ANOVA using geometric mean

[ggao]

I was wondering if I could do ANOVA with geometric mean instead of LS means?

[Rick_SAS]

There was a similar discussion a few months ago: See the thread "Questions about geometric mean and SAS procedures" and see if that helps.

[ggao]

Don't quite really get that. I have three treatments A, B and C. For each treatment I have measurement for day1, day 15 and 30. I basically create indexes for day15 ( day15/day1) and day30 (day30/day1). Want to see the difference in index day15 and day30 between the groups. Since some of the index are much bigger than the rest, I want to use geometric mean to compare. Can you give more detailed explaination please?

[PGStats]

Can you explain why you want to use "geometric mean ANOVA" ?

[ggao]

The ranges of the ratios are from less than 0 to over 100. That's why I think that geometric mean may be more appropriate. But correct me if you think I am wrong. And please advise what technics should I use?

[PGStats]

You cannot take the geometric mean of negative values. 

One common misunderstanding about ordinary least squares estimation is about normality. Your values do not have to be distributed normally for OLS inference to be valid. Your values can be the sum of some fixed effects and random errors (noise). It is the random error that must obey a normal distribution. Thus the residuals from OLS must be normal, not the raw data.

Geometric mean ANOVA is actually ordinary ANOVA of log-transformed data. It should be considered for heteroscedastic data, when the standard deviation of errors seems to be proportional to the mean.

So, the first question is: Why is ANOVA not appropriate for your data?

[mariko5797]

I have a follow up question to this response. I want to compare the geometric mean concentrations of three BMI groups at two different time points (after start of treatment and after end of treatment). Would I simply be running an ANOVA on the (natural) log-transformed concentrations, then taking that p-value? My main goal is to see if at least one BMI group is different using Welch's ANOVA.

[PGStats]

I can't be definite without seeing your data and knowing more about the context, but your proposed approach sounds like a good idea to me.

[ggao]

Sorry, I made a typo. It should be less than 1 not less than 0. So no negative numbers.

[Ksharp]

That is really interesting question. I also want know.

But I think you can't. Using algorithm mean you can get chi-square value,

ChiSq1/ChiSq2 ~ F distribution, so you can do ANOVA , but I am afraid you can use geometric mean to get that chi-square value.

Maybe someone can develop some other quasi-ANOVA for geometric mean.

[Rick_SAS]

If you post your data and state the research question that you want to answer, we might be able to recommend how to analyze the data.

[ggao]

The study design is for three product A, B and C. 130 people approximately in each group. So total is 390 panelists. There is no crossover. Each panelist only use one product. Bacterial has been measured in baseline, day15 and day30. What I want to know is if there is significant difference between three products. And also want to compare baseline, day15 and day30 to see which product is acting quicker. Any feedback is appreciate.

[SteveDenham]

Product has 3 levels A, B and C. Time has 3 points 0, 15 and 30.

proc glimmix data=yourdata;

class subjid product time;

model bacterial_counts=product|time/link=log;

random time/residual subject=subjid type=un;

lsmeans product time product*time/diff adjdfe=row adjust=simulate ilink;

run;

Steve Denham