This section describes a normal-scores test for testing the null hypothesis that a set of scores
is distributed symmetrically around a specified value *C*. If the null hypothesis is
true, then *C* is both the mean and the median of the distribution.

Elsewhere I describe two methods for transforming scores to a normal distribution. Here we use the simpler of those two methods--the equal-area method. The use of the more complex median-scores method is illustrated below.

Consider a set of normal scores applying to *just the right half* of a
normal
distribution. For instance, if *N* = 9, we want the scores that divide the right
half of a
normal distribution into 10 equal areas. Thus we want the scores whose right-hand areas are
.45, .40, .35, .30, .25, 20, .15, .10, and .05. A normal-curve table shows these 9 scores
are

.126 .253 .385 .524 .674 .842 1.036 1.282 1.645

In this test you rank all *N* scores by the absolute values of their deviations
from the
hypothesized center *C*, with the score closest to *C* getting a rank of
1. As usual, assign
mean ranks to ties. Then use those ranks to assign normal scores. If any scores exactly
equal *C*, assign them a normal score of 0. Then attach minus signs to the scores
below *C*,
leaving the scores above *C* as positive. Define* M* as the mean of
these scores. Then use an ordinary* z* test to test the null hypothesis that the
scores have a mean of zero; use 1/sqrt(*N*) for the standard error of the mean.
Thus *z* = *M* × sqrt(*N*).

The logic behind this test is that if the scores are in fact distributed symmetrically
around *C*, then positive and negative signs will be assigned with roughly equal
frequency to
the normal scores, and their expected mean will be zero and their standard deviation will be
approximately 1. But if, say, positive deviations from *C* are larger and more
numerous than
negative deviations, then *M* will tend to be positive. The logic is rather similar
to the logic
of the Wilcoxon signed-ranks test. The major
advantage of the normal-scores test is that there is little or no tendency for the various
possible values of *M* to be tied with each other--unlike the Wilcoxon rank-sum
*S*. That raises power, for reasons explained elsewhere.

New left-tail area = .5 + Original left-tail area/2

For instance, in the previous example with *N* = 9, the 9 left-tail areas
were

.074 .180 .286 .393 .500 .607 .714 .820 .926

The last equation transforms these values to

.537 .590 .643 .697 .750 .803 .857 .910 .963

The* z*-values corresponding to these areas are then the desired normal scores.
They are:

.093 .227 .367 .514 .674 .854 1.066 1.342 1.786

In other ways the method is like the equal-area method.