2014 Final Exam KEY
Question 1
1.1
[25 points]
[2 points] Describe in detail the design of this experiment [see appendix].
RCBD Split-split plot with a 3x2x3 factorial treatment structure, with 3 reps per
block*factorial combination
Weight (in ounces)
a) Mainplot: a group of six piglets; b) Subplot: one piglet. c) Sub-subplot: Repeated
measures of weight per piglet
Design:
Response Variable:
Experimental Unit:
Class
Variable
1
2
3
4
Block or
Treatment
Block
Treatment
Treatment
Treatment
Number of
Levels
2
3
2
3
Subsamples?
Covariable?
Fixed or
Random
Random
Fixed
Fixed
Fixed
Description
Farm
Breeds of pigs
Diets
Time (weeks)
NO
NO
Data Pigs;
Do Block = 1 to 2;
Do Breed = 1 to 3;
Do Rep = 1 to 3;
Do Diets = 1 to 2;
Do Time = 1 to 3;
Input Weight @@;
Output;
End;
End;
End;
End;
End;
Cards;
75
77
73
81
82
77
92
91
93
59
50
59
69
89
74
83
91
89
82
102
100
89
79
76
74
73
91
88
89
97
88
86
105
103
107
83
71
79
77
59
76
68
81
76
74
83
81
83
56
71
61
65
74
75
76
78
84
82
96
89
97
62
53
74
77
72
72
76
86
82
80
95
103
93
65
68
71
73
59
71
78
66
59
71
71
72
82
68
62
78
77
89
96
71
80
84
84
91
100
71
95
89
80
82
87
77
77
81
;
Proc GLM Data = Pigs Order = Data;
Class Block Breed Diets Time;
Model Weight = Block Breed Block*Breed
Diets Breed*Diets Block*Breed*Diets
Time Breed*Time Diets*Time Breed*Diets*Time;
test h=Breed e= Block*Breed;
test h=Diets e= Block*Breed*Diets;
test h=Breed*Diets e= Block*Breed*Diets;
Means Breed / tukey e= Block*Breed;
Means Diets;
Means Time;
Contrast "Time lineal" Time 1 0 -1;
Contrast "Time quadratic" Time 1 -2 1;
Output out = PigsPR R=Res P=Pred;
Proc Univariate Data = PigsPR Normal;
Var Res;
Proc GLM Data = Pigs;
* Levene's for Breed;
Class Breed;
Model Weight = Breed;
Means Breed / hovtest = Levene;
Proc GLM Data = Pigs;
* Levene's for Diets;
Class Diets;
Model Weight = Diets;
Means Diets / hovtest = Levene;
Proc GLM Data = Pigs;
* Levene's for Time;
Class Time;
Model Weight = Time;
Means Time / hovtest = Levene;
Run;
Quit;
1.2
[3 points] Show that the data meet the assumptions of normality of residuals and homogeneity of
variances among Diets, among Breeds and among times.
Tests for Normality
Test
Statistic
p Value
Shapiro-Wilk W 0.989571 Pr < W 0.5753
Levene's Test for Homogeneity of Weight Variance
ANOVA of Squared Deviations from Group Means
Source
DF
Sum of Squares
Mean Square
Breed
2
10679.4
5339.7
Error
105
1284041
12229.0
Levene's Test for Homogeneity of Weight Variance
ANOVA of Squared Deviations from Group Means
Source
DF
Sum of Squares
Mean Square
Diets
1
19180.3
19180.3
Error
106
3693421
34843.6
Levene's Test for Homogeneity of Weight Variance
ANOVA of Squared Deviations from Group Means
F Value
0.44
Pr > F
0.6474
F Value
0.55
Pr > F
0.4598
Source
Time
Error
DF
2
105
Sum of Squares
10020.7
2306986
Mean Square
5010.3
21971.3
F Value
0.23
Pr > F
0.7965
All assumptions are met. We fail to reject the null hypothesis that the residuals are normally distributed
(p = 0.5753). We fail to reject the null hypotheses that Time, Diets and Breed are homogeneous
(p=0.7965, p = 0.4598, and p=0.6474, respectively).
1.3
[7 points] Run the appropriate ANOVA and answer the following questions (with p-values):
a. Is there a significant effect of breed on weight?
b. Is there a significant effect of diet on weight?
c. Is there a significant effect of time (weeks) on weight?
d. Are there any significant interactions between Diets and Breeds?
Dependent Variable: Weight
Source DF Sum of Squares Mean Square F Value Pr > F
Model 23 12954.85185 563.25443 22.20 <.0001
Error 84 2131.55556 25.37566
Corrected Total 107 15086.40741
R-Square Coeff Var Root MSE Weight Mean
0.858710 6.342294 5.037426 79.42593
Source
Block
Block*Breed
Diets
Breed*Diets
Block*Breed*Diets
Time
Breed*Time
Diets*Time
Breed*Diets*Time
DF
1
2
1
2
3
2
4
2
4
Type III SS
3245.037037
35.018519
1108.481481
47.574074
57.055556
2390.740741
98.148148
156.074074
208.370370
Mean Square
3245.037037
17.509259
1108.481481
23.787037
19.018519
1195.370370
24.537037
78.037037
52.092593
F Value
127.88
0.69
43.68
0.94
0.75
47.11
0.97
3.08
2.05
Pr > F
<.0001
0.5044
<.0001
0.3957
0.5257
<.0001
0.4301
0.0514
0.0943
***
NS
NS
NS
Tests of Hypotheses Using the Type III MS for Block*Breed as an Error Term
Source
DF
Type III SS Mean Square
F Value
Pr > F
Breed
2
5608.351852 2804.175926
160.15
0.0062 ***
Tests of Hypotheses Using the Type III MS for
Source
DF
Type III SS Mean Square
Diets
1
1108.481481 1108.481481
Breed*Diets 2
47.574074
23.787037
Block*Breed*Diets as an Error Term
F Value
Pr > F
58.28
0.0047 ***
1.25
0.4027 NS
a) Yes, there is a significant effect of Breed on weight (p=0.0062).
b) Yes, there is a significant effect of Diet on weight (p = 0.0047)
c) Yes, there is a significant effect of Time on weight (p = <0.0001), but due to the fact that it is
significant, we have to proceed with the conservative degrees of freedom test.
d) No, the interaction between Diet and Breed is not significant.
