Equivalence:Effects Representation & Cell Means Model

Parameter constraints i

iij

ij . . . = = . . . = ( ) 0ijk

ijk( )

Means and mean effects

i i

ij i j ij

( )

i i

ij ij i j

ij j i

( )

( ) ( )

Page 43: Statistical Analysis

43

Contrasts

k

1jj

k

1jjj a with a

Zero toSum tsCoefficien whoseParameters ofn CombinatioLinear A Contrast

Contrasts often eliminatenuisance parameters; e.g.,

Page 44: Statistical Analysis

44

Contrasts

i i i i

ij i j ij

ij il kj kl ij il kj kl

( )

( ) ( ) ( ) ( )

Main Effects

Interactions

4 2 2cr ijij

ij il kj klijkl

( ) {( ) ( ) ( ) ( ) }

Show


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