CROP 590
Name:
KEY
Experimental Design in Agriculture
Second Midterm
Winter 2016
10 pts
1) You have just been hired by a seed company to evaluate the effectiveness of five
fungicides treatments on the yield of wheat. Last year, they conducted a similar
experiment and obtained a Mean Square Error for yield of 600 (measured in bu/acre).
They would like to be able to detect differences of 40 bushels/acre 75% of the time
using an alpha level of 0.05.
If you were to use 4 replications in a Randomized Complete Block Design, could you
expect to detect differences of this magnitude with the specified power and Type I error
levels? You may assume that this Seed Company always uses a standard plot size (X=1).
Use the t tables at the end of this exam and show your calculations to support your
conclusion. (Hint: there is more than one way to solve this problem)
t=
r=
dfe = (r-1)(t-1)
t(0.05, 12 df) =
t(0.50, 12 df) =
𝜎 2=
d=
X=
5
4
12
2.179
0.695
600
40
1
d2 
d2 
2(t1  t 2 )2 s2
r
2(2.179  0.695)2 600 2 * 8.259 * 600

 2477.963
4
4
d = sqrt(2477.9628) = 49.78
49.78>40, so the level of precision is not sufficient.
No, they cannot expect to achieve the desired power with the conditions specified.
Alternatively, you could solve for r using d=40 and the same t values as above with
12 df, to see if the result for r is less than or equal to 4.
r
2(2.179  0.695)2 600 2 * 8.2599 * 600

 6.19
402
1600
Another approach is to fix r and d and to see if the plot size required exceeds the
standard size of X=1. The estimated plot size required is 2.40 standard units, so we
don’t have sufficient precision using X=1.
1
2) Nitrification inhibitors (NI) have been used to delay conversion of the ammonium forms
of nitrogen in fertilizers into the more soluble nitrate form, thereby reducing losses of N
due to leaching. An experiment was conducted to determine if the benefit of NI in sweet
corn fields depends on the time of fertilizer application. 15N was applied in a drip
irrigation system at an early, optimum, and late date of application, and uptake by the
sweet corn was determined (% of total N fertilizer applied that was taken up by the
crop). These treatments were applied both in the presence and absence of an NI.
Results from the ANOVA using SAS PROC GLM are shown below:
Dependent Variable: N uptake
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
7
2893.478889 413.354127 19.49 <.0001
Error
10
212.038889
21.203889
Corrected Total 17
3105.517778
R-Square Coeff Var Root MSE Nuptake Mean
0.931722 9.631167 4.604768
47.81111
Source
DF
Type III SS Mean Square F Value Pr > F
block
2 459.954444 229.977222 10.85 0.0031
Time of Application 2 1333.341111 666.670556 31.44 <.0001
NI
1 591.680000 591.680000 27.90 0.0004
Time*NI
2 508.503333 254.251667 11.99 0.0022
Table of means for all treatment combinations:
Time of Application
6 pts
6 pts
Nitrification Inhibitor
Early
Optimum
Late
Mean
No Inhibitor (No NI)
22.53
51.77
51.93
42.08
With Inhibitor (+ NI)
48.87
53.87
57.90
53.54
Mean
35.70
52.82
54.92
47.81
a) Briefly interpret the results of the ANOVA (just the conclusions from the F tests for now)
There is a significant Time*NI interaction (P=0.0022), so results for the main effects
should be interpreted with caution. Blocking was effective.
b) Calculate an LSD value (using an alpha level of 0.05) that could be used to compare any
two levels of the six treatment combinations in this experiment.
2 ∗ 𝑀𝑆𝐸
2 ∗ 21.204
𝐿𝑆𝐷 = 𝑡𝛼=0.05,10 𝑑𝑓 √
= 2.228√
= 2.228 ∗ 3.760 = 8.38
𝑟
3
2
8 pts
c) Briefly interpret the results from this experiment based on the trends apparent in the
table of means and considering your answers to questions a) and b) above.
The effect of a Nitrification inhibitor
(NI) on 15N uptake depends on the
time of fertilizer application. When
fertilizer is applied early, addition of
an NI increases N uptake. When
fertilizer is applied at optimum or late
time periods, the effects on N uptake
are not significant.
3) An entomologist wishes to determine the effects of the food source (red clover, canola, or
almonds) and hive design (standard or enhanced ventillation) on survival of honeybees. He
conducts an experiment that includes all possible combinations of these two treatment
factors in a Completely Randomized Design. Write orthogonal contrast coefficients that
would address the following questions:
1. Does the level of ventillation in the hives affect honeybee survival?
2. Is there a difference between the orchard crop (almonds) and the field crops (clover
and canola)?
3. Does the species of field crop affect honey bee survival?
4. Does the difference between the orchard and field crops depend on the level of
ventillation?
5. Does the difference between clover and canola depend on the level of ventillation?
Fill in the appropriate coefficients below the corresponding treatment combinations:
10 pts
6 pts
Ventillation:
Crop
Standard
Clover
Standard
Canola
Standard
Almond
Enhanced
Clover
Enhanced
Canola
Enhanced
Almond
Contrast #
1
2
3
4
5
-1
-1
-1
1
1
-1
-1
1
1
-1
-1
2
0
-2
0
1
-1
-1
-1
-1
1
-1
1
-1
1
1
2
0
2
0
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)(-1) + (-1)(-1) + (-1)(2) + (1)(-1) + (1)(-1) + (1)(2) = 0
6 pts
b) Is this a complete set of orthogonal contrasts? If not, how many additional contrasts
would be required to make a complete set?
Yes. There are six treatment combinations and t-1=5 orthogonal contrasts.
3
4) A corn breeder has developed some new varieties of corn that are high in iron content.
She wishes to study the effects of feeding six varieties of corn that differ in iron content
on plasma iron levels of dairy cows. She intends to use a Latin Square Design with
animals and stage of lactation (time after calving) as blocking factors.
4 pts
a) How many cows will she need to conduct this experiment? 6
4 pts
b) How many total degrees of freedom will there be in the ANOVA?
8 pts
c) Given the following estimates of Mean Squares from the ANOVA, would you say that
it was important to use the stage of lactation as a blocking factor to control
experimental error? Use an estimate of Relative Efficiency to justify your answer.
35
Mean Square for Cows = 5.6
Mean Square for Stage of Lactation = 2.1
Mean Square Error = 2.3
To answer this question you need to consider a model that does not use lactation as a
blocking factor, and compare that to the Latin Square design. We therefore want to
know the Relative Efficiency of the Latin Square compared to an RBD with cows as
blocks. If cows are the rows and stage of lactation is the columns:
RE 
MSC  (t  1)MSE 2.1  (6  1) *2.3

