Name:
KEY
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.
You should consider using a CRD, since it is a small experiment and there was no
benefit to blocking in the previous year. Being a “small” experiment, you probably do not
have many treatments or replications, and the area utilized in the field may be fairly
uniform. Using a CRD will provide the maximum degrees of freedom for error, which
may improve the power of the test.
There may also be good reasons for continuing to use an RBD. If it is not possible to
collect a particular type of data on the whole experiment in a reasonable time frame,
blocking could permit you to divide the workload across several days, or to have different
individuals collect data in different blocks. If you are planning to combine results with the
previous year, it may be preferable to use the same design in both years.
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? The ¼-acre section planted for each variety (a plot).
6 pts
c) Are there any blocks in this design? If so, what are they? The six farms are the
blocks.
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
Arrange the plots so that
each variety is exposed to
the same variation due to
moisture from the 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
9.32153
20
17.4750
0.87375
Corrected Total 23
45.4396
Error
10.67 0.0002
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
8 pts
3
27.9646
9.32153
10.67 0.0002
b) Explain how you would use the Pr>F value to determine if there are significant
differences among the treatments (using  = 0.05).
The observed P value of 0.0002 is much smaller than 0.05, so we can reject the null
hypothesis and conclude that there are significant differences among the treatments.
6 pts
c) What is the meaning of the R-Square value?
The R-Square value tells us what proportion of the total sums of squares is explained
by the model (the treatments in this case). A low value of R-Square would suggest
that the treatments have little effect on the variable that is measured, or that a better
model is needed (for example, blocking may be needed to control unwanted variation
in the experiment).
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
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.
Plant height is significantly greater for Mix-1 than for Mix-2 and the Control (No amendment).
The difference between Mix-1 and Mix-3 is not significant.
This interpretation can be made because Mix-1 and Mix-3 share the common letter ‘A’,
whereas Mix-2 and the Control have different letters than Mix-1. You could also reach the
same conclusions by calculating the differences between the mean of Mix-1 and the other
treatments and comparing those differences to the LSD value of 1.126.
8 pts
Mix-1 vs Mix-3
38.700 – 37.633 = 1.067 1.067<1.126, so the difference is not significant
Mix-1 vs Mix-2
38.700 – 36.917 = 1.783 1.783>1.126 so the difference is significant
Mix-1 vs None
38.700 – 35.733 = 2.967 2.967>1.126 so the difference is significant
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.
Critical t(=0.05, 20 df) = 2.086
sY 
MSE
0.87375

 0.3816
r
6
2.086*0.3816 = 0.7960
Upper limit = 38.7+0.796 = 39.496
Lower limit = 38.7-0.796 = 37.904
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