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I have a continuous variable which is not normally distributed i want to transform it to normal using the two step approach method in the link below:

Abstract This article describes and demonstrates a two-step approach for transforming non-normally distributed continuous variables to become normally distributed. Step 1 involves transforming the variable into a percentile rank, which will result in uniformly distributed probabilities. The second step applies the inverse-normal transformation to the results of Step 1 to form a variable consisting of normally distributed z-scores. The approach is little-known outside the statistics literature, has been scarcely used in the social sciences, and has not been used in any IS study. The article illustrates how to implement the approach in Excel, SPSS, and SAS and explains implications and recommendations for IS research.

https://aisel.aisnet.org/cais/vol28/iss1/4/

Is this method going to reorder the the observations?

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    $\begingroup$ What approach? Please describe what you mean to make the question self-contained. $\endgroup$
    – Tim
    Mar 22, 2022 at 14:11
  • $\begingroup$ @Tim thank you, I edited the question. $\endgroup$
    – Stats34
    Mar 22, 2022 at 14:17
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    $\begingroup$ Why do you want to do this? There is no need for any variable to be normally distributed in your model. That is not an assumption of linear regression. $\endgroup$
    – Noah
    Mar 22, 2022 at 14:26
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    $\begingroup$ The residuals don't need to be normally distributed. And even if they did, making the outcome variable normally distributed would not accomplish this. $\endgroup$
    – Noah
    Mar 22, 2022 at 14:40
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    $\begingroup$ A similar question appeared a few days ago here as an isolated problem, and I gave an answer. I have no idea where, if anywhere, this method would be useful. // My implementation does not change the order of the data. $\endgroup$
    – BruceET
    Mar 22, 2022 at 15:50

2 Answers 2

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First, as noticed in the comment by Noah, almost never in statistics do you need to "transform" the data to be normally distributed. That is not the case for linear regression, nor for most of the other statistical methods. In the comment, you say that you are doing that for the residuals to be normally distributed. This approach would not make them normally distributed because it changes the marginal distribution, while in the case of linear regression we are talking about the conditional distribution. There is no simple way how you could "transform" the data to meet this assumption and the solution would be problem-specific.

Answering your main question: the method is called equipercentile equating and uses the standard normal quantile function to transform the empirical quantiles. It does not change anything about the ordering of the observations, it just transforms the values. The transformation is as follows:

  • for each value, $x_i$ find it's rank $r_i$, i.e. the index of this value if $x_i$ values were sorted in increasing order,
  • transform the ranks to quantile ranks by dividing the ranks by sample size $q_i = r_i / N$,
  • use standard normal quantile function $\Phi^{-1}$ to calculate the $z$-scoares $z_i = \Phi^{-1}(q_i)$.

As you can see, this doesn't change anything about the ordering of the observations. The order would only change if you actually sorted the $x_i$ values, but this is not needed.

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  • $\begingroup$ thank you very much for your response, however I have another question my data are violating other assumption as linearity and Homoscedasticity cook's value and other assumptions, in this case can I still use simple linear regression, I was supposed to use multivariate multiple regression because I have 7 independent and 2 dependent variables, but I can't find this method online using spss software. $\endgroup$
    – Stats34
    Mar 22, 2022 at 14:52
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    $\begingroup$ @Stats34 if you have another question, please ask it by opening a separate question. If it is about the actual results of your regression analysis, it would be helpful to describe your data in detail, show the plots and the results you got. $\endgroup$
    – Tim
    Mar 22, 2022 at 14:54
  • $\begingroup$ I will add a new question thank you. $\endgroup$
    – Stats34
    Mar 22, 2022 at 14:56
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    $\begingroup$ @Stats34 if either of the answers answered your question, please mark it as accepted so the question is "solved". $\endgroup$
    – Tim
    Mar 22, 2022 at 15:06
  • $\begingroup$ I added the new question sir stats.stackexchange.com/questions/568700/… $\endgroup$
    – Stats34
    Mar 22, 2022 at 15:32
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No, this does not change the order of the variables. It is a one-to-one transformation from any range of values (ideally when no values are ever exactly the same, but in practice as long as there are not too many ties) to $(-\infty, \infty)$ (but with 95% of values between -1.96 and 1.96).

However, note that this may often not be something that you want to do. The author of the paper appears to be confused on why such an approach would be needed. Any assumption with linear regression models are about the residuals of the model (additionally, depending on the situation, methods may be quite robust to deviations from normality in residuals: see all the literature and previous questions about assessing normality of residuals). What the author proposes will not work on the residuals of a model, but rather on the raw variables. In fact, the approach could make residual normality worse, e.g. when normality of residuals is already perfect, but there are two explanatory variable levels that have very different means (then you would never ever want to do a transformation of the outcome variable).

Oddly, the author also does not seem to say whether they intend this just for dependent or independent variables, when for the latter it would often not make much sense for a linear model. One setting, in which this approach applied to independent variables is known to be potentially helpful, is when training neural networks for prediction purposes or denoising autoencoders. In that setting, it has been used in Kaggle competitions (first in the Porto Seguro challenge) and lead to improved model performance.

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    $\begingroup$ thank you very much for your elaborated response. $\endgroup$
    – Stats34
    Mar 22, 2022 at 14:57

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