ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Final TAKE HOME EXAM – FALL 08
Analysis - Disease Progress Curve :
Controlling papaya ring spot virus
Data kindly provided by Pedro Torres, graduate student in Statistics and TA for our course.
Data was used for a study that is summarized below.
 Title: Nonlinear models for analyzing disease progress
 Authors: Raúl Macchiavelli, Wilfredo Robles, Edwin Abreu and Alberto Pantoja.
College of Agricultural Sciences,Univ. of Puerto Rico – Mayagüez
Monitoring plant diseases
 Diseases are normally monitored over time,
assessing the amount of disease present in a
population of plants:
 “Disease Progress Curve”
 Represents an interpretation of all host, pathogen
and environmental effects occurring during an
epidemic (Campbell and Madden, 1990)
Disease Progress Curve - Models
Y
amount of disease
Proportion of diseased trees (out of 20)
t
time
 dY/dt absolute rate of disease increase (or decrease)
 Quantitative description of epidemics:
dY/dt vs. Y,
dY/dt vs. t
Logistic Disease Progress Curve
Gompertz Disease Progress Curve
dY
 rl Y (1  Y )
dt
1
Y
1  exp(  B  rl t )
 Y
log 
 1 Y
dY
 rg Y   log Y 
dt
Y  exp   B exp(rg t ) 
 log   log Y    log   log Y0   rg t
 Y0 

  rl t
  log 

 1  Y0 
Treatment is a Factor: categorical variable.
 Need to create 0/1 variables to identify
each treatment level.
Day is a quantitative explanatory variable.
 Effect of each Treatment level: Differences
in intercept of curve
 Day effect: slope of curve is not zero
 Treatment*Day: changes in slope for some
treatments.
Controlling papaya ring spot virus
Twenty plots were planted, each with 20 papaya plants
Controlling papaya ring spot virus
Twenty plots were planted, each with 20 papaya plants

There were 4 different treatments for the control of
certain insects (aphids) which are vectors of the virus
o
Control (no weeds) T
o
Plastic (black)
PC
o
Plastic (silver)
PP
o
Weeds
M

Each treatment was randomly assigned to 5 plots (CRD)




The experiment was monitored weekly
(8 weeks)
Each week, every plant was checked to see whether it
showed symptoms
Once the plant showed symptoms, it was classified as
diseased for the rest of the experiment
Analyzing the Data

The variable of interest is the disease index for
treatment i, time j and plot k:
number of plants with symptoms
Yijk 
20
1
ST 524

Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Problems with traditional approach
o Non normal distribution
o Non linear models
o Non constant Variances
o Observations of the same tree in different weeks are
dependent
o Observations on the same plot may not be
independent (contagion)
Traditional analysis
o Fit separate curves for each plot using linear
regression
o Compare slopes for different treatments
Generalized linear models
o
The linear component is defined like in traditional linear models:
i  xi '
o
A monotonic differentiable link function g describes how the expected value of y , E Y    , related to
the linear predictor:
g ( i )  xi '
o
The response variables y are independent and have a probability distribution from an exponential family.
This implies that the variance of the response depends on the mean through a variance function V:
var(Yi )   V ( i )
o
The dispersion parameter  is either assumed known (for example, for the binomial distribution,  = 1) or it
must be estimated to account for overdispersion.
a) Nonlinear fitting of observed proportion of diseased plants (out of 20 plants per plot)
using PROC NLIN in SAS

Need to set up dummy variables for treatments

Needs initial values of parameters

Logistic Fit
Yijk 
number trees diseased
20
 E Yijk  
     o   i  1  day j 90    2i  day j  90  * i 
log 
 1  E Yijk   ij
 

Yijk 
exp ij 
1  exp ij 
 eijk
ij   o   i  1  day j 90    2i  day j  90  * i 
  oi   1i   day ij 90 
Normal  0,  2 
eijk

Gompertz Fit
Yijk 


number trees diseased
20
 log   log E Yijk    ij   o   i  1  day j 90    2i  day j  90  * i 




