Name:
KEY
CSS 590 – Experimental Design in Agriculture
Second Midterm
Winter 2008
14 pts
1) Name a statistical test or briefly describe a diagnostic test that can be used to determine
if the assumptions for the ANOVA listed on the left have been met (use a different
answer for each of the assumptions).
Assumption
Statistical or Diagnostic Test
Homogeneity of Variance
Levene’s Test or Bartlett’s test
Effects in the model are
additive
Tukey’s Test for Additivity
Residuals are randomly
and independently
distributed
Residual plots – the points should be evenly scattered
above and below zero, without any obvious patterns in
relation to the predicted values
Residuals are normally
distributed
Normal probability plots – the residuals should graph on
a straight line
1
2) You are studying the effect of three cultivation methods on fresh weight of spinach. You
decide to use a Randomized Block Design with 5 Blocks. This is your first experience
collecting data of this sort, and you are not sure about the optimum harvest time nor the
appropriate plot size needed to obtain an acceptable level of precision. On one harvest
date you collect samples from two quadrats in each plot, and you then ask your assistant
to enter the data and calculate the ANOVA. He provides you with the output below:
The GLM Procedure
Dependent Variable: weight
8 pts
Source
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
6
16.35733333
2.72622222
7.03
0.0002
Error
23
8.92266667
0.38794203
Corrected Total
29
25.28000000
R-Square
Coeff Var
Root MSE
weight Mean
0.647046
11.12232
0.622850
5.600000
Source
DF
Type III SS
Mean Square
F Value
Pr > F
Block
Method
4
2
4.16333333
12.19400000
1.04083333
6.09700000
2.68
15.72
0.0570
<.0001
a) Looking at the degrees of freedom and F ratios, you realize that the analysis has been
done incorrectly. Draw a skeleton ANOVA with sources of variation and degrees of
freedom for the correct analysis, and indicate what F ratios would be required to make
an appropriate test for cultivation methods (hint: there may be more than one correct
answer – choose one).
Option 1 – analyze means
Source
df
SS
Block
4
Method
2
SSM
Error
8
SSE
Total
14
Option 2 – include subsamples
F ratio
Source
MSM/MSE
SS
F ratio
Block
4
Method
2
SSM
MSM/MSE
Block*Method
8
SSE
MSE/MSS
Sampling Error 15
SSS
Total
2
df
29
6 pts
b) How would you justify your answer to your assistant, and explain why his analysis
was not valid?
The residual in his analysis includes variation among plots treated alike (true
experimental error) and variation among quadrats within each plot. Pooling the two
sources of variation together will give you too many degrees of freedom in the error
term, and will probably provide an estimate of error that is too small, thereby inflating the
Type I error rate. One approach is to calculate the means for each plot and perform
ANOVA on the means. The other option is to keep the data for individual quadrats in the
data set, but specify that the appropriate error for testing methods is the block*method
interaction. In an RBD, this term represents the error among experimental units to which
the treatments were randomly applied.
3
3) A researcher is evaluating the effect of 5 soil mixes on growth of strawberries in the
greenhouse. A heater on the north wall blows warm air into the greenhouse, and the
windows to the outside on the east wall provide extra light in the morning. Consequently,
she decides to use the benches, and the position of pots on each bench as blocking
factors in a Latin Square Design.
North
Heater
1
(Numbers indicate
the position of pots
on a bench)
2
East
3
Windows
4
5
3 pts
a) Which environmental factor is accounted for by the benches?
The variation due to light from the East Windows.
6 pts
b) Calculate the F statistics for Position and Bench in the ANOVA. Are the effects of
Position and Bench significant? (Use the tables at the end of the exams and indicate
how you reached your conclusion)
Source
df
SS
Total
24
338.16
Treatment
4
Position
Bench
Error
MS
F
180.56
45.14
15.53
4
80.56
20.14
6.93
4
42.16
10.54
3.63
12
34.88
2.91
The critical F with 4 and 12 df (=0.05) is 3.26. Both of the observed F values are
greater than 3.26, so we conclude that the effects of the bench and position on the
bench were significant.
c) Calculate the relative efficiency of the Latin Square Design compared to a
Randomized Block Design with Benches as Blocks.
RE 
MSP  (t  1)MSE 20.14  (5  1) * 2.91

