Name:
KEY
CSS 590 – Experimental Design in Agriculture
Second Midterm
Winter 2007
10 pts
15 pts
1) While entering yield data from an experiment onto a spreadsheet, the value of 17796
g/plot was entered rather than the true value of 1796. Draw a picture showing how this
might appear on a residual plot. Include titles for the X and Y axes of your diagram and
the zero mark on the Y axis.
2) What is the purpose of conducting a Levene Test? If the test is significant, how would
you proceed with your data analysis and summarization?
A Levene test is conducted to determine if the assumption of homogeneity of variance
among groups is valid. If the test is significant, then the assumption is not valid, and
some form of transformation may be needed. To determine the appropriate
transformation, you would first consider the type of data that has been collected, which
might suggest an underlying distribution that would apply to this data set. For example,
you could determine if the mean is proportional to the standard deviation, or to the
variance. In the former case we would consider using a log transformation, and in the
latter, a square root transformation. Finally, we would want to recheck the residual plot
after analyzing the transformed data. For data summary, means would be back
transformed to the original scale. Alternatively, one could use a Generalized Linear
Model, or a nonparametric test, if no suitable transformation can be identified.
1
3) An animal scientist wishes to test the effects of 5 different feeding regimens on the
weight gain of pigs. Only five pigs are available for this research.
15 pts
a) Can you suggest a design that would permit her to address the research question of
interest given this limitation? Show the sources of variation and degrees of freedom for
the ANOVA for this design.
She could use a Latin Square Design and give all five feeding regimens to each pig. The
pigs would be one blocking factor (rows) and the sequence of the feeding regimen could
be a second blocking factor (columns).
Source
Total
p -1
24
Pigs
p-1
4
Sequence
p-1
4
Feeding Regimen
p-1
4
Error
10 pts
df
2
(p-1)(p-2)
12
b) Briefly describe how the feeding regimens would be assigned to the experimental units
in this experiment (you do not need to describe the randomization process step by step,
but describe the features of an appropriate randomization).

Each animal would receive each regimen once

The order in which the regimens are given would be unique for each pig

At any point in time, each of the five regimens would be assigned to one and only
one pig

