A must have three pivot columns. (See Exercise 30 in Section 1.7, or realize that the equation Ax = 0
has only the trivial solution and so there can be no free variables in the system of equations.) Since
A is 33, the pivot positions are exactly where a,
478
Exercises
Table 14.17
Correct analysis of variance for the dye experiment
The SAS System
General Linear Models Procedure
Dependent Variable: Y
Source
Model
Error
Corrected Total
Source
BLK
A
B
C
A*B
A*C
B*C
A*B*C
Table 14.18
DF
5
2
2
2
4
4
4
8
DF
31
2
14.5
Table 14.12
Design d1 for a 23 experiment confounding F1 F2 G1 and design d2 for a
2
32 experiment confounding G2 H; G2 H 2
Design
d1
Table 14.13
Block
I1
II1
Block
I
I1 II2
II
I1 III2
III
II1 I2
IV
II1 II2
V
II1 III2
VI
Design
d2
Block
I2
II2
III2
T
14.4
473
Designing Confounded Asymmetrical Experiments
We can use this idea only when the numbers of levels of all factors are powers of the same
prime number. For all of the other examples mentioned above, the use of pseudofactors
would transform the exp
14.3
14.3
Designing Using Pseudofactors
471
Designing Using Pseudofactors
14.3.1
Confounding in 4p Experiments
A treatment factor F with four levels coded 0, 1, 2, 3 can be represented by two factors F1
and F2 each having two levels coded 0, 1. The levels
472
Chapter 14
Table 14.8
Confounding in General Factorial Experiments
42 experiment in 4 blocks of 4, confounding three
degrees of freedom (F1 G1 , F1 F2 G2 , F2 G1 G2 ) from FG
Block
I
II
III
IV
14.3.2
L1 , L2
0,0
0,1
1,0
1,1
Pseudofactors F1 , F2 , G1
470
Chapter 14
Table 14.7
Confounding in General Factorial Experiments
Analysis of variance for the dye experiment
Source of
Variation
Block
A
AL
AQ
B
BL
BQ
C
CL
CQ
AB
AC
BC
Error
Total
Degrees of
Freedom
2
2
1
1
2
1
1
2
1
1
4
4
4
6
26
Sum of
Squares
182.
14.2
469
Confounding with Factors at Three Levels
To normalize the contrasts, one would divide AL by
ci2 /(rbc)
2/9 and divide AQ
by
ci2 /(rbc)
6/9.
The sum of squares for testing the hypothesis that the linear contrast for A is negligible
is the square o
14.2
Confounding with Factors at Three Levels
467
confound either a main effect or a two-factor interaction. However, the three-factor interactions are also thought to be negligible, so one possible choice is to confound (ABD;
A2 B 2 D 2 ) together with (
480
Exercises
Table 14.20
Data for the sugar beet experiment
Block I
Levels of
N, P, K
Yield
211
2575
120
2472
200
2411
002
2403
010
2220
021
2252
101
2295
112
2362
222
2434
Block II
Levels of
N, P, K
Yield
121
2599
220
2517
022
2411
110
2252
212
2381
201
479
Exercises
(b) Calculate the normalized contrast estimate for Linear A Linear B, using the
method outlined in Section 14.2.4.
(c) Compute the sum of squares for testing the hypothesis that the Linear A Linear
B contrast is negligible, using the method
481
Exercises
8. Consider a 22 32 design confounding AB, (CD 2 ; C 2 D), and (ABCD 2 ; ABC 2 D).
(a) Give the designnamely, list the treatment combinations block by block.
(b) Describe how to randomize the design.
(c) Give a set of ve orthogonal treatment
15.2
Table 15.8
493
Fractions from Block Designs; Factors with 2 Levels
1
fraction
4
of a 25 experiment and data from the sludge
experiment.
