CROP 590
Name:
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)
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)
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
Question 2 continued
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.
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
Ventillation:
Standard
Standard
Standard
Enhanced
Enhanced
Enhanced
Crop:
Clover
Canola
Almond
Clover
Canola
Almond
Contrast #
1
2
3
4
5
6 pts
a) Describe how you would verify that these contrasts are orthogonal to each other (give
one example).
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?
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?
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.
Mean Square for Cows = 5.6
Mean Square for Stage of Lactation = 2.1
Mean Square Error = 2.3
8 pts
5) What is meant by experimentwise (or family wise) Type I error?
8 pts
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.
5
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
6
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