1.4
[3 points] If appropriate, use conservative degrees of freedom to test the Time and all 2 and 3 way
Diet and Breed interactions with Time. State whether or not you should continue the analysis past
this step. If so, carry out the appropriate next analysis and run any other assumptions tests. Do
your conclusions about Time or Time interactions change using regular and conservative degrees
of freedom?
ADJUSTED:
Error
DF
84
ADJ DF
42
Source
DF
Time
2
Breed*Time
4
Diets*Time
2
Breed*Diets*Time 4
Adj df
1
2
1
2
F Value
47.11
0.97
3.08
2.05
p-value
<.0001 ***
0.4301 NS
0.0514 NS
0.0943 NS
adj. p-value
0.0001 ***
0.3874 NS
0.0866 NS
0.1414 NS
All of the factors and interactions have the same significance or non-significance as in the split-split plot
analysis, so we stop here, and do not proceed to do a repeated measures analysis. None of the conclusions
change from the previous analysis (Time is significant, all interactions are NS).
1.5
[3 points] Carry out the appropriate means separation tests to compare all pairs of Breeds
(controlling the maximum experiment wise error rate). Which breed weighed the most (on
average)?
Tukey's Studentized Range (HSD) Test for Weight
Note: This test controls the Type I experimentwise error rate, but it generally has a
higher Type II error rate than REGWQ.
Alpha 0.05
Error Degrees of Freedom 2
Error Mean Square 17.50926
Critical Value of Studentized Range 8.33078
Minimum Significant Difference 5.8099
Means with the same letter
are not significantly different.
Tukey Grouping Mean N Breed
A 89.2500 36 3
B 76.8611 36 2
B 72.1667 36 1
Breed 3 had the greatest average weight over the course of the study (89.25 ounces), which was
significantly different from the average weight of Breeds 1 and 2 (which were not significantly different
from one another).
Level of
Diets N Weight
Mean Std Dev
1 54 82.6296296 12.0600640
2 54 76.2222222 10.8760888
Diet 1 had the greatest average weight over the course of the study (82.630 ounces), which was
significantly different from the average weight of Diet 2 (p = 0.0047), from the ANOVA.
1.6
[3 points] Use contrasts to determine if the response in time is lineal.
Dependent Variable: Weight
Contrast
DF Contrast SS Mean Square F Value Pr > F
Time lineal
1 2334.722222 2334.722222
Time quadratic
1
56.018519
56.018519
92.01 <.0001
2.21 0.1411
The response in time is lineal (p<.0001).
1.7
[3 points] Graph the main effects or the appropriate interactions if significant and comment.
proc gplot data=Pigs ;
** Main effect plots **;
axis1 offset=(5 pct,5 pct);
axis2 offset = (5 pct,5 pct);
symbol1 i=std1mtj v=none color=BLUE;
plot Weight * Breed = 1 / ;
run;
axis1 offset=(5 pct,5 pct);
axis2 offset = (5 pct,5 pct);
symbol1 i=std1mtj v=none color=RED;
plot Weight * Diets = 1 / ;
run;
axis1 offset=(5 pct,5 pct);
axis2 offset = (5 pct,5 pct);
symbol1 i=std1mtj v=none color=RED;
plot Weight * Time = 1 / ;
run;
The graphs confirm what we know from the ANOVA and means separation above. The highest weight
was found at Time 3, Breed 3, and Diet 1. Since the interactions are not significant, the appropriate graphs
are the main effects
1.8
[1 points] What other measurement could the researchers have taken before starting the experiment
that could have been used as covariable to reduce the variability of the experiment?
The researchers could have taken a measurement of initial pig weight (at birth), in order to determine if
the differences in average weight were due to diet or breed, or if they were due to differences in initial
weight.
Question 2
2.1
[25 points]
[2 points] Describe in detail the design of this experiment [see appendix].
Response Variable:
Experimental Unit:
RCBD with a 3 x 4 factorial treatment structure nested in locations with 1 rep per
block*factorial combination
Healthy Fruit per tree
Tree
Class
Variable
1
2
3
4
Number of
Levels
10
2
3
3
Design:
Block or
Treatment
Treatment
Block
Treatment
Treatment
Subsamples?
Covariable?
Fixed or
Random
Random
Random
Fixed
Fixed
Description
Locations/Farmers
Greenhouses
Varieties (2 new and 1 old)
Pesticides (0, 2, and 4 qts/A)
NO
NO
Data Lemon;
Do Block = 1 to 2;
Do Trt = 1 to 9;
Do Location = 1 to 10;
Input Y @@;
Output;
End;
End;
End;
Cards;
74
65
55
64
57
56
64
56
67
76
56
56
54
76
61
55
50
61
67
59
50
82
68
57
71
84
72
81
80
64
65
69
51
71
68
64
59
51
63
70
68
52
78
77
60
83
66
74
72
73
50
67
63
67
68
64
80
85
64
82
62
78
58
61
66
50
52
60
56
61
56
50
50
50
50
56
75
66
57
57
55
50
50
66
63
60
55
64
64
64
75
67
60
58
77
50
65
59
56
63
62
72
50
59
76
80
73
73
81
58
67
75
73
67
74
64
64
56
70
81
62
50
81
72
68
79
85
81
77
50
71
61
57
81
68
83
56
71
51
67
56
66
87
72
58
57
58
51
62
55
64
57
65
65
57
57
69
66
62
66
67
69
60
66
60
79
63
51
68
58
73
66
73
67
60
62
55
74
73
60
;
Proc GLM Order = Data;
Class Location Block Trt;
Model Y = Location Block(Location) Trt Trt*Location;
Random Location Block(Location) Trt*Location / test;
Contrast 'MvsO' Trt 1 1 -2
1 1 -2
1 1 -2 / e = Trt*Location;
Contrast 'MvsM' Trt 1 -1 0
1 -1 0
1 -1 0 / e = Trt*Location;
Contrast 'Lin' Trt 1 1 1
0 0 0 -1 -1 -1 / e = Trt*Location;
Contrast 'Quad' Trt 1 1 1 -2 -2 -2
1 1 1 / e = Trt*Location;
Contrast 'MOxL' Trt 1 1 -2
0 0 0 -1 -1 2 / e = Trt*Location;
Contrast 'MOxQ' Trt 1 1 -2 -2 -2 4
1 1 -2 / e = Trt*Location;
Contrast 'MMxL' Trt 1 -1 0
0 0 0 -1 1 0 / e = Trt*Location;
Contrast 'MMxQ' Trt 1 -1 0 -2 2 0
1 -1 0 / e = Trt*Location;
LSMeans Trt / pdiff adjust=tukey lines e = trt*Location;
Output out = LemonPR p = Pred r =Res;
Proc Univariate Data = LemonPR normal;
Var Res;
Proc Plot Data = LemonPR;
Plot Res*Pred = Trt;
Proc GLM;
Class Location;
Model Y = Location;
Means Location / hovtest = Levene;
Proc GLM;
Class Trt;
Model Y = Trt;
Means Trt / hovtest = Levene;
Proc VarComp Method = Type1;
Class Location Block Trt;
Model Y = Trt Location Block(Location) Trt*Location / Fixed = 1;
Run;
Quit;
2.2
[3 points] Show that the data meet the assumptions of normality of residuals and homogeneity of
variances among treatment groups.
