Field Plot Technique CSS 590
Second Midterm Exam
Winter, 2010
1) The residual plot below was obtained from a yield trial of 112 barley varieties. The
experimental design was an RBD with 2 blocks.
Heading Date in Barley
8
6
4
Residual
10 pts
Name____KEY______________
2
0
-2
-4
-6
-8
156
158
160
162
164
166
168
170
172
Predicted
How would you interpret this graph? If this were your own trial, what steps would you
take to address any concerns you have about the data?
There appear to be two outliers on the graph, which probably indicates one
observation was recorded incorrectly (which gives the other replication a residual of
equal magnitude that is opposite in sign). I would start with the field book to see if
there was an error with data input into the computer. I would also check my notes
and talk to anyone familiar with the trial to find out if there was something peculiar
about one of those plots (e.g., a planting error, extreme stress, or mechanical
damage caused an atypical flowering time). If there is no plausible explanation for
the discrepancy I would consider dropping the data point as if it were a missing plot.
The likelihood of obtaining such an observation due to chance is extremely small,
and the variation among the two outliers would greatly inflate the estimate of
experimental error.
Common transformations are not likely to help in this example because the
heterogeneity of variance does not follow a pattern typical of known distributions
(e.g. a binomial or Poisson distribution).
The pattern that is observed among the rest of the residuals in this data set is not
really a problem. Because the data are measured in days, observed values must be
1
recorded in whole integers. With only two replications, genotypic predictions will
either be in whole units or in half units. Block effects will also be included in predicted
values, but that will only add or subtract a single constant from the averages for the
treatments (genotypes). The pattern results from the fact that the predicted values
can only take on a limited number of possible values in this experiment. This is fairly
common for treatments that are class variables.
2) An animal scientist would like to determine if three different species of pasture grass
affect milk yield of Jersey cows in Australia. She would like to use the individual cows
as blocks to control variation among animals. She also knows that milk yield varies
throughout the year, so she decides to use time of year as an additional blocking
factor. She intends to use a Latin Square Design. Each cow is individually fed equal
quantities of pasture grass.
a) Show one possible randomization for a Latin Square Design by assigning the
pasture grasses (A,B, and C) to the experimental units below.
Cow
Period
1
2
3
Sept-Oct
1
B
C
A
Nov-Dec
2
A
B
C
Jan-Feb
3
C
A
B
6 pts
8 pts
b) Provide a skeleton ANOVA for this experiment, showing sources of variation and
degrees of freedom.
Source
Total
Cows
Period
Pasture
Error
t2-1
t-1
t-1
t-1
(t-1)(t-2)
df
8
2
2
2
2
c) Assume that the means for the pastures are A=16, B=30, and C=26 liters of milk
per cow per day. Calculate the Sums of Squares for Pastures from these means.
6 pts
Average = 24
SS = 3*[(16-24)2 + (30-24)2 + (26-24)2] = 64 + 36 + 4 = 3*104= 312
d) Do you think there will be adequate power in this experiment to detect
differences among the pastures? Can you suggest a way to increase power
without including additional treatments in a Latin Square Design?
She could replicate the squares using additional cows with the same three
pasture grasses.
2
3) You wish to evaluate the effect of three methods for pruning grapes (no pruning,
standard method, new method) and two fertilizer levels (low and high) on fruit yield.
Your experiment consists of all possible combinations of these two treatment factors
in a Randomized Complete Block Design. Write orthogonal contrast coefficients that
would address the following questions:
1. Does fertilizer level affect fruit yield?
2. Does pruning affect fruit yield?
3. Are yields with the New pruning method the same as with the Standard
method?
4. Is the difference between the New and Standard methods the same at both
levels of fertilizer?
Fill in the appropriate coefficients below the corresponding treatment combinations:
12 pts
Fertilizer:
Low
Low
Low
High
High
High
Pruning
None
Standard
New
None
Standard
New
1
-1
-1
-1
1
1
1
2
-2
1
1
-2
1
1
3
0
-1
1
0
-1
1
4
0
1
-1
0
-1
1
Contrast #
5 pts
a) Describe how you would verify that these contrasts are orthogonal to each other
(give one example).
