NCSU ST512
Quiz 4 –Practice Sum 2 2011
1. I have 2 oil based paints (O1 and O2) and a latex paint (L). To
for water permeability, I paint 5 boards with each paint, spray
boards with water and then measure the amount of water absorbed
grams. Here are the resulting means and totals for each set of
boards:
Paint
Mean
Total
Contrast Coefficients
L
26
130
_2_
O1
21
105
_-1_
O2
19
95
_-1_
test
all 15
in
5
Q
DIV
SS
60
5*6=30 120
(a) Fill in the coefficients for a contrast to compare latex to the average
of the oil based paints.
(b) Compute the sum of squares for the contrast in (a). 120
(c) Compute a sum of squares for testing the null hypothesis that there are
no differences in water permeability among the three paints.
HO: MU1=MU2=MU3=MU
OVERALL MEAN = (130 + 105 + 95)/15 = 22
SS= 5(26 - 22)^2 + 5(21-22)^2 + 5(19-22)^2 = 130
(d) Assuming the error sum of squares is 300, finish the test in question
(c) by computing the calculated F statistic.
MS(Treatment) = SS(Treatment)/(3-1) = 130/2 = 65
Error MS = Error SS / dfe = 300/3*(5-1) = 300/12 = 25
F(Treatment) = MS(Treatment)/Error MS = 65/25 = 2.6
2. Here is a plot like that in the class notes of the 4 means for a two
factor factorial experiment. The factors are fertilizer ingredients N
and P each at 2 levels and the design is a randomized complete block
with 6 blocks, each block being a bench in a greenhouse and each
observation Y being growth of a flower.
Heights of the plotted points are labeled on the vertical axis. Each
flower is in its own pot on one of the greenhouse benches.
19
15
14
12
|
+
p1
|
|
|
+
p2
+
p2
|
+
p1
|
|______________________________________
|
|
n1
n2
N1
N2
mean
N2-N1
August 2, 2011
P1
12
19
15.5
7
P2
14
15
14.5
1
mean
13
17
15
P2-P1
2
-4
-6
-6 = N*P
Page 1
NCSU ST512
Quiz 4 –Practice Sum 2 2011
Compute the following if possible, including the right sign:
(a) The main effect of N (7+(1))/2= 4 its sum of squares
6*2*(13-15)^2 + 6*2*(17-15)^2 = 96
and the degrees of freedom (2-1) = 1 for this sum of squares.
(b) The simple effect of N at the low level of P (19-12) = 7 and its sum
of squares 6*(19-15.5)^2 + 6*(12-15.5)^2 = 147.
(c) The simple effect of P at the high level of N
is (15-19) = -4
6*(15-17)^2 + 6*(19-17)^2 = 48.
(d) The interaction between N and P = -6 and its sum of squares
mean
total
N
P
N1P1
12
72
-1
-1
N1P2
14
84
-1
1
N2P1
19
114
1
-1
N2P2
15
90
1
1
NP
1
-1
-1
P at
N2
P at
N1
N at
P1
N at
P2
Sum of (SS
0
0
-1
Q
54
divisor
SS(Q)
48
-12
6*4=24
24
1
-36
24
-1
1
24
6*2=12
1
0
0
12
12
-1
0
1
0
42
12
0
-1
0
1
6
12
48^2/24=96
(-12)^2/24
= 6
(-36)^2/24
= 54
(-24)^2/12
= 48
(12)^2/12
= 12
(42)^2/24
= 147
(6)^2/24
= 3
Simple effects for N)= 147+3=150=54+96
Sum of (SS Simple effects for P)= 48+12=102 = 96+6
Grand Total = 360
overall mean = 360/24 = 15
(e) The error degrees of freedom 2*2*(6-1) = 20 .
(f) Your friend asks you if he should put the source "REPS" in his ANOVA
table as he has seen in some published papers. What do you say? (yes or
no with a brief explanation)
YES, since it is a Randomized Complete Block Design.
3. In a 2x2 factorial with factors A and B and 10 replications the main
effect of A was 10, the simple effect of A at the high level of B was 2.
B1
A1
A2
A2-A1
B2
B2-B1
x
18
= 10*2-2
2
(18+2)/2=10
A Main eff
A*B
(2-18)/2=-9
r=10
August 2, 2011
Page 2
NCSU ST512
Quiz 4 –Practice Sum 2 2011
Find, if possible, (if not, put "NP")
(a) the simple effect of A at the low level of B = 18
(b) the mean _NP___ of all observations that had both A and B at the low
level.
(c) Interaction effect A*B :
(2 – 18)/2 = - 8
4. Suppose a 3 x 2 factorial with factors A (3 levels: al, a2, a3)
and B (2 levels: bl, b2) is run with 4 replications in randomized complete
blocks design. List the standard ANOVA breakdown for this factorial
experiment including sources for each main effect and interaction (give
sources and degrees of freedom only).