Tests for Normality
Test Statistic p Value
Shapiro-Wilk W 0.992939 Pr < W 0.5360
Levene's Test for Homogeneity of Y Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
Location 9 71816.0 7979.6 0.80 0.6153
Error 170 1692292 9954.7
Levene's Test for Homogeneity of Y Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
Trt 8 20972.5 2621.6 0.41 0.9166
Error 171 1106866 6472.9
All assumptions are met. We fail to reject the null hypothesis that the residuals are normally distributed (p
= 0.5360). We fail to reject the null hypotheses that Location and Trt Variances are homogeneous
(p=0.6153, and p=0.9166, respectively).
2.3
[6 points] Using a single treatment factor with 8 levels representing the 8 combinations of
pesticide rate and variety, carry out the appropriate Proc GLM for this experiment and answer
the following questions based on the overall ANOVA (report F and p-values):
a.
b.
c.
d.
Does treatment significantly affect fruit number?
Is the effect of treatment location specific?
Is there significant variation in fruit number within locations?
Is there significant variation in fruit number among locations?
Source
DF
Type III SS Mean Square F Value Pr > F
Location
9
1963.466667 218.162963
1.86
0.1553
Error
12.272 1441.755706 117.485185
Error: MS(Block(Location)) + MS(Location*Trt) - MS(Error)
Source
DF
Type III SS Mean Square
Block(Location)
10
1015.222222 101.522222
Location*Trt
72
4804.933333 66.735185
Error: MS(Error) 80 4061.777778 50.772222
F Value Pr > F
2.00
0.0441
1.31
0.1168
Source
DF
Trt
8
Error
72
Error: MS(Location*Trt)
F Value Pr > F
7.27
<.0001
Type III SS Mean Square
3880.844444 485.105556
4804.933333 66.735185
Source
Type III Expected Mean Square
Location
Var(Error) + 2 Var(Location*Trt) + 9 Var(Block(Location)) + 18 Var(Location)
Block(Location) Var(Error) + 9 Var(Block(Location))
Trt
Var(Error) + 2 Var(Location*Trt) + Q(Trt)
Location*Trt
Var(Error) + 2 Var(Location*Trt)
a) Yes, there is a significant effect of Treatment on fruit number (p=<0.0001).
b) No, the effect of treatment is not location specific (interaction between location and treatment) (p =
0.1168).
c) Yes, there is significant variation in fruit number between greenhouses (p = 0.0441).
d) No, there is not significant variation in fruit number among locations (p = 0.1553).
2.4
[6 points] Use contrasts to partition the sums of squares associated with 2.3 in order to help
you answer the following questions (report p-values):
a.
Is there a difference between the new varieties and the old variety?
b.
Is there a difference between the two new varieties?
c.
Characterize the response of fruit number to pesticide rate.
d.
Are the dosage responses different between old and new varieties?
e.
Are the dosage responses different between the two new cultivars?
Contrast Coefficients:
Dosage
Variety
Mvs old
M vs M
L
Q
MOxL
MOxQ
MMxL
MMxQ
0
1 2 3
1 1 -2
1 -1 0
1 1 1
1 1 1
1 1 -2
1 1 -2
1 -1 0
1 -1 0
2
1 2 3
1 1 -2
1 -1 0
0 0 0
-2 -2 -2
0 0 0
-2 -2 4
0 0 0
-2 2 0
1
1
1
-1
1
-1
1
-1
1
Tests of Hypotheses Using the Type III MS for
Contrast
DF Contrast SS Mean Square F
MvsO
1 268.669444 268.669444 4.03
MvsM
1 492.075000 492.075000 7.37
Lin
1 2990.008333 2990.008333 44.80
Quad
1
14.802778
14.802778 0.22
MOxL
1
49.504167
49.504167 0.74
MOxQ
1
14.734722
14.734722 0.22
MMxL
1
27.612500
27.612500 0.41
MMxQ
1
23.437500
23.437500 0.35
4 .
2 3
1 -2
-1 0
-1 -1
1 1
-1 2
1 -2
1 0
-1 0
Location*Trt as an Error Term
Pr > F
0.0486
0.0083
<.0001
0.6391
0.3919
0.6399
0.5221
0.5553
a) Yes, there is a difference in fruit number between the new varieties and the old variety (p = 0.0486).
b) Yes, there is a difference in fruit number between the two new varieties (p = 0.0083).
c) The pesticide rates can be characterized by a linear response (p = <0.0001), but not a quadratic
response (p = 0.6391).
d) No, there is not an interaction between the new varieties and the old variety and the different pesticide
rates. The new and old varieties do not have different linear or quadratic responses (p = .6391, and
0.6399, respectively).
e) No, the responses of the two new varieties are not different at the different pesticide rates (p = 0.5221
and 0.5553, respectively).
2.5 [3 points] Rank the 9 treatment combinations in terms of mean fruit number and assign them to
significance groups using Tukey’s method of means separation.
Tukey Comparison Lines for Least Squares Means of Trt
LS-means with the same
letter are not significantly
different.
Y LSMEAN Trt LSMEAN Number
A
72.10 7 7
V1R4
B A
69.00 9 9
V3R4
B A
68.95 4 4
V1R2
B A C 67.50 8 8
V2R4
B D C 63.65 5 5
V2R2
B D C 62.85 6 6
V3R2
B D C 61.85 1 1
V1R0
D C 59.60 2 2
V2R0
D
57.20 3 3
V3R0
2.6 [1 points] Among the group of varieties with the highest yield (NS differences in yield
among each other in Tukey), which variety would you recommend to minimize the use of
pesticide?