Levels of
A, B, C, D, E
00010
00111
01001
01100
10001
10100
11010
11111
yijklm
195
496
87
1371
102
1001
354
775
A
492
Chapter 15
Fractional Factorial Experiments
1
design obtained by confounding ABD, ACE, and BCDE will give a 4 fraction in which
CD is not aliased with main effects.
1
A list of useful 4 fractions is given in Table 15.55 at the end of the chapter.
Exam
15.2
491
Fractions from Block Designs; Factors with 2 Levels
1
one block for the 4 fraction, specically the block that satises
a1 + a 2 + a 4
1 (mod 2) and a3 + a4 + a5
0 (mod 2).
The treatment combinations in this fraction are
00011 00110 01000 01101 100
490
Chapter 15
Fractional Factorial Experiments
Contrast Estimate
T
0.4
0.2
0.0
b
-0.2
Figure 15.1
Soup experiment:
normal probability
plot of contrast
estimates
b
-0.4
1.8 T
1.4
1.0
b b
Labels: levels of B
Interaction plots for
the soup experiment
0
1
y.
15.2
489
Fractions from Block Designs; Factors with 2 Levels
This was a screening experiment, since the researchers had little idea of which factors
were going to turn out to be important in affecting the variability of the soup mix weight.
1
They decided
488
Chapter 15
Table 15.4
Fractional Factorial Experiments
Resolution numbers of fractional factorial experiments
Resolution
III
IV
V
Table 15.5
Main effects aliased with:
2-factor interactions or higher
3-factor interactions or higher
4-factor interactio
486
Chapter 15
Table 15.3
Fractional Factorial Experiments
Aliasing
scheme for the
1
-fraction (001,
2
010, 100, 111)
of a 23
experiment
I
A
B
C
ABC
BC
AC
AB
The list of aliased contrasts is called the aliasing scheme for the design. We generally
write th
15.2
487
Fractions from Block Designs; Factors with 2 Levels
fraction satises either
a1 + a2 + a3 + a4
0 (mod 2)
and has dening relation I
a 1 + a2 + a3 + a4
ABCD, or it satises
1 (mod 2)
and has dening relation I
a1 + a2 + a3 + a4 + a5
and has dening rel
15.2
Fractions from Block Designs; Factors with 2 Levels
485
Table 15.2 shows several interesting features. First, the column corresponding to ABC
is not a contrast. The coefcients are the coefcients that one would use in obtaining the
sum of the four obs
484
Chapter 15
Fractional Factorial Experiments
some of the ideas introduced in Section 7.6 to reduce the sensitivity to noise variables of a
product or manufacturing process.
The use of SAS in analyzing fractional factorial experiments is explored in Sec
466
Chapter 14
Confounding in General Factorial Experiments
to be confounded. If the pair (Az1 B z2 P zp ; A2z1 B 2z2 P 2zp ) is chosen for confounding
together with the pair (Ay1 B y2 P yp ; A2y1 B 2y2 P 2yp ), the b 9 blocks are produced
from the the ni
PSTAT 120A
Winter 2016
Midterm 1
1/28/16
Time Limit: 75 Minutes
Name (Print):
Instructor:
Teaching Assistant:
Wade Herndon
This exam contains 6 pages (including this cover page) and 7 problems. Check to see if any pages
are missing. Enter all requested in
PSTAT 120A
More Counting Problems
Winter 2016
Problem 1: Each year starts on 1 of 7 days and each year is either a leap year or not a leap
year. How many possible calendars are there?
Multiplication rule for counting:
7(2) = 14.
Problem 2: Three different
PSTAT 120A
Midterm 2 Formulas
Binomial distribution
n x
f (x) =
p (1 p)nx , x = 0, 1, . . . , n
x
E [X] = np
Poisson distribution
f (x) =
e x
, x = 0, 1, 2, 3 . . .
x!
E [X] =
Geometric distribution
f (x) = (1 p)x1 p, x = 1, 2, 3, . . .
E [X] =
1
p
Exp