A randomization process would be used to assign the regimens to pigs and
sequences
2
4) A Sugar Company conducted an experiment to compare two varieties of sugar cane in
combination with three levels of nitrogen (150, 210, and 270 lbs. N per acre
respectively). The experiment was run in 4 complete blocks.
a) Fill in the shaded cells to complete the following ANOVA. (There is an F table at the
end of this exam)
16 pts
ANOVA
Source
Blocks
Variety
Nitrogen
VxN
Error
Total
df
3
1
2
2
15
23
SS
249
216
16
400
672
1553
MS
83.0
216.0
8.0
200.0
44.8
F
F crit
4.82
0.18
4.46
4.54
3.68
3.68
Mean Yield (tons)
Variety 1
Variety 2
Mean
5 pts
Mean
69
63
66
b) Interpret the results from this experiment. What can you say about the response of
sugarcane to Nitrogen fertilizer? (you do not need to calculate contrasts, but discuss
trends).
c) Is there a significant difference between the varieties at the highest N level?
There is a significant interaction between variety and nitrogen, so conclusions should be
based on the two-way table of means. The yield of sugar cane in this study depended
upon the variety grown and the rate of applied nitrogen. The yield of Variety 2
decreased as the rate of application increased, while variety 1 showed a positive
response to the higher application rates.
Variety 1
Variety 2
sed = sqrt(2*44.8/4) = 4.73
75
t(0.05,15 df) = 2.131
70
LSD for interactions = 2.131*4.73
Sugar Yield (tons)
9 pts
N-150 N-210 N-270
65
69
73
69
63
57
67
66
65
= 10.09
Mean1-Mean2 = 73-57 = 16
16 > 10.09
65
60
55
Yes, the varieties differ at the highest N
level.
50
N-150
(could also compare a calculated t to the
critical t, or compare confidence intervals)
3
N-210
Nitrogen Application
N-270
5) You wish to compare the tolerance of 5 mint varieties to a new herbicide. A Randomized
Block Design is used, with 3 blocks. Initial plant densities are the same for all varieties,
and herbicide is applied to all plots in the experiment. After two months, numbers of mint
plants are counted in 4 quadrats in each plot.
The GLM Procedure
Dependent Variable: Count
Source
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
14
10409.83333
743.55952
11.16
<.0001
Error
45
2997.50000
66.61111
Corrected Total
59
13407.33333
Source
Block
Variety
Block*Variety
R-Square
Coeff Var
Root MSE
count Mean
0.776428
16.43268
8.161563
49.66667
DF
Type III SS
Mean Square
F Value
Pr > F
2
4
8
470.233333
9035.333333
904.266667
235.116667
2258.833333
113.033333
3.53
33.91
1.70
0.0377
<.0001
0.1255
Tests of Hypotheses Using the Type III MS for Block*Variety as an Error Term
Source
Block
Variety
10 pts
DF
Type III SS
Mean Square
F Value
Pr > F
2
4
470.233333
9035.333333
235.116667
2258.833333
2.08
19.98
0.1873
0.0003
a) What is the appropriate F value for comparing the varieties? Explain your answer.
The F value for comparing varieties is 19.98. This ratio is obtained when we use the
experimental error (block*variety) as the error term. The default analysis uses the
sampling error in the denominator of the F test, which gives too large a value and
inflates the level of significance.
5 pts
b) Is the F test for blocks significant? what is your proof?
No. The observed Probability of 0.1873 is > than 0.05, so we fail to reject the null
hypothesis.
5 pts
c) Calculate the standard error for a variety mean.
standard error of a mean = sqrt(113.03/(3*4) = 3.07
4
sY 
MSE
r *n
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
4.6
15
4.54
16
4.49
17
4.45
18
4.41
19
4.38
20
4.35
21
4.32
22
4.3
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
7
199.5 215.71 224.58 230.16 233.99 236.77
19 19.16 19.25
19.3 19.33 19.36
9.55
9.28
9.12
9.01
8.94
8.89
6.94
6.59
6.39
6.26
6.16
6.08
5.79
5.41
5.19
5.05
4.95
5.88
5.14
4.76
4.53
4.39
4.28
4.21
4.74
4.35
4.12
3.97
3.87
3.79
4.46
4.07
3.84
3.69
3.58
3.5
4.26
3.86
3.63
3.48
3.37
3.29
4.1
3.71
3.48
3.32
3.22
3.13
3.98
3.59
3.36
3.2
3.09
3.01
3.88
3.49
3.26
3.1
3
2.91
3.8
3.41
3.18
3.02
2.92
2.83
3.74
3.34
3.11
2.96
2.85
2.76
3.68
3.29
3.06
2.9
2.79
2.71
3.63
3.24
3.01
2.85
2.74
2.66
3.59
3.2
2.96
2.81
2.7
2.61
3.55
3.16
2.93
2.77
2.66
2.58
3.52
3.13
2.9
2.74
2.63
2.54
3.49
3.1
2.87
2.71
2.6
2.51
3.47
3.07
2.84
2.68
2.57
2.49
3.44
3.05
2.82
2.66
2.55
2.46
3.42
3.03
2.8
2.64
2.53
2.44
3.4
3
2.78
2.62
2.51
2.42
3.38
2.99
2.76
2.6
2.49
2.4
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
0.05
0.01
1.376 12.706 63.667
1.061 4.303 9.925
0.978 3.182 5.841
0.941 2.776 4.604
0.920 2.571 4.032
0.906 2.447 3.707
0.896 2.365 3.499
0.889 2.306 3.355
0.883 2.262
3.25
0.879 2.228 3.169
0.876 2.201 3.106
0.873 2.179 3.055
0.870
2.16 3.012
0.868 2.145 2.977
0.866 2.131 2.947
0.865
2.12 2.921
0.863
2.11 2.898
0.862 2.101 2.878
0.861 2.093 2.861
0.860 2.086 2.845
0.859 2.080 2.831
0.858 2.074 2.819
0.858 2.069 2.807
0.857 2.064 2.797
0.856 2.060 2.787
0.856 2.056 2.779
0.855 2.052 2.771
0.855 2.048 2.763
0.854 2.045 2.756
0.854 2.042 2.750