# ANOVA: contrast to ratio of adjusted geometric means

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!

• 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?
– whuber
Jun 13, 2023 at 21:35
• 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. Jun 13, 2023 at 21:46
• Also added a couple of intermediate steps. Jun 13, 2023 at 21:50
• 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. Jun 13, 2023 at 22:09
• I indeed estimate a ratio in the end. Small y is the logarithm of the big Y. Jun 13, 2023 at 22:41