Yijk  exp  exp  ij   eijk
ij   o   i  1  day j 90    2i  day j  90  * i 
  oi   1i   day ij 90 
eijk
Normal  0,  2 
2
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
b) Estimation of parameters and treatment effect of Generalized Linear Model with PROC
GENMOD in SAS and
o
o
o
o
No need for initial estimates
Use of CLASS statement sets up dummy variables (treatment effect) directly
Use of CONTRAST statement allows comparing treatment effect, equality of slopes, etc.
Use of ESTIMATE statement allows predictions, confidence intervals .
Yijk  number of diseased trees out of 20 per plot
binomial  20,  ij 
Yijk
 
log  ij
 1 
ij


     i  1day j   2i  day j  i 

E Yijk    ij
Var Yijk   20 ij 1   ij 
i = 1, 2, 3, 4 treatments
j=1, 2, 3, 4, 5, 6, 7, 8 timepoints
k = 1, 2, 3, 4, 5 blocks
c)
Note
Non linear models with normal residuals (NLIN) do not take into account actual
distribution or longitudinal nature.
 Because of contagion, Number of diseased trees (out of 20) is not a binomial
random variable, variance may not correspond to a binomial random variable.

E Yijk    ij
Var Yijk   20 ij 1   ij  
  overdispersion parameter

Non linear models fitting a binomial distribution with possibly overdispersion do not
take into account longitudinal nature.
 Overdispersion parameter  may be estimated as the square root of deviance
divided by its degrees of freedom. If the ratio (deviance/d.f) is greater than 1
indicates that overdispersion is present. Use scale= deviance option in PROC
GENMOD, MODEL statement, to fit a binomial with overdispersion. Standard errors
and tests are adjusted to account for extra variation
d) Estimation of the parameters of Generalized Linear Model with PROC NLINMIXED in SAS
 Repeated observations from the same plot are correlated, same random plot effect.
Yijk  number of diseased trees out of 20 per plot
Yijk | uik
binomial  20,  ij 
  ij
log 
 1 
ij

uik
E Yijk    ij

     i  1day j   2i  day j  i   uik

Normal  0,  2 
i = 1, 2, 3, 4 treatments
j = 1, 2, 3, 4, 5, 6, 7, 8 timepoints
k = 1, 2, 3, 4 blocks
Var Yijk   20 ij 1   ij 
 ij 

exp int   i   1day j   2i  day j  i 


1  exp int   i   1day j   2i  day j  i 
 No accounting for correlation between measurements within same tree.
3

ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Questions
PROC NLIN is used to fit two disease curves.
Q1.a.
Write down both estimated equations.
Logistic Fitting
Yijk  number of diseased trees out of 20 per plot
Yijk
binomial  20,  ij 
 
log  ij
 1 
ij


  0.2073  0.3215* id _ m  1.7844* id _ pc  2.3447 * id _ pp

 0.0593* day j  0.00231* day j * id _ m  0.0425* day j * id _ pc  0.0535* day j * id _ pp
Treatment=C
 
log  ij
 1 
ij


  0.2073  0.0593* day j

Treatment= M
  ij
log 
 1 
ij


  0.2073  0.3215* id _ m  0.0593* day j  0.00231* day j * id _ m

 0.2073  0.3215  0.0593* day j  0.00231* day j
 .1142  0.06161day j
Treatment = PC
 
log  ij
 1 
ij


  0.2073  1.7844* id _ pc  0.0593* day j  0.0425* day j * id _ pc

 0.2073  1.7844  0.0593* day j  0.0425* day j
 1.5771  0.1018* day j
Treatment=PP
 
log  ij
 1 
ij


  0.2073  2.3447 * id _ pp  0.0593* day j  0.0535* day j * id _ pp

 0.2073  2.3447  0.0593* day j  0.0535* day j
 2.1374  0.1128* day j
Gompertz Fitting
Yijk 


number trees diseased
20
 log   log E Yijk     o   i  1  day j 90    2i  day j  90  * i 


