When a predicted mean response on a transformed scale is converted back to its
original scale, the predicted mean becomes the predicted median. A correction
is applied to the back-transformed prediction so it correctly predicts the
mean, rather than the median. This correction may be toggled on/off by clicking
**Edit** > **Preferences** > **Math Analysis** > **Correct biased predictions of mean**.

Use the predicted mean for making inferences and prediction about your process.

Consider the case where we transform the response data, \(y_i\), using some transformation \(f\) to ensure the normality of the residuals,

\[z_i = f(y_i)\]

We fit this new, transformed data to a model, resulting in *transformed scale*, least squares estimates
for the variance, \(\sigma_z^2\), and response mean function, \(z(A,B,...)\), as a function of the
factors in the model, \(A, B,...\),

\[\begin{split}\begin{eqnarray}
\hat{\sigma}_{z}^2 &=&~\text{transformed scale mean square error}~=~\text{constant},\\
\hat{z} &=& \hat{z}(A,B,...) = \hat{\beta}_0+\hat{\beta}_A A + \hat{\beta}_B B + \hat{\beta}_{AB} A B+...\\
\end{eqnarray}\end{split}\]

Because the normal distribution is a symmetric distribution, the response mean function is also the median function – the mean and median are the same on the transformed scale.

Note

It is important to verify that the model assumptions – normality of residuals, constant variance, and independence – have been achieved after transformation, and perhaps even more so when relying on original scale mean bias correction.

To obtain the *median* on the *original scale*, apply the inverse transformation \(f^{-1}\) to the mean
function on the transformed scale,

\[\hat{m}_y(A,B,...) =~\text{original scale median}~= f^{-1}(\hat{z})\]

On the original scale, the mean and median are not the same. To obtain an estimate of the *mean* on the
*original scale*, we can use an approximation involving the second derivative (curvature) of the inverse transformation
and the mean squared error from the regression. To first order in the mean square error, \(\hat{\sigma}_{z}^2\),

\[\hat{y}(A,B,...) =~\text{original scale mean}~\approx f^{-1}(\hat{z}) + \frac{1}{2} \left(\left. \frac{d^2 f^{-1}}{d z^2}\right\rvert_{\hat{z}}\right) \hat{\sigma}^2_z\]

Consider the transformation,

\[z = f(y) = \frac{1}{y}, y > 0\]

In that case, the inverse transformation and its derivatives are,

\[\begin{split}\begin{eqnarray}
f^{-1}(z) &=& +\frac{1}{z}, z > 0, \\
\frac{d f^{-1}}{d z} &=& -\frac{1}{z^2},\\
\frac{d^2f^{-1}}{d z^2} &=& +\frac{2}{z^3}
\end{eqnarray}\end{split}\]

Therefore, on the original scale,

\[\begin{split}\begin{eqnarray}
\hat{m}_y &=& f^{1}(z) = \frac{1}{\hat{z}},\\
\hat{y} &\approx& \frac{1}{\hat{z}}+ \frac{1}{2} \left(\frac{2}{\hat{z}^3}\right)\hat{\sigma}^2_z = \frac{1}{\hat{z}}\left(1+ \frac{1}{\hat{z}^2} \hat{\sigma}^2_z\right)
\end{eqnarray}\end{split}\]

Note

Remember that \(\hat{z} = \hat{z}(A,B,...)\), \(\hat{m}_y =\hat{m}_y(A,B,...)\), and \(\hat{y} = \hat{y}(A,B,...)\) are functions of the factor settings and change with the different values for A, B,…

Logarithmic tranformations for log normal data are a special case where the mean function on the original scale can be computed exactly (Finney 1941). For these transformations,

\[\hat{y} = \exp{\left[a \left(\hat{z}+ a \frac{1}{2} \hat{\sigma}^2_z \right)\right]}\]

Here, \(a\) controls the base of the logarithm. For the natural log transformation, \(a = \log_e(e) = \ln{e} = 1\). For the log base-10 transformation, \(a = \ln{10}\). This gives two sets of equations, depending on the type of logarithm transformation used,

For natural logarithm transformation the median and mean are,

\[\begin{split}\begin{eqnarray}
\hat{m}_y &=& e^\hat{z},\\
\hat{y} &=& e^\hat{z} e^{\frac{1}{2} \hat{\sigma}^2_z} = \hat{m}_y e^{\frac{1}{2} \hat{\sigma}^2_z}\\
\end{eqnarray}\end{split}\]

For log base 10 logarithm transformations,

\[\begin{split}\begin{eqnarray}
\hat{m}_y &=& 10^\hat{z},\\
\hat{y} &=& 10^\hat{z} 10^{\frac{1}{2} \ln{(10)} \hat{\sigma}^2_z} = \hat{m}_y 10^{\frac{1}{2} \ln{(10)} \hat{\sigma}^2_z}
\end{eqnarray}\end{split}\]

Suppose that the transformation is the base-10 logarithm, \(z = f(y) = log_{10}(y)\), as in the image displayed at the top of this help entry. We wish to calculate the mean on the original scale at the point (B=1,C=1,D=1) in the design (the factor A was not in the model for this example).The least square estimates for the mean square error and mean function on the transformed scale are,

\[\begin{split}\begin{eqnarray}
\hat{\sigma}_z^2 &=& (0.45)^2 = 0.002025,\\
\hat{z}(B,C,D) &=& 0.683549 + 0.126689 B+0.239201 C+ 0.072841 D \\
\end{eqnarray}\end{split}\]

So the mean (and median) on the transformed scale at the desired point is,

\[\begin{split}\begin{align}
\hat{z}(B=1,C=1,D=1)~=~& 0.683549~+\\
& 0.126689(1)~+\\
& 0.239201(1)~+\\
& 0.072841(1) = 1.122279
\end{align}\end{split}\]

The original transformation is the base-10 logarithm so we use,

\[\begin{split}\begin{eqnarray}
\hat{m}_y(B=1,C=1,D=1) &=& 10^\hat{z} = 10^{1.122279} \approx 13.25,\\
\hat{y}(B=1,C=1,D=1) &=& \hat{m}_y 10^{\frac{1}{2} \ln(10)\hat{\sigma}^2_z} = (13.25)[10^{0.5 (2.302585) (0.002025)}] \approx 13.32
\end{eqnarray}\end{split}\]

The mean bias correction is approximately \(13.32-13.25 = 0.07\) in this case.

References

D. J. Finney. On the distribution of a variate whose logarithm is normally distributed.

*Journal of the Royal Statistical Society Supplement*, 7:155–161, 1941.D. Miller. Reducing transformation bias in curve fitting.

*The American Statistician*, 38(2):124–126, 1984.G.W. Oehlert. A note on the delta method.

*American Statistician*, 46:27–29, 1992.