Unbalanced 2-Factor Studies
KNNL Chapter 23
Unequal Sample Sizes
When sample sizes are unequal, calculations and
parameter interpretations (especially marginal ones)
become messier
Observational stu
4.2. SOLVING SYSTEMS BY SUBSTITUTION 237
33. We begin by solving the rst equation for 3:.
33: + 83; = 6 First equation.
3:1: 2 6 83; Subtract 83; from both sides
at = 6 38y Divide both sides by 3.
:
240 CHAPTER 4. SYSTEMS
g3: 8, is already solved for y, so lets substitute
a: 8 for y in the rst equation.
37. The second equation, y =
2
9m + 63; = 9 First Equation.
9:r |- 6 (gm 8) = 9 Substitute gs:
246 CHAPTER 4. SYSTEMS
Well concentrate on eliminating 3:. Multiply the rst equation by 5, then add
the results.
511: + 40y = 205
53;' 93; = 50
31y = 155
Divide both sides by 31.
1
3.1 = g Divide both
238 CHAPTER 4. SYSTEMS
III-H
Clear the calculator screen by pressing the CLEAR button, then enter the left-
hand side of the rst equation with the following keystrokes. The result is
shown in the seco
4.3. SOLVING SYSTEMS BY ELIMINATION 245
Because each of the last tvvo statements are true, this guarantees that (3:, y) =
(2, 4) is a solution of the system.
5. Start with the given system.
56
56
8:rl
244 CHAPTER 4. SYSTEMS
Check: We must show that the solution (3:, y) = (4., 1) satises both equa-
tions.
:1:+4y:0 9x7y243
(4) + 4(1) = 0 9(4) 7(1) = 43
4+4:0 367:43
0 = 0 43 = 43
Because each of the l
4.3. SOLVING SYSTEMS BY ELIMINATION 249
Take the ansvver y = 2 and substitute 2 for y in the rst equation.
32:r -| 7y : 238 First equation.
32:,B l 7(2) = 238 Substitute 2 for y.
32:1: 14 : 238 Multip
250 CHAPTER 4. SYSTEMS
19. Start with the given system.
9:1393;
2:1363;
63
34
Well rst concentrate on eliminating the variable 3:. Multiply the rst equation
by 2, the second equation by 9. then add th
HW 61
Due 3/24
Spring 2014
Math 2568
Oguz Kurt
Problem 1
TRUE/FALSE: Suppose A and B are n n matrices. Then (A + B) (A + B) = A2 + 2 AB + B 2 .
Solution
.
1 Please,
submit only problems 4, 5, 16, 18,
Homework 51 Due Friday, 2/28
Spring 2014
Problem 1
Find a vector !v R3 so that !v (1, 2, 3) = (1, 0, 0).
1 Please,
submit only the problems 1,2,3,4,6,8,12,15,17,20,22,25,27.
Math 2568
Oguz Kurt
Proble
HW71
Due 4/4
Spring 2014
Math 2568
Oguz Kurt
Problem 1
Let V = P3 be the vector space of polynomials of degree 3. Define a linear transformation f : V R1
by the rule that f (p) = p(1) for p(x) P3 . Le
HW81
Due 4/21
Math 2568
Spring 2014
Oguz Kurt
Determinants
Problem 1
2
k
Find values of k for which det(B) = 0 for B = 40
k
1 Please,
submit problems 1, 3, 6, 8, 12, 13, 14, 15, 16, 17
k
k+1
8
3
3
1 5
Homework 3 Due Monday, 2/10
Math 2568
Spring 2014
Oguz Kurt
Problem 1
True/False/Why?: If !v , w,
! !u are linearly independent, then !v + !u, w
! + !u, !u are linearly independent.
Problem 2
True/Fal
HW91
Due
Math 2568
Spring 2014
Oguz Kurt
Similarity and Diagonalization
Problem 1
1
Show that A = 1
2
1 Please,
0
1
1
4
1
3 and B = 0
3
7
do NOT submit any
0
1
1
1
1 are not similar. (by finding an i
4.3. SOLVING SYSTEMS BY ELIMINATION 247
Divide both sides by 17.
17y = 68
58 . . .
y = Dw1de both Sldes by 17.
'9' = 4 Simplify.
Take the answer y = 4 and substitute 4 for y in the rst equation.
12:13
234 CHAPTER 4. SYSTEMS
29. We begin by solving the rst equation for m.
3:3 5y = 3 First equation.
3:1: 2 3 + 5y Add 53; to both sides
:3 = 3 25y Divide both sides by 3.
3: : 1 + 3:; Divide both term
Randomized Complete Block and Repeated
Measures (Each Subject Receives Each
Treatment) Designs
KNNL Chapters 21,27.1-2
Block Designs
Prior to treatment assignment to experimental units, we
may have i
STA 4211 Exam 1 Spring 2015 PRINT Name _
For all significance tests, use = 0.05 significance level. Show work for any partial credit!
Q.1. In the broiler chicken study, with factor A (base diet: Sorgh
STA 4211 Exam 1 Take-Home Portion
Due at 10:40 AM 2/9/18
A manufacturer buys chemicals from three vendors (A,B,and C). You sample nI=6 units from each vendor and
make an assessment of the purity of th
2-Factor ANOVA
Beauty and Judgment Halo Effect
Landy and Sigall (1974). Beauty is Talent: Task Evaluation as a Function of the Performers
Physical Attraction, Journal of Personality and Social Psychol
4.3. SOLVING SYSTEMS BY ELIMINATION 243
47. The second equation, '9 = 3 23:, is already solved for y. Substitute
3 2:1: for y in the rst equation and solve for :5.
9:1: -I- 4y : 73 First Equation.
9:.
4.2. SOLVING SYSTEMS BY SUBSTITUTION 241
intercept form so that we can compare them.
14:i.: 7y 2 112 Second Equation.
Ty = 14:1: + 112 Add 14:3 to both sides.
3; : 143312112 Divide both sides by 7.
y
248 CHAPTER 4. SYSTEMS
13. Start with the given system.
28
60
29: 63;
3113 + 183
Well concentrate on eliminating y. Multiply the rst equation by 3. then add
the results.
611: 18y = 84
33: + 18y = 60
3
4.3. SOLVING SYSTEMS BY ELIMINATION 253
Store 20/61 in Y with the following keystrokes. The result is shown in the rst
image below.
EEEEI mmlww
Clear the calculator screen by pressing the CLEAR button
236 CHAPTER 4. SYSTEMS
Finally, substitute 16/7 for y in 3: = 2 3y.
3
=2_
:1: 4y
1
a: = 2 g (76) Substitute 16/7 for y.
12
a: = 2 + ? Multiply.
1-4 12
:r = ? -| ? Equivalent fractions with.
a common d
4.3. SOLVING SYSTEMS BY ELIMINATION 251
Divide both sidee by 37.
373; = 148
y = % Divide both sides by 37.
y = 4 Simplify.
Take the answer 3; = 4 and substitute 4 for y in the rst equation (you could
4.2. SOLVING SYSTEMS BY SUBSTITUTION 235
HEB!
Clear the calculator screen by pressing the CLEAR button, then enter the left-
hand side of the rst equation with the following keystrokes. The result is
252 CHAPTER 4. SYSTEMS
Take the answer 3; : 5 and substitute 5 for y in the rst equation (you could
also make the substitution in the second equation).
3:c 5y = 34 First equation.
3:1: 5(5) 2 34 Subst