1 / 36

Orthogonality

Orthogonality. 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.

reilly
Télécharger la présentation

Orthogonality

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Orthogonality 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 havingC levels, or C columns ), the SSBcolcan be broken up into (C-1) individual SSQ values, each with a singledegree of freedom, each addressing a different inquiry into the data’s message.

  2. Orthogonality 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”.

  3. Orthogonality 2. The (C-1) SSQ values are guaranteed to add up exactly to the total SSBcol you started with. What? How? -Watch! 1 2 3 4 Consider 4 column means: 6 4 1 -3 Grand Mean = 2

  4. Call these values: Y1, Y2, Y3, Y4, 4 and define 1 = a1j Yj , j=1 4 2 = aj Yj , j=1 4 and= aj Yj , j=1

  5. 4 (3.) ai1j . ai2j = 0 for all i1, i2, j=1 i1 = i2. 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 A linear combination of treatment means satisfying (2) is called a contrast. 2 orthogonal

  6. 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 Y1 Y2 Y3 Y4 Y= 2 6 4 1 -3 2 1/2 1/2 -1/2 -1/2 6 36 1 = 2 = 1/2 -1/2 1/2 -1/2 3 9 3= 1/2 -1/2 -1/2 1/2 -1 1 46

  7. Yj -Y)2 = (6-2)2 + (4-2)2 + (1-2)2 + (-3-2)2 = 16 + 4 + 1 + 25 = 46 OK! How does this help us?

  8. Consider the following data, which, let’s say, are the column means of a one factor ANOVA, with the one factor being “DRUG”: Y.1 Y.2 Y.3 Y.4 5 6 7 10 Y.. = 7 and (Y.j - Y..)2 = 14. (SSBc = 14.R, where R = # rows)

  9. Consider the following two examples: Example 1 1 3 4 2 Sulfa Type S1 Sulfa Type S2 Anti-biotic Type A Placebo Suppose the questions of interest are (1) Placebo vs. Non-placebo (2) S1 vs. S2 (3) (Average) S vs. A

  10. How would you combine columns to analyze the question? P S1 S2 A 1234 -3 1 1 1 P vs. P: S1 vs. S2: S vs. A: 0 -1 1 0 0 -1 -1 2 Note Conditions 2 & 3 Satisfied

  11. divide top row by middle row by bottom row by    (to satisfy condition 1)

  12. 5 6 7 10 Y.1 Y.2 Y.3 Y.4 Zi2 PS1 S2 A Placebo vs. drugs S1 vs. S2 Average S vs. A 1 12 1 12 1 12 3 12 5.33 1 2 1 2 0.50 0 0 2 6 8.17 1 6 1 6 0 14.00

  13. Example 2: Y.1 Y.2 Y.3 Y.4 sulfa type sulfa type antibiotic type antibiotic type S1 S2 A1 A2

  14. Exercise: • Suppose the questions of interest are: • The difference between sulfa types • The difference between antibiotic types • 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?

  15. Y.1 Y.2 Y.3 Y.4 Zi2 S1 S2 A1 A2 1 2 1 2 0 0.5 S1 vs. S2 A1 vs. A2 Ave. S vs. Ave. A 0 1 2 1 2 0 4.5 0 1 4 1 4 1 4 1 4 9.0 (5)(6)(7) (10) 14.00 OK! Now to the analysis:

  16. Example: ASP1 . . . . . 6 Placebo . . . . . 5 ASP2 . . . . . 7 Buff . . . . . 10 { R=8 Y..= 7

  17. ANOVA F.05(3,28)=2.95

  18. Now, an orthogonal breakdown: Placebo ASP1 ASP2 Buff Z +5 +6 +7 10 Placebo vs. others ASP1 vs. ASP2 ASP vs. Buff - 3 12 1 12 1 12 8 12 1 12 1 2 1 2 -1 2 0 0 -1 6 7 6 -1 6 2  6 0 1/2 -1/2 5/2

