We have
observational units, which we divide into 2 blocks (say,
and
) of
and
experimental units respectively, in advance of an experiment. Thus,
.
We conduct the experiment consisting in applying a different treatment to each block. We measure a non-categorical response
, resulting in measurements
.
Let
and
be the group means of the two groups.
Let
![]() |
be the mean difference.
We would like to test whether treatment
has significantly larger mean than treatment
. If there is no difference (the null hypothesis, or
), the assignment of the observational units to the blocks has no effect, and
and
are only labels. Thus, any label assignment should produce the same result as the selected one.
There are
![]() |
ways to assign labels
and
to
observational units, and under null hypothesis, all of them are equally probable. Thus, we may consider a pure thought experiment in which we consider all possible label assignments. For every such assignment, we may compute the difference of the means
.
The proportion of label assignments for which the diference
is larger than that of the original assignment is the significance level of the test.
It can be shown that the normal theory can be applied to analyze the resulting test from the point of view of significance level and power. Below we outline the corresponding argument.
We see that
![]() |
where
![]() |
with probability
each and
is the value of the response of an observational unit.
Since all random variables
and
are jointly independent, the conditional distribution of
conditioned on some particular set of values of variables
is exactly the same as the marginal distribution with all
discarded.
Thus, the distribution of
is that of a difference of two random variables
, where
is an average of
independent copies of
and
is an average of
independent copies of
. Thus, if
and
are approximately
and thus large, by the Central Limit Theorem both these variables are approximately normally distributed with the same mean and variance as
. Hence, we may assume that the distributions are approximately
and
.
We observe that we would come to exactly the same conclusion if the distribution of the response variable were
.
![\[<br />
\delta = \overline{x}_A-\overline{x}_B<br />
\]](/sites/default/files/tex/bee5176fc72c39e2b0ac0b8fe20ae2653df1f2f3.png)
![\[<br />
{n\choose k} = \frac{n!}{k! l!}<br />
\]](/sites/default/files/tex/00cb89c15ee0e4011eb6311b5541146c7cd67e97.png)
![\[<br />
\delta = \sum_{i=1}^n U_i X_i<br />
\]](/sites/default/files/tex/c1a46152a9b6b67d64ce518e9749abf90adb8ef4.png)
![\[<br />
U_i\in\{1/k,-1/l\}<br />
\]](/sites/default/files/tex/c605413a0e530639fba124550cd268db00df314f.png)