. Variety 1 with Rate 2
Means separation: The means separation test indicates that Variety 1 at 4 qt/A, Variety 3 at 4 qt/A,
Variety 1 at 2 qt/A and Variety 2 and 4 qt/A are all not significantly different from one another, in
terms of healthy fruit /tree. So, at the high rate of pesticide, all varieties are the same, but at a
lower pesticide rate (2 qt/A), Variety 1 is recommended.
75
70
65
V1
V2
60
55
50
R0
R2
R4
2.7 [1 points] Would you feel comfortable in extending your recommendation to all greenhouse
locations in Southern California?
YES, because there are no significant interaction between location and treatments, so the
response is the same across locations (in addition there are no significant differences across
locations)
2.8. [3 points] What is the major source of variation in this experiment (use Proc VarComp)?
Type 1 Estimates
Variance Component Estimate Percent
Var(Location)
5.59321
0.079919212
Var(Block(Location))
5.63889
0.080571916
Var(Location*Trt)
7.98148
0.114044278
Var(Error)
50.77222
0.725464594
Total =69.9858
What this table shows is that the component of variation due to differences among locations (7.9% of the
total) is small compared to the component of variation due to error (72.5% of the total). The residual
error is the largest component of variation. This explains the relatively low proportion of variation
explained by our model (R2=0.742)
Question 3
data final_2014_2;
Input Block Gene $ Cultivar ID $ yield;
*yield = badyield**(-1.98323);
*yield = log10(badyield); *;
*yield = sqrt(badyield); *;
Cards;
1
T
1
A
138
1
T
2
B
114
1
T
3
C
89
1
T
4
D
73
1
S
1
E
97
1
S
2
F
115
1
S
3
G
48
1
S
4
H
49
2
T
1
A
124
2
T
2
B
99
2
T
3
C
74
2
T
4
D
49
2
S
1
E
.
2
S
2
F
86
2
S
3
G
46
2
S
4
H
46
3
T
1
A
88
3
T
2
B
67
3
T
3
C
70
3
T
4
D
47
3
S
1
E
58
3
S
2
F
74
3
S
3
G
44
3
S
4
H
44
4
T
1
A
60
4
T
2
B
62
4
T
3
C
60
4
T
4
D
43
4
S
1
E
50
4
S
2
F
55
4
S
3
G
41
4
S
4
H
40
;
Proc Print data = final_2014_2;
ID Block Gene Cultivar ID;
var yield;
Proc GLM data = final_2014_2;
title 'Exploratory model 2';
Class Block Gene Cultivar;
model yield = Gene|Cultivar|Block @2;
Proc GLM data = final_2014_2;
title 'ANOVA';
Class Block Gene Cultivar;
model yield = Gene|Cultivar Block;
Proc GLM data = final_2014_2;
title 'TRTMT ANOVA';
Class Block ID;
model yield = Block ID;
means ID;
Output out = fPR r = fres p = fpred;
Contrast 'diff. btw. African'
ID
Contrast 'diff. Btw. C. American'
ID
Contrast 'diff. btw Af and Am'
ID
Contrast 'Gene'
ID
Contrast 'int. gene by African'
ID
Contrast 'int. gene by C. American' ID
Contrast 'int. gene by btw Af and Am' ID
Proc Gplot data = fPR;
title 'res*pred';
plot fres*fpred = ID;
[20 points]
1 -1 0 0 1 -1 0 0;
0 0 1 -1 0 0 1 -1;
1 1 -1 -1 1 1 -1 -1;
1 1 1 1 -1 -1 -1 -1;
1 -1 0 0 -1 1 0 0;
0 0 1 -1 0 0 -1 1;
1 1 -1 -1 -1 -1 1 1;
Proc GLM data = final_2014_2;
title 'levene';
Class ID;
model yield = ID;
means ID / Hovtest = Levene;
Proc Univariate data = fPR normal;
title 'Normality test';
var fres;
qqplot;
histogram;
proc gplot data = final_2014_2;
title 'Gene1*Cultivar interaction TRANSFORMED DATA/UNTRANFORME DATA';
** Two-way Plots **;
axis1 offset=(5 pct,5 pct);
axis2 offset = (5 pct,5 pct);
symbol1 i=std1mtj v=none color=BLUE;
symbol2 i=std1mtj v=none color=BLACK;
symbol3 i=std1mtj v=none color=GREEN;
symbol4 i=std1mtj v=none color=ORANGE;
symbol5 i=std1mtj v=none color=RED;
plot yield * Gene = Cultivar/
description = "Plot of Yield by Gene2 and cultivar genetic background";
run; quit;
3.1 [2 points] Use the table in the appendix to describe the design of the experiment.
Design:
Response Variable:
Experimental Unit:
Class
Variable
1
2
3
RCBD factorial
Yield in bushels / Acre
A 30 m2 Field Plot
Block or
Treatment
Block
Treatment
Treatment
No. of
Levels
4
4
2
Subsamples?
Covariable
NO
NO
Description
240 m2 field segment
Genetic background of one of four cultivars
On of two alleles of the same gene
3.2 [3 points] Check all of the assumptions of your model.
Tests for Normality
Test
Statistic
p Value
Shapiro-Wilk W 0.967291 Pr < W 0.4478
The Shapiro-Wilk test suggest that the data is normality distributed (P = 0.4478).
Levene's Test for Homogeneity of yield Variance
Levene's Test for Homogeneity of yield Variance
ANOVA of Squared Deviations from Group Means
Source
ID
Error
DF Sum of Squares Mean Square F Value Pr > F
7
2816790
402399
23
3037586
132069
3.05 0.0202
Levene's Test for Homogeneity of yield Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
Gene
1
487771
487771
Error
29
22348814
770649
0.63 0.4327
Levene's Test for Homogeneity of yield Variance
ANOVA of Squared Deviations from Group Means
Source
DF Sum of Squares Mean Square F Value Pr > F
Cultivar
Error
3
3440166
1146722
27
6296861
233217
4.92 0.0075
The Levene’s test suggest that the variance between treatments is not homogeneous (ID P =
0.020 or if run separately Cultivar P=0.0075). Both analyses are considered correct
Source
DF Type III SS Mean Square F Value Pr > F
Gene
1 2033.472222 2033.472222
47.12 0.0001
Cultivar
3 7999.003472 2666.334491
61.78 <.0001
Gene*Cultivar
3
317.601852
7.36 0.0109
Block
3 6970.836806 2323.612269
53.84 <.0001
Block*Gene
3
309.597222
103.199074
2.39 0.1442
Block*Cultivar
9 2290.284722
254.476080
5.90 0.0101
952.805556
The exploratory model suggest that the block effects are multiplicative (P = 0.0101).
