Lecture16 Module3 Anova 1

8/11/2019 Lecture16 Module3 Anova 1

1/10

Desi nDesi n of Ex erimentsof Ex eriments--II

MODULEMODULE IIIIII

LECTURELECTURE -- 1616

MODELSMODELS

Dr. Shalabh

Department of Mathematics and Statistics

Indian Insti tute of Technology Kanpur


2/10

Two-way classification with interactions

2

Consider the two-way classification with an equal number, say K observations per cell. Let : kth observation in

(i, j)th cell , i.e., receiving the treatments ith level of factorA andjth level of factor B,

and are inde endentl drawn from so that the linear model under consideration is

ijky

1,2,..., ; 1, 2,..., ; 1, 2,...,i I j J k K = = =

2N

ij i

ijk ij ijk y = +

where are identically and independently distributed following Thusijk 2(0, ).N

( )

( ) ( ) ( )

ijk ij

oo io oo oj oo ij io oj oo

i j ij

E y

=

= + + + +

= + + +

oo

i io oo

j oj oo

=

=

=

1 1 1 1

0, 0, 0, 0.

with

ij ij io oj oo

I J I J

i j ij ij

i i

= = = =

= +

= = = =

Assume that the design matrixX

is of full rank so that all the parametric functions of are estimable.ij


3/10

The null hypothesis are

3

0 1 2

0 1 2

0

...

: ... 0

: 0 , .allAll for

=

I

J

ij

H

H i j

= = =

=

1 : ,

: ,

The corresponding alternative hypothesis is

At least one for

At least one for

i jH i j

H i j

1 : , .At least one forij ik H j k

Minimizing the error sum of squares

2

1 1 1

( ) ,ijk i j ij

i j k

E y = = =

=

the normal equations are obtained as

E=

,

0 ,for all

i

Ei

E

=

0 .

for all and

j

ij

E i j

=


4/10

The least squares estimates are obtained as

I J K

4

1 1 1

1

1

ooo ijk i j k

I

i ioo ooo ijk ooo

i

y yIJK

y y y yJK

= = =

=

= =

= =

1

1

J

j ojo ooo ijk ooo

j

ij ijo ioo ojo ooo

y y y yIK

y y y y

=

= =

= +

1 1

1

I J

ijk ioo ojo ooo

i j

y y y yK = =

= + .

2

, , ,1 1 1

2

( )

i j

I J K

ijk i j ij

ij i j k

I J K

SSE Min y

= = =

=

1 1 1

2

1 1 1

( )

ijk i j iji j k

I J K

ijk ijo

i j k

y

y y

= = =

= = =

=

=

2

2 ~ ( ( 1)).with

SSEIJ K


5/10

Now minimizing the error sum of squares under , i.e., minimizing0 1 2 ... 0 = = = = =IH

5

2

1

1 1 1

( ) = = =

= I J K

ijk j ij

i j k

E y

with respect to and and solving the normal equations,j ij

1 1 10, 0 0for all and for all andj ij

E E Ej i j

= = =

ooo

j ojo ooo

y

y y

=

=

.ij ijo ioo ojo oooy y y y = +

The sum of squares due to is0H

2

, ,1 1 1

( )j ij

I J K

ijk j ij

i j k

Min y

= = =

2

1 1 1

2 2

( )

( ) ( )

I J K

ijk j ij

i j k

I J K I

i k i o ioo ooo

y

y y JK y y

= = =

=

= +

1 1 1 1

2

1 ( ) .

i j k i

I

ioo ooo

iSSE JK y y

= = = =

== +


6/10

Thus the sum of squares due to deviation from or the sum of squares due to effectA is0H

6

2

0

1

2

2

( )

~ ( 1).

