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The PairedComparison: A Good Design
for
FarmerManaged Trials
revised
August, 2003
Rick Exner, ISU Extension PFI
Coordinator and
Richard Thompson, Practical Farmers of Iowa
THERE IS A PRESSING NEED FOR
AN ONFARM RESEARCH METHODOLOGY THAT:
 EMPOWERS FARMERS TO QUANTITATIVELY EVALUATE ALTERNATIVE AGRICULTURAL PRACTICES;
 PRODUCES RESULTS CONSIDERED CREDIBLE BY BOTH AGRONOMISTS AND OTHER FARMERS;
 IS “FARMABLE” USING EXISTING EQUIPMENT AND WITH A MINIMUM OF ADDITIONAL EFFORT

The pairedcomparison
trial can meet many of the needs of onfarm research. If treatment strips of one or two
equipmentwidths are run completely across the field, no extra stopping, turning or
adjustments are required. Because pairs of these narrow strips lie close to one another,
field variability between treatment strips is minimized. The pair of treatments is
replicated at least six times in the field to overcome chance field differences. The order
of the two practices in each pair is chosen at random, avoiding a source of unintentional
bias. Each data pair yields one difference. These differences can be analyzed
using a simple, singlesample technique. Farmers, themselves, if they have a calculator or
spreadsheet that does square roots, can generate the L.S.D. (least significant difference)
with which to evaluate a twotreatment trial.
RANDOMIZATION
AND REPLICATION
In this trial the practices, or
“treatments,” are paired six times, making six “replications,” or
“reps.” Replication is crucial because it gives you a
“second opinion” (and a third, fourth, ...sixth) on your question. With only one
pair in the trial, you would have to base your conclusions on only one observation 
without knowing how representative that single result really was.
Notice that in this trial the practices or
“treatments” are paired in random order. That is, the strips don’t just go
“AB AB AB...” across the field. Randomization is important to
avoid the biasing effects that you often don’t even anticipate. For instance, if
practice A were always downhill or were always downwind from practice B, the results could
reflect the position of the strips as well as the practices in them. So to avoid
any question of conscious or unconscious bias: randomize!
Work out a new randomization for each trial. (A
field layout is only random the first time you use it.) If the trial will have
two treatments, start with two pieces of paper and write the name of a treatment on each
one. Put them in a milk bucket or a hat and give it a shake. Now pick one piece of paper
out without looking to see which it is. That will be the treatment for the first strip of
the first pair. Obviously the second strip in the pair gets the other treatment. Repeat
this procedure for each pair of strips (each replication). Write down your results as you
go. When you finish, take a look at your randomization. If you don’t see any really
good reason why that arrangement of treatments would prejudice trial results then keep it.
Otherwise repeat the whole process. Write down your field diagram in two places 
one safe place in your office and one convenient place like a pocket notebook. The last
thing you want is to ruin your work by losing track of where you did what in the field.
This is the first step in your record keeping. As you see things in the field, write them
down in a pocket notebook, then transfer the information to a more permanent record. Your
recorded observations can lead to insights months  even years later.

AN EXAMPLE OF A PAIREDCOMPARISON
TRIAL AND ITS ANALYSIS

Fertilizer
Type




Pair number 
None 
Starter 
Difference
(X) 
(XX) 
(XX)^{2} 
(Yield in bu. /acre) 
1 
117.9 
118.9 
1.0 
6.62 
43.78 
2 
125.2 
112.0 
+13.2 
7.58 
57.51 
3 
119.9 
100.7 
+19.2 
13.58 
184.51 
4 
114.9 
110.8 
+4.1 
1.52 
2.30 
5 
116.9 
119.9 
3.0 
8.62 
74.25 
6 
119.7 
118.5 
+1.2 
4.42 
19.51 
n = 6 