Data will be transformed but to help with the later analysis of the interactions, the ANOVA is
also run on the untransformed data
Source
DF Type III SS Mean Square F Value Pr > F
Gene
1 2482.370400 2482.370400
16.43 0.0006
Cultivar
3 7854.252976 2618.084325
17.33 <.0001
Gene*Cultivar
3 1205.018601
401.672867
2.66 0.0761
Block
3 6824.190476 2274.730159
15.06 <.0001
Contrast
diff. btw. African
DF Contrast SS Mean Square F Value Pr > F
1
19.580875
19.580875
0.13 0.7226
Contrast
DF Contrast SS Mean Square F Value Pr > F
diff. Btw. C. American
1
410.062500
2.71 0.1151
diff. btw Af and Am
1 7443.188582 7443.188582
49.26 <.0001
Gene
1 2482.370400 2482.370400
16.43 0.0006
int. gene by African
1
787.537396
787.537396
5.21 0.0335
int. gene by C. American
1
410.062500
410.062500
2.71 0.1151
int. gene by btw Af and Am
1
0.910173
0.910173
0.01 0.9389
410.062500
No overall significant interaction, but one of the contrast for interactions is significant even
in the untransformed data
3.3 [6 points] If necessary transform the data using a power transformation and report the results
demonstrating that all the assumptions are met. Run the appropriate ANOVA model to test for
effects of the gene and cultivar genetic background and the interactions between them. Report the
results of your ANOVA and describe which effects are significant.
data final_power_trans;
input means stddev;
logmeans = log10(means);
logvar = log10(stddev*stddev);
cards;
102.50 35.30
85.50 25.09
73.25 12.04
53.00 13.56
68.33 25.15
82.50 25.15
44.75 2.99
44.75 3.77
;
Proc GLM data = final_power_trans;
title 'power trans';
model logvar = logmeans;
run; quit;
Parameter
Estimate Standard Error t Value Pr > |t|
Intercept -7.580524919
1.91249374
-3.96
0.0074
logmeans
1.04656856
5.15
0.0021
5.394158485
a = 1 – (5.394158485 / 2) = -1.98323
The data will be transformed by raising them to the power of -1.98323.
TRANSFORMED DATA
Test for Normality OK
Test
Statistic
p Value
Shapiro-Wilk W 0.952353 Pr < W 0.1812
The Shapiro-Wilk test suggest that the data is normally distributed (P = 0.1812).
Test of homogeneity of variances
Levene's Test for Homogeneity of yield Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
ID
Error
7
5.11E-14
7.3E-15
23
1.8E-13
7.84E-15
0.93 0.5015
Levene's Test for Homogeneity of yield Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
Gene
1
4.11E-14
4.11E-14
Error
29
1.63E-12
5.63E-14
0.73 0.4000
Levene's Test for Homogeneity of yield Variance
ANOVA of Squared Deviations from Group Means
Source
DF Sum of Squares Mean Square F Value Pr > F
Cultivar
Error
3
9.07E-14
3.02E-14
27
5.87E-13
2.17E-14
1.39 0.2670
The transformation eliminated the heterogeneity of variances. Levene’s test suggest that the
variances are homogeneous (ID P = 0.5015, also works separately for Cultivar P=0.367 and
Gene P=0.40).
Source
DF
Type III SS
Mean Square F Value Pr > F
Gene
1 1.0995647E-6 1.0995647E-6
64.06 <.0001
Cultivar
3 3.5023752E-6 1.1674584E-6
68.02 <.0001
Gene*Cultivar
3 7.2635596E-7 2.4211865E-7
14.11 0.0015
Block
3 1.9891237E-6 6.6304123E-7
38.63 <.0001
Block*Gene
3
3.266445E-9
0.19 0.9001
Block*Cultivar
9 1.4429678E-7 1.6032976E-8
0.93 0.5438
9.799335E-9
The problem with non-additivity has disappeared as all Block*treatment interaction are no longer
significant (P = 0.9001 and P = 0.5438).
Now time for the ANOVA:
Source
DF
Type III SS
Mean Square F Value Pr > F
Gene
1 1.2443457E-6 1.2443457E-6
86.45 <.0001
Cultivar
3 3.5229784E-6 1.1743261E-6
81.59 <.0001
Gene*Cultivar
3
7.211971E-7 2.4039903E-7
16.70 <.0001
Block
3 1.9685005E-6 6.5616684E-7
45.59 <.0001
The data suggest that there is a significant effect on yield due to the gene (P < 0.0001) and the
cultivar genetic background (P < 0.0001). There is significant gene by cultivar genetic
background interactions (P < 0.0001) and there were significant block effects (P < 0.0001).
3.4 [6 points] Before starting the experiment the investigator formulated questions of interest about the
significant genetic background effects and interactions that may arise. Please test the following:
Contrast
DF
Contrast SS
Mean Square F Value Pr > F
diff. btw. African
1
1.124603E-8
1.124603E-8
0.78 0.3872
diff. Btw. C. American
1 3.4573416E-7 3.4573416E-7
24.02 <.0001
diff. btw Af and Am
1 3.1298497E-6 3.1298497E-6
217.45 <.0001
Gene
1 1.2443457E-6 1.2443457E-6
86.45 <.0001
int. gene by African
1 1.0701751E-7 1.0701751E-7
7.44 0.0130
int. gene by C. American
1 3.2820811E-7 3.2820811E-7
22.80 0.0001
int. gene by btw Af and Am
1 2.6331091E-7 2.6331091E-7
18.29 0.0004
3.4.1. Is there a difference in yield between the African and C. American cultivars?
Yes P<0.0001
3.4.2
Is there a difference in yield between the two African cultivars?
No P=0.3872 (because the effects are crossed for S and T and they cancel each other)
3.4.3
Is there a difference in yield between the two C. American cultivars?
Yes P<0.0001
3.4.4
Is the effect of the gene on yield different between the African and C. American cultivars?
Yes P=0.0004
3.4.5
Is the effect of the gene on yield different between the African cultivars (i.e. between cultivars 1 and 2)?