 0.6125  0.2169* id _ m  1.3102* id _ pc  1.7687 * id _ pp
0.0404*  day j 90 
0.000993*  day j  90  * id _ m 
0.0332*  day j  90  * id _ pc 
0.0439*  day j  90  * id _ pp 
4
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Treatment C


 log  log E Yijk    0.6125  0.0404*  day j 90 


Treatment M


 log   log E Yijk    0.6125  0.2169* id _ m  0.0404*  day j 90   0.000993*  day j  90  * id _ m 


 0.6125  0.2169  0.0404*  day j 90   0.000993  day j  90 
 0.3956  0.041393*  day j  90 
Treatment PC


 log   log E Yijk    0.6125  1.3102* id _ pc  0.0404*  day j 90   0.0332*  day j  90  * id _ pc 


 0.6125  1.3102  0.0404*  day j 90   0.0332*  day j 90 
 0.6977  0.0736*  day j 90 
Treatment PP


 log   log E Yijk    0.6125  1.7687 * id _ pp  0.0404*  day j 90   0.0439*  day j  90  * id _ pp 


 0.6125  1.7687  0.0404*  day j 90   0.0439*  day j 90 
 1.1562  0.0843*  day j 90 
5
ST 524
Solution Final Take Home Exam
Q1.b.
NCSU - Fall 2008
Due: 12/09/08
Calculate R2, measure of goodness of fit,

