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I have a between-group independent variable with two level (A and B) and a dependent variable Y that I transformed in order to normalize the distribution of the residual. I used a Box Cox transformation with the following formula : ((Y^3,169833)-1)/3,169833

With this transformation I observed the following results (Y' refers to the transformed variable) enter image description here

I used this formula to back transform my data:(Y'*3,169833+1)^(1/3,169833) and I obtained the means and CI in red in the figure.

My question is how can I get the difference between my two means and the 95% CI for that difference? I tried to calculate the difference between the trasnformed means, to get its 95% CI and then to transform back. However I obtained an improbable result. I guess that maybe this is because I transformed Y and not the difference between my to groups. Is there a formula to compute the CI of a difference between means from the two CI of the two means?

Thank you very much in advance

Best regards

Update after Alexy's answer

I have a question about the fact that "inferences about f(x) are not the same thing as inferences about x". I agree with that. However I thought that it was for that reason that we have to back transform the transformed data in order to interpret the result more easily. For example, Bland & Altman (1996)* have suggested that we can back-transformed the confidence limits of a log transform data using the anti-log in order interpret the results in the original unit. Do you agree with this possibility? Or does the fact that "inferences about f(x) are not the same thing as inferences about x" preclude this possibility? I you agree with Bland & Altman (1996), do you think that we can approximate the sd for x by following the following steps? enter image description here

Does it make a little sense or do I am totally wrong in doing that?

*Of course I have forgotten that Bland & Altman (1996) have underlined the impossibility to back transform the standard deviation.

2nd Update

So I tried to compute the sd for back-transformed Y from the back-transformed CI and the critical t value and the square root of the sample size (see the results above). I wanted to find a method in order to check the validity of this method. My assumption was that if the method is valid, the inference made with transformed data would have to lead to the same conclusion as the inference made with the back-transformed data. If so the t-test with the two sets of data should lead to the same results and a standardized version of the effect size estimate (point and CI) should also be similar with the two sets of data. So I performed a t-test for independent samples with the transformed data and another t-test for independent samples with the back transformed data (using the sd computed from the above-mentioned formula). I also computed the Cohen's d and their 95% CI from the two sets of data. The following table presents the results of my analyses.

enter image description here

From that results it seems that the inference made with the back-transformed data and the approximate sd is consistent with those made with the transformed data. Of course the inference made with the back-transformed data is very very slightly liberal compared with the inference made with the transformed data. A more central issue with the formula is that as Alexy said, "inferences about f(x) are not the same thing as inferences about x". However, while keeping this fact in mind, it is easier to interpret the back-transformed data than the transformed data and since the MOE is similar for the two sets of data why we could not use the back-transformed data?

I aknowledge that the obtained back-transformed mean difference is not equivalent to the arithmetic mean difference. So I think that if we decide to use this method to compute the back-transformed mean difference and its CI, we have to remind that fact to the reader.

What do you think about my solution to the problem? Do you see other limits of my solution? Do you think that I underestimate the limits that I mentioned previously?

Thanks again for your time and for your answers.

Reference

Bland, J. M., & Altman, D. G. (1996). Statistics notes: Transformations, means, and confidence intervals. BMJ, 312(7032), 1079–1079. doi:10.1136/bmj.312.7038.1079.

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  • $\begingroup$ Hi Psychokwak. There is something, I don't unterstand from your question: I assume that you chose your $\lambda$ for the Box-Cox transformation with respect to some model (e.g. regression)? But if you have fit such a model, why are you trying to infer a CI for the difference of the means from simple summary statistics? Normally, the software will give you a CI for the difference of the means which you can then back-transform. $\endgroup$ – COOLSerdash Jul 6 '14 at 8:09
  • $\begingroup$ Hi COOLSerdash, thanks for your comment. I used statistica which provided the best lambda to normalize the data. Then I ran my analysis using the transformed data. Statistica provided the t statistic and the transformed means with their standard deviations and the corresponding CI for the transformed means. Statistica did not provide the mean difference and its CI in the original unit. So I tried to compute them from the individual means. I try to get the mean difference and its CI in the original unit in order to have an effect size estimate readily interpretable. Do you see what I mean? $\endgroup$ – Psychokwak Jul 6 '14 at 8:23
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    $\begingroup$ So you transform the unconditional $Y$ (i.e. just the plain $Y$)? But the $t$-test only requires that the conditional $Y$ is approx. normally distributed (i.e. the residuals). Unfortunately, I am not familiar with Statistica. Another problem is that you can't back-transform any values $<-\frac{1}{\lambda}$. Try to back-transform a value of $-0.32$ with your $\lambda=3.169833$, for example. $\endgroup$ – COOLSerdash Jul 6 '14 at 9:25
  • $\begingroup$ I performed my Box-Cox transformation with my IV selected as a grouping variable. The Statistica's help indicates that "when one or more grouping variables are selected for the Box-Cox analysis/transformation, the transformation will be computed with the goal of transforming the within group distributions to normality with homogenous variances; this option is useful when analyses (e.g. ANOVA) are to be performed with categorical predictors." $\endgroup$ – Psychokwak Jul 6 '14 at 12:31
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    $\begingroup$ I do not agree that in general one can perform the inverse transformation on the CI of of a transformed estimate in order to obtain the CI on the untransformed estimate. For the reasons I have already written. That said, there are transformations of the variance of a function of $x$ that can provide more or less accurate estimates of the variance of $x$ (from there to standard errors to CIs). If you read Altman and Bland's third paragraph, they are pointing this out. $\endgroup$ – Alexis Jul 6 '14 at 15:59
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The first thing you need to understand is that $f\left(\sigma^{2}_{x}\right) \ne \sigma^{2}_{f\left(x\right)}$. Put in words a function of the variance of a variable does not equal the variance of the same function of that variable.

This is substantively important when performing inference (e.g. hypothesis tests, or interval estimation) on transformed data. Your statistical inferences were based on a Box Cox transformation, great. But an inverse Box Cox transformation of your results (e.g. your interval estimates) does not provide you with a valid estimate of the same kind of interval for untransformed data.

So what can you do?

First, you can accept that inferences about $f\left(x\right)$ are not the same thing as inferences about $x$.

Second, you can simply perform inferences on $x$.

Third, in case where you must perform inference on $f\left(x\right)$, but want to say something about $x$, you can use the delta method to create an approximation of $\sigma_{f\left(x\right)}$.

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  • $\begingroup$ Hi Alexis, thank you very much for your answer. I will look at the delta method. However I have a question about your answer (please see the update of my previous question). $\endgroup$ – Psychokwak Jul 6 '14 at 7:21

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