Source
BLOCK
A
B
A*B
Error
Df
4-1 = 3
3-1 = 2
2-1 = 1
2*1 = 1
(3*2-1)*(4-1)=15
(a) Suppose the totals of the observations in the experiment in
question 2 are listed in our standard form as
[albl] = 30,
[a2bl]
[alb2] = 40,
mean
total
C2
(iii)
C3 (iv)
totals
means
C1 (i)
a1b1
30/4
30
1
a1b2
10
40
-1
a2b1
20
80
-2
a2b2
25
100
2
a3b1
50/4
50
1
a3b2
20
80
-1
1
180
180/8
2
[a3bl] = 50
[a2b2] = l00,
-1
70
70/8
-1
= 80,
130
130/8
-1
[a3b2] = 80
Q
divisor
0
4*12
40
2*4
380
GM=380/24
160
6*8
SS(Q)
0
200
533.33
and that the error sum of squares is l20. Compute sums of
squares and associated degrees of freedom for these tests:
r=4
Error Sum of Squares = 120.
MSE = 120/(15) = 8
Error df = (3*2-1)*(4-1) = 15
i) Test for main effects of A.
Sum of squares
> 4*3[(70/8-380/24)^2+(180/8-380/24)^2+(130/8-380/24)^2]
[1] 758.3333
Degrees of freedom = 2
MS(A) = 758.3333/2 = 379.1667
F = 379.1667/8 = 47.39584
ii) Test for comparing the effect of level 2 of A to the average
of the effects of levels, l and 3 of A.
Sum of squares
> (-70+2*180-130)^2/(8*(1+4+1))
[1] 533.3333
Degrees of Freedom = 1
MS(C1) = 533.3333
F-533.3333/8
August 2, 2011
Page 3
NCSU ST512
iii)
iv)
Quiz 4 –Practice Sum 2 2011
Test to see if the comparison in part (b) is the same within both
levels of factor B.
Sum of squares 0
Degrees of freedom
1
Also, compute the t-statistics for testing that the low and high levels
of A, when combined with the high (b2) level of B, produce the same
effect. In other words,
test Ho: alb2 - a3b2 = 0
F = (200/1)/(120/8) = 25
F= t^2 , t = sqrt(F) = sqrt(25) = -5 (note sign) with dfe= 15.
Also find s.e(“a1b2-a3b2”) = sqrt(MSE*(2/4)) = 2
t= (10-20)/2 = -5
t-statistics = -5
with
15 degrees of freedom.
5. I have a 5x2 factorial with factors W (weight at levels 20, 40, 60, 80,
100) and D (distance 40 yards, 80 yards) in which I study the effect on
some heart measurement of carrying the weight over the given distance. I
do all 10 (W,D) combinations for each of 5 people so I have 50
observations. Here is my (W, D) table of totals.
W
D
20
100
sum
mean
24
24
26
27
100
110
4
4.4
sum
23 40 46 48
Mean 2.3 4.0 4.6 4.8
53
5.3
40
80
10
13
40
18
22
60
22
24
80
Grand Mean = 210/50 = 4.2
My book shows -2 -1 0 1 2 as the linear orthogonal polynomial coefficients
for a factor at 5 equally spaced levels, like W.
D
40
80
total
20
10
13
23
40
18
22
40
W
60
22
24
46
80
24
24
48
100
26
27
53
Wlin
-2
-1
0
1
2
Q
68
divisor
SS(Q)
5*2*10=100
(a) Compute the sum of squares for the linear effect of W 46.24
degrees of freedom 1 df
46.24
and its
(b) I want to see if the linear effect of W is the same for both distances.
Compute sum of squares for the interaction of the W linear effect with D
H o :  2  * 11   1 * 12   0  * 13  1 * 14   2  * 15   2  * 21   1 * 22   0  * 23  1 * 24   2  * 25
or
H o :  211  12  0* 13  14  215    221  22  0* 23  24  225   0
H1 :  211  12  0* 13  14  215    221  22  0* 23  24  225   0
Contrast
Wlin in D1 vs D2=
(-2*10-1*18+0*22+1*24+2*26)-( -2*13-1*22+0*24+1*24+2*27)=8
August 2, 2011
Page 4
NCSU ST512
Quiz 4 –Practice Sum 2 2011
SS (Wlin in D1 vs D2) = 8^2/(5*(10*2)) = 0.64
((-2*10-1*18+0*22+1*24+2*26)-( -2*13-1*22+0*24+1*24+2*27))^2/(5*(10*2))
[=
0.64
(c) If I had obtained the table of totals above without having blocked on
people, how would that change your computations for parts (a) and (b) of this
question?
It would not have changed since we are using totals for W and D over people.
(d) Suppose I take the 4 degree of freedom sum of squares for W and from it I
subtract your sum of squares in part (a). I form an F test using this and it
is significant (I reject H0). Explain, as you would to the experimenter, how
to interpret this result. In other words, what is it that I am testing here?
W has 5 levels, df(W) = 5-1= 4
Full SS for W – SS for W Linear
F = [(Full SS for W – SS for W Linear ) / 3]/Error MS
SS(W) = 5*2*((2.3-4.2)^2+(4.0-4.2)^2+(4.6-4.2)^2+(4.8-4.2)^2+(5.3-4.2)^2)
= 53.8
F = [(Full SS for W – SS for W Linear ) / 3]/Error MS
= [(53.8-46.24)/3]/8= 0.315
We are testing whether the higher degree polynomial coefficients(quadratic,
cubic, quartic) are all simultaneously equal to 0, i.e., whether we need a
higher degree polynomial than linear to explain effect of W.
August 2, 2011
Page 5