Why couldn't we use x / average for standardization instead of z scores?

Question

As I was reviewing data standardization and z score theory, i had this intuition. Suppose you have the results of people who took two different tests:

TEST A (mean=70%; std.dev=6%)
+--------------+-------+---------+-------+
| Participant# | score | z-score | x/avg |
+--------------+-------+---------+-------+
| 1            |    60 | -1.66   | 0.85  |
| 2            |    65 | -0.83   | 0.92  |
| 3            |    80 | 1.66    | 1.14  |
| 4            |    90 | 3.33    | 1.28  |
| 5            |    40 | -5.00   | 0.57  |
| ...          |       |         |       |
+--------------+-------+---------+-------+

TEST B (mean=75%;std.dev=7%)
+--------------+-------+---------+-------+
| Participant# | score | z-score | x/avg |
+--------------+-------+---------+-------+
| 1            |    60 | -2.14   | 0.8   |
| 2            |    70 | -0.71   | 0.93  |
| 3            |    80 | 0.71    | 1.06  |
| 4            |    90 | 2.14    | 1.2   |
| ...          |       |         |       |
+--------------+-------+---------+-------+

We can see that participant #3 in test A has a higher z score than participant #3 in test B and so that he's relatively better than his counterpart.

I can't find any info on the name of the measurement x/avg, but I have the intuition that it could be used as a proxy for standardized data.

I am surely wrong as it is mentioned nowhere, but why?

Answer 1

This may make sense for test scores, which are always scored on a $0–100$ scale. But in more general contexts it does not really make sense. Consider for example the these two very small data sets:

$$A=\{-990, 1000\}\\ B=\{-999.9, 1000\}$$

Surely these two are almost the same, and so their normalized versions should also be the same. But the first has a mean of $10$ and the second has a mean of only $0.1$. With your idea, these normalize quite differently:

$$A\div\bar A=\{-99, 100\}\\ B\div{\bar B}=\{-9999, 10000\}$$

This does not really make sense. They were almost the same before, so they should normalize the same. Normalization is supposed to reveal a hidden similarity. It is not supposed to obscure an obvious similarity.

But because the data sets almost the same, their standard deviations are also nearly the same, and when converted to Z-scores they are still nearly the same.

---

Another way to see the difficulty: Suppose we were to increase every participant's score on a test by 1000 points. Surely this shouldn't change the normalized scores? The test was the same. And in fact if you do this, the Z-scores won't change.

But your $x\div\bar x$ score will change, perhaps drastically. Consider participant 5's score on test A, which goes up from $\frac{40}{70} = 0.57$ to $\frac{1040}{1070} = 0.97$. And consider participant 4, whose score goes down from $\frac{90}{70} = 1.29$ to $\frac{1090}{1070} = 1.02$. Does this make sense?