Yes P= 0.013 (this shows the crossed effect of the gene in S and T alleles)
3.4.6
Is the effect of the gene on yield different between the C. American cultivars (i.e. between cultivars 3 & 4)?
Yes P=0.0001
[3 points] Please present line plots of Gene by cultivar interaction for both the original and
transformed data. Based on these graph, and the significance of the ANOVAs and contrasts
of transformed and untransformed data indicate which of the following statements reflect the
reality and why:
The Tolerant allele increases yield relative to the susceptible allele in at least some cultivars
CORRECT ONE
The Tolerant allele decreases yield relative to the susceptible allele in at least some cultivars
The Tolerant allele increases yield relative to the susceptible allele in all cultivars
The Tolerant allele decreases yield relative to the susceptible allele in all cultivars
Original and transformed data interaction plots
T
he interactions are observed in both transformed and untransformed data, so there is a different
response of gene by cultivar (and at least one of the interactions is significant in the untransformed
data and all the interactions are significantly different in the transformed data).
These graphs show the interactions between varieties within each the African and Central American
varieties
Question 4
Data Gibberellin;
Input Chamber GA $ Light SeedWgt X;
Z = SeedWgt + 2.38*(X-17.5);
Cards;
1
C
8
111.28
15
2
C
8
111.12
14
3
C
8
115.47
13
1
C
10
115.89
14
2
C
10
112.04
15
3
C
10
115.57
14
1
C
12
118.89
13
2
C
12
115.17
14
3
C
12
113.99
15
1
G3
8
102.18
19
2
G3
8
98.35
20
3
G3
8
97.41
20
1
G3
10
103.71
20
[30 points]
2
G3
10
101.21
19
3
G3
10
102.88
20
1
G3
12
99.16
21
2
G3
12
99.47
20
3
G3
12
103.37
19
1
G4
8
117.09
13
2
G4
8
109.19
15
3
G4
8
111.10
14
1
G4
10
119.23
13
2
G4
10
115.30
14
3
G4
10
118.30
13
1
G4
12
113.31
15
2
G4
12
110.86
15
3
G4
12
117.75
13
1
G4&3 8
92.40
23
2
G4&3 8
91.89
22
3
G4&3 8
96.94
21
1
G4&3 10
98.45
22
2
G4&3 10
92.23
23
3
G4&3 10
95.46
22
1
G4&3 12
94.50
23
2
G4&3 12
94.76
22
3
G4&3 12
97.72
22
;
Proc GLM;
Title 'One-way ANOVAs for X and SEEDWGT';
Class Chamber Light GA;
Model X SeedWgt = Chamber Light Chamber*Light GA GA*Light;
Test h=Chamber e= Chamber*Light;
Test h=Light
e= Chamber*Light;
Contrast 'G3'
GA -1 1 -1 1;
Contrast 'G4'
GA -1 -1 1 1;
Contrast 'G3*G4'
GA 1 -1 -1 1;
Proc GLM;
Title 'General regression';
Model SeedWgt = X;
Proc GLM Order = Data;
Title 'The ANCOVA';
Class Chamber Light GA;
Model SeedWgt = Chamber Light Chamber*Light GA GA*Light X/ solution;
Test h=Chamber e= Chamber*Light;
Test h=Light
e= Chamber*Light;
LSMeans GA
Contrast
Contrast
Contrast
Contrast
Contrast
Light/ StdErr;
'Light Lineal'
'Light Quadratic'
'G3'
'G4'
'G3*G4'
Light -1 0 1 / e=Chamber*Light;
Light 1 -2 1 / e=Chamber*Light;
GA -1 1 -1 1;
GA -1 -1 1 1;
GA 1 -1 -1 1;
Proc GLM;
Title 'Homogeneity of slopes for light';
Class Chamber Light;
Model SeedWgt = Chamber Light X Light*X;
Proc GLM;
Title 'Homogeneity of slopes for GA';
Class Chamber Light GA;
Model SeedWgt = Chamber Light Chamber*Light GA GA*Light X GA*X;
Proc GLM Order = Data;
Title 'ANOVA on Z';
Class Chamber Light GA;
Model Z = Chamber Light Chamber*Light GA GA*Light;
Test h=Chamber e= Chamber*Light;
Test h=Light
e= Chamber*Light;
Output out = PRz p = Pred r = Res;
Proc Univariate Data = PRz normal;
Title 'Normality of residuals';
Var Res;
Proc Plot Data = PRz;
Plot Res*Pred = GA;
Proc GLM;
Title
Class
Model
Means
Proc GLM;
Title
Class
Model
Means
Run;Quit;
'Homogeneity of variances';
GA;
Z = GA;
GA / hovtest = Levene;
'Homogeneity of variances';
Light;
Z = Light;
Light / hovtest = Levene;
1) [2 points] Describe in detail the design of the experiment.
Design:
RCBD with 3 blocks (Chamber models). Split-plot with photoperiod as main plot and GA
levels as subplot (GA levels are itself organized as a 2x2 factorial).
Response Variable:
Weight of seeds from one complete tray (in grams)
Experimental Unit:
a) Mainplot: photoperiod (assigned to a chamber of a specific model). b) Subplot: one tray
of 100 Arabidopsis plants.
Class
Variable
1
2
3
Number of
Levels
3
3
4
Block or
Treatment
Block
Treatment
Treatment
Subsamples?
Covariable?
NO
Yes
Fixed or
Random
Random
Fixed
Fixed
Description
Chamber model
Photoperiod (main plot)
Combinations of GA3 and GA4
Level of disease
2) [5 points] Run the ANOVA and ANCOVA designs and compare the results. Use the
correct experimental design and correct error terms in both of them!
