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OrthogonalityOrthogonalityOrthogonalityOrthogonalityOne way to delve further into the impact a factor has on the yield is to break the Sum of Squares (SSQ) into “orthogonal” components.

One way to delve further into the impact a factor has on the yield is to break the Sum of Squares (SSQ) into “orthogonal” components.

If SSBcol has (C-1) df (which corresponds with having C levels, or C columns ), the SSBcol can be broken up into (C-1) individual SSQ values, each with a single degree of freedom, each addressing a different inquiry into the data’s message.

2

If each “question” asked of the data is orthogonal to all the other “questions”, two generally desirable properties result:

1. Each “question” is independent of each other one; the probabilities of Type I and Type II errors in the ensuing hypothesis tests are independent, and “stand alone”.

OrthogonalityOrthogonalityOrthogonalityOrthogonality

3

Consider 4 column means: 6 4 1 -3

Grand Mean = 2

2. The (C-1) SSQ values are guaranteed to add up exactly to the total SSBcol you started with.

What? How? -Watch!

OrthogonalityOrthogonalityOrthogonalityOrthogonality

11 22 33 44

4

Call these values: Y1, Y2, Y3, Y4, 4

and define 1 = a1j Yj , j=1

4

2 = aj Yj ,

j=1

4

and = aj Yj , j=1

5

Under what conditions will

3 4

2i = Yj - Y)2 ?

i=1 j=1

one answer: 4 (1) aij = 1 for all i

j=1 (i=1,2,3)

4

(2) aij = 0 for all i j=1

2

4

(3.) ai1j . ai2j = 0 for all i1, i2,

j=1 i1 = i2.

A linear combination of treatment means satisfying (2)

is called a contrast.

orthogonal

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Writing the aij’s as a “matrix”, one possibility among many:

1/2 1/2 -1/2 -1/2 1/2 -1/2 1/2 -1/2 1/2 -1/2 -1/2 1/2

6 4 1 -3 2

Y1 Y2 Y3 Y4 Y= 2

1 = 1/2 1/2 -1/2 -1/2 6 362 = 1/2 -1/2 1/2 -1/2 3 93= 1/2 -1/2 -1/2 1/2 -1 1

4646

7

Yj -Y)2 = (6-2)2 + (4-2)2 + (1-2)2 + (-3-2)2

= 16 + 4 + 1 + 25 =

OK! How does this help us?

4646

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Consider the following data, which,let’s say, are the column means of a one factor ANOVA, with the one factorbeing “DRUG”:

Y.1 Y.2 Y.3 Y.4

Y.. = 7 5 6 7 10

and (Y.j - Y..)2 = 14.

(SSBc = 14.R, where R = # rows)

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Consider the following two examples:

Example 1

1 2 3 4

Placebo

Sulfa

Type

S1

Sulfa

Type

S2

Anti-biotic

Type A

Suppose the questions of interest are

(1) Placebo vs. Non-placebo

(2) S1 vs. S2

(3) (Average) S vs. A

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How would you combine columns to analyze the question?

P vs. P:

S1 vs. S2:

S vs. A:

1 2 3 4

-3 1 1 1

0 -1 1 0

0 -1 -1 2

Note Conditions 2 & 3 Satisfied

P S1 S2 A

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divide top row bymiddle row bybottom row by

(to satisfy condition 1)

12

Y.1 Y.2 Y.3 Y.4 Zi2

312

112

112

112

5.33

12

12

0.50

16

16

26

8.17

0

0

0

Placebovs. drugs

S1 vs. S2

Average Svs. A

P S1 S2 A

14.0014.00

5 6 7 10

13

Example 2:

Y.1 Y.2 Y.3 Y.4

antibiotictype

antibiotictype

sulfatype

sulfatype

S1 S2 A1 A2

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Exercise:

Suppose the questions of interest are:1. The difference between sulfa types2. The difference between antibiotic types3. The difference between sulfa and antibiotic types, on

average.

Write down the three corresponding contrasts. Are they orthogonal? If not, can we make them orthogonal?

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OK! Now to the analysis:

Y.1 Y.2 Y.3 Y.4 Zi2

S1 vs. S2

A1 vs. A2

Ave. Svs. Ave. A

(5) (6) (7) (10)

S1 S2 A1 A2

0.5

4.5

9.0

0

12

12 0 0

012

12

14

14

14

14

14.0014.00

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Example:

{R=8

Placebo....

. 5

ASP1

.

.

.

. .

6

ASP2

.

.

.

. .

7

Buff....