 2.1858
t * MSE
5* 2.91
The Latin Square Design which accounts for benches and for position on the bench is
118% more efficient than the RBD with benches as the only blocking factor.
4
4) A Sugar Company conducted an experiment to compare two varieties of sugar cane in
combination with three levels of nitrogen (150, 210, and 270 lbs. N per acre
respectively). The experiment was run in 4 complete blocks.
a) Fill in the shaded cells to complete the following ANOVA. (There is an F table at the
end of this exam)
16 pts
ANOVA
Source
Blocks
Variety
Nitrogen
VxN
Error
df
3
1
2
2
15
SS
249
600
532
28
672
Total
23
2081
MS
83
600
266
14
44.8
F
13.393
5.938
0.313
F crit
4.54
3.68
3.68
Mean Yield (tons)
4 pts
N-150
N-210
N-270
Mean
Variety 1
66
72
75
71
Variety 2
54
61
68
61
Mean
60
66.5
71.5
66
b) Show how you could calculate the Sums of Squares for Varieties from the table of
means above (the value you obtain should agree with the number in the ANOVA
table)

SST  r *b j Yj  Y
4 pts

2
2
2
 4*3*  71  66    61  66    600


c) Is there a significant difference in yield among the varieties? Justify your answer.
The interaction between variety and nitrogen is not significant, therefore we can interpret
the main effects. The calculated F for varieties is greater than the critical value of 4.54,
so we reject the null hypothesis and conclude that there are differences among the
varieties. Those differences do not depend on the amount of nitrogen applied.
3 pts
d) Calculate a standard error for a variety mean.
sY 
MSE

rb
44.8
 1.93
4 *3
5
5) A plant breeder made selections for seed size in a population of pigeon peas, and
developed two experimental varieties. One had large seeds and the other had small
seeds. She wished to determine if seed size showed any association with early vigor,
growth and development of seedlings, and if the level of soil compaction would influence
that relationship. She applied 3 levels of compression (0, 10 and 20 kJm-3) to soils in
PVC tubes. The two varieties were planted in tubes at all three levels of compression.
Each treatment combination was replicated in 6 tubes, which were randomly arranged in
the greenhouse (making a total of 36 experimental units). The weights of seedlings in
each tube were measured at 3 weeks after planting.
10 pts
a) Write orthogonal contrast coefficients that would address the following questions,
and fill in the appropriate coefficients below the corresponding treatment
combinations in the table (there is a table of orthogonal polynomial coefficients at the
end of this exam).
1) Does seed size affect seedling weight?
2) Is there a linear relationship between soil compaction and seedling weight?
3) Is there a curvilinear relationship between soil compaction and seedling weight?
4) Do the varieties show a similar linear response to soil compaction?
5) Do the varieties show a similar curvilinear response to soil compaction?
Small
Seed Size
Large
Soil
Compaction
0
10
20
0
10
20
Means
81.3
44.5
38.5
86.1
65.9
52.4
L
1
-1
-1
-1
1
1
1
2
-1
0
1
-1
0
3
1
-2
1
1
4
1
0
-1
5
-1
2
-1
k2
SS(L)
40.1
6
1608.01
30.23
1
-76.5
4
8778.37
165.01
-2
1
37.5
12
703.12
13.22
-1
0
1
9.1
4
124.22
2.33
1
-2
1
-24.1
12
290.41
5.46
F
Contrast #
4 pts
b) Describe how you would verify that these contrasts are orthogonal to each other
(give one example).
The sums of crossproducts of the corresponding coefficients must equal zero. For
contrasts 1 and 2:
(-1)(-1)+ (-1)(0)+ (-1)(1)+ (1)(-1)+ (1)(0)+ (1)(1) = 1-1-1+1=0
6
(Problem 5 cont’d. – means are repeated here for ease of reference)
Small
Seed Size
9 pts
Large
Soil Compaction
0
10
20
0
10
20
Means
81.3
44.5
38.5
86.1
65.9
52.4
c)
Fill in the values for L, 2, and Sums of Squares for contrast #4 using the means in
the table (show your work here).
L = 81.3 – 38.5 - 86.1 + 52.4 = 9.1
2
2
2
2
2
   (1)  (1)  (1)  (1)  4
r*L2 6*(9.1)2
SSL 

