It is often useful to be able to transform a set of scores into a particular distribution--most often into a standard normal distribution. That allows you to then apply statistical methods designed for that distribution. It's difficult for a critic to object to use of a technique that assumes a normal distribution if you have transformed the data into a normal shape.

This section discusses two major ways to transform data. Both are practical with a wide variety of statistical packages. One very inexpensive package that includes these capabilities is Mystat for Business; the "regular" Mystat includes just the first of the two methods. We'll call the two methods the equal-area method and the median-score method. The equal-area method is easier to understand and to execute, while the median-scores method seems somewhat more satisfying intuitively. In most practical applications it probably makes little difference which method you use. We'll assume here that you want to transform scores into a normal distribution, though both methods can be applied to other distributions as well.

The central problem is that the normal distribution, or other distributions you might use, are distributions of infinitely large populations. Thus it is impossible for any finite number of scores to form exactly a normal distribution. Rather you must transform the scores so you can say that the transformed scores come "as close to a normal distribution as a set of that many scores can come". The problem is that the quoted phrase has no single meaning, so we must choose some particular meaning. The equal-area and median-score methods assign two different meanings to the quoted phrase.

-1.282 -.842 -.524 -.253 0 .253 .524 .842 1.282

Therefore these 9 scores have a certain claim to being called the 9 scores that come as close to forming a standard normal distribution as any 9 scores can come.

More generally, if you have a sample of *N* cases, you divide a standard
normal
distribution into* N*+1 equal areas, and take the *N* cutting points
between adjacent areas.
Those* N* values are the *N* transformed scores. Assign the highest
transformed score to the
highest raw score, the second-highest transformed score to the second-highest raw score, and
so on.

If you rank the raw scores from low to high, then the area to the left of each score is
Area = rank/(*N*+1). For instance, in a set of 9 scores the 8th lowest score
(which is also
the second highest score) gets a rank of 8, so Area = rank/(*N*+1) = 8/10 = .8.
Therefore
the appropriate transformed score is the score chosen so that 80% of a standard normal
distribution is to the left of that score. A standard normal table shows that value is
.842.

If some of the raw scores are tied, use mean ranks. For instance, if the 7th and 8th
ranked scores were tied when *N* = 9, then you would use Area =
rank/(*N*+1) = 7.5/10 =
.75.

Like the 9 scores derived by the equal-area method, these 9 scores have a reasonable claim to being called the 9 scores that come as close to forming a standard normal distribution as any 9 scores can come. The difference is a matter of intuition, but most statisticians think that median scores are intuitively more appealing than equal-area scores.Rank 1 2 3 4 5 6 7 8 9 Median -1.446 -.917 -.564 -.271 0 .271 .564 .917 1.446

One might well ask why not literally use expected values? That is, why not use the means instead of the medians of the sampling distributions of ranked scores? There are two practical answers. First, expected values are far more difficult to calculate than median scores. Second, it is much easier to generalize the median-score method to distributions other than normal distributions.

The median-score method is like the equal-area method in that you first find an Area
to the left of each score, then use a standard normal table (or other table if you're
transforming to some other distribution) to find the scores corresponding to those areas. In
the equal-area method with* N* = 9, the 9 areas were found by the formula Area
=
rank/(*N*+1) and were simply .1 .2 .3 .4 .5 .6 .7 .8 .9. In the median-score
method the
areas are

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

To see why these particular values are used, consider the second-highest value of
.820. It can be shown that in any continuous distribution (not just a normal distribution), if
you draw infinitely many random samples of size 9 from the distribution, and take the
second-highest score in each sample, the median of all those second-highest scores will be
the point chosen so that .820 of the population distribution's total area falls to its left.
Similar statements can be made about .607, .714, and the other scores in this list of 9 scores.
Therefore the second step in the median-score method is just like the second step in
the equal-area method; find the scores corresponding to particular areas. The two methods
differ only in the first step. Where the equal-area method uses the formula Area =
rank/(*N*+1), the median-score method uses the beta distribution. It can be
shown that the
desired Area equals the Inverse Beta Function (what Systat and Mystat for Business call BIF)
for the 3 values .5, rank, and *N*+1 - rank. Thus for instance if *N* =
9 and you want the
Area for the second-highest score (whose rank is 8), you want BIF(.5, 8, 2). In those
programs, if* N* = 9 then all the Areas can be found at once by starting with a
column of
ranks (titled Rank), and then writing the command

LET AREA = BIF(.5, RANK, 10 - RANK)

Whatever *N* is, replace 10 in this command by *N*+1. Ties are
handled the same way as in
the equal-area method; assign the mean rank to tied values.

A table of median-score normal scores, for N up to 100 but with no allowance for ties, is available.