Error SS 
R 2  100 1 
%
USS
 Total  

For Logistic Fitting
1.6916 
R 2  100 1 
 96.97%
 55.88 
For Logistic Fitting
1.9131
R 2  100 1 
 96.58%
 55.88 
Q1.c.
which model, GOMPERTZ or Logistic, shows better fit?
Logistic fitting have a higher R-square, thus Logistic model is better fit for proportion of
diseased plants. Note that model may be improved since the approximate 95% Confidence
limits for parameters bm and rm includes value 0, which would indicate that there is no
differences between the logistic curves for treatments M and Control. Also, it should be of
interest to test whether both plastic treatments respond to same logistic curve, i.e., no
differences between these two treatments; and whether there are differences between
treatments M and C against the “plastic” treatments PP and PC.
PROC GENMOD is used for a fitting the number of diseased trees within each plot as a binomial random
variable with n=20 (trees) and the probability for a tree being diseased as a function of Treatment and
Day. Full model fits four slopes (for linear time effect), one for each treatment and four separate
intercepts (treatment effects).
Q1.d.
Would you recommend to adjust for overdispersion?.
Yes, Deviance/DF = 280.1112/152 = 1.8428 is greater than 1, indicating
overdispersion.
Contrasts test
☼ whether slopes for plastic covers have same effects
☼ whether slopes for control and weedy condition are the same.
☼ whether average slope for “plastic” is equal to average slope for “nonplastic”
☼ whether effects of plastic covers are the same
☼ whether effects of control and weedy treatments are the same.
☼ whether average effect for “plastic” treatments is equal to average effect for “nonplastic”
treatments
Q1.e.
Which model do you select, based on above results (PROC GENMOD)? Make reference
to contrasts. Write down conclusions. Indicate limitations.
After adjusting for overdispersion, Type III likelihood ratio test, in Table LR Statistics
for Type 3 Analysis indicates that Treatments, Day and their interaction are highly
significant (p-value < 0.0001). Contrasts show that there are not differences between
main effects of the two plastic treatments (p=0.3434), No differences between between
main effects of the M and C treatments (p =0.5173), significant differences between
these two groups (p<0.0001) , and similarly for the day effect: no differences in linear
slope between the two “plastic” treatments (p=0.4793), nor the two M and C
treaments (p=0.8031), but significant differences between the linear day effect of these
two groups (p<0.001). The recommended model should fit a common binomial model
(curve) for the two “plastic” treatments, and a common binomial model (curve) for the
two M and C treatments, with correction for overdispersion.
6
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Note: The scale parameter was estimated by the square root of DEVIANCE/DOF.
LR Statistics For Type 3 Analysis
Source
Num DF
Den DF
F Value
Pr > F
Chi-Square
Pr > ChiSq
treatid
3
152
14.72
<.0001
44.15
<.0001
day
1
152
1113.09
<.0001
1113.09
<.0001
day*treatid
3
152
7.21
0.0001
21.62
<.0001
Contrast Results
Contrast
Num DF
Den DF
F Value
Pr > F
Chi-Square
Pr > ChiSq
Type
PP=PC
1
152
0.90
0.3449
0.90
0.3434
LR
T=M
1
152
0.42
0.5183
0.42
0.5173
LR
(PPandPC= TandM)
1
152
43.18
<.0001
43.18
<.0001
LR
rpc=rpp
1
152
0.50
0.4804
0.50
0.4793
LR
rm=rt
1
152
0.06
0.8034
0.06
0.8031
LR
(rpc and rpp) = (rm and rt)
1
152
21.32
<.0001
21.32
<.0001
LR
PROC NLMIXED is used to fit a model taking into account the distribution of the number of diseased trees
within a plot as a binomial random variable with parameter  ij depending on the treatment and time of
measurement. Random block effects are also included in model. Full model fits separate slopes and
intercepts for each treatment group, while the second model fits two models with common intercept and
slope for treatments T and M and separate common intercept and slope for treatment PP and PC.
Q1.f.
Which model do you select, make reference to contrasts. Limitations.
Results from NLMIXED are similar to what we found in PROC GENMOD, significance of
contrasts tested indicate that there should be two models for the logits, one modeling logits
from the “plastic” treatments, and the other modeling logits from the M and C treatments. A
limitation of this modeling is that the repeated measures structure of the residual variance has
been modeled, although random effect associated with repetitions is included in the model,
while GENMOD does not include either random effects or the repeated measures structure for
the residual variation.
7
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Parameter Estimates
Parameter
Estimate
Standard
Error
DF
t Value
Pr > |t|
Alpha
Lower
Upper
Gradient
b0
0.3505
0.1498
159
2.34
0.0205
0.05
0.05468
0.6463
-0.0013
bm
-0.2966
0.2102
159
-1.41
0.1602
0.05
-0.7119
0.1186
-0.00035
bpc
-1.7488
0.2503
159
-6.99
<.0001
0.05
-2.2432
-1.2544
-0.00049
bpp
-2.2418
0.2734
159
-8.20
<.0001
0.05
-2.7818
-1.7018
0.000131
r
0.06896
0.005835
159
11.82
<.0001
0.05
0.05744
0.08049
0.030399
rm
0.001911
0.008116
159
0.24
0.8141
0.05
-0.01412
0.01794
0.029448
rpc
0.03107
0.01009
159
3.08
0.0024
0.05
0.01114
0.05099
-0.00962
rpp
0.03934
0.01088
159
3.61
0.0004
0.05
0.01785
0.06083
0.006952
sigma2
0.3813
0.1122
159
3.40
0.0009
0.05
0.1597
0.6030
0.000135
Contrasts
Label
Num DF
Den DF
F Value
Pr > F
PP=PC
1
159
2.63
0.1065
T=M
1
159
3.77
0.0539
rpc=rpp
1
159
0.44
0.5077
PPandPC = TandM
1
159
79.45
<.0001
slopes: (T,M) = (PP,PC)
1
159
98.96
<.0001
Final model is
Parameter Estimates
Parameter
Estimate
Standard
Error
DF
t Value
Pr > |t|
Alpha
Lower
Upper
Gradient
b0
0.2077
0.1088
159
1.91
0.0582
0.05
-0.00730
0.4226
-0.00242
bpp
-1.8373
0.1883
159
-9.75
<.0001
0.05
-2.2093
-1.4653
-0.0008
r
0.07018
0.004330
159
16.21
<.0001
0.05
0.06163
0.07873
-0.03194
rpp
0.03384
0.007520
159
4.50
<.0001
0.05
0.01899
0.04869
-0.02421
sigma2
0.4212
0.1183
159
3.56
0.0005
0.05
0.1876
0.6548
0.000523
8
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Yijk  number of diseased trees out of 20 per plot
Yijk
binomial  20,  ij 
 
log  ij
 1 
ij


  0.2077  1.8373* id _ pp +0.0718*  day j  90   0.03384*  day j  90  * id _ pp

Q1.g.
Write down the model for the proportion of diseased trees in a plot receiving a silver
plastic cover at day t.
 
