CROP 590 ecampus
Name:
KEY
Experimental Design in Agriculture
Midterm Exam
Instructions: For all problems that involve calculations (not multiple choice), please
remember to include the steps in your calculations to show how you got your answer.
For multiple choice questions, choose the one best answer.
1) A chemist recently discovered a new seed treatment that can effectively control
important soil‐borne pathogens in soybeans. Shortly after making this valid discovery,
he decides to take a job with another company. A pathologist is hired to replace him
who has no experience in monitoring pathogens of soybeans. 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). He uses
a Completely Randomized Design with four replications. The observed F value for
treatments from this experiment is 2.36.
5 pts
1a) What is the critical F value for treatments at the =0.05 probability level (refer to
the F table at the back of this exam)?
i.) 3.06
ii.) 3.26
iii.) 3.68
iv.) None of the above
5 pts
We are told that there really are differences among the
treatments, but the observed F of 2.36 is less than the
critical F with 4 and 15 df, so we accept the null
1b) The new pathologists prepares a report on his findings.
What is the likely outcome of this experiment?
i.) The null hypothesis is correctly accepted
ii.) The null hypothesis is correctly rejected
iii.) A Type I error is committed
iv.) A Type II error is committed
2) Which of the treatments underlined below is most likely to be a random effect?
5 pts
a) Eight experimental diets used to rear insects in the laboratory
b) Collections of an invasive weed species taken from 20 farmers’ fields in a region
c) Investigation of effects of five tillage methods on crop growth
d) Three levels of irrigation applied to determine optimum amount for tomato quality
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
5 pts
3a) What type of experimental design is being used in this experiment?
i.) Completely Randomized Design (CRD)
ii.) Randomized Complete Block Design (RBD)
iii.) There is no design because the experiment is unreplicated
iv.) None of the above
5 pts
3b) What is the experimental unit? The ¼‐acre section planted for each variety (a plot).
5 pts
3c) Are there any blocks in this design? If so, what are they? The six farms are the
blocks.
6 pts
3d) 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.
4a) Fill in the missing values in the shaded cells below to complete the ANOVA.
Dependent Variable: Height
8 pts
Source
DF Sum of Squares Mean Square F Value Pr > F
Total
23
45.4396
3
27.9646
9.32153
20
17.4750
0.87375
Mixture
Error
Mixture
Mix‐1
Mix‐2
Mix‐3
None
6 pts
10.67 0.0002
Mean
38.700
36.917
37.633
35.733
4b) 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.
8 pts
4c) Use the results from the ANOVA and the t‐table at the back of this exam to calculate
a 95% confidence interval for the mean of the Mix‐1 amendment. Include the upper
and lower limits for the confidence interval.
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
6 pts
4d) The LSD value is 1.126. Is the difference between the Mix‐1 and Mix‐2 treatments
significant?
38.700 – 36.917 = 1.783
1.783 > 1.126 Therefore, yes, Mix‐1 and Mix‐2 are significantly different.
3
5) You wish to compare eight varieties of peas in a field that was determined to have a soil
variability index of b=0.6. The last time you conducted a trial in this field you obtained a
CV of 16% using a plot size of 12 m2 and four replications in a Randomized Complete
Block Design.
9 pts
5a) Determine the size of plot you would you need to have an 80% probability of
detecting differences of 30% of the mean, using an RBD with four blocks and a
significance level of 5%. Use the tables at the end of this exam and show your work.
r=4
dfe = (r‐1)(t‐1) = (4‐1)(8‐1) = 21
t1(0.05, 21 df) = 2.08
2
2 2.08 0.859 16
t2(0.40, 21 df) = 0.859
4 ∗ 30
CV% = 16
d = 30%
b = 0.6
Xb = [2*(2.08+0.859)2*162]/(4*302)
Xb = 1.228
(Xb)1/0.6 = (1.228) 1/0.6
X = 1.409
1.409 x 12m2 = 16.91 m2 (could round up to an even number)
3 pts
1.228
5b) Assuming that your planter and harvest equipment are designed for a plot that is 2m
wide, how long will your plot need to be to attain the optimum plot size?
Length = 16.91/2 = 8.45 m
(or round up to a length that is easier to measure in the field)
6 pts
6) A fellow graduate student is planning an experiment. He would like to have a 90%
probability of detecting differences between treatment means that are less than or
equal to 20 units, using a Type I error rate of 0.05. He is not sure how to determine if his
experimental design has the desired level of power. Using the estimate of variance and
other parameters that he provides, you calculate a detectable difference of 25 units for
his experiment. What advice would you give to him?
Because the d value is larger than 20 units, the experiment does not have the desired
level of power or precision (confidence intervals around the treatment means are too
wide). He will need to increase the number of replications or possibly the plot size, or
find other ways to reduce experimental error.
