ST512
HOMEWORK 4 SSII10
1)
Experimental Layout. A study will be carried out in the greenhouse on the effects of 2
methods of obtaining cuttings (M1, M2) on the growth of 5 cultivars of an ornamental shrub
(V1, V2, V3, V4, V5). Identify uniquely each method by cultivar combination from T1
through T10. Use PROC PLAN to obtain:
a. a randomization plan for an CRD with 4 pots per cutting method and cultivar
combination, with three cuttings per pot. These pots for 10 methods by cultivar
treatment combinations will be randomly distributed along four selected benches in
the greenhouse. Sketch a plan for the layout in the greenhouse indicating the position
of each treatment combination. Assume each bench will contain a single row of 10
pots.
b. a randomization plan for an RCBD with 4 blocks corresponding to 4 benches in the
greenhouse, and all 10 methods by cultivar treatment combinations represented once
in each block. Sketch a plan for the layout in the greenhouse indicating the position of
each treatment combination in each block. Assume each block will contain a single
row of 10 pots ( 'plots').
2) Ex. 7.4.5 (STD), Unequal number of repetitions in a CRD.
R.T. Claussen, Cornell University, gives leaf lengths (sum of three leaves in millimeters) at
the time of flowering for Sedum oxypetalum from six locations in the trans-mexican volcanic
belt. The following data are reconstructed to match his totals and sum of squares.
Location
H
Length
147
Within each group
0
df
Totals
147
Corrected SS for
0
each group
Overall mean= 103.0714

ri y i.  y..
r  y
t
i `
i
i.

2
 y..
1929.719

LA
70
R
49, 104, 140, 113, 105, 123
SN
75
Tep
105, 116, 126
Tis
80, 90
0
5
0
2
1
70
634
75
347
170
0
4747.333
0
220.6667
50
Total sum = 1443
Corrected total SS = 9998.929
40.41156
788.0051
1093.719
475.9201
653.153
2
a. Construct the Analysis of variance table,
b. Test the null hypothesis of no location effect
c. Calculate the standard error of the least square mean for location R,
d. Calculate the standard error of the least squares mean for location LA.
3) Ex. 5.7.5 (STD) Paired data and a RCBD. Shuel (1952) gives the sugar concentration of
nectar in half-heads of red clover kept at different vapor pressures for 8 hours.
HEAD
Pressure 4.4 mm Hg.
Pressure 9.9 mm Hg.
1
62.5
51.7
2
65.2
54.2
3
67.6
53.3
4
69.9
57.0
5
69.4
56.4
6
70.1
61.5
7
67.8
57.2
8
67.0
56.2
9
68.5
58.4
10
62.4
55.8
a. Test the hypothesis that there is no difference in mean sugar content at the different
vapor pressures, using the fact that the two half-heads of each clover were stored at
different pressures. Also compute a 99 percent confidence interval for the difference in
mean sugar content.
b. Note that each column is a block involving two halves of the same red clover head.
Run an analysis of variance, following a randomized complete block design.
c. Test the null hypothesis of no treatment effect.
d. Is the MSError the same as in part a.?
July 28, 2010
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ST512
HOMEWORK 4 SSII10
e. Is one error and integral multiple of the other? How do you account for this?
4) Example 9.2 (STD book). Missing observations in a RCBD. Data is from
the oil content of Redwing flaxseed study discussed in class. Response
is Oil content, given in percentage. Experimental design follows a
randomized complete block design with 4 blocks and 6 treatments. Plots
were inoculated with spores suspensions of Septoria linicola, which
causes pasmo in flax. Treatments are stage of the plant when receiving
the inoculation. Assume that due to reasons not related to the
treatment applied, two of the plots were ruined and no measurement was
taken in any of these two plots, resulting in two missing observation,
represented with a dot in the following table.
Missing observations: Full Bloom in block 2 and Uninoculated in block 3.
1)
treat
block_1
block_2
block_3
block_4
Mean
Early_Bloom
33.3
31.9
34.9
37.1
34.300
Full_Bloom
34.4
.
34.5
33.1
34.000
Full_Bloom_P
36.8
36.6
37.0
36.4
36.700
Ripening
36.3
34.9
35.9
37.1
36.050
Seedling
34.4
35.9
36.0
34.1
35.100
Uninoculated
36.4
37.3
.
36.7
36.80
Mean
35.2667
35.3200
35.6600
35.7500
35.5000
Please fill the blanks
The GLM Procedure
Simple Means
July 28, 2010
Level of
treat
N
Early_Bloom
Full_Bloom
Full_Bloom_P
Ripening
Seedling
uninoculated
4
_
4
4
4
_
Level of
block
N
1
2
3
4
6
_
_
6
--------------y-------------Mean
Std Dev
34.3000000
__________
36.7000000
36.0500000
35.1000000
__________
2.23308456
0.78102497
0.25819889
0.91469485
0.98994949
0.45825757
--------------y-------------Mean
Std Dev
35.2666667
__________
__________
35.7500000
1.41938954
2.10760528
0.98640762
1.71551741
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ST512
HOMEWORK 4 SSII10
Analysis of Variance Table
The GLM Procedure
Dependent Variable: y
Sum of
DF
Source
SS Treat | Block ,  
SS  Block | Treat 
SS Treat | Block 
Mean Square
Model
8
28.06797619
3.50849702
Error
__
19.37202381
__________
Corrected Total
__
47.44000000
Source
SS  Block |  
Squares
block
treat
Source
block
treat
R-Square
Coeff Var
Root MSE
y Mean
0.591652
3.438646
1.220719
35.50000
F Value
___
Pr > F
0.0820
DF
Type I SS
Mean Square
F Value
Pr > F
3
5
0.99166667
27.07630952
0.33055556
5.41526190
0.22
3.63
0.8795
0.0282
DF
Type III SS
Mean Square
F Value
Pr > F
3
5
__________
___________
__________
__________
____
____
______
______
Missing observations:
2. Choose:
Type I SS
=