The sum of cross-products of the coefficients for all pairs of contrasts should be
zero. For example, for contrast 1 vs contrast 2:
(-1)(-2) + (-1)(1) + (-1)(1) + (1)(-2) + (1)(1) + (1)(1) = 0
5 pts
b) Is this a complete set of orthogonal contrasts? If not, how many additional
contrasts would be required to make a complete set?
No, a complete set would consist of t-1 = 5 contrasts. We would need one more to
make a complete set.
3
4) A trial was conducted to determine oil yields of three varieties of meadowfoam at four
levels of nitrogen fertility. The trial was conducted in a Randomized Block Design
with four blocks. The results of a SAS analysis and means for all treatment
combinations are summarized below.
The GLM Procedure
Dependent Variable: oilyield
Source
DF
Sum of
Squares
Model
Error
Corrected Total
14
33
47
51468
20721
72189
Source
DF
Rep
NRate
Variety
Variety*NRate
Contrast
NRate linear
NRate quadratic
NRate cubic
Mean Square
F Value
Pr > F
3676
628
5.85
<.0001
Type III SS
Mean Square
F Value
3
3
2
6
7218
30929
10684
2637
2406
10310
5342
440
DF
Contrast SS
Mean Square
1
1
1
11546
19322
61
11546
19322
61
Oil yield of meadowfoam (lbs/acre)
Nitrogen lbs/acre
0
20
40
Variety
MF183
330
403
406
Ross
337
400
411
Starlight
326
357
377
Average
331
387
398
(See questions on next page)
4
Pr > F
3.83
16.42
8.51
0.70
0.0185
<.0001
0.0010
0.6515
F Value
Pr > F
18.39
30.77
0.10
0.0001
<.0001
0.7585
60
380
395
346
374
Average
380
386
351
372
4) (cont’d from previous page)
Give a brief interpretation of the results from this experiment.
How would you summarize and report your findings?
There is no interaction between variety and nitrogen fertility, so we can focus on the
main effects of variety and nitrogen.
There are significant differences among varieties, so I would report the means for
each variety and provide a standard error.
sY 
MSE

rn
628
 6.26
4*4
You could also determine which varieties are different using an LSD test (you would
have to guess at the t value because our tables only go up to 30 df)
t  2.03 (for 33 df)
sed = sqrt(2*628/16)
LSD = t*sed = 18.0
Ross and MF183 are both higher yielding than Starlight. Ross and MF183 are not
different from each other.
For nitrogen, the response is quadratic (based on the polynomial contrasts). Oil yield
increases to a maximum at 60 lbs N/acre and then declines. I would use Excel to
determine the appropriate quadratic equation using the 4 nitrogen means (averaged
across the 3 varieties.
Average nitrogen response for
meadowfoam varieties
450
400
oil yield lbs/acre
12 pts
350
2
y = -0.0502x + 3.7031x + 331.41
300
250
200
150
100
50
0
0
20
40
nitrogen lbs/acre
5
60
5) A forester wished to know the effects of vegetation management in the first four
years after planting on the subsequent growth of Douglas Fir trees. Four herbicide
treatments were applied to 400 m2 plots in five Randomized Complete Blocks. The
treatments were no herbicide, early vegetation control (years 1&2), late vegetation
control (years 3&4), and yearly herbicide sprays. In year 7, cores were taken from
the stems of three trees in each plot to measure ring segment length.
5 pts
a) What is the experimental unit in this study?
The experimental unit is the 400 m2 plot, because that is the unit to which each
treatment is applied. However, for the variable ring segment length, one could
argue that the group of 3 trees in each plot represents the experimental unit, so
either answer was acceptable.
The individual trees are sampling units (subsamples) within each plot.
12 pts
6 pts
b) Fill in the appropriate values for the degrees of freedom and Mean Squares in
the ANOVA table for this experiment.
Mean
Square
df
SS
Total
59
2110
Block
4
560
140
Herbicide
3
666
222
Block x Herbicide
12
324
27
Residual
40
560
14
c) Conduct an F test to determine if there are significant differences among the
Herbicide treatments. What are your conclusions?