One-way ANOVA for covariable X
Class
Levels Values
Chamber
3 1 2 3
Light
3 8 10 12
GA
4 C G3 G4 G4&3
Number of Observations Read 36
Number of Observations Used 36
Dependent Variable: X
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
17
473.3333333
27.8431373
Error
18
13.6666667
0.7592593
Corrected Total 35
487.0000000
R-Square Coeff Var Root MSE
0.971937
Source
36.67 <.0001
X Mean
4.979171 0.871355 17.50000
DF Type III SS Mean Square F Value Pr > F
Chamber
2
2.1666667
1.0833333
1.43 0.2660
Light
2
0.5000000
0.2500000
0.33 0.7237
Chamber*Light
4
0.8333333
0.2083333
0.27 0.8906
GA
3 468.1111111 156.0370370
205.51 <.0001
Light*GA
6
1.7222222
0.2870370
0.38 0.8834
Contrast DF Contrast SS Mean Square F Value Pr > F
G3
1 441.0000000 441.0000000
G4
1
11.1111111
11.1111111
14.63 0.0012
G3*G4
1
16.0000000
16.0000000
21.07 0.0002
580.83 <.0001
Tests of Hypotheses Using the Type III MS for Chamber*Light as an Error Term
Source
DF
Type III SS
Mean Square
F Value
Pr > F
Chamber
2
2.16666667
1.08333333
5.20
0.0772
Light
2
0.50000000
0.25000000
1.20
0.3906
Level of GA
N
Mean
C
SeedWgt
X
Std Dev
Mean
Std Dev
9 14.1111111 0.78173596 114.380000 2.54248402
Level of GA
N
SeedWgt
X
Mean
Mean
Std Dev
Std Dev
G3
9 19.7777778 0.66666667 100.860000 2.32896436
G4
9 13.8888889 0.92796073 114.681111 3.68866658
G4&3
9 22.2222222 0.66666667
94.927778 2.44033183
GA affects the disease (no effect of photoperiod). G3 increases the disease relative to the control
and the effect of G3 is enhanced by the presence of G4 (significant interaction G3xG4 on X).
Therefore, we need to be careful in the interpretation of the effect of GA on seed-weight, since
the effect can be mediated by the effect of GA on the disease and the effect of the disease on
seed-weight.
One-way ANOVAs for SEEDWGT
Dependent Variable: SeedWgt
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
17
2809.617439
165.271614
Error
18
91.757583
5.097644
Corrected Total 35
2901.375022
32.42 <.0001
R-Square Coeff Var Root MSE SeedWgt Mean
0.968374
Source
2.125740 2.257796
106.2122
DF Type III SS Mean Square F Value Pr > F
Chamber
2
65.876772
32.938386
6.46 0.0077
Light
2
55.974606
27.987303
5.49 0.0138
Chamber*Light
4
11.786844
2.946711
0.58 0.6823
GA
3 2649.776778
883.258926
173.27 <.0001
Light*GA
6
4.367073
0.86 0.5442
26.202439
Contrast DF Contrast SS Mean Square F Value Pr > F
G3
1 2491.008100 2491.008100
G4
1
71.346178
71.346178
14.00 0.0015
G3*G4
1
87.422500
87.422500
17.15 0.0006
488.66 <.0001
Tests of Hypotheses Using the Type III MS for Chamber*Light as an Error Term
Source
DF
Type III SS
Mean Square
F Value
Pr > F
Chamber
2
65.87677222
32.93838611
11.18
0.0230
Light
2
55.97460556
27.98730278
9.50
0.0303
The ANOVA indicates significant effects of Chamber (block), light and GA with no interaction
The ANCOVA
Dependent Variable: SeedWgt
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
18
2887.078513
160.393251
Error
17
14.296509
0.840971
Corrected Total 35
2901.375022
190.72 <.0001
R-Square Coeff Var Root MSE SeedWgt Mean
0.995073
Source
0.863408 0.917045
106.2122
DF Type III SS Mean Square F Value Pr > F
Chamber
2 36.55569866 18.27784933
21.73 <.0001
Light
2 63.67956537 31.83978269
37.86 <.0001
Chamber*Light
4
3.46479180
0.86619795
1.03 0.4202
GA
3
0.29693662
0.09897887
0.12 0.9485
Light*GA
6
4.35938952
0.72656492
0.86 0.5404
X
1 77.46107398 77.46107398
92.11 <.0001
Tests of Hypotheses Using the Type III MS for Chamber*Light as an Error Term
Source
DF
Type III SS
Mean Square
F Value
Pr > F
Chamber
2
36.55569866
18.27784933
21.10
0.0075
Light
2
63.67956537
31.83978269
36.76
0.0027
Parameter
Estimate
Standard Error t Value Pr > |t|
Intercept 149.4660366 B
5.47521430
27.30
<.0001
X
0.24806143
-9.60
<.0001
-2.3807317
Slope -2.38. A negative slope indicates that the increase of the disease level decreases the see-weight.
Least Squares Means
GA
SeedWgt LSMEAN Standard Error Pr > |t|
C
106.311965
0.894504
<.0001
G3
106.282778
0.642416
<.0001
G4
106.084024
0.946498
<.0001
G4&3
106.170122
1.210629
<.0001
Interpret the results of the ANOVA and ANCOVA. Explain any difference in the two analyses.
Based on the ANCOVA results answer the following questions
Once the effect of the disease is removed the effect of GA on seed-weight disappears. This
indicates that the observed effect of GA on seed-weight was an indirect effect of the effect of GA
on the disease!
The effect of light is highly significant. Light does not affect the disease but still affects seed
weight confirming that photoperiods of 10 or 12 h of light increase seed yield relative to 8 h. The
response is not lineal (based on the means it seems to saturate it effects at 10h.
3.1. [2 points] Are there significant differences in seed-weigh among different GA levels?
No. They disappear in the ANCOVA model P=0.9485
3.2. [2 points] Are there significant interactions for seed-weight between the GA3 and GA4
effects?
Not in the ANCOVA model. P=0.8997
Dependent Variable: SeedWgt
Contrast DF Contrast SS Mean Square F Value Pr > F
G3
1
0.00021905
0.00021905
0.00 0.9873
G4
1
0.14396681
0.14396681
0.17 0.6842
G3*G4
1
0.01377586
0.01377586
0.02 0.8997
3.3. [2 points Are there significant differences in seed-weigh among different photoperiods?
Indicate P
Yes P=0.0027
3.4. [2 points Are the differences in photoperiod lineal?
Light SeedWgt LSMEAN Standard Error Pr > |t|
8
104.336606
0.265534
<.0001
10
107.324106
0.265534
<.0001
12
106.975955
0.267937
<.0001
Tests of Hypotheses Using the Type III MS for Chamber*Light as an Error Term
Contrast
DF
Contrast SS
Mean Square
F Value
Pr > F
Light Lineal
1
40.68075737
40.68075737
46.96
0.0024
Light Quadratic
1
22.05143729
22.05143729
25.46
0.0073
The differences in photoperiod are significant but not
lineal (significant quadratic response P=0.0073).
