ST 524
NCSU - Fall 2008
Homework 2
Due: Thursday September 25, 2008
I. An evaluation of sample size is carried on paper “Sample Size for Single Double and Three-Way Hybrid
Corn Ear Traits”
1. Statistical Design
2. Response variables
Write down the statistical linear model for yield of grains at 13% humidity, in words if preferred.
Yij     i   j  eij , where
Yij , is the yield of grains at 13% humidity for ith hybrid in jth block, i = 1, ..., 6 , j = 1, 2, 3
 is the overall mean
 i , is the fixed effect associated with ith hybrid. ,
t

i 1
i
0
 j is the fixed effect corresponding to jth block,   j  0
r
j 1
eij is the random effect for i hybrid and j block., eij ~ iidN  0,  2 
th
th
3. The Analysis of variance table for Yield is presented next,
How many blocks should be used if the minimum detectable difference (  ) in any direction
to be found between any two hybrids is 5 ton h-1 at a significance level of 0.05 and power equal
0.9
Tuesday September 15, 2008
1
ST 524
NCSU - Fall 2008
2
 2.23 
r  2 1.96  1.28  
 ,
Use
 5 
r2
Two replications are needed to detect a minimum
difference between pairs of treatment equal to 5 ton-1
at a significance level equal 0.05, and
with power=0.90
2
4. Subsampling
Write down the statistical linear model for response Weight of 100 grains.
Yijk     i   j  eij   ijk , where
Yij , is the weight of 100 grains for the kth ear of
 is the overall mean
ith hybrid selected in jth block, i = 1, ..., 6 , k = 1, …, 5, , j = 1, 2, 3,
 i , is the fixed effect associated with ith hybrid. ,
t

i 1
i
0
 j is the fixed effect corresponding to jth block,   j  0
r
j 1
eij is the random effect for i hybrid and j block., eij ~ iidN  0,  2 
th
 ijk
th
is the random effect for he kth ear of ith hybrid selected in jth block,.
 ijk ~ iidN  0,  2 
5. Analysis of variance table for Weight of 100 grains is presented next,
Fcalc, Hybrids = 26.9/ 61.5  0.4374 ,
Fcalc, Blocks = 106.2 / 61.5  1.7268
a) Explain differences between Experimental and Sampling Error.
Experimental error represent the variation due to uncontrolled random effects associated to
experimental units (plots) that received a given hybrid: plot-to-plot variation.
Tuesday September 15, 2008
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ST 524
NCSU - Fall 2008
Sampling error represents the variation due to uncontrolled random effects associated to
observational units (ears), a number of which were selected randomly within each
experimental unit (plot): ear-to-ear variation.
b) Experimental Coefficient of variation (Exp. CV) is calculated by 24.2% 
19.6
 100 . How would you interpret these two
32.4
while Am CV corresponds to 13.6% 
c)
61.5
 100 ,
32.4
values?
24.2% represent the relative experimental variation with respect the overall mean weight of
100 grains, ie., 61.5  7.8421 g (experimental error) represents 24.2% of 32.4 g (mean
weight) at an ear basis.
13.6% represent the relative sampling variation with respect the overall mean weight of 100
granis, ie., 19.6  4.4272 g (sampling error) represents 19.6% of 32.4 g (mean weight), at an ear
basis.
Calculate the variance components in the Expected Experimental Error MS.
E(Experimental Error MS) =
E(Sampling Error MS) =
Estimated value for
 2
 2  s e2
 2
is 19.6 = MS(Sampling Error) =
ˆ2
MS  Experimental Error   ˆ2 61.5  19.6
2

