11
$\begingroup$

Suppose you have a data set that doesn't appear to be normal when its distribution is first plotted (e.g., it's qqplot is curved). If after some kind of transformation is applied (e.g., log, square root, etc.) it seems to follow normality (e.g., qqplot is more straight), does that mean that the dataset was actually normal in the first place and just needed to be transformed properly, or is that an incorrect assumption to make?

$\endgroup$
7
  • 12
    $\begingroup$ I'd say it means that the transformed data are approximately normal, not the original data. $\endgroup$
    – BruceET
    Feb 28, 2022 at 19:31
  • 13
    $\begingroup$ Nonlinear functions like logs and roots are usually not considered to be "scaling." The latter term is conventionally reserved for linear changes only: that is, multiplication by a constant followed by adding another constant. The phrase "actually normal" is quite interesting, because there is a philosophy of data analysis that says not to let the form in which the data are expressed determine how you express them for analysis. From this perspective, such data very well might be conceived of as "actually normal" values that were given to us in the form of the exponentials or squares, e.g. $\endgroup$
    – whuber
    Feb 28, 2022 at 19:55
  • $\begingroup$ @whuber why aren't they considered to be scaling exactly? I thought that the monotinicity of the log function is what makes it suitable for scaling. $\endgroup$
    – mesllo
    Feb 28, 2022 at 23:12
  • 9
    $\begingroup$ I suspect different communities might have different concepts of "scaling" in various contexts, but overall, this term generally is used in math and stats to refer to a proportionate change in size ("scale"), often with the option to change the origin of a unit of measurement. Thus, some effort is usually made to distinguish the kinds of non-linear transformations or re-expressions you mention from mere "scaling." It's a useful distinction. $\endgroup$
    – whuber
    Feb 28, 2022 at 23:45
  • 2
    $\begingroup$ Any random variable with a continuous distribution on $\mathbb R$ can be transformed monotonically into any other distribution (with a minor caveat about values which combined have probability $0$ of occurring). That does not mean they have the same distribution $\endgroup$
    – Henry
    Mar 2, 2022 at 14:16

3 Answers 3

24
$\begingroup$

NO

It means that the transformed distribution is normal (at least roughly). Depending on the transformation, it might suggest a lack of normality of the original distribution. For instance, if a log-transformed distribution is normal, then the original distribution was log-normal, which certainly is not normal.

$\endgroup$
9
$\begingroup$

_Comment continued: Consider lognormal data x, which does become exactly normal when transformed by taking logs. In this case (with $n=1000),$ Q-Q plots and the Shapiro-Wilk normality test agree for original and transformed data.

set.seed(2022)
x = rlnorm(100, 50, 7)
y = log(x)
par(mfrow = c(1,2))
 hdr1 = "Lognormal Sample: Norm Q-Q Plot"
 qqnorm(x, main=hdr1)
  abline(a=mean(x),  b=sd(x), col="blue")
 hdr2 = "Normal Sample: Norm Q-Q Plot"
 qqnorm(y, main=hdr2) 
  abline(a=mean(y), b=sd(y), col="blue")
par(mfrow = c(1,1))

enter image description here

shapiro.test(x)

        Shapiro-Wilk normality test

data:  x
W = 0.1143, p-value < 2.2e-16     # Normality strongly rejected

shapiro.test(y)

        Shapiro-Wilk normality test

data:  y
W = 0.99017, p-value = 0.678     # Does not rejece null hyp: normal
$\endgroup$
0
$\begingroup$

In general, the answer is no. It will be normal only if it was generated by the back transformation that corresponds to your (series of) transformations (see edit below). Nevertheless, there are good chances that the distribution of the transformed data is approximately normal, but keep in mind that not every bell-shaped distribution is a normal distribution. You need more than eye bowling before you start your analyses.

Edit regarding the first sentence in my answer: the transformation must be monotonic. For example, if you take data that was generated by a normal distribution, square it and then apply square root - you will not end with a normal distribution.

$\endgroup$
1
  • $\begingroup$ In practice, your second paragraph itself needs a pedantic qualifier. For example, human heights are often regarded as approximately normal, but squaring and then square rooting -- although a silly thing to do -- would get you back where you started, and the same applies whenever all the data are positive. What you have in mind, no doubt, is that the support of a normal distribution is the entire real line, so there is always some fraction of the probability for negative values, and then in principle there is the difficulty you allude to. $\endgroup$
    – Nick Cox
    Mar 4, 2022 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.