Experimental Design in Agriculture
CROP 590
Second Midterm Exam
Winter, 2014
Name______KEY______________
Please show your work.
1) You have completed an experiment with five treatments and four replicates per
treatment using a Completely Randomized Design. You intend to conduct an
Analysis of Variance to see if there are significant differences among treatments.
8 pts
a) You use SAS to perform Levene’s test on this data set and obtain a probability
value (Pr>F) of 0.2893. What does this result tell you about the assumptions
needed for a valid ANOVA?
Levene’s Test is used to test the assumption of homogeneous variances. The
null hypothesis in this case is that the variances for the five treatment groups are
the same. Because the P value is greater than 0.05, we can accept the null
hypothesis. Therefore the assumption of homogeneity of variance needed for a
valid ANOVA is met. If the other assumptions needed for ANOVA are also
satisfied, then we can proceed with the analysis.
8 pts
b) Describe at least one approach that you could use to determine if the assumption
of normality required for a valid ANOVA has been met. Outline the steps that you
would take to complete your diagnosis.
The assumption of normality refers to the residuals after the model has been
fitted. The distribution of the original observations could be skewed by effects
due to particular treatments. Because there are relatively few observations in
each group (four replicates per treatment), it would be best to obtain residuals
and evaluate them collectively to see if they meet the assumption of normality.
To do this you would run the ANOVA and output the residuals to a new data set.
You could then use various graphical procedures and statistical tests for
normality.

Stem and leaf plots and box plots could be used to determine if the tails of
the distribution are fairly symmetric (not skewed), and if the median is close
to the mean as expected for a normal distribution. There should be few, if
any, extreme values.

A Q-Q plot or normal probability plot could be used. If residuals fall close to
the line then they fit the expectations for a normal distribution.

The number of observations in this trial is limited, but in some experiments it
may be possible to plot a histogram of residuals to see if they approximate a
normal, bell-shaped curve.

A Shapiro-Wilk test could be used to test the null hypothesis that residuals
are normally distributed.
1
2) Match the mean comparison tests (A-F) with the features that best describe them in
the table below. For full credit, each option should be used once.
A
B
C
D
E
F
Bonferroni Adjustment
Dunnett’s Test
LSD Test
Student-Newman-Keuls Test
Tukey’s Test
Waller-Duncan Bayes (BLSD)
12 pts
Features
Uses a single critical value that is calculated
from a table of studentized range values.
Effectively controls Type I experimentwise
error.
This test can be used to compare all treatments
with a control treatment. It is widely accepted
and provides good control of experimentwise
error.
The critical value is adjusted depending on the
magnitude of the F test among treatments.
Aims to reduce Type I experimentwise error as
well as Type II error.
This test has a high power to detect differences
but only controls the comparisonwise Type I
error rate. It should only be used to make
preplanned comparisons and should be limited
to a number of comparisons that do not exceed
the degrees of freedom for treatments.
Uses a single critical probability value that is
calculated by dividing the desired
experimentwise error rate by the number of
pairwise comparisons that are to be made.
Provides very strict control of Type I error but
may have very low power to detect differences
among treatments.
A multiple range test that uses more than one
critical value, depending on how far apart the
means are in the ranking.
2
Name of Test
E) Tukey’s Test
B) Dunnett’s Test
F) Waller-Duncan Bayes (BLSD)
C) LSD Test
A) Bonferroni Adjustment
D) Student-Newman-Keuls Test
3) A group of gardening enthusiasts wish to evaluate the eating quality of 5 heirloom
varieties of tomato. They expect that quality may be influenced by the growing
environment (farm) and the personal preferences of the evaluators. They decide to
use a Latin Square Design to account for this variability. Five farmers volunteer to
grow all of the varieties, and 5 panelists participate in a sensory evaluation of the
varieties. They evaluate each variety for a number of attributes (appearance, flavor,
texture, etc.) and combine their assessments into an overall score.
a) Complete the ANOVA by filling in the shaded cells below.
13 pts
8 pts
Source
Total
Farm
Panelist
Variety
Error
df
24
4
4
4
12
SS
66.256
27.120
20.872
13.992
4.272
MS
6.780
5.218
3.498
0.356
F
19.054
14.657
9.826
b) What is the relative efficiency of the Latin Square when compared to an RBD
using only the Farms as Blocks?
RE 
MS(panelist)  (t  1)MSE 5.218  (5  1)0.356