X = 5.62


§ = 381.85

Sample Variance: s_{x}^{2}
= §/(n1)
= 381.85/5 = 76.37 Variance of the Mean: s_{x}^{2} = s_{x}^{2}/n
= 76.37/6 = 12.73
Standard Error of the Mean: s_{x} = (12.73)^{˝} = 3.57 bu./acre
(the ^{“˝”} means take the square root of 12.73)
Student’s t_{.05, 5} = 2.571
Get t from a “t table.” A
high school statistics book would have a t table. “.05” (sometimes called “a”)
is the chance of an error at the 95% confidence level. “5” is the number “n” minus 1, or (61) here.
The table may use the term “degrees of freedom” for the number (n1). The
correct number for t is in the table at the intersection of the “.05”
column and the “5” row. You might want to demand a higher level of confidence
(99%), or you might settle for a less stringent burden of proof (90%, 80%). Least Significant Difference at 95% confidence level:
LSD_{.05} = s_{x}
× t_{.05, 5} = 3.57 × 2.571
= 9.17 bushels per acre
So here the observed treatment difference (X=5.62 bushels) was less than the difference
(9.17 bushels) that would occur simply by chance 5 times out of 100. That is, the
difference observed was not statistically significant at the commonly used 95% confidence
level. So in this case there was no significant yield difference between the corn that got
starter fertilizer and the corn that received none. However, the result comes from just
one field and one year. It’s a good idea to repeat a trial elsewhere to see if the
outcome is reproducible.
Properly conducted pairedcomparison trials on Iowa
farms in 1987 were capable of detecting finer treatment differences than some experiment
station research (P. Rzewnicki, et al. Fall 1988. American Journal of Alternative
Agriculture, Vol. 3, No. 4). Onfarm research does not replace experiment station work,
which often uses more complex designs. The point is that for what these simple onfarm
trials set out to accomplish, they do a very credible job.

STUDENT’S tDISTRIBUTION
CRITICAL POINTS
“2TAILED” TEST OF
DIFFERENCE (EITHER GREATER OR LESS)
Number of
Pairs

Degrees of
Freedom

a = .20

a = .10

a = .05

a = .01

1 
N.A. 
N.A. 
N.A. 
N.A. 
N.A. 
2

1

3.078

6.314

12.706

63.657

3

2

1.886

2.920

4.303

9.925

4

3

1.638

2.353

3.182

5.841

5

4

1.533

2.132

2.776

4.604

6

5

1.476

2.015

2.571

4.032

7

6

1.440

1.943

2.447

3.707

8

7

1.415

1.895

2.365

3.499

9

8

1.397

1.860

2.306

3.355

10

9

1.383

1.833

2.262

3.250

11

10

1.372

1.812

2.228

3.169

12

11

1.363

1.796

2.201

3.106

13

12

1.356

1.782

2.179

3.055

14

13

1.350

1.771

2.160

3.012

15

14

1.345

1.761

2.145

2.977

When a =
.20, the confidence level = (1.00  .20), or 80%. (Chance of wrong conclusion = 20%.)
When a = .10,
the confidence level = (1.00  .10), or 90%. (Chance of wrong conclusion = 10%.)
When a = .05,
the confidence level = (1.00  .05), or 95%. (Chance of wrong conclusion = 5%.)
When a = .01,
the confidence level = (1.00  .01), or 99%. (Chance of wrong conclusion = 1%.)
A “Student’s t” table can
be found in the back of most statistics textbooks. If the experimental question is:
“Is practice ‘A’ different from practice ‘B’ (less or
more)?” then you want a “twotailed” t table that divides the
chance for error () between the lower and the upper tails of the bell curve. If your
question is specifically “Is practice ‘A’ greater than practice
‘B,’?” then you would use a onetailed ttable. Most t
tables are twotailed and don’t even bother to say so. Notice how t
diminishes as the number of replications increases? Other things being equal, a smaller t
means a smaller L.S.D., and a smaller L.S.D. means the trial is more sensitive to the
treatment differences you are looking for.