log  ij
 1 
ij


 0.2077  1.8373* id _ pp+0.07018*  day j  90   0.03384*  day j  90  * id _ pp

 silver plastic
  0.2077  1.8373   0.07018  0.03384  *  day j  90 
 1.6296  0.10402*  day j  90 
Q1.h.
Interpret coefficients of model.
-1.6296 is the predicted value for logit at t=0, which means that at t=0, i.e., at day 90, the predicted
e1.6296
proportion of diseased plants per plot is
 0.1639 , 16.39% at day = 90
1 e1.6296
0.10402 is the increase in the logit when day increases one unit, i.e., the daily rate of increase on logit
is given by 0.10402, which indicates that the odds ratio is 1.1096, and a daily increase of 10.96%
Q1.i.
Write down the equation for the prediction of response in a plot with a silver plastic cover
at day 85. And similarly for a weedy plot at day 85.
  
log  ij 
 0.2077  1.8373* id _ pp+0.07018* day j  90  0.03384* day j  90 * id _ pp
 1  
ij  silver plastic ,t 85





  0.2077  1.8373   0.07018  0.03384  * 85  90 
 1.6296  0.10402* 85  90 
 2.1497
 silverplastict day 85
 
log  ij
 1 
ij

e2.1497

 0.1044 , 10.44%
1 e2.1497

 0.2077  0.07018*  day j  90 

 silver plastic ,t 85
 0.2077  0.07018*  85  90 
 0.2077  0.07018*  85  90 
 weedy day 85 
0.1432
 0.1432
e
 0.4543 , 45.43%
1 e0.1432
9
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Question 2
Scientific paper:
Effect of Plant Species and Environmental Conditions on Epiphytic Population Sizes of
Pseudomonas syringae and other Bacteria. R. D. O’Brien and S. E. Lindow. 1989. the
American Phytopathological Society. V. 79, No. 5, 1989,
Question 2 will Refer only to Experiment 1 in the above paper.
Please answer the following
Description
Q2.1.
Objective
9 strains * 4 plant species
10
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
dry + high light
72 hrs
Humid + low light
48 hrs
Uncontrolled outdoor
environment
12 days
Sampling unit: 15-40 individual leaves per plant (15-20 g. fresh weigh per plant)
Sampling periods:
☼ Before Inoculation
☼ Immediately Before Inoculation
☼ Immediately after inoculation
☼ After wet and dry growth chamber incubation
☼ Periodically during incubation of plants under field conditions.
o
11
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Response Variable
Bacterial Population size per gram fresh weight of leaves.
Log-transformed population size
Q2.3.
Indicate What Are The Different Experimental Units,
a. Main-Unit :
a run of the experiment at each environment
b. Sub-Unit:
a 20-cm diameter pot , one plant per pot
c. Sub-Sub Unit: a sample of 15-20 g fresh weigh leaves
Q2.4.
How Is A Block Defined?
a. There were 3-4 replicates for each experimental condition, a block can be
assumed a complete run of the experiment, thus there will be three to four
runs (?) No information about blocks in text.
Q2.5.
What Are The Factors And Their Type: Random Fixed,
a. Main-plot Factor: Environment: Humid+Dry; Humid+uncontrolled : Fixed
b. Sub-plot Factor:
i. Strain
: Random (random sampled selected)
ii. and Plant Species
: Fixed
c. Sub-sub-sub unit factor: Removal method
: Fixed
All Factors were considered fixed-effect factors.
Counting
Q2.6.
Number Of Blocks:
3-4
Q2.7.
Total Number Of Main-Units:
2 * (3-4) = 6-8
Q2.8.
Total Number Of Sub-Units:
