Could you, please, help me with the following problem?

Suppose we have a one-way ANOVA with a single 2-level factor. The dependent variable is a logarithmized value: $y_i = log(Y_i)$.

$y_{iz} = \mu + a_i + \epsilon_{iz}$

$z$ is the number of observations in each $i$ group, different for different $i$.

We want to estimate the contrast:

$C = a_1 - a_2$

If we exponentiate this contrast we get a ratio of geometric means of the dependent variable in the two treatment groups on the original scale.

$C = \frac{\sum_{h=1}^{z} log(Y_{1h})}{z} - \frac{\sum_{h=1}^{z} log(Y_{2h})}{z}$

$C = log(\prod_{h=1}^{z} Y_{1h}^{\frac{1}{z}})- log(\prod_{h=1}^{z} Y_{2h}^{\frac{1}{z}})$

$C = log(\frac{(\prod_{h=1}^{z} Y_{1h})^{\frac{1}{z}}}{(\prod_{h=1}^{z} Y_{2h})^{\frac{1}{z}}})$

Now, suppose we add additional factors in the model. Importantly, we do not add interaction terms in the model.

What do we get if we exponentiate the same contrast from the model with additional variables? Do we still get an adjusted ratio of geometric means of some sort? Have you seen any literature on this?

I would appreciate any insights!

  • $\begingroup$ I have a hard time making sense of using the group counts as subscripts. Indeed, this makes nonsense out of your second expression for $C,$ where "$z$" appears in at least three distinct roles: index, power, and terminal value in each product! Could you clarify how this works, perhaps with a small concrete example? $\endgroup$
    – whuber
    Jun 13, 2023 at 21:35
  • $\begingroup$ z is the number of observations in each group i. I have corrected the formula: replaced the index with h. The terminal value and the power should both be z. $\endgroup$
    – dfgh19283
    Jun 13, 2023 at 21:46
  • $\begingroup$ Also added a couple of intermediate steps. $\endgroup$
    – dfgh19283
    Jun 13, 2023 at 21:50
  • $\begingroup$ are you sure you want to estimate the difference between the two group means, and not the ratio? With lognormal distributions, ratios usually make the most sense. $\endgroup$ Jun 13, 2023 at 22:09
  • $\begingroup$ I indeed estimate a ratio in the end. Small y is the logarithm of the big Y. $\endgroup$
    – dfgh19283
    Jun 13, 2023 at 22:41

1 Answer 1


We call a mean adjusted when we include other covariates in the model with the treatment dummy variable. You can read about it here. Then we exponentiate the adjusted means to obtain the adjusted geometric means. I have not been able to find a definition of an adjusted geometric mean as such, however, I have found a few articles where they do what I need to do: exponentiate the adjusted mean to obtain an adjusted geometric mean. For example, you can refer to this article.


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