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Noise is often the villain in the analysis of an experimental design. There
is just too much noise to see the target. The design we introduce next
is an effective noise reducer . The price is a loss of information (nothing
comes free). Also, as you will see, often it is not possible to do.
The setup is the same as the completely randomized design. We have two
treatments, T_{1} and T_{2}, applied to a response. We still want to test an
estimate the effect, .
The difference is that we can select
a pair of experimental units. For example, identical twins on a study involving
humans, the same house for a study on two house paints (halve the North
wall), the same field for a study on two varieties of wheat, etc. As I
said, "this may be impossible to do.''

Randomized Paired Design: Randomly select n paired experimental
units from the reference population. Within a pair, randomly assign one
of the pair to Treatment 1 and the other to Treatment 2. The experiment
(study) is run for a pre assigned time and during this time all other variables
are kept under control. At the end of the assigned time, we measure the
responses for the
n paired experimental units. Letting X
and Y denote the responses for Treatments 1 and 2, respectively
the data are in the paired form:
(X_{1},Y_{1}), ..., (X_{n},Y_{n}).
Although, the pairs are independent, within a pair there is dependency.
In fact, the more dependency within a pair, the more the noise reduction.
Hence, the twoindependentsample analysis of Chapter 9 is out. The key
is that
is still a typical Y minus a typical X,
i.e., read that as Y_{i} minus a typical X_{i}
where the subscript i refers to the ith pair. Thus the sample
of interest IS THE DIFFERENCES. That is,
D_{1} = Y_{1}  X_{1}, D_{2} = Y_{2}  X_{2}, ... , D_{n} = Y_{n}  X_{n}
Too wordy! Lets have an example . This is taken from Siegel, Nonparametric
Statistics. Ten pairs of identical twins, age 4, were randomly selected
for an experiment to investigate how nursery school affects the the social
awareness of a 4 year old. For each pair, one twin was randomly assigned
to go to nursery school while the other stayed home. At the end of the
time period, all 20 took the same test and their scores were recorded (bigger
means more socially aware). The data are, pair number in column 1, nursery
school twin in column 2, response of stayathome twin in column 3, difference in responses in column
4:
pair N H D
1 74 63 11
2 43 33 10
3 61 41 20
4 79 67 12
5 80 65 15
6 73 80 7
7 56 43 13
8 98 84 14
9 84 74 10
10 52 48 4
There are two immediate observations form this data set:
 1.
 The twin who went to nursery school seems to be more socially aware. A
dotplot on the differences is given next and a formal analysis is discussed
below.
. . : . . . . . .
++++++
5.0 0.0 5.0 10.0 15.0 20.0
 2.
 The pairing has really cut the noise. The range of the nursery school scores
is
98 43 = 55, the range of the stayathome scores is
84 33 = 51, but the range of the differences is
20  (7) = 27. Hence the noise level has been cut by about 1/2. The reason this reduction in noise takes place here is that four year olds are all over the map on social
relationships. Some are ready for school, some are far from ready, some
talk continuously while others are very shy, etc. And the scores reflect
this, (note the scores 98 and 43 in column 1). But identical twins are alike in social awareness (before the experiment). So if one twin scores high then so does the other while if one twin scores low so does the other. This certainly makes sense for these our identical twins.
Within a pair the scores are much more similar and, hence, the differences
are smaller.
Alright! I hear you clamoring. This is ad hoc. We want pvalues.
We WANT estimates and confidence intervals. Put up or shut up.
We can't use Chapter 9 but since we have a single sample, the D's,
we can use Chapter 7 for estimates and confidence intervals. For example,
we can estimate
by the median of the paireddifferences which
is 11.5. A confidence interval for the median is (10, 14.5)
which can be obtained using the class code (One sample bootstrap confidence intervals for the population mean and median) and typing in the paired differences in the big data box. Selecting median and submiting produces the bootstrap confidence interval for the median.
Exercise 11.3.1
 1.
 Finish the example for the twin data. Recall the paired differences were:
pair N H D
1 74 63 11
2 43 33 10
3 61 41 20
4 79 67 12
5 80 65 15
6 73 80 7
7 56 43 13
8 98 84 14
9 84 74 10
10 52 48 4
 (a)
 Obtain the value of the Wilcoxon test statistic. (Actually determine the number of negative averages (2) and subtract it for 10(11)/2.
 (b)
 Obtain the pvalue for a two sidedtest. Use the class code of course (Wilcoxon for paired designs). Conclude in terms of the problem.
 (c)
 Obtain (from class code) the estimated effect and the associated confidence interval. Conclude in terms of the problem.
 2.
 From Cushney and Peebles (1905)a, J. of Phisiology: Ten patients were selected for a study. The average number of hours that they slept was deterimed.
There were two parts to the study. In Part 1, they were given by a flip of the coin one of two drugs, Laevo and Dextro, and the average (over a week)
number of excess hours (over their usual average) was recorded. In Part 2 (after a wash out period), they were given the other drug, and the average (over
a week) number of excess hours (over their usual average) was recorded. The data are:
Patient Dextro Laevo
1 0.7 1.9
2 1.6 0.8
3 0.2 1.1
4 1.2 0.1
5 0.1 0.1
6 3.4 4.4
7 3.7 5.5
8 0.8 1.6
9 0.0 4.6
10 2.0 3.4
 (a)
 Obtain the value of the Wilcoxon test statistic, (diff = D  L).
 (b)
 Compare it what you would expect under H_{0}.
 (c)
 Obtain the pvalue for a two sidedtest. Use the class code (Wilcoxon for paired designs) of course. Conclude in terms of the problem.
 (d)
 Obtain (from class code) the estimated effect and the associated confidence interval. Conclude in terms of the problem.
 3.
 The data below are some measurements recorded by Charles Darwin in 1878. They consist of 15 pairs of heights in inches of crossfertilized plants and
selffertilized plants, Zea mays, each pair grown in the same pot.
POT CROSS SELF
1 23.500 17.375
2 12.000 20.375
3 21.000 20.000
4 22.000 20.000
5 19.125 18.375
6 21.550 18.625
7 22.125 18.625
8 20.375 15.250
9 18.250 16.500
10 21.625 18.000
11 23.250 16.250
12 21.000 18.000
13 22.125 12.750
14 23.000 15.500
15 12.000 18.000
 (a)
 Obtain the value of the Wilcoxon test statistic, (diff = C  S).
 (b)
 Compare it what you would expect under H_{0}.
 (c)
 Obtain the pvalue for a two sidedtest. Use the class code (Wilcoxon for paired designs). Conclude in terms of the problem.
 (d)
 Obtain (from class code) the estimated effect and the associated confidence interval. Conclude in terms of the problem.
Next: SignedRank Wilcoxon
Up: Design of Experiments
Previous: Completely Randomized Designs
20010101