 124.22
2
k
4

d) Interpret the results from this experiment. What can you say about the effects of
seed size and soil compaction on seedling growth? (Use the F table at the end of this
test and refer to specific contrasts to justify your answer)
There was no indication that blocks were used, so assume a CRD (or state that you
have assumed an RBD). The error df = 6*(6-1) = 30, so Fcritical for each contrast is 4.17.
All of the contrasts are significant except for #4.
Contrast #5 indicates that the response to soil compaction depends on seed size (which
is represented by the two varieties). Both varieties showed a linear decrease in seedling
weight with increased compaction levels, but the response for the small-seeded variety
was slightly curvilinear. On the average, the small-seeded variety had lower seedling
weight than the large-seeded variety, but there was little difference between the varieties
in uncompacted soils.
100
Seedling Weight
8 pts
80
60
small
40
large
20
0
0
5
10
15
Compaction kJm
7
20
-3
25
F Distribution 5% Points
Denominator
df
1
1 161.45
2 18.51
3 10.13
4
7.71
5
6.61
6
5.99
7
5.59
8
5.32
9
5.12
10
4.96
11
4.84
12
4.75
13
4.67
14
4.60
15
4.54
16
4.49
17
4.45
18
4.41
19
4.38
20
4.35
21
4.32
22
4.30
23
4.28
24
4.26
25
4.24
26
4.23
27
4.21
28
4.20
29
4.18
30
4.17
Student's t Distribution
Numerator
(2-tailed probability)
2
3
4
5
6
7
199.5 215.71 224.58 230.16 233.99 236.77
19.00 19.16 19.25 19.30 19.33 19.36
9.55
9.28
9.12
9.01
8.94
8.89
6.94
6.59
6.39
6.26
6.16
6.08
5.79
5.41
5.19
5.05
4.95
5.88
5.14
4.76
4.53
4.39
4.28
4.21
4.74
4.35
4.12
3.97
3.87
3.79
4.46
4.07
3.84
3.69
3.58
3.50
4.26
3.86
3.63
3.48
3.37
3.29
4.10
3.71
3.48
3.32
3.22
3.13
3.98
3.59
3.36
3.20
3.09
3.01
3.88
3.49
3.26
3.10
3.00
2.91
3.80
3.41
3.18
3.02
2.92
2.83
3.74
3.34
3.11
2.96
2.85
2.76
3.68
3.29
3.06
2.90
2.79
2.71
3.63
3.24
3.01
2.85
2.74
2.66
3.59
3.20
2.96
2.81
2.70
2.61
3.55
3.16
2.93
2.77
2.66
2.58
3.52
3.13
2.90
2.74
2.63
2.54
3.49
3.10
2.87
2.71
2.60
2.51
3.47
3.07
2.84
2.68
2.57
2.49
3.44
3.05
2.82
2.66
2.55
2.46
3.42
3.03
2.80
2.64
2.53
2.44
3.40
3.00
2.78
2.62
2.51
2.42
3.38
2.99
2.76
2.60
2.49
2.40
3.37
2.98
2.74
2.59
2.47
2.39
3.35
2.96
2.73
2.57
2.46
2.37
3.34
2.95
2.71
2.56
2.45
2.36
3.33
2.93
2.70
2.55
2.43
2.35
3.32
2.92
2.69
2.53
2.42
2.33
8
df
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0.4
0.05
0.01
1.376 12.706 63.667
1.061 4.303 9.925
0.978 3.182 5.841
0.941 2.776 4.604
0.920 2.571 4.032
0.906 2.447 3.707
0.896 2.365 3.499
0.889 2.306 3.355
0.883 2.262 3.250
0.879 2.228 3.169
0.876 2.201 3.106
0.873 2.179 3.055
0.870 2.160 3.012
0.868 2.145 2.977
0.866 2.131 2.947
0.865 2.120 2.921
0.863 2.110 2.898
0.862 2.101 2.878
0.861 2.093 2.861
0.860 2.086 2.845
0.859 2.080 2.831
0.858 2.074 2.819
0.858 2.069 2.807
0.857 2.064 2.797
0.856 2.060 2.787
0.856 2.056 2.779
0.855 2.052 2.771
0.855 2.048 2.763
0.854 2.045 2.756
0.854 2.042 2.750
9