If the problem had stated that the true difference between means was 25 units, then
the answer would be “yes”, he has sufficient power. But here we are told that he can
only detect a difference that is 25 units, which means that he needs more power to
detect a difference of 20 units.
4
500
400
300
200
100
0
‐100
‐200
‐300
‐400
‐500
Residual
Residual
7) An experiment was conducted to evaluate the seed yield of 270 meadowfoam
genotypes. Each genotype was replicated twice. After conducting the ANOVA, two
residual plots were produced, as shown below. The chart on the left shows the residuals
vs the predicted values. The residuals were then sorted in the order that the
experimental units occurred in the field. The resulting diagram is shown on the right.
500
1000
1500
0
2000
8 pts
200
400
600
Plot Order
Predicted Values
5 pts
500
400
300
200
100
0
‐100
‐200
‐300
‐400
‐500
7a) List the assumptions that are necessary for a valid ANOVA.
The residuals are random, independent and normally distributed with a mean of zero
and common variance. The effects in the model are additive.
7b) Based on the evidence available in these charts, discuss any of the assumptions that
you think might have been violated, and justify your answer. Is there any evidence
that some assumptions might have been met?
The distribution of residuals is not random, because there is a clear pattern based on the
position of plots in the field. There is the potential for some bias due to these trends. They also
are not independent. Plots that are close together are similar, and plots that are far apart are
increasingly different from each other. There is no obvious heterogeneity of variance – the
spread of residuals is fairly consistent across the range of the X axis in both graphs. However,
since the residual variance is estimated based on deviations from an average (zero), it seems
likely that the residual variance will be larger in the beginning and towards the end of data
collection. The normality assumption is a little difficult to assess without a normality plot or
frequency distribution of residuals, but there is no obvious cause for concern about normality
from these graphs. Note that the clustering on the graph on the left is consistent with a normal
distribution where observations are more common near the mean. The plot is very dense
because there were so many treatments in the experiment (270).
5 pts
7c) Do you have any ideas about how these results might have occurred? (there are
many possibilities – just give one plausible explanation.)
For a trait like yield, a field gradient that was not accounted for in the experimental design
would be the most likely explanation. It is conceivable that there was some sort of mechanical
factor that was changing over time. For example, if the blower on the harvesting or threshing
equipment was reduced as harvesting operations proceeded through the field. We would need
some more information about how the plots were laid out in the field to give a more definitive
answer.
5
Denominator 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
1
161.45
18.51
10.13
7.71
6.61
5.99
5.59
5.32
5.12
4.96
4.84
4.75
4.67
4.60
4.54
4.49
4.45
4.41
4.38
4.35
4.32
4.30
4.28
4.26
4.24
4.23
4.21
4.20
4.18
4.17
F Distribution
Fv1,v2 for P(F ≥ Fv1,v2) = 5%
Student's t Distribution
tdf for P(|t| ≥ tdf) = 
Numerator df
( for 2-tailed probability)
2
3
4
5
6
199.5 215.71 224.58 230.16 233.99
19.00 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
4.10
3.71
3.48
3.32 3.22
3.98
3.59
3.36
3.20 3.09
3.89
3.49
3.26
3.10 3.00
3.81
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
3.01
2.78
2.62 2.51
3.39
2.99
2.76
2.60 2.49
3.37
2.98
2.74
2.59 2.47
3.35
2.96
2.73
2.57 2.46
3.34
2.95
2.71
2.56 2.45
3.33
2.93
2.70
2.55 2.43
3.32
2.92
2.69
2.53 2.42
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
0.40
0.50
0.05
0.01
1.376 1.000 12.706 63.667
1.061 0.816 4.303 9.925
0.978 0.765 3.182 5.841
0.941 0.741 2.776 4.604
0.920 0.727 2.571 4.032
0.906 0.718 2.447 3.707
0.896 0.711 2.365 3.499
0.889 0.706 2.306 3.355
0.883 0.703 2.262 3.250
0.879 0.700 2.228 3.169
0.876 0.697 2.201 3.106
0.873 0.695 2.179 3.055
0.870 0.694 2.160 3.012
0.868 0.692 2.145 2.977
0.866 0.691 2.131 2.947
0.865 0.690 2.120 2.921
0.863 0.689 2.110 2.898
0.862 0.688 2.101 2.878
0.861 0.688 2.093 2.861
0.860 0.687 2.086 2.845
0.859 0.686 2.080 2.831
0.858 0.686 2.074 2.819
0.858 0.685 2.069 2.807
0.857 0.685 2.064 2.797
0.856 0.684 2.060 2.787
0.856 0.684 2.056 2.779
0.855 0.684 2.052 2.771
0.855 0.683 2.048 2.763
0.854 0.683 2.045 2.756
0.854 0.683 2.042 2.750