Experimental Design in Agriculture
Name:
KEY
CROP 590
First Midterm
Winter 2016
Please show your work!
1a) In just a few sentences, describe the objectives of a planned experiment that you have
encountered in your field of study.
6 pts
Answers will vary
1b) Using your example, define the following:
Answers will vary
8 pts
The treatments
The experimental units
Blocking factors (if any), and explain the reason that they are used (or not used)
1c)
9 pts
Continuing with your example, describe three specific things that the researcher could do
to reduce the chances of making a Type II error.
- choose treatment levels that are farther apart to increase differences among the
responses that are being measured.
- reduce experimental error by increasing the plot size or number of replications
- increase degrees of freedom by using more treatment levels or more replication
- control experimental error by refining the experimental technique
- use blocking to remove a known source of experimental error
- select a higher alpha level (Type I error rate)
1
2)
A plant breeder in Zambia would like to compare three varieties of maize for grain yield: a
hybrid, and open-pollinated variety (OPV) and a local (landrace) variety. The experiment is
conducted as a Completely Randomized Design with 4 replications.
The mean grain yields for the varieties in Mg/ha are as follows:
Varieties
Means
Hybrid
OPV
Local
3.35
2.90
2.15
a) Calculate the Sum of Squares for Treatments:
8 pts
Mean = 2.8
SSvarieties = 4*[3.35-2.8)2+(2.90-2.8)2+(2.15-2.8)2]= 4*[(0.55)2+(0.1)2+(-0.65)2]=2.94
b) If the LSD (=0.05) value is 0.55 for this experiment, then the hybrid is significantly
different from the OPV (circle the correct answer).
4 pts
TRUE
FALSE
3)
9 pts
Discuss the potential advantages of using long, narrow plots in field experiments. Describe
at least one scenario where another plot shape might be preferred.
Long, narrow plots are often more convenient for machine harvest. Experimental units can
be positioned in close proximity within a square block, allowing precise comparisons within
a relatively homogeneous area (the block). If there is a field gradient that is not completely
accounted for by the arrangement of blocks, plots can be oriented with the long dimension
parallel to the gradient, so that each plot is exposed to the same conditions.
Long, narrow plots may not be optimal when border effects are large, or when there is the
potential for treatment applications to drift from one plot to another (e.g., irrigation,
chemical sprays, fertilizer).
Different plot shapes may also be needed for specific conditions (such as plots grown under
pivot irrigation) or to meet particular experimental objectives (such as plant density trials).
4)
5 pts
When there are missing plots in a Randomized Complete Block Design (choose one
answer):
a) A degree of freedom must be subtracted from the Error df for each missing plot
b) We should not rely on the Analysis Tool Pak in Excel to perform the ANOVA
c) Adjusted means in SAS (lsmeans) may be different than the arithmetic means
d) All of the above
2
5) An animal scientist wished to test the effects of 5 different feeding regimens on the
weight gain of pigs. Each treatment was randomly assigned to five pigs.
a) Fill in the boxes to complete the ANOVA.
14 pts
6 pts
Source
DF
SS
MS
F
Total
24
1081
Feeding Regimens
4
272
68
1.68
Error
20
809
40.45
b) Do the results indicate that there were differences among the feeding regimens at
the  = 0.05 probability level? What is your proof?
(Use the tables at the back of this exam)
F critical at  = 0.05 with 4 and 20 df is 2.87
F observed (1.68) is less than 2.87, so we fail to reject the null hypothesis and
conclude that there are no differences among the feeding regimens.
6 pts
c) The grand mean for the experiment was 55 pounds. What was the Coefficient of
Variation (CV%)?
CV% 
MSE
*100 
Y
8 pts
40.45
*100  11.56%
55
d) Regimen #1 has a mean of 60 pounds. Calculate the upper and lower limits for a 95%
confidence around this mean. (Use the tables at the back of this exam)
half-width  t 0.05,df 20
MSE
40.45
 2.086
 5.93
r
5
Lower limit = 60 – 5.93 = 54.07
Upper limit = 60 + 5.93 = 65.93
3
6) The response of soybeans to four seed treatments was measured in a Randomized
Complete Block Design. The output of the analysis from SAS PROC GLM is shown below.
Dependent Variable: response
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
8
878.0000000
109.7500000
Error
15
83.9583333
5.5972222
Corrected Total 23
961.9583333
19.61 <.0001
R-Square Coeff Var Root MSE response Mean
0.912721
Source
14.08940
2.365845
16.79167
DF Type III SS Mean Square F Value Pr > F
block
5 435.2083333
87.0416667
15.55 <.0001
treatment
3 442.7916667 147.5972222
26.37 <.0001
a) Do the results indicate that there were differences among the treatments? Support
your statement with evidence from the output above.
5 pts
F for treatments = 26.37, and Pr>F is <.0001
This p value is much smaller than the critical p of 0.05, so we reject the null
hypothesis and conclude that there are differences among the seed treatments.
b) Calculate the standard error of a difference between treatment means for this
experiment.
6 pts
6 pts
2MSE r  2 *5.597 6  1.36
c) The relative efficiency for this experiment compared to a CRD is 4.16. Circle the
correct statement below:
i.
The value of 4.16 > 1, so a CRD would have been more efficient than the RBD
ii.
Blocking increased efficiency compared to a CRD by 416%
iii.
25 replications in a CRD would be needed to obtain the same level of
efficiency as the RBD.
iv.
None of the above
4
F Distribution 5% Points
Denominator
Student's t Distribution
Numerator
(2-tailed probability)
df
1
2
3
4
5
6
7
1 161.45 199.50 215.71 224.58 230.16 233.99 236.77
2 18.51 19.00 19.16 19.25 19.30 19.33 19.35
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
6.09
5
6.61
5.79
5.41
5.19
5.05
4.95
4.88
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
3.33
3.22
3.14
11
4.84
3.98
3.59
3.36
3.20
3.09
3.01
12
4.75
3.89
3.49
3.26
3.11
3.00
2.91
13
4.67
3.81
3.41
3.18
3.03
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
3.01
2.78
2.62
2.51
2.42
25
4.24
3.39
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
5
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
1.376
1.061
0.978
0.941
0.920
0.906
0.896
0.889
0.883
0.879
0.876
0.873
0.870
0.868
0.866
0.865
0.863
0.862
0.861
0.860
0.859
0.858
0.858
0.857
0.856
0.856
0.855
0.855
0.854
0.854
0.2
0.05
3.078 12.706
1.886 4.303
1.638 3.182
1.533 2.776
1.476 2.571
1.440 2.447
1.415 2.365
1.397 2.306
1.383 2.262
1.372 2.228
1.363 2.201
1.356 2.179
1.350 2.160
1.345 2.145
1.341 2.131
1.337 2.120
1.333 2.110
1.330 2.101
1.328 2.093
1.325 2.086
1.323 2.080
1.321 2.074
1.319 2.069
1.318 2.064
1.316 2.060
1.315 2.056
1.314 2.052
1.313 2.048
1.311 2.045
1.310 2.042