. 10

Y..= 7

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Source SSQ df MSQ FDrugs

Error

112

140

3

28

37.33

5

7.47

ANOVA

F.05(3,28)=2.95

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Now, an orthogonal breakdown:

Placebovs. others

ASP1

vs. ASP2

ASP vs. Buff

0

- 312

112

12

-16

-16

2 6

112

112

0

-12

0 12

76

812

Placebo ASP1 ASP2 Buff Z +5 +6 +7 10

Pl’bo vs Bff 0 02 2 2

19

Z 8

12

Z2 Z2 x 8

42.64

1

7

.50 4.00

8.17 65.36

14.00 112.00

5.33

5

10025/2

20

ANOVA

Source SSQ df MSQ F

Drugs

Error

{Z1

Z2

Z3

112{42.64 4.00 65.36

3{ 111111

42.6442.64 4.004.0065.3665.36

8.538.53 .80.8013.0713.07

140140 28 5 28 5

FF1-.051-.05(1,28)=4.20(1,28)=4.20

Z4 100 1 100 20

FF1-.05/31-.05/3(1,28)<7.64(1,28)<7.64

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Another Example:

The variable (coded) is mileage per gallon.

GasolineIIIIIIIVV

YIELD-419211018

Standard GasolineStandard, plus additive A made by PStandard, plus additive B made by PStandard, plus additive A made by QStandard, plus additive B made by Q

A significant difference between Placebo and the rest, and between ASP’s and BUFF, but not between the two different ASP’s.

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Questions actually chosen:

Standard gasoline vs gasoline with an additiveP vs. QBetween the two additives of PBetween the two additives of Q

(Z1)(Z1)(Z2)(Z2)(Z3)(Z3)(Z4)(Z4)

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With appropriate orthogonal matrix and Z 2 values:

12

422.8

II III IV V Z2i

120

120

352.8

36.0

2.0

32.0

12

12

Z1

Z2

Z3

Z4

I

+ 420

0

0

0

14

0

14

0

120

14

0

120

14

0

12

By far, the largest part of the total variability in yields is associated with standard gasoline vs. gasoline with an additive.

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Let n=4. Let the four observed yields be thefour yields of a 22 factorial experiment:

Orthogonal Breakdowns In 2k and 2k-p designs

Y1 = 1Y2 = aY3 = b Y4 = ab

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Example: Miles per Gallon by Gas Type and Auto Make

1 2 3 4

16 28 16 2822 27 25 3016 17 16 1910 20 16 1818 23 19 24

8 23 16 25

15 23 18 24

Group

26

Suppose: 1 1 152 a 233 b 184 ab 24

A = Gas Type = 0, 1B = Auto Make = 0, 1

(22 design)

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Earlier we formed

estimate of 2A

estimate of 2B

estimate of 2AB

1 a b ab

-1

-1

1

1

-1

-1

-1

1

-1

1

1

1

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Which for present purposes we replace by:

1 a b ab Z

1 2

1 2

1 2

.

.

.

.

.

.

.

.

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

--

--

--

--

----

Now we can see that these coefficients of the yields are elements of the orthogonal matrix.So, A, B and AB constitute orthogonal estimates.

A

B

AB

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Standard one-way ANOVA:

F.95 (3,20)3.1

Source SSQ df MSQ F

Col 324 3 108 5.4

Error 400 20 20

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Then, A = (-15+23-18+24) =

B = (-15-23+18+24) =

AB = (15-23-18+24) =

A2 = 49, B2 = 4 , AB2 = 1

1

1

1

14

4

-2

4

4

4

4

4

4

31

Multiply each of these by the numberof data points in each column:

A2 6(49) = 294B2 6(4) = 24AB2 6(1) = 6

TOTAL : 324TOTAL : 324

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And:And:

ANOVA:

F.95 (1,20)4.3

Source SSQ df MS Fcalc

Col 324 3

Error 400 20 20

AABB

ABAB{294

246{

111{

29424

6

14.71.2

.3

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If: 1 c 152 a 233 b 184 abc 24

A = Gas TypeB = Auto MakeC = Highway

(23-1 design)

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ANOVA:

We’d get the same breakdown of the SSQ, but being the + block of I = ABC,

Source SSQ df

A+BC 294 1B+AC 24 3 1AB+C 6 1

Error 400 20

{ { e t

c .

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What if contrasts of interest are not orthogonal?

*Bonferroni Method: The same F test (SSQ = RxZi^2) but using = a/k, where a is the overall error rate.*Scheffe Method: p.108, skipped.Reference: Statistical Principles of Research Design and Analysis by Robert O. Kuehl.

Let k be the number of contrasts of interest.1. If k <= c-1 Bonferroni method2. If k > c-1 Bonferroni or Scheffe method

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1. Can k be larger than c-1?2. Can k be smaller than c-1?

If k contrasts are orthogonal,

No.

Yes.

For case 2, do the same F test (but the sum of SSQ will not be equal to SSB). See Slide 16.