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I have run a series of mixed linear regressions in Stata, some with inverse-square-root ($1/\sqrt{x}$) transformations and others with square root ($\sqrt{x}$) transformations.

How do I calculate untransformed confidence intervals from the transformed results??

Glen_b
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Nadia Brown
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1 Answers1

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You calculate the full transformed confidence interval and then transform it back.

Let's say the transformed confidence interval is 5 ± 3, or CI95% = [2, 8]. You would take the 2, and 8 values and transform them back. You do NOT transform the 3 (the width of the confidence interval). The result in this example CI95% = [0.25, 0.016].

Be careful of interpretation because you're switching around what is larger and what is smaller. Also, the CI inferences only hold in the transformed space and these back transformed values would just be referece / orientation values for the reader rather than primary values for inference. If this is something like a beta coefficient you can't even do the back transform because the line is no longer straight and you wouldn't be able to interpret it (you could calculate the whole line and error curves and back transform them though). There may be other interpretation issues.

Generally, it's best to interpret in the transformed space and present back transformed values only for convenience and where they aid that interpretation. It would be best if you can ascribe some meaning to the transformed value.

John
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  • This answer is overlooking something crucial: the back-transformed confidence interval is no longer a confidence interval with the same nominal coverage. In extreme cases, it would be unlikely for the back-transformed interval even to include a reasonable estimate of the statistic it's supposed to cover! So, pending an adequate answer to @Glen_b's request for information about that target of the confidence interval, this advice seems premature. – whuber May 16 '13 at 14:17
  • You're right but I always thought of that as an over interpretation issue, not an issue of working out the CI. The transform sounds like a skew correction so the mean and CI will be quite different from the original space. So, one might report the CI of the transformed values and then say, for reference, that this is what an untransformed value looks like. Perhaps I should have been more specific about interpretation in the answer. And you're also right that a much better answer could be supplied if Glen_b were answered. – John May 16 '13 at 16:53
  • To expand that further, I've reported a very severe transform in logistic regression in the past. I typically show both the logit function and the function in probability space both containing confidence intervals with very different looking coverage patterns. Some people have a difficult time interpreting the logit but I always emphasize that the interpretation goes back to there. Are you suggesting this is wrong? – John May 16 '13 at 16:57
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    But it's more than interpretation: my point is that your transformation ruins the defining property of the interval. For instance, suppose $\log(x)$ has a standard Normal distribution and there are $100$ iid draws. The usual symmetric $5$% two-sided (Student t) CI for the mean of $\sqrt{x}$ has coverage of about $94.6$%, close to what we want, but your back-transformed CI for the mean of $x$ only has coverage around $32$%. That's a big error, regardless of how you choose to interpret it. – whuber May 16 '13 at 17:03
  • I'm not sure why you switched between log, sqrt, and then the original. I don't think I'm following this comment. If I gather x, analyze the log and get the student T based CI and mean then the coverage of that CI for those log values is about 95%. Now, if I simulate x, and generate means of the transformed values, then back transform those values, the new distribution of back transformed means won't fall in that initial range? – John May 16 '13 at 19:46
  • Comments have to be brief, but all the information is there: *log* of $x$ has a normal distribution, so $x$ is lognormal. The objective is a CI for the mean of $x$, but all you have is a CI for the mean of $\sqrt{x}$: that's the context of the question. Simulate that and see what happens. – whuber May 16 '13 at 19:50
  • See, I would never have interpreted the back transformed values as a CI for means of x, only a CI for back transforms of means of transformed samples of X... which simulates just fine. Of course it can't be a CI for something which it is not. – John May 16 '13 at 19:52
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    Thanks: that clarifies what you are doing. It leaves us back with @Glen_b's question for the OP: CIs for *what*? – whuber May 16 '13 at 20:00