Sum of squares due to

with

I

ioo ooo

i

SSA H SSE JK y y

SSAI

=

= =

Minimizing the error sum of squares under i.e., minimizing0 1 2: ... 0JH = = = =

2

2

1 1 1

( )I J K

ijk i ij

i j k

E y = = =

=

and solving the normal equations

2 2 2

0, 0 0for all and for all andi ij

E E E

j i j

= = =

yields the least squares estimators

ooo

i iooo ooo

y

y y

=

=

.ij ijo ioo ojo oooy y y y = +

The minimum error sum of squares is

I J K J2 2

1 1 1 1

( ) ( )ijk i ij ojo oooi j k j

y SSE IK y y = = = =

= +


7/10

and the sum of squares due to deviation from or the sum of squares due to effect B is0H

7

0

2

1

( )

Sum of squares due to

J

ojo ooo

j

SSB H SSE

IK y y

SSB

=

=

=

Next, minimizing the error sum of squares under for all i, j, i.e., minimizing

2 ~ .w

0 : 0ijH all =

2

3

1 1 1

( )I J K

ijk i j

i j k

E y = = =

=

with respect to and and solving the normal equations, j

yields the least squares estimators as

3 3 30, 0 0for all and for alli j

E E Ei j

= = =

.

ooo

i ioo ooo

y

y y

=

=

= o o ooo


8/10

The sum of squares due to is0H

8

2

, ,1 1 1

2

( )

( )

i j

I J K

ijk i j

i j k

I J K

ijk i j

Min y

y

= = =

=

1 1 1

2

1 1

( ) .

i j k

I J

ijo ioo ojo ooo

i j

SSE K y y y y

= = =

= =

= + +

Thus the sum of squares due to deviation from or the sum of squares due to interaction effect AB is0H

0Sum of squares due to

I J

SSAB H SSE =

1 1

2

2 ~ (( 1) 1)).with

ijo ioo ojo ooo

i j

y y y y

SSABI J

= =

= +

The total sum of squares can be partitioned as

TSS = SSA + SSB + SSAB + SSE

where SSA, SSB, SSAB and SSE are mutually orthogonal. So either using the independence of SSA, SSB, SSAB and

SSE as well as their respective distributions or using the likelihood ratio test approach, the decision rules for the

null hypothesis at level of significance are based on F - statistics as follows:


9/10

9

[ ]1 0. ~ ( 1, ( 1) ,1

( 1). ~ 1 1

under

under

F F I IJ K H I SSE

IJ K SSBF F J IJ K H

=

=

[ ]3 0

1

( 1). ~ ( 1)( 1), ( 1) .

( 1)( 1)

and

under

J SSE

IJ K SSABF F I J IJ K H

I J SSE

=

[ ]0 1 1

0 2 1

( 1), ( 1)

So

Reject

Reject

if

if

H F F I IJ K

H F F

>

> [ ]( 1), ( 1)J IJ K

If or is re ected one can use t -test or multi le com arison test to find which airs of are

[ ]0 3 1 ( 1)( 1), ( 1) .Reject ifH F F I J IJ K >

' 's s or

significantly different.

If is rejected, one would not usually explore it further but theoretically t- test or multiple comparison tests can be

0 0

0H

used.


10/10

It can also be shown that 2 2

1

( )1

I

i

i

JKE SSA

I

=

= +

10

2 2

1

2 2

1 1

( )1

( )( 1)( 1)

J

j

j

I J

ij

i j

IKE SSB

J

KE SSAB

I J

=

= =

= +

= +

2( ) .E SSE =

The analysis of variance table is as follows:

-

variation

freedom

FactorA ( I - 1) SSA1

MSAF

MSE

=

1

MSAMSA

I

=

1 ( , )C F p n p=

Factor B (J - 1) SSB2

MSBF

MSE=

1

MSBMSB

J=

00:H =Interaction AB (I - 1) (J - 1) SSAB

3

MSABF

MSE=

( )( 1)

SSABMSAB

I I J=

Error I J (K - 1) SSE

Total (I J K 1) TSS

( 1)

SSEMSE

IJ K=

Lecture16 Module3 Anova 1

Documents

Transcript of Lecture16 Module3 Anova 1