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When a regression is calculated with a simple linear model that returns intercept and slope for an equation like this $y=a + bx$ one can predict $y$, the response variable, based on that equation.

Equally one could rearrange for $x$: $x=\frac{\left(y-a\right)}{b}$ and calculate the value of $x$. This isn't available in R's predict() function but can easily be done. Can one do this calculation and still be statistically sound?

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  • $\begingroup$ For monotonic prediction function, this should be straightforward. Non-monotonic functions do not necessarily have a single solution for x={y-a}/b. What do you mean by statistically sound? $\endgroup$
    – Kees Mulder
    Commented Nov 6, 2014 at 15:25
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    $\begingroup$ This is common in analytical chemistry (where known values of the standards are used as x-values and the instrument response as y-values for calibration). Thus, package chemCal provides a function inverse.predict. I don't understand your last sentence. $\endgroup$
    – Roland
    Commented Nov 6, 2014 at 15:26
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    $\begingroup$ No, you need to estimate the standard error of the inverse prediction. chemCal::inverse.predict does this. Look at the function's code. $\endgroup$
    – Roland
    Commented Nov 6, 2014 at 15:30
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    $\begingroup$ This is sometimes called "inverse regression", though unfortunately the recent rise of 'sliced inverse regression' makes the use of the term in relation to calibration-type applications rather harder to search for (-sliced helps). it looks like the invest function of the investr package in R does what you seek. $\endgroup$
    – Glen_b
    Commented Nov 6, 2014 at 21:41
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    $\begingroup$ There is now a dedicated package called investr for this problem. cran.r-project.org/web/packages/investr/investr.pdf I get what you mean by "statistically sound", looking at the paper it doesn't seem a trivial problem to just invert the analysis. journal.r-project.org/archive/2014-1/greenwell-kabban.pdf $\endgroup$
    – Andy74
    Commented Jan 27, 2016 at 19:47

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