2*(9*4)*(3-4) = 216-288
Q2.9.
Total Number Of Sub-Sub-Units:
2*2*(9*4)*(3-4) =432-576
Q2.10.
How Many Main-Unit Within Each Block:
2
Q2.11.
How Many Sub-Units Within Each Main-Unit:
9*4=36
Q2.2.
12
ST 524
Q2.12.
Solution Final Take Home Exam
How Many Sub-Sub-Units Within Each Sub-Unit:
Each block should have
2 main units
2*9*4 = 72 sub units
2*2*9*4 = 144 sub-sub units
NCSU - Fall 2008
Due: 12/09/08
2
Statistical Analysis
Q2.13.
Linear Model, based on above information.
Yijk  log  colony size 
Yij    Ei  Bm  aim  S j  Pk   S * P  jk   E * S ij   E * P ik   E * S * P ijk  b jkmi 
 Rl   E * R il   S * R  jl   P * R kl   S * P * R  jkl   E * S * R ijl   E * P * R ikl   E * S * P * R ijkl  eijklm
aim
Normal  0,  a2 
b jkm i 
Normal  0,  b2 
Normal  0,  e2 
eijklm
Q2.14.
Present The ANOVA Table, Sources Of Variation, Df, Ms If Possible,
Number of blocks = 4
MS
Block
DF
r=4
3
4.51
E(MS)
Fixed Effect
2
 e2  2r b2  72r a2  144 block
Environment
1
195.41
 e2  2 b2  72 a2  Q  E 
Error(a)
3
0.71
 e2  2 b2  72 a2
Strain
8
6.51
 e2  2 b2  Q  S 
Plant Species
3
6.53
 e2  2 b2  Q  P 
Strain*Plant Sp
24
0.65
 e2  2 b2  Q  S * P 
Env*Strain
8
2.52
 e2  2 b2  Q  E * S 
Env*PlantSp
3
2.75
 e2  2 b2  Q  E * P 
Env*Strain*PlantSp
24
0.50
 e2  2 b2  Q  E * S * P 
Error(b)
210
Removal
1
12.23
 e2  Q  R 
Env*Rem
1
4.75
 e2  Q  E * R 
Strain*Rem
8
 e2  Q  S * R 
Plant Sp*Rem
3
 e2  Q  P * R 
Strain*Plant Sp*Rem
24
 e2  Q  S * P * R 
Env*Strain*Rem
8
 e2  Q  E * S * R 
Env*Plant Sp*Rem
3
 e2  Q  E * P * R 
Env*Strain*Plant Sp*Rem
24
 e2  Q  E * S * P * R 
Residual
216
 e2
Total
575
Source
 e2  2 b2
13
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Random Effect: Strain and its interactions
Source
MS
E(MS)
Random Effects
Block
DF
r=4
3
4.51
2
 e2  2r b2  72r a2  144 block
Environment
1
195.41
 e2  2 b2  72 a2  Q  E 
Error(a)
3
0.71
 e2  2 b2  72 a2
Strain
8
6.51
 e2  2 b2  64 S2
Plant Species
3
6.53
 e2  2 b2  Q  P 
Strain*Plant Sp
24
0.65
 e2  2 b2  16 S3*P
Env*Strain
8
2.52
 e2  2 b2  32 E2*S
Env*PlantSp
3
2.75
 e2  2 b2  Q  E * P 
Env*Strain*PlantSp
24
0.50
 e2  2 b2  8 E2*S*P
Error(b)
210
Removal
1
12.23
 e2  Q  R 
Env*Rem
1
4.75
 e2  Q  E * R 
Strain*Rem
8
 e2  32 S2*R
Plant Sp*Rem
3
 e2  Q  P * R 
Strain*Plant Sp*Rem
24
 e2  8 S2*P*R
Env*Strain*Rem
8
 e2  16 E2*S*R
Env*Plant Sp*Rem
3
 e2  Q  E * P * R 
Env*Strain*Plant Sp*Rem
24
 e2  4 E2*S*P*R
Residual
216
 e2
Total
575
 e2  2 b2
14
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Yijk  log  colony size 
Yij    Ei  Bm  aim  s j  Pk   s * P  jk   E * s ij   E * P ik   E * s * P ijk  b jkmi 
 Rl   E * R il   s * R  jl   P * R kl   s * P * R  jkl   E * s * R ijl   E * P * R ikl   E * s * P * R ijkl  eijklm
aim
Normal  0,  a2 
b jkmi 
Normal  0,  b2 
eijklm
Normal  0,  e2 
sj
Normal  0,  s2 
 s * P  jk
Normal  0,  s2*P 
 E * s ij
Normal  0,  E2*s 
 E * s * P ijk
 s * R  jl
 s * P * R  jkl
 E * s * P * R ijkl
Normal  0,  E2*s*P 
Normal  0,  s2*R 
Normal  0,  s2*P*R 
Normal  0,  E2*s*P*R 
Sub-Sub-Plot Factor And Interactions Was Analyzed As A Randomized Complete Block
Design. Indicate The Number Of Blocks That Should Be Considered.
Each combination of BLOCK*E*S*P is considered a block when analyzing