Fcalc = 222/27 = 8.22
Fcritical (0.05, 3, 12) = 3.49
8.22>3.49 so reject H0 and conclude that there are differences among the
herbicide treatments
6
8 pts
6) Match each scenario described below with the data transformation that is most likely
to address anticipated problems with heterogeneity of variance. Choose from the
following:
i. square root transformation
ii. arcsin transformation
iii. log transformation
iv. no transformation required – expect homogeneity of variance
Variable Measured
a. Disease incidence (% infected plants) of crop
varieties that have been uniformly inoculated with
the pathogen. Some varieties are highly resistant
and others are highly susceptible to the disease.
Transformation
arcsin
b. Weed counts with varying herbicide treatments.
Values range from 2 to 250. The means are
proportional to the standard deviation.
log
c. Grain protein content with different fertilizer
treatments. Values range from 9.5 to 13.1
percent.
none
d. Insects caught in traps at varying elevations.
Results vary from 10 to 50 insects per week. The
means are proportional to the variance.
square root
7
F Distribution 5% Points
Denominator
Numerator
df
1
2
3
4
5
6
7
1 161.45 199.5 215.71 224.58 230.16 233.99 236.77
2 18.51 19.00 19.16 19.25 19.30 19.33 19.36
3 10.13
9.55
9.28
9.12
9.01
8.94
8.89
4
7.71
6.94
6.59
6.39
6.26
6.16
6.08
5
6.61
5.79
5.41
5.19
5.05
4.95
5.88
6
5.99
5.14
4.76
4.53
4.39
4.28
4.21
7
5.59
4.74
4.35
4.12
3.97
3.87
3.79
8
5.32
4.46
4.07
3.84
3.69
3.58
3.50
9
5.12
4.26
3.86
3.63
3.48
3.37
3.29
10
4.96
4.10
3.71
3.48
3.32
3.22
3.13
11
4.84
3.98
3.59
3.36
3.20
3.09
3.01
12
4.75
3.88
3.49
3.26
3.10
3.00
2.91
13
4.67
3.80
3.41
3.18
3.02
2.92
2.83
14
4.60
3.74
3.34
3.11
2.96
2.85
2.76
15
4.54
3.68
3.29
3.06
2.90
2.79
2.71
16
4.49
3.63
3.24
3.01
2.85
2.74
2.66
17
4.45
3.59
3.20
2.96
2.81
2.70
2.61
18
4.41
3.55
3.16
2.93
2.77
2.66
2.58
19
4.38
3.52
3.13
2.90
2.74
2.63
2.54
20
4.35
3.49
3.10
2.87
2.71
2.60
2.51
21
4.32
3.47
3.07
2.84
2.68
2.57
2.49
22
4.30
3.44
3.05
2.82
2.66
2.55
2.46
23
4.28
3.42
3.03
2.80
2.64
2.53
2.44
24
4.26
3.40
3.00
2.78
2.62
2.51
2.42
25
4.24
3.38
2.99
2.76
2.60
2.49
2.40
26
4.23
3.37
2.98
2.74
2.59
2.47
2.39
27
4.21
3.35
2.96
2.73
2.57
2.46
2.37
28
4.20
3.34
2.95
2.71
2.56
2.45
2.36
29
4.18
3.33
2.93
2.70
2.55
2.43
2.35
30
4.17
3.32
2.92
2.69
2.53
2.42
2.33
8
Student's t Distribution
(2-tailed probability)
df
0.40
0.05
0.01
1 1.376 12.706 63.667
2 1.061 4.303 9.925
3 0.978 3.182 5.841
4 0.941 2.776 4.604
5 0.920 2.571 4.032
6 0.906 2.447 3.707
7 0.896 2.365 3.499
8 0.889 2.306 3.355
9 0.883 2.262 3.250
10 0.879 2.228 3.169
11 0.876 2.201 3.106
12 0.873 2.179 3.055
13 0.870 2.160 3.012
14 0.868 2.145 2.977
15 0.866 2.131 2.947
16 0.865 2.120 2.921
17 0.863 2.110 2.898
18 0.862 2.101 2.878
19 0.861 2.093 2.861
20 0.860 2.086 2.845
21 0.859 2.080 2.831
22 0.858 2.074 2.819
23 0.858 2.069 2.807
24 0.857 2.064 2.797
25 0.856 2.060 2.787
26 0.856 2.056 2.779
27 0.855 2.052 2.771
28 0.855 2.048 2.763
29 0.854 2.045 2.756
30 0.854 2.042 2.750