 8.38 , thus ˆ e  8.38
5
5
d) Researchers want to analyze the increase on the number of blocks and/or number of ears per
plot, necessary to find significance for the set of effects similar to the observed in Table 3.
What combinations of r and s would you suggest? Consider power = 0.80
Estimated value for
 e2 is given by
Type
Hybrid
Mean
Single
Single
Three-way
Three-way
Double
Double
P30F33
P Flex
AG 8021
DG 501
AG 2060
DKB 701
Overall
33.2
33.5
33.8
31.1
30.5
32.3
32.4
Obs
Effect
Label
1
2
3
4
hybrid
hybrid
hybrid
hybrid
hybrid
hybrid
hybrid
hybrid
Weight of 100 grains (g)
Variance
CV
Standard
Deviation
35.10
17.8
5.9245
12.47
10.5
3.5313
17.57
12.4
4.1917
14.10
12.1
3.7550
16.83
13.4
4.1024
21.43
14.3
4.6293
19.6
13.6
4.4272
Num
DF
Den
DF
t
r
s
alpha
5
5
5
5
10
10
10
10
6
6
6
6
5
6
7
9
23
18
15
11
0.05
0.05
0.05
0.05
dendf2
fcrit
ncparm
power
20
25
30
40
2.71089
2.60299
2.53355
2.44947
17.0537
16.0156
15.5707
14.6810
0.80707
0.80255
0.80584
0.79937
Possible combinations: (5 blocks, 23 ears/plot), (6 blocks, 18 ears/plot), (7 blocks, 15 ears/plot), (9 blocks, 11 ears/plot)
2
2
Calculate the Standard Error of a hybrid mean for each case, where s.e y i..      e . Which
rs
r
combination of r and s in d) will result in the smallest standard error for the mean?
For original combination of r and s
 
e)
 
s.e y i.. 
ˆ2
rs

ˆ e2
r
 
Note that s.e y i.. 
Tuesday September 15, 2008

19.6 8.38

 4.1  2.0248 .
3 5
3
Experimental Error MS
61.5

 2.0248
rs
3 5
3
ST 524
NCSU - Fall 2008
 
r
s
Power
5
23
s.e y i.. 
 
19.6 5  8.38

 1.3588
5  23 5  23
0.80707
6
18
s.e y i.. 
 
19.6 18*8.38

 1.2562
6 18
6 18
0.80255
7
15
s.e y i.. 
 
19.6 15*8.38

 1.1763
7 15
7 15
0.80584
9
11
s.e y i..
 
s.e y i.. 
0.79937
19.6 11*8.38

 1.0626
9 11
9 11
 
Smallest standard error of the mean is given by r = 9 and s = 11, with s.e yi..  1.0626
f)
The optimum allocation solution for number of ears selected within each plot, when
minimizing the total cost, is given by s

c1s 2
, where s is the number of ears selected
c2 se2
within each plot, c1 is the management cost for each plot and c2 is the cost per ear sampled,
find the optimum allocation number of ears per plot for a relative cost c1 = 3.0 and c2 = 0.1,
with s 2  ˆ 2  19.6 and se2  ˆ e2  8.38 . Do any of the suggested combinations of (r, s) for
the desired power correspond with the optimum number of ears to be sampled within each
plot?
s
c1s 2
3.0 19.6

 8.37  9 , optimum number of ears is 9 ears per plot.
2
c2 se
0.1 8.38
No, there is not any combination with s = 9, but the combination of r = 9 blocks and s = 11
is the nearest to the optimum allocation of ears per plot.
g) Through independent calculations of sample size, researchers found that three blocks and 21
ears per block were adequate to attain a CV = 5%, an improvement over the observed CV
average (Am CV) of 13.6%.
Comment the following
AmCV is dependent on the Sampling Error and a
reduction of the AMCV is directly linked to the
number of ears sampled within each plot. Researchers
estimated the sample size required within each hybrid
and ‘averaged’ them to get the value of 21 ears to be
sampled within each plot. Their final suggestion may
go against the initial objective of reducing AmCV. A
sample of 9 ears may reduce the CV to a value about
10% , although the variance of a hybrid mean is
reduced. See below.
10  1.96 ^ 2*17.8 ^ 2 /(9 /(1  9 /10)) for PF0F33
Weight of 100 grains (g)
Type
Hybrid
Single
Single
Three-way
Three-way
Double
Double
P30F33
P Flex
AG 8021
DG 501
AG 2060
DKB 701
Overall
Mean
33.2
33.5
33.8
31.1
30.5
32.3
32.4
Variance
35.10
12.47
17.57
14.10
16.83
21.43
19.6
CV
17.8
10.5
12.4
12.1
13.4
14.3
13.6
Standard
Deviation
5.9245
3.5313
4.1917
3.7550
4.1024
4.6293
4.4272
Number of
ears/plot (s)
for desired
CV=5%
22
13
15
15
17
18
17
Use the standard error of the mean to compare a design with three blocks and 21 ears
subsampled within each plot, with a design with 7 blocks and 9 ears per plot. What power is
associated with each scenario?
Tuesday September 15, 2008
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ST 524
NCSU - Fall 2008
r
s
3
21
7
9
Obs t ro so Effect
Num
DF
 
