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For a normally distributed random variable $X$, we can "standardize" $X$ by defining a new random variable $\frac{X-\mu}{\sigma}$ where $\mu$ is a mean of $X$, and $\sigma$ is a standard deviation of $X$.

Now, why do we do this? A textbook said that it was to make the calculation of probability easier since we can refer to the standard normal table once we have the standard normal. However, I feel like this reason is outdated as I believe we no longer use the standard normal table.

One reason I can think of is that many statistical methods rely on the chi-squared distribution, which is the distribution of the sum of the squared standard normal variables. Are there any others?

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  • $\begingroup$ If you are aware about this chatroom created by Sextus Empiricus to reach you on your recently deleted post, then please communicate back (if you want). $\endgroup$ Apr 6, 2023 at 3:13

1 Answer 1

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Comparison of Scales

A very common reason to use z-scores is to compare variables with completely different scales. Consider these two variables I simulated in R, IQ and salary.

enter image description here

For simplicity's sake (because salary isn't normally distributed in reality), lets just say they both have a Gaussian distribution. However, they are not very comparable. Telling somebody they make $30,000 USD per year and have an IQ of 130 tells you something, but it says nothing about the location of their score, nor if it's considered several deviations below or above average. However, if we transform these variables into z-scores...

enter image description here

We now have something comparable, as they are both on the same scale. We know for example that a person with a z-score of +2 on both measures is way above average compared to the distribution.

Fixing Messy Regressions

Sometimes complicated regressions need some legwork to get them to work. A common issue in mixed model regressions is that interactions with measures on totally different scales often leads to buggy behavior. Standardizing scores between two variables in an interaction is a common fix. For example, if I fit these same variables above into an interaction in lmer, R will kick back a warning that it's going to explode.

#### Load Libraries ####
library(lmerTest)
set.seed(123)

#### Simulate Data ####
money <- rnorm(n=1000, mean = 50000, sd = 10000)
iq <- rnorm(n=1000, mean = 120, sd = 10)
subject <- factor(rbinom(n=1000,size=50,prob=.5))
response <- rnorm(n=1000)

fit <- lmer(
  response
  ~ iq*money
  + (1|subject)
)

Shown in this error when I save fit:

boundary (singular) fit: see help('isSingular')
Warning messages:
1: Some predictor variables are on very different scales: consider rescaling 
2: Some predictor variables are on very different scales: consider rescaling

Refitting with scaled data:

fit <- lmer(
  response
  ~ scale(iq)*scale(money)
  + (1|subject)
)

We don't fix the singular matrix (this is just because of the lazy way I simulated the data), but it has now removed the error regarding the scale invariance, which improves the chances the regression will fit correctly:

boundary (singular) fit: see help('isSingular')

General Statistical Inference

Many inferential tests in statistics are based off CLT and z-scores give useful heuristics for understanding this sorta thing.

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  • $\begingroup$ The only issue with this is that salaries tend not to be normally distributed. $\endgroup$
    – John
    Mar 20, 2023 at 15:46
  • $\begingroup$ Yeah perhaps that is a bad example. I've made a minor edit to make that caveat. $\endgroup$ Mar 20, 2023 at 15:47

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