## Even in 2x2... [General Statistics]

» ... But we see a similar effect with the EMA’s Method A above, which is a bloody ANOVA and all effects fixed. I don’t get it.

Dear all,

the reason of the difference is that sequences were unbalanced (not equal number of subjects in each of the sequences).

As you know (for ln-transformed data) the

The way of calculation of the

Personally I would not report these least squares means at all. If you are calculating ANOVA (e.g. by GLM) you get point estimate directly, i.e. without calculation of least squares means.

For unbalanced sequences there might be more questions ... together with misleading terminology - e.g. geometric means ratio - for unbalanced sequences, the ratio of geometric means reported in descriptive statistics differ from geometric means ratio reported with 90% confidence interval which is also very, very strange. As these ratios are different, somoone wants to have the "correct" geometric means of T and R for which the ratio T/R is equal to point estimate (i.e. somoone wants "geometric least squares means"). Nevertheless geometric means are geometric means! Behind geometric least squares means I see only values of "T" and "R", for which (by dividing) we get the point estimate. But as it was pointed out, least squares mean of T can be affected by R, and vice versa.

Best regards,

zizou

Dear all,

the reason of the difference is that sequences were unbalanced (not equal number of subjects in each of the sequences).

As you know (for ln-transformed data) the

*estimated marginal means*(i.e.*least squares means*in SAS terminology) differ from*arithmetic means*of T and R if sequences are unbalanced.The way of calculation of the

*estimated marginal means*involves some modification of "standard"*marginal means*. I'm lazy to go through the matrix algebra (which is used by most of the softwares), so simply:*marginal means*are "corrected" to*estimated marginal means*by using a difference between*mean of all data*and*mean of marginal means of T and R*(when the sequences are balanced then this difference is zero).Personally I would not report these least squares means at all. If you are calculating ANOVA (e.g. by GLM) you get point estimate directly, i.e. without calculation of least squares means.

For unbalanced sequences there might be more questions ... together with misleading terminology - e.g. geometric means ratio - for unbalanced sequences, the ratio of geometric means reported in descriptive statistics differ from geometric means ratio reported with 90% confidence interval which is also very, very strange. As these ratios are different, somoone wants to have the "correct" geometric means of T and R for which the ratio T/R is equal to point estimate (i.e. somoone wants "geometric least squares means"). Nevertheless geometric means are geometric means! Behind geometric least squares means I see only values of "T" and "R", for which (by dividing) we get the point estimate. But as it was pointed out, least squares mean of T can be affected by R, and vice versa.

Best regards,

zizou

### Complete thread:

- Least square mean calculation for the fully replicate design Jaimik Patel 2019-10-31 11:19 [General Statistics]
- Very, very strange! Helmut 2019-10-31 14:36
- Very, very strange! Shuanghe 2019-10-31 19:43
- Very, very strange! Helmut 2019-10-31 19:50

- Very, very strange! Jaimik Patel 2019-11-02 12:26
- G matrix mittyri 2019-11-03 01:38

- Very, very strange! Shuanghe 2019-10-31 19:43
- Inner workings of REML ElMaestro 2019-10-31 19:48
- Even in ANOVA… Helmut 2019-10-31 19:53
- Even in 2x2... zizou 2019-11-01 21:05
- trying to understand emmeans mittyri 2019-11-02 02:26

- Even in 2x2... zizou 2019-11-01 21:05

- Even in ANOVA… Helmut 2019-10-31 19:53
- Least square mean calculation for the fully replicate design PharmCat 2019-11-03 00:46

- Very, very strange! Helmut 2019-10-31 14:36