Power
s.e y i..
 
s.e y i.. 
19.6 21 8.38

 1.7619
3  21
3  21
0.40995
19.6 9*8.38

 1.2281
79
79
0.55294
 
s.e y i.. 
Den
DF
FValue
ProbF Label
alpha r
s dendf2
ncparm
fcrit
power
44 6 3 5 hybrid
5
10
0.44 0.8078 hybrid 0.05 3 21
10
9.34244 3.32583 0.40995
128 6 3 5 hybrid
5
10
0.44 0.8078 hybrid 0.05 7 9
30
9.34244 2.53355 0.55294
While keeping the total number of observational units constant, an increase on the number of blocks
reduces the standard error of the mean, and increases the power.
h) Calculate the sample size needed to estimate the observed difference between Single and
Double Hybrids with a power = 0.80.
Single
Single
Hybrid
P30F33
P Flex
Mean
Csingle vs double
Csingle vs three-way
33.2
1
1
33.5
1
1
Note that F
calc ,C1 
Threeway
AG
8021
Threeway
Double
Double
DG 501
AG 2060
DKB 701
33.8
31.1
30.5
-1
32.3
-1
-1
-1
estimated
contrast
3.9
1.8
Fcalc
0.9274
0.1976
SS  C2  1
SS  C1  1
3  5 1.82
3  5  3.92

 0.1976

 0.9274 , and Fcalc ,C2 
Exp Error MS
61.5
Exp Error MS
61.5
Standard Error of contrast is given by

Estimate statement in Proc Mixed uses
coefficients ½, ½, -½ , -½ , and s.e (contrast) is
2.0248

Exp Error MS
4  61.5
2
2
2
2
s.e. Cˆ1 
 4.0497
1  1   1   1 
3 5
15
 
From output,
Obs
Effect
1
2
3
4
5
contrast
contrast
contrast
contrast
contrast
i)
FValue
0.93
0.93
0.93
0.93
0.93
Label
Single
Single
Single
Single
Single
vs
vs
vs
vs
vs
alpha
r
s
ncparm
power
EXP_
Error_MS
se_contrast
0.05
0.05
0.05
0.05
0.05
6
7
8
9
10
23
19
17
15
13
8.53244
8.22329
8.40878
8.34695
8.03780
0.80158
0.79241
0.80495
0.80487
0.79223
212.34
178.82
162.06
145.30
128.54
1.24044
1.15953
1.09161
1.03745
0.99437
Dual
Dual
Dual
Dual
Dual
Calculate the sample size needed to estimate the observed difference between Single and
Three-way Hybrids with a power = 0.80.
For Contrast “single vs Three-way”, the power for any of the combinations of r and s considered is
below 0.80, the largest values for power is given by
Obs
Effect
215
216
FValue
contrast
contrast
0.20
0.20
Label
Single
Single
alpha
vs Triple
vs Triple
0.05
0.05
r
s
10
10
ncparm
24
25
3.16098
3.29268
power
0.41304
0.42705
Increasing r and s in calculation of power, we get power = 0.8 for the following combination of r and s
Obs
585
672
1520
1714
Single
Single
Single
Single
Label
FValue
vs
vs
vs
vs
0.20
0.20
0.20
0.20
Triple
Triple
Triple
Triple
effect
c
c
c
c
alpha
0.05
0.05
0.05
0.05
r
7
8
17
19
s
91
79
36
32
ncparm
8.38976
8.32390
8.06049
8.00780
power
0.80031
0.80100
0.80092
0.79941
j) Does any of the suggested sample size agree with the needed number of blocks found in 3?
No, since the minimum detectable difference between a pair of means is larger than any
difference observed between mean, pairwise-difference range is 0.1 to 3.8 in absolute value.
Tuesday September 15, 2008
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