WORK SHEET PROBLEM
DEEPBANDED P & K FERTILIZER
VERSUS NONE
HARLAN AND SHARON GRAU FARM, 1990

Treatment Type




Pair number

Deep Band

None

Difference
(X) 
(XX)

(XX)^{2}

(Yield in Bu./acre) 
1 
98.8 
85.8 
_______ 
____.____ 
_____.____

2 
96.9 
85.8 
_______ 
____.____ 
_____.____

3 
97.4 
87.0 
_______ 
____.____ 
_____.____

4 
101.2 
89.9 
_______ 
____.____ 
_____.____

5 
103.5 
93.2 
_______ 
____.____ 
_____.____

6 
85.8 
87.8 
_______ 
____.____ 
_____.____

n = _____ 

X =
______.____
(Mean Difference)

§ = _________.____
(Sum of Squares)

n–1 = _____
Sample Variance: s_{x}^{2} = §/(n1) =
______.____
Variance of the Mean: s_{x}^{2} = s_{x}^{2}/n
= ______.____
Standard Error of the Mean: s = (s_{x}^{2})^{˝} = ______.____
bu./acre
(the ^{“˝”} means take the square root of s_{x}^{2})
Student’s t_{.05, n1} =
______.______
t_{.05, n1} comes from a “t
table.” A high school statistics book would have a t table. “.05”
(sometimes called “a”) is the chance, at
the 95% confidence level, of incorrectly concluding there is a real yield difference
between treatments. The number (n1) is one less than the number of pairs in the field.
The table may use the term “degrees of freedom” for the number (n1). The
correct number for t is in the table at the intersection of the “.05”
column and the “n1” row.
Least Significant Difference at 95% confidence
level:
LSD_{.05} = s_{x} × t_{.05, n1} = _____._____ ×
_____._____
= _____._____ bushels per acre
Was the observed treatment difference (X) more or less than the difference (LSD_{.05}
in bushels) that would occur simply by chance 5 times out of 100? If it was greater, then
the yield difference observed was statistically significant at the commonly used 95%
confidence level.

BLANK WORK SHEET
YOUR EXAMPLE HERE

Treatment Type




Pair
number

Deep Band

None

Difference
(X) 
(XX)

(XX)^{2}

(Yield in
Bu./acre) 
1 
_______ 
_______ 
_______ 
____.____ 
_____.____

2 
_______ 
_______ 
_______ 
____.____ 
_____.____

3 
_______ 
_______ 
_______ 
____.____ 
_____.____

4 
_______ 
_______ 
_______ 
____.____ 
_____.____

5 
_______ 
_______ 
_______ 
____.____ 
_____.____

6 
_______ 
_______ 
_______ 
____.____ 
_____.____

7 
_______ 
_______ 
_______ 
____.____ 
_____.____

8 
_______ 
_______ 
_______ 
____.____ 
_____.____

n = _____ 

X =
______.____
(Mean Difference)

§ = _________.____
(Sum of Squares)

n–1 = _____
Sample Variance: s_{x}^{2} = §/(n1) =
______.____
Variance of the Mean: s_{x}^{2} = s_{x}^{2}/n
= ______.____
Standard Error of the Mean: s = (s_{x}^{2})^{˝} = ______.____
bu./acre
(the ^{“˝”} means take the square root of s_{x}^{2})
Student’s t_{.05, n1} =
______.______
Least Significant Difference at 95% confidence
level:
LSD_{.05} = s_{x} × t_{.05, n1} = _____._____ ×
_____._____
= _____._____ bushels per acre
Was the observed treatment difference (X) more or less than the difference (LSD_{.05}
in bushels) that would occur simply by chance 5 times out of 100? If it was greater, then
the yield difference observed was statistically significant at the commonly used 95%
confidence level.

BLANK FIELD PLOT LAYOUT
