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Date:         Mon, 15 Nov 1999 16:41:44 +0000
Reply-To:     John Whittington <John.W@MEDISCIENCE.CO.UK>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         John Whittington <John.W@MEDISCIENCE.CO.UK>
Subject:      Re: Statistic
Comments: To: dwall@QBLM.QUINTILES.COM
In-Reply-To:  <199911151436.OAA22628@vicar.netnames.net>
Content-Type: text/plain; charset="us-ascii"

At 16:20 15/11/99 +0200, dwall@QBLM.QUINTILES.COM wrote:

>Has anyone heard of a geometric CV (co-efficient of variation). >I know what a geometric mean and SD are. >I have some test data and it appears that the geometric CV is >the SD of log-transformed data (without anti-logging again). >This is then multiplied by 100 to convert to a percentage. >Is this correct???

I can't say I've heard of a 'geometric CV' but I would be rather surprised if it were what you have described. By analogy with a normal 'CV', one would imagine that a geometric one would have to be expressing the variability in terms of the mean in some fashion - whereas what you describe is simply a function of the SD.

If you had said that it was the SD of log-transformed data DIVIDED BY the mean of log-transformed data (both without anti-logging), then multiplied by 100, then that would have made much more sense.

Another possibility, of course, would be to take it as SD/mean*100, using the back-transformed (i.e. anti-logged) SD and mean - perhaps reasonable in some senses, in as much as it is only really after back-transformation that the SD and mean *are* really 'geometric'. Otherwise, what you'd be talking about would be 'the CV of log-transformed data'.

I'm sure someone around will know the true/conventional answer!

Regards

John

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