Solution for
Problem 13.3.4
Prob 13.3.4
Test Hyp: First 4 Steps
a1) Ho : 1 =  2 =  0
All means are equal
a2) Ha : i   o
for at least one i
At least one mean is significantly
different from others
b1) Ho : 1 = 2 =  0
All slopes are equal
b2) Ha : i  o
for at least one i
At least one slope is significantly
different from others
Test all species of Drosophlia have same
response to increasing levels of insecticide
Perform ANCOVA
Means
X Trt 1
Trt 2
Trt 3
0
0.3
0.6
0.9
89
77
12
2
87
43
22
8
45
40
Tyx= 0
Txx= 0
91
71
23
5
0.45 47.5
Tyy=116.7
Eyy= 14315 Eyx= -133.5
Exx= 1.35
Syy= 14431 Syx= -133.5
Sxx= 1.35
44.2
Survival of Drosophlia
Find Adjusted SS for Common
Estimate from ANOVA Table
Source
Trt
df
Adjusted SS
t-1
T' yy = S' yy - E' yy
MST
MSE
Error
t(r-1)-1
E' yy
Total
(sum)
tr-2
S'yy
Common Error 8
MS
Eyy = 1113.3
Fobs
MST / MSE
Survival of Drosophlia
Find SSQ & X-Product for Common Slope
Adjust Error Sum of Squares of Y
for Common Slope
Tyy=
116.7 Tyx= 0
Eyy= 14315
Txx= 0
Eyx= -124.5 Exx= 1.35
Syy= 14431.7 Syx= -124.5 Sxx= 1.35
2
Exy
Exy
=1113.3 where b 
Eyy  Eyy = -98.89
Exx
Exx
Survival of Drosophlia
Setup table to
Compare Common vs Separate
S.V.
Difference
df
Adjusted SS
MS
t-1
Separate Error t(r-2)  Syyi= ?
Common Error t(r-1)-1
Eyy = 1113.33
Fobs
Survival of Drosophlia
For each Treatment
Get SSQ’s and X-product
Where
bi = Sxyi
Sxxi
X Trt 1
Trt 2
Trt 3
0
0.3
0.6
0.9
89
77
12
2
87
43
22
8
91
71
23
5
Syy=4851
Syy=5898
Syy=3566
Sxy= -45.9
Sxx= 0.45
Sxy= -48.9
Sxx= 0.45
Sxy= -38.7
Sxx= 0.45
b1 = -102
b2 = -108.7
b3 = -86
Survival of Drosophlia
Calculate Separate Slopes
and sum to get Separate Residual Error SS
S.V.
df
Adjusted SS
Syyi = Syyi - bi Sxyi
Trt 1
4-2
4851 - (-102) (-45.9) = 169.2
Trt 2
4-2
5898 - (-108.7) (-48.9)= 584.2
Trt 3
4-2
3566 - (-86) (-38.7)
Separate Error 6
= 237.8
 Syyi=991.2
Survival of Drosophlia
Compare Common vs Separate
S.V.
Difference
df
2
Adjusted SS
122.13
Separate Error 6  Syyi= 991.2
Common Error 8
MS
Fobs
MSD= 61.07 .37
MSE=165.2
Eyy = 1113.33
Conclude Slopes are Equal
b` = -98.88
F.05,2,6 = 5.14
Proceed with ANCOVA
Source
Trt
df
t-1
Adjusted SS
MS
T' yy = S' yy - E' yy
MST
MSE
Error
t(r-1)-1
E' yy
Total
(sum)
tr-2
S'yy
Fobs
MST / MSE
Sxy2 =1603.33
Exy2
=1113.33 S' yy  Syy Eyy  Eyy Sxx
Exx
Syy= 14431.7 Syx= -133.5
Sxx= 1.35
Proceed with ANCOVA
Source
df
Adjusted SS
Trt
2
Error
8
Total
(sum)
10
T' yy =
116.7
E' yy= 1113.3
MS
Fobs
58.3 0.42
139.2
S'yy= 14431.7
Reject H0
Conclude Adjusted means are equal
F.05,2,8 = 4.459
Compare at Mean of X
Series1
Series2
Series3
Gp Adjusted
Mean
A 47.500
A 45.000
A 40.000
100
80
60
40
20
0
0
0.2
0.4
0.6
0.8
1
• Which species has greatest relative best (best advantage)?
Can’t tell since neither slopes nor means are significantly different
NOTE: ADJUSTED = Unadjusted since X is same for all trts
N trt
4
4
4
1
2
3
SAS: Program to test
HOMOGENEITY of Slopes
proc glm data=a;
class trt;
model y = trt x trt*x;
estimate "slope for trt 1" x 1 trt*x 1 0 0;
estimate "slope for trt 2" x 1 trt*x 0 1 0;
estimate "slope for trt 3" x 1 trt*x 0 0 1;
run;
SAS Output
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
5
13440.46667 2688.09333 16.27
0.0020
Error
6
991.20000 165.20000
Corrected Total 11
14431.66667
R-Square
Coeff Var
Root MSE
y Mean
0.931318
29.10117
12.85302
44.16667
Source DF
trt
2
x
1
x*trt
2
Type I SS
116.66667
13201.66667
122.13333
Source DF
trt
2
x
1
x*trt
2
Type III SS
213.03810
13201.66667
122.13333
Parameter
slope for trt 1
slope for trt 2
slope for trt 3
Mean Square
58.33333
13201.66667
61.06667
F Value
0.35
79.91
0.37
Pr > F
0.7162
0.0001
0.7057
Mean Square
106.51905
13201.66667
61.06667
Standard
Estimate
Error t
-102.000000 19.1601438
-108.666667 19.1601438
-86.000000 19.1601438
F Value
0.64
79.91
0.37
Pr > F
0.5576
0.0001
0.7057
Value Pr > |t|
-5.32
0.0018
-5.67
0.0013
-4.49
0.0042
END