 3.73
t * MSE
5*0.356
A relative efficiency of 3.73 represents a 273% increase in efficiency by including
the panelists as blocks in a Latin Square Design.
You would need 3.73 * 5 =18.65  19 blocks to achieve the same level of
efficiency using an RBD with farms as the only blocking factor.
8 pts
c) Do you think the use of a Latin Square was justified in this case? Give evidence
to support your answer.
Yes. The MS (and F value) for farms is larger than for the panelists, so we can
assume that the RE estimate compared to an RBD with panelists as blocks
would be a little larger than the 3.73 RE calculated above. The critical F for
testing Farms and Panelists with 4 and 12 df is 3.26. The calculated F values for
both blocking factors are much greater than 3.26, so the use of a Latin Square
Design was justified.
d) Would the farms be a fixed or a random effect? Justify your answer.
6 pts
Farms would be random because they represent a sample of farms that are
owned by members of the group of gardening enthusiasts. Since the farmers are
volunteers, it is clear that there is no specific interest in determining how varieties
perform on their farms. The intent is to remove any farm to farm variation that
would increase error if the varieties were not evaluated in a common
environment. Another objective may be to estimate how much variation can be
attributed to farms. This would still be a random effect because we are more
interested in estimating the variance among farms than in knowing the effect of
specific farms on quality.
3
4) An experiment was conducted to compare the tolerance of 5 mint varieties to a new
herbicide. A Randomized Block Design was used, with 3 blocks. Initial plant densities
were the same for all varieties, and herbicide was applied to all plots in the
experiment. After two months, numbers of mint plants were counted in 4 quadrats in
each plot.
The data were analyzed with SAS PROC GLM. Some of the output is shown below.
The GLM Procedure
Dependent Variable: count
Source
DF Sum of Squares Mean Square
Model
14
10409
743.5
Error
45
2997
66.6
Corrected Total 59
13406
Source
5 pts
DF Type III SS Mean Square
Block
2
470
235
Variety
4
9035
2259
Block*Variety
8
904
113
a) What value from this output provides an estimate of the variance among plants
within each plot?
The MSE of 66.6
8 pts
b) Calculate the appropriate F value for testing the null hypothesis that the means of
all varieties are equal. Use the F table from the back of this exam to determine
whether to accept or reject the null hypothesis, using =0.05.
F = MSVariety / MSBlock*Variety = 2259 / 113 = 19.99
Critical F with 4 and 8 df = 3.84
19.99 > 3.84, so we reject the null hypothesis and conclude that there are
differences among the varieties.
8 pts
c) Calculate the standard error for a variety mean for this experiment.
se 
MSE
113

 3.07
r *n
3* 4
4
5) A researcher wished to know how soil type and a seed treatment (fungicide)
influenced the emergence of red clover seedlings. She utilized factorial combinations
of three soil types (Sand, Silt Loam, and Clay) and two levels of the fungicide (None
and Treated). She grew three pots of each treatment combination in a Completely
Randomized Design in the greenhouse. She recorded the number of emerged
seedlings in each pot. The ANOVA for this is experiment is shown below.
Source
Total
Fungicide (F)
Soil Type (S)
FxS
Error
df
17
1
2
2
12
SS
7566
1301
4589
741
936
MS
445
1301
2294
371
78
F
Prob.>F
16.67
29.42
4.75
0.0015
0.0000
0.0302
Soil Type
Fungicide
None
Treated
Mean
8 pts
Silt Loam
82
92
87
Clay
42
77
60
Mean
73
90
82
a) Use the ANOVA and table of means provided above to interpret the results from
this experiment.
b) Calculate the appropriate standard error for the means that you discuss.
There are significant interactions between fungicide and soil type, so the main
effects should be interpreted with caution. This is a noncrossover interaction,
because the fungicide treatment consistently has higher emergence than
untreated clover.
In general, emergence of
clover is greatest on sandy
soil, and decreases as the
clay content increases. This
effect is less severe when
fungicide is applied.
Emergence is reduced to 42
when there is no fungicide
applied on a clay soil.
se 
MSE
78

 5.10
r
3
No Fungicide
emerged seedlings
8 pts
Sand
95
101
98
Treated
120
100
80
60
40
20
0
Sand
5
Silt Loam
Clay
F Distribution 5% Points
Denominator
Numerator
df
1
2
3
4
5
6
7
1 161.45 199.5 215.71 224.58 230.16 233.99 236.77
2 18.51 19.00 19.16 19.25 19.30 19.33 19.36
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
6.08
5
6.61
5.79
5.41
5.19
5.05
4.95
5.88
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
3.32
3.22
3.13
11
4.84
3.98
3.59
3.36
3.20
3.09
3.01
12
4.75
3.88
3.49
3.26
3.10
3.00
2.91
13
4.67
3.80
3.41
3.18
3.02
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
3.00
2.78
2.62
2.51
2.42
25
4.24
3.38
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
6
Student's t Distribution
(2-tailed probability)
df
0.40
0.05
0.01
1 1.376 12.706 63.667
2 1.061 4.303 9.925
3 0.978 3.182 5.841
4 0.941 2.776 4.604
5 0.920 2.571 4.032
6 0.906 2.447 3.707
7 0.896 2.365 3.499
8 0.889 2.306 3.355
9 0.883 2.262 3.250
10 0.879 2.228 3.169
11 0.876 2.201 3.106
12 0.873 2.179 3.055
13 0.870 2.160 3.012
14 0.868 2.145 2.977
15 0.866 2.131 2.947
16 0.865 2.120 2.921
17 0.863 2.110 2.898
18 0.862 2.101 2.878
19 0.861 2.093 2.861
20 0.860 2.086 2.845
21 0.859 2.080 2.831
22 0.858 2.074 2.819
23 0.858 2.069 2.807
24 0.857 2.064 2.797
25 0.856 2.060 2.787
26 0.856 2.056 2.779
27 0.855 2.052 2.771
28 0.855 2.048 2.763
29 0.854 2.045 2.756
30 0.854 2.042 2.750