Name:
CSS 590 – Experimental Design in Agriculture
First Midterm
Winter 2012
Please remember to show your calculations.
6 pts
1) A chemist recently discovered a new seed treatment that can effectively control
important soil-borne pathogens in soybeans. Shortly after making this discovery, he
decides to take a job with another company. Another pathologist with less experience is
hired to replace him. The new pathologist conducts a field experiment to compare
disease levels of soybeans treated with the new product, three commercial seed
treatments, and a control (no seed treatment). The observed F value for treatments from
this experiment is 2.36. He compares this value to a critical F with 4 and 12 df and
=0.05.
The new pathologist reports his findings to his supervisor. What is the likely outcome of
this experiment? (refer to the F table at the back of this exam)
a) The null hypothesis is correctly accepted
b) The null hypothesis is correctly rejected
c) A Type I error is committed
d) A Type II error is committed
10 pts
2) Last year you conducted a small field experiment at the OSU Vegetable Farm. You used
a Randomized Complete Block Design. When you analyzed the results, you noted that
the P-value (PR>F) for Blocks was 0.4089. You calculated the Relative Efficiency
compared to a CRD to be 0.96. You will be conducting a similar experiment again this
year in an adjacent field with comparable soil conditions. Will you use the same
experimental design as you did last year? Explain your answer.
1
3) An extension agent wishes to identify the best variety of wheat for production in the
Willamette Valley. The wheat breeder has given him seed of three new varieties that
performed well in on-station trials. Six growers in the Willamette Valley have agreed to
collaborate and test the new varieties on their farms. Each grower has a one-acre field
that will be divided into four sections. The three new varieties and an older standard
variety will be randomly assigned to the four ¼-acre sections on each farm. The field
layout on Mr. Smith’s farm looks like this:
New Variety 2
Standard Variety
New Variety 1
New Variety 3
6 pts
a) What type of experimental design is being used in this experiment?
i.)
ii.)
iii.)
iv.)
Completely Randomized Design (CRD)
Randomized Complete Block Design (RBD)
There is no design because the experiment is unreplicated
None of the above
6 pts
b) What is the experimental unit?
6 pts
c) Are there any blocks in this design? If so, what are they?
8 pts
d) On Mr. Brown’s farm, there is a stream at the north end of his field. On the diagram
below, show how you would arrange the four ¼-acre sections of the field to best
account for potential variation due to the stream.
Stream
2
4) An experiment was conducted to evaluate three soil amendments (Mixtures) on the
growth of collard greens. A control with no amendment was also included for
comparison. Each of the treatment levels was randomly assigned to six plots in the field.
Plant height was measured in each plot.
a) Fill in the missing values in the shaded cells below to complete the ANOVA.
The GLM Procedure
24 pts
Dependent Variable: Height
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
3
27.9646
Corrected Total 23
45.4396
9.32153
0.0002
Error
R-Square Coeff Var Root MSE Height Mean
0.6154
2.5097
0.93475
37.246
Source DF Type III SS Mean Square F Value Pr > F
Mixture
3
27.9646
9.32153
0.0002
8 pts
b) Explain how you would use the Pr>F value to determine if there are significant
differences among the treatments (using  = 0.05).
6 pts
c) What is the meaning of the R-Square value?
3
5) The SAS output for an LSD test for the experiment described in question 4 is shown
below.
t Tests (LSD) for Height
Alpha
0.05
Least Significant Difference 1.126
t Grouping
A
Mean
N
Mixture
38.700
6
Mix-1
37.633
6
Mix-3
36.917
6
Mix-2
35.733
6
None
A
B
A
B
B
C
12 pts
8 pts
a) Assume that Mix-1 is a new soil amendment that we wish to compare to each of the
other treatments. Use the output to interpret results for the three comparisons of
interest.
b) Use the results from the ANOVA in question 4 and the t-table at the back of this
exam to calculate a 95% confidence interval for the mean of the Mix-1 amendment.
4
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
27
28
29
30
Student's t Distribution
Numerator
(2-tailed probability)
2
3
4
5
6
199.5 215.71 224.58 230.16 233.99
19 19.16 19.25 19.30 19.33
9.55
9.28
9.12
9.01
8.94
6.94
6.59
6.39
6.26
6.16
5.79
5.41
5.19
5.05
4.95
5.14
4.76
4.53
4.39
4.28
4.74
4.35
4.12
3.97
3.87
4.46
4.07
3.84
3.69
3.58
4.26
3.86
3.63
3.48
3.37
4.1
3.71
3.48
3.32
3.22
3.98
3.59
3.36
3.20
3.09
3.88
3.49
3.26
3.10
3.00
3.8
3.41
3.18
3.02
2.92
3.74
3.34
3.11
2.96
2.85
3.68
3.29
3.06
2.90
2.79
3.63
3.24
3.01
2.85
2.74
3.59
3.20
2.96
2.81
2.70
3.55
3.16
2.93
2.77
2.66
3.52
3.13
2.90
2.74
2.63
3.49
3.10
2.87
2.71
2.60
3.47
3.07
2.84
2.68
2.57
3.44
3.05
2.82
2.66
2.55
3.42
3.03
2.80
2.64
2.53
3.40
3.00
2.78
2.62
2.51
3.38
2.99
2.76
2.60
2.49
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
0.5
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