3. Are Treatment and Block orthogonal effects? Explain.
Type III SS
Solution
Parameter
Intercept
block
block
block
block
treat
treat
treat
treat
treat
treat
Estimate
1
2
3
4
Early_Bloom
Full_Bloom
Full_Bloom_P
Ripening
Seedling
uninoculated
37.21488095
-0.48333333
-0.76130952
0.20297619
0.00000000
-2.65446429
-3.12142857
-0.25446429
-0.90446429
-1.85446429
0.00000000
B
B
B
B
B
B
B
B
B
B
B
Standard
Error
t Value
Pr > |t|
0.81834165
0.70478263
0.75049541
0.75049541
.
0.94591683
1.03169613
0.94591683
0.94591683
0.94591683
.
45.48
-0.69
-1.01
0.27
.
-2.81
-3.03
-0.27
-0.96
-1.96
.
<.0001
0.5049
0.3289
0.7911
.
0.0149
0.0097
0.7921
0.3564
0.0717
.
Least Squares Mean for Treatment - Treatment LSMEAN
Average for each treatment level over the block effects
July 28, 2010
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ST512
HOMEWORK 4 SSII10
 37.2149 
 0.4833


 0.7613


 0.2030 
 ˆ1.  1 1 4 1 4 1 4 1 4 1 0 0 0 0 0   0  37.2149  1 4   0.4833  0.7613  0.2030  0    2.6545    34.3 
 ˆ   1 1 4 1 4 1 4 1 4 0 1 0 0 0 0   2.6545  37.2149  1 4 0.4833  0.7613  0.2030  0  3.1214   33.8331
 
 
 
 
 2.  


 36.9545 
 ˆ 6.  1 1 4 1 4 1 4 1 4 0 0 0 0 0 1   3.1214  37.2149  1 4  0.4833  0.7613  0.2030  0   0


 0.2545
 0.9045


 1.8545


 0 
y LSMEAN
Standard
Error
Pr > |t|
34.3000000
__________
36.7000000
36.0500000
35.1000000
__________
0.6103597
_________
0.6103597
0.6103597
0.6103597
_________
<.0001
<.0001
<.0001
<.0001
<.0001
<.0001
treat
Early_Bloom
Full_Bloom
Full_Bloom_P
Ripening
Seedling
uninoculated
Balanced Design with missing observations:
4. Choose:
Simple Arithmetic Mean =
Least squares means are estimates of

Least Squares Mean
 i  
LSMEAN Standard Error =
se( LSMEAN, r=4)
=
se(LSMEAN, r=3) =
ErrorMS
1.49016

 0.6104,
ri
4
ErrorMS
1.49016

 0.7048
ri
3
5. Compare se(Full Bloom LSMEAN) with the value calculated above, 0.7048
6. Compare se(Full Bloom LSMEAN - Early_Bloom LSMEAN ) to se (Ripening LSMEAN - Early_Bloom LSMEAN) .
July 28, 2010
Page 4