 0.9855
t * MSE
6*2.3
(0.9855-1)*100 = -1.45%
There is a slight loss in efficiency (1.45%) from including Stage of Lactation as a blocking
factor.
These questions can be confusing, but a simple check that you have done this correctly
would be to look at the F ratio for stage of lactation (2.1/2.3). It is less than one, and
therefore clearly not significant.
5) What is meant by experimentwise (or family wise) Type I error?
8 pts
8 pts
Experimentwise error is the risk of making at least one Type I error among the set
(family) of comparisons in the experiment. It indicates the proportion of experiments in
which one or more differences are falsely declared to be significant. Type I
experimentwise error increases as the number of mean comparison tests increases.
6) Which of the following multiple comparison procedures does not provide good control
of experimentwise error (circle one)?
a) Tukey’s Test
b) Bonferroni Correction
c) LSD Test
d) Scheffés Method
4
7) A scientist is studying seed dormancy in a newly domesticated crop. She plants 100
seeds of four varieties in separate germination trays and counts the number of
germinated seedlings after two weeks. Each variety is replicated four times. The 16
germination trays are completely randomized in the germination chamber under
standard conditions. She runs an ANOVA and the residual plot is shown below.
16 pts
Discuss any concerns that you might have about the ANOVA assumptions for this
experiment. How would you determine if the assumptions have been violated?
What possible solutions could you suggest to the researcher to address potential
violations of the ANOVA assumptions?
Your comments should be specific for this experiment.
The assumptions for the ANOVA are that the error terms are random, independent and
normally distributed with a mean of zero and common variance. The effects in the model
are also assumed to be additive.
This is a CRD, so the four predicted values along the X axis represent the means for each of
the varieties. In other words, the fact that the data are grouped in four discrete lines is not
in itself a cause for concern. The lack of an even distribution of residuals above and below
the mean of zero on the Y axis suggests that there might be a problem with heterogeneity
of variance.
There are two possible outcomes here – a seed may germinate or not – and counts are
made of the number of germinated seedlings. The data are discrete rather than continuous,
and the response of interest could be expressed as a percent of the total number of seeds
planted, so a binomial distribution is expected. In some cases, binomial data may
approximately follow a normal distribution. In this experiments there are two varieties that
have average germination percentages close to the extremes of 0 and 100, which raises
more concern about the assumptions of normality and homogeneity of variance. Values at
the extremes will have smaller variances than those at intermediate levels. This is evident in
5
the graph, where the residuals at the extremes are clustered closer to the mean of zero
compared to the residuals for the two varieties with intermediate germination percentages.
A Levene’s test could be conducted to see if there is significant heterogeneity of variance.
Tests for normality of residuals could also be conducted, and Q-Q plots could be used to see
if the normality assumption is met. There is only one factor in the model, so there is no
reason to conduct a Tukey’s test for additivity.
The simplest solution might be to use an arcsin transformation of the data. The ANOVA is
conducted on the transformed data, and then means are back-transformed to the original
scale. It is also necessary to check that all of the ANOVA assumptions for the transformed
data set are met before proceeding with the analysis.
A more sophisticated solution would be to use a generalized linear mixed model with the
appropriate link function for binomial data. In this way the model would be fit to the data,
rather than trying to fit the data to the ANOVA model.
6
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
19.3 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
7
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
0.5
0.05
0.01
1.000 12.706 63.667
0.816 4.303 9.925
0.765 3.182 5.841
0.741 2.776 4.604
0.727 2.571 4.032
0.718 2.447 3.707
0.711 2.365 3.499
0.706 2.306 3.355
0.703 2.262 3.250
0.700 2.228 3.169
0.697 2.201 3.106
0.695 2.179 3.055
0.694 2.160 3.012
0.692 2.145 2.977
0.691 2.131 2.947
0.690 2.120 2.921
0.689 2.110 2.898
0.688 2.101 2.878
0.688 2.093 2.861
0.687 2.086 2.845
0.686 2.080 2.831
0.686 2.074 2.819
0.685 2.069 2.807
0.685 2.064 2.797
0.684 2.060 2.787
0.684 2.056 2.779
0.684 2.052 2.771
0.683 2.048 2.763
0.683 2.045 2.756
0.683 2.042 2.750