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Is there a symbol to indicate that variables have been standardized?

For example, if I have 2 different scoring functions Score1 and Score2. Let's say I want to form a combo score and show that the scores have been standardized, e.g.,

\[ \text{ComboScore1} = \frac{\text{std. Score1} + \text{std. Score2}}{2} \]

two differentiate from a flavor of a ComboScore where the individual scores have not been standardized

\[ \text{ComboScore2} = \frac{\text{Score1} + \text{Score2}}{2} \]

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  • $\begingroup$ You are right, I removed the tag $\endgroup$
    – user39663
    Oct 13, 2014 at 2:48
  • $\begingroup$ How are you standardizing exactly? Are these standardized for sample mean and standard deviation, or in some other way? $\endgroup$
    – Glen_b
    Oct 13, 2014 at 2:50

1 Answer 1

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Standards can vary from one application area to another.

Speaking for myself, I'd probably do something this:

Let $z_1 = \frac{\text{score}_1-\text{?}}{\text{??}}$ (depending on how you're standardizing, I'd fill in the missing parts differently) and similarly for $z_2$. Then

\[ \text{ComboScore}_1 = \frac{z_1 + z_2}{2} \]

Which is to say I'd explicitly (algebraically) define the z-scores, then explicitly algebraically define the combo score.

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  • $\begingroup$ Yes, I would mention in the prepending text something like "after standardization (z-score normalization) of the individual scores, the combo score is calculated as follows [ComboScore equation]." That's a good idea, I will just add also something like $Score1_z = \frac{Score1_i - \bar{Score1}}{\sigma_{Score1}}$ $\endgroup$
    – user39663
    Oct 13, 2014 at 2:52
  • $\begingroup$ Conventially, $\overline{x}$ indicates a sample mean of $x_i$'s, while $\sigma_x$ indicates a population standard deviation. If you know the population s.d. how is it you don't know the mean? $\endgroup$
    – Glen_b
    Oct 13, 2014 at 2:55
  • $\begingroup$ I am confused now, where did I say that I don't know the mean? $\endgroup$
    – user39663
    Oct 13, 2014 at 2:58
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    $\begingroup$ The fact that you use a sample mean to standardize rather than the population mean implies you don't know the population mean. If you knew the population parameters, you'd use $\frac{x_i-\mu_x}{\sigma_x}$. If you only had sample quantities you'd used $\frac{x_i-\bar x}{s_x}$. You mixed the two (sample mean, population sd), which leads to the question. $\endgroup$
    – Glen_b
    Oct 13, 2014 at 2:59
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    $\begingroup$ The answer to that should be fairly clear from the fact that I used $s_x$ for the sample standard deviation of $x$ in my previous comment... it would be $s_{\text{score}_1}$. The use of $s$ for sample standard deviations is pretty conventional. See here for example. $\endgroup$
    – Glen_b
    Oct 13, 2014 at 3:06

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