  19. Z 8 12 Z2 Z2 x 8 5.33 42.64 1  7  .50 4.00 8.17 65.36 5  100 25/2 14.00 112.00

  20. ANOVA Source SSQ df MSQ F Z1 Z2 Z3 42.64 4.00 65.36 8.53 .80 13.07 42.64 4.00 65.36 { { { 1 1 1 3 112 Drugs Error 100 100 20 Z4 1 140 28 5 F1-.05/3(1,28)<7.64 F1-.05(1,28)=4.20

  21. A significant difference between Placebo and the rest, and between ASP’s and BUFF, but not between the two different ASP’s. Another Example: The variable (coded) is mileage per gallon. Gasoline I II III IV V YIELD -4 19 21 10 18 Standard Gasoline Standard, plus additive A made by P Standard, plus additive B made by P Standard, plus additive A made by Q Standard, plus additive B made by Q

  22. Questions actually chosen: Standard gasoline vs gasoline with an additive P vs. Q Between the two additives of P Between the two additives of Q (Z1) (Z2) (Z3) (Z4)

  23. With appropriate orthogonal matrix and Z 2values: I IIIII IV V Z2i 1 20 1 20 1 20 1 20 + 4 Z1 Z2 Z3 Z4 352.8 20 1 4 1 4 1 4 1 4 0 36.0 2.0 32.0 1 2 1 2 0 0 0 0 1 2 1 2 0 0 422.8 By far, the largest part of the total variability in yields is associated with standard gasoline vs. gasoline with an additive.

  24. Orthogonal Breakdowns In 2k and 2k-p designs Let n=4. Let the four observed yields be the four yields of a 22 factorial experiment: Y1 = 1 Y2 = a Y3 = b Y4 = ab

  25. Example:Miles per Gallon by Gas Type and Auto Make 1234 16 28 16 28 22 27 25 30 16 17 16 19 10 20 16 18 18 23 19 24 8 23 16 25 15 23 18 24 Group

  26. Suppose: 1  1 15 2  a 23 3  b 18 4  ab 24 A = Gas Type = 0, 1 B = Auto Make = 0, 1 (22 design)

  27. 1 a b ab -1 -1 1 1 -1 -1 -1 1 -1 1 1 1 Earlier we formed estimate of 2A estimate of 2B estimate of 2AB

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

  29. Source SSQ df MSQ F Col 324 3 108 5.4 Error 400 20 20 F.95 (3,20) =3.1 Standard one-way ANOVA:

  30. 1 1 1 14 Then, A = (-15+23-18+24) = B = (-15-23+18+24) = AB = (15-23-18+24) = A2 = 49, B2 = 4 , AB2 = 1 4 4 4 4 4 -2 4 4

  31. Multiply each of these by the number of data points in each column: A26(49) = 294 B26(4) = 24 AB26(1) = 6 TOTAL : 324

  32. ANOVA: Source SSQ df MS Fcalc Col 324 3 Error 400 20 20 14.7 1.2 .3 294 24 6 A B AB 1 1 1 294 24 6 F.95 (1,20) =4.3 { { { And:

  33. If: 1  c 15 2  a 23 3  b 18 4  abc 24 A = Gas Type B = Auto Make C = Highway (23-1 design)

  34. Source SSQ df { { A+BC 294 1 B+AC 24 3 1 AB+C 6 1 Error 400 20 e t c . We’d get the same breakdown of the SSQ, but being the + block of I = ABC, ANOVA:

  35. What if contrasts of interest are not orthogonal? • Let k be the number of contrasts of interest. • If k <= c-1  Bonferroni method • If k > c-1  Bonferroni or Scheffe method *Bonferroni Method: The same F test (SSQ = RxZi^2) but using a = 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.

  36. If k contrasts are orthogonal, • Can k be larger than c-1? • Can k be smaller than c-1? No. Yes. For case 2, do the same F test (but the sum of SSQ will not be equal to SSB). See Slide 16.

More Related