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how does one calculate confidence values for geometric means

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Dr. Lester M. Shulman

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Aug 25, 2004, 4:30:05 AM8/25/04
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I use Excell to calculate the geometric mean of a series of antibody titers
using GEOMEAN. How can I calculate the 95% confidence level for this
geometric mean? Thank you. Dr Lester M. Shulman


Example:
512
1024
1024
1024
1024
1024
1024
1024
1024
2048
2048
2048
2048
4096
4096
4096
4096
4096
4096
4096
4096
8192
8192
8192
8192
8192
8192
8192

GEOMEAN=2756.404

JE McGimpsey

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Aug 25, 2004, 5:26:30 AM8/25/04
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I'm cross-posting this to the microsoft.public.excel.worksheet.functions
group (since it isn't Mac specific), in the hope that one of the stats
gurus can do a better job answering this than I can...

In article <BD5228BC.210D%cvls...@netvision.net.il>,

Jerry W. Lewis

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Aug 25, 2004, 6:55:59 AM8/25/04
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Depends on what you assume about the distribution of the data. If you
assume the data are lognormal, then you should take logs of the X values
calculate the arithmetic mean and usual Student's t confidence intervals
(say LCL to UCL) on the log scale, then back transform the results to
the original scale. Thus exp(LCL) to exp(UCL) is a 95% confidence
interval for exp(average(ln(data))) = geomean(data) for lognormal data.

Jerry W. Lewis, PhD

Jerry W. Lewis

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Aug 25, 2004, 7:39:47 AM8/25/04
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Sorry, I read the question, but didn't look at the data. The data are
interval censored. That is a much more complicated problem.

You could do a maximum likelihood analysis. If you assume that the
underlying continuous variable is from a lognormal distribution, then
the contribution to the log likelihood of the observation in A1 is
=LN(NORMDIST(LN(A1),$D$1,$D$2,TRUE)-NORMDIST(LN(A1/2),$D$1,$D$2,TRUE))
assuming that a titer of 512 is interpreted as "the true value lies
between 256 and 512". Here D1 initially contains average(ln(data)) and
D2 initially contains stdev(ln(data)). In another cell (say D3) use a
formula to sum the individual contributions to the log likelihood. Then
use Solver to maximize D3 by changing D1:D2, to get the maximum
likelihood estimates for the mean and standard deviation of the assumed
continuous (unobserved) underlying log data. With logs of the posted
data, I get an MLE of 7.575 for mu and 0.833 for sigma vs. 7.922 and
0.873 ignoring the censoring

Jerry

Michael R Middleton

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Aug 25, 2004, 3:47:59 PM8/25/04
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Dr. Lester M. Shulman -

Since your question isn't Macintosh-specific nor Excel-specific, it's more
likely that you'll get an answer from a statistics newsgroup or from the
Excel statisticians who frequent the microsoft.public.excel.misc newsgroup.

- Mike

www.mikemiddleton.com

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