Seed-weight is higher at 10-12 h than at 8h, but there
are no differences between 10 and 12 h suggesting
that 10 h is sufficient to saturate the photoperiod response.
3.5. [2 points Are the differences in photoperiod different at the different GA treatments?
No. there is no significant interaction between light and GA treatments in the ANOVA.
P=0.5404
4) Answer the following questions:
4.1. [2 points] Are the slopes homogeneous within GA treatments and within photoperiods?
Homogeneity of slopes for light
Dependent Variable: SeedWgt
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
7
2878.691338
411.241620
Error
28
22.683684
0.810132
Corrected Total 35
2901.375022
507.62 <.0001
R-Square Coeff Var Root MSE SeedWgt Mean
0.992182
Source
0.847429 0.900073
106.2122
DF Type III SS Mean Square F Value Pr > F
Chamber
2
37.267363
18.633681
23.00 <.0001
Light
2
4.015414
2.007707
2.48 0.1021
X
1 2751.801118 2751.801118 3396.73 <.0001
X*Light
2
0.153351
0.076676
0.09 0.9100
Slopes of seed-weigh vs disease are homogeneous for the three photoperiods
Homogeneity of slopes for GA
Dependent Variable: SeedWgt
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
21
2891.512949
137.691093
Error
14
9.862074
0.704434
Corrected Total 35
2901.375022
195.46 <.0001
R-Square Coeff Var Root MSE SeedWgt Mean
0.996601
Source
0.790216 0.839306
106.2122
DF Type III SS Mean Square F Value Pr > F
Chamber
2 32.93884880 16.46942440
23.38 <.0001
Light
2 65.17865884 32.58932942
46.26 <.0001
Chamber*Light
4
6.16141440
1.54035360
2.19 0.1235
GA
3
4.42364106
1.47454702
2.09 0.1471
Light*GA
6
5.44555607
0.90759268
1.29 0.3240
X
1 72.00482116 72.00482116
102.22 <.0001
X*GA
3
4.43443572
1.47814524
2.10 0.1464
Slopes of seed-weigh vs disease are homogeneous for the four GA treatments
4.2 [2 points] Are the residuals from the ANCOVA model normally distributed?
ANOVA on Z
Dependent Variable: Z
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
17
109.9041056
6.4649474
Error
18
14.2965167
0.7942509
8.14 <.0001
Source
DF Sum of Squares Mean Square F Value Pr > F
Corrected Total 35
124.2006222
R-Square Coeff Var Root MSE
0.884892
Source
Z Mean
0.839082 0.891208 106.2122
DF Type III SS Mean Square F Value Pr > F
Chamber
2 37.30090556 18.65045278
23.48 <.0001
Light
2 64.04677222 32.02338611
40.32 <.0001
Chamber*Light
4
3.50444444
0.87611111
1.10 0.3853
GA
3
0.29731111
0.09910370
0.12 0.9442
Light*GA
6
4.75467222
0.79244537
1.00 0.4565
Tests of Hypotheses Using the Type III MS for Chamber*Light as an Error Term
Source
DF
Type III SS
Mean Square
F Value
Pr > F
Chamber
2
37.30090556
18.65045278
21.29
0.0074
Light
2
64.04677222
32.02338611
36.55
0.0027
Almost identical to the ANCOVA! This indicates that the adjusted values are correct.
Normality of residuals
The UNIVARIATE Procedure
Tests for Normality
Test
Statistic
p Value
Shapiro-Wilk W 0.970054 Pr < W 0.4270
Plot of Res*Pred.
1.5 ˆ
‚
‚
‚
‚
‚
‚
1.0 ˆ
‚
‚
‚
‚
‚
‚
0.5 ˆ
‚
‚
Res ‚
‚G
‚
‚
0.0 ˆ
Symbol is value of GA.
G
G
G
G
C
G
G
C
G
C
G
G
G
G G
C
G
‚
‚
‚
‚
‚
‚
-0.5 ˆ
‚
‚
‚
‚
‚
‚
-1.0 ˆ
‚
‚
‚
‚
‚
‚
-1.5 ˆ
G
G
G
G
G
C
C
C G
C
G
G
G
C
G
G
G
Šˆƒƒƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒƒƒˆƒ
103
104
105
106
107
108
109
110
Pred
Residuals of the ANOVA are normally distributed
4.3 [2 points] Are the variances for photoperiod and for GA treatments homogeneous?
Homogeneity of variances
Dependent Variable: Z
Levene's Test for Homogeneity of Z Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
GA
Error
3
6.4302
2.1434
32
528.1
16.5027
0.13 0.9416
Levene's Test for Homogeneity of Z Variance
ANOVA of Squared Deviations from Group Means
Source
DF Sum of Squares Mean Square F Value Pr > F
Light
Error
2
5.6965
2.8482
33
89.7145
2.7186
Level of
Light
N
1.05 0.3621
Z
Mean
Std Dev
8
12 104.336667 1.33350075
10
12 107.324167 1.54117699
12
12 106.975833 1.14676984
Both light and GA show homogeneous variances.
4.4 [2 points] Is the regression between seed-weight and disease significant? What is the
average slope? Explain what does that slope and its sign mean in terms of seed-weight and
disease levels.
General regression
Dependent Variable: SeedWgt
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
1
Error
34
124.169230
Corrected Total 35
2901.375022
2777.205792 2777.205792
760.45 <.0001
3.652036
R-Square Coeff Var Root MSE SeedWgt Mean
0.957203
Source DF
X
1.799256 1.911030
106.2122
Type I SS Mean Square F Value Pr > F
1 2777.205792 2777.205792
Parame
ter
Estimate
Standar
d Error
t Val
ue
Pr > |
t|
Interc
ept
148.0027
253
1.54855
700
95.57
<.0001
X
2.388028
7
0.08659
704
27.58
<.0001
Significant regression suggests that it might
be worth adjusting for this covariable. The
increase of the disease is associated with a
decrease in seed-weight
760.45 <.0001
4.5 [2 points] Is the disease independent of the treatments?
No. The ANOVA of the covariable indicate a significant GA effect
4.6 [3 points] What factors affect significantly the level of disease? Is there any significant
interaction? How can this result help you to understand the differences between the ANOVA
and ANCOVA results?
GA3 presence results in an increase in disease severity, which is further increased in the presence
of the two GA forms (GA4&3). This is an interesting discovery that can generate a new grant!