Sample Solution to Assignment 2
Your solution to the random assignment will be different.
1. Convenience Assignment
It is easiest to plant one variety on 18 plots on one side of the field and the other
variety on the 18 plots on the other side. Below is a picture of such and assignment
and the yields, in bushels per acre, for each plot. The yields and a summary of those
yields are given below.
A
130
A
149
A
141
A
150
A
139
A
155
a) Variety n
A
18
B
18
sp 
mean
144.9
141.8
A
149
A
133
A
156
A
142
A
155
A
138
A
139
A
152
A
137
A
155
A
139
A
150
B
155
B
131
B
146
B
136
B
147
B
137
B
137
B
147
B
132
B
152
B
137
B
145
B
145
B
136
B
148
B
133
B
153
B
136
std. dev
8.29
7.65
nA  1s 2A  nB  1sB2
nA  nB  2
std. error of difference  s p
178.292  177.652

 63.6233  7.976
34
1
1
1
1

 7.976

 2.659
n A nB
18 18
Using a two-sample t-test, the value of the test statistic is
t
144.9  141.8  1.17
2.659
with a two-sided P-value of 0.2499. Because the P-value is not small (> 0.05),
this indicates that although variety A has a slightly higher sample mean yield
than variety B, the two varieties are not statistically different.
The
population means could be the same.
b) This is probably not a fair comparison. One side of the field may be more
fertile than the other or get more water than the other. There could be many
lurking variables that could be affecting the yields besides that variety.
1
2. Systematic Assignment (Alternating)
Many people think that an alternating sequence is a random, or at least an unbiased,
sequence. Below is a picture of an alternating pattern, like a checkerboard, and the
yields, in bushels per acre, for each plot. The yields and a summary of those yields
are given below.
A
130
B
137
A
141
B
138
A
139
B
143
a) Variety n
A
18
B
18
sp 
mean
142.3
144.5
B
137
A
133
B
144
A
142
B
143
A
138
A
139
B
140
A
137
B
143
A
139
B
138
B
155
A
143
B
146
A
148
B
147
A
149
A
149
B
147
A
144
B
152
A
149
B
145
B
145
A
148
B
148
A
145
B
153
A
148
std. dev
5.75
5.37
nA  1s 2A  nB  1sB2
nA  nB  2
std. error of difference  s p

175.752  175.37 2
 30.9497  5.563
34
1
1
1
1

 5.563

 1.854
n A nB
18 18
Using a two-sample t-test, the value of the test statistic is
t
142.3  144.5  1.19
1.854
with a two-sided P-value of 0.2390. Because the P-value is not small (> 0.05),
this indicates that although variety B has a slightly higher sample mean yield
than variety A, the two varieties are not statistically different. The
population means could be the same.
2
3. Random Assignment
a) Below is one possible random assignment of varieties to plots. This
randomization was accomplished using a six-sided die. Rolling the die once
gives a row number, rolling the die again gives column number thus
identifying a unique plot. For example if the two rolls were 3, 6, then row 3,
column 6 would be planted with A. This was done until 18 unique plots were
assigned variety A. The remaining plots were assigned variety B. Below are
the yields and a summary of those yields.
b) and c)
1
2
3
4
5
6
d) Variety n
A
18
B
18
sp 
1
A
130
A
149
A
141
A
150
B
127
B
143
mean
150.4
136.3
2
A
149
B
121
A
156
B
130
A
155
B
126
3
A
139
B
140
A
137
B
143
B
127
B
138
4
A
167
B
131
A
158
B
148
A
159
A
149
5
B
137
A
159
A
144
B
152
A
149
B
145
6
A
157
B
136
A
160
B
133
B
153
B
136
std. dev
9.49
8.74
nA  1s 2A  nB  1sB2
nA  nB  2
std. error of difference  s p
179.492  178.742

 83.22385  9.123
34
1
1
1
1

 9.123

 3.041
n A nB
18 18
Using a two-sample t-test, the value of the test statistic is
t
150.4  136.3  4.64
3.041
A two-sample t-test on these data has a test statistic of 4.64 with a P-value
less than 0.0001. Because the P-value is so small (< 0.05), there is a
statistically significant difference between the sample mean yields for the two
varieties. With this randomization variety A is about 14 bushels per acre
better than variety B, on average.
e) Closer examination of the “Truth” reveals that in each plot variety A is 12
bushels higher than variety B.
3
JMP Output
1. Yield using Convenience Assignment
170
Convenience
160
150
140
130
120
A
B
Variety
t Test
B-A
Assuming equal variances
–3.1111
2.6577
2.2901
–8.5123
0.95
Difference
Std Err Dif
Upper CL Dif
Lower CL Dif
Confidence
-10
-5
0
5
t Ratio
DF
Prob > |t|
Prob > t
Prob < t
–1.17059
34
0.2499
0.8750
0.1250
10
Means and Std Deviations
Level
A
B
Number
18
18
Mean
144.944
141.833
Std Dev
8.28515
7.64853
Std Err Mean
1.9528
1.8028
4
2. Yield using Systematic Assignment
170
Systematic
160
150
140
130
120
A
B
Variety
t Test
B-A
Assuming equal variances
Difference
Std Err Dif
Upper CL Dif
Lower CL Dif
Confidence
2.2222
1.8543
5.9905
–1.5461
0.95
t Ratio
DF
Prob > |t|
Prob > t
Prob < t
1.198443
34
0.2390
0.1195
0.8805
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Means and Std Deviations
Level
A
B
Number
18
18
Mean
142.278
144.500
Std Dev
5.74769
5.37149
Std Err Mean
1.3547
1.2661
5
3. Yield using Random Assignment
170
Random
160
150
140
130
120
A
B
Variety
t Test
B-A
Assuming equal variances
–14.111
3.042
–7.928
–20.294
0.95
Difference
Std Err Dif
Upper CL Dif
Lower CL Dif
Confidence
-15 -10
-5
0
5
10
t Ratio
DF
Prob > |t|
Prob > t
Prob < t
–4.63811
34
<.0001
1.0000
<.0001
15
Means and Std Deviations
Level
A
B
Number
18
18
Mean
150.444
136.333
Std Dev
9.49441
8.74475
Std Err Mean
2.2379
2.0612
6