the sub-sub-plot factor. Thus there are r*2*9*4=72r (216 for r=3, or 288
for r=4)
Q2.15.
Compare Your ANOVA Table With Table 2. Experiment 1. If any discrepancies
are observed, please explain them.
15
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
Whole plots = 2*4 = 8
df=8-1 = 7
Sub plots = 2*9*4 = 72
df=72-1=71-1missing=70
Error (b) = 423
From table: 210+216 = 426-3=423; Error (b) and Residual Error were
combined
Total = 575 – missing obs = 500 (75 obs missing)
In second anova, for Removal, Error= 494 = 423 + 3 + 2*(8+3+24) –(1 + 1) , Strain, Plant
and their interactions was pooled with Error
Q2.16.
Describe An Alternative Plan Of Statistical Analysis For The Described Model.
Why the full model can not run?
16
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
1. Run a separate model for each Removal method
Source
DF
r=4
MS
E(MS)
Block
3
2
 e2  2r b2  72r a2  144 block
Environment
1
 e2  2 b2  72 a2  Q  E 
Error(a)
3
 e2  2 b2  72 a2
Strain
8
 e2  2 b2  Q  S 
Plant Species
3
 e2  2 b2  Q  P 
Strain*Plant Sp
24
 e2  2 b2  Q  S * P 
Env*Strain
8
 e2  2 b2  Q  E * S 
Env*PlantSp
3
 e2  2 b2  Q  E * P 
Env*Strain*PlantSp
24
 e2  2 b2  Q  E * S * P 
Residual
210
 e2  2 b2
Total
287
2. The two removal methods are correlated, since they are measures taken from the same 1540 fresh individual leaves, create a new variable defined as the difference in population size
between these two methods, and run model in 1.
3. Run analysis for each ENVIROMENT separate, include REMOVAL METHOD as repeated
measures (pseudo-replication).
17
ST 524
Solution Final Take Home Exam
NCSU - Fall 2008
Due: 12/09/08
4. Q3. This question ask you to write down a description of your research project,
indicating
Q3.1.
Objective
Q3.2.
Response Variable
Q3.3.
Experimental design. Detailed description
a. Indicate What Are The Different Experimental Units,
i. Main-Unit
ii. Sub-Unit (if any)
iii. Sub-Sub Unit (if any)
b. How Is A Block Defined?
c. What Are The Factors And Their Type: Random Fixed,
Q3.4.
Present the Analysis of Variance table
a. Sources of Variation (SOV)
b. Degrees of Freedom
c. Expected Mean Squares
d. F test for each SOV
Q3.5.
What type of statistical tests do you plan to carried on results to answer
your research questions: pairwise mean comparisons, contrasts, orthogonal
polynomial contrasts, curve fitting, etc
Q3.6.
Do you have repeated measures, how do you plan to analyze them?
References
https://www.crops.org/publications/pdfs/CESGuide.pdf
https://www.crops.org/publications/pdfs/cinstauthmans.pdf
https://www.crops.org/publications/pdfs/jpr-instructions.pdf
From North Dakota Agricultural Exp Station- Research Project Guidelines
Procedures:
This section is to provide a general design of the project. To begin, re-state each of
the objective statements followed by a description of the procedures/methods for that
objective. The procedure statements should show that the research needs and plans have
been considered carefully and the proposed work has the potential to provide data and
information which will permit accomplishing the objectives.
While the details of the experimental design do not need to be specified, provide
sufficient information to indicate that an appropriate design is planned.
18