Jaume
If you assume that Y1 = log(X1),..., Yn=log(Xn) come from N(mu,
sigma-squared)
then with Ybar = sum(Y)/n and SE = SD(y)/sqrt(n) and large n,
Ybar - 1.96SE to Ybar + 1.96SE is a 95%CI for mu.
so
exp(Ybar-1.96 SE) to exp(Ybar+1.96 SE) is a 95% CI for exp(mu).
But if X is log-normal, the mean of X, is not exp(mu).
It is exp(mu + sigma-squared/2).
If you are happy with the log-normal assumption, multiply the exponentiated
CI endpoints by exp(s^2/2), where s^2 is the sample variance
A useful paper is Duan, JASA, September 1983: Smearing Estimate: A
nonparametric retransformation method.
George
----- Original Message -----
From: "Bob Baskin" <BobBaskin@westat.com>
To: "'Jaume'" <jaguado@medicina.ub.es>; "S-News (E-mail)"
<s-news@lists.biostat.wustl.edu>
Sent: Friday, May 24, 2002 12:33 PM
Subject: Re: geometric mean
> <I know that this question is not new but I haven't seen the answer yet.>
> the correct answer probably is 'that depends...'
>
> <Is it correct to work with the log-data and then make the antilog of the
> confidence interval limits?.>
> in general, under log and-or exponential transforms, variances do not
> transform correctly, but percentiles do.
>
> under different assumptions about how the errors arise you can derive
> different types of asymptotic confidence intervals.
>
> typical approaches for finding c.i.s for geometric means are
>
> 1) using some distributional assumption such as the data has a log-normal
> distribution.
> logging the data (as you said), finding a normal c.i. for the mean of the
> logged data, and
> 1a)transforming the percentiles. in general, this gives no estimate of
> standard error but you may not need it. also, if you are transforming the
> empirical distribution to get the percentiles,
> you will be limited in the alpha you can use (or have to form a random
> c.i.).
> 1b) making assumptions that allows you to transform the variance, such as
> the data has a log-normal distribution.
>
> note: if you assume that your data, X, is log-normal(mean=mu, standard
> deviation-sigma), then sigma, the standard deviation of ln(X), provides a
> measure of dispersion of X/exp(mu) in the sense that
> [exp(-1.96*sigma),exp(1.96*sigma)] is a 95% c.i. for X/exp(mu). so under
> the assumption of log-normal X, the anti-log of the standard normal c.i.
> provides a c.i. for X.
> (actually, this note is probably all you wanted)
>
> 2) using a taylor linearization.
> if X is your random variable, and GM(X) is the geometric mean of X, then:
> GM(X) approximately= E(X) - var(X)/[2E(X)]
>
>
> your english seems fine, but i would recommend that you talk to a spanish
> speaking statistician at your university.
>
> do you have a theory as to how your errors arise?
> do you need an estimate of variance or only the confidence interval?
> can you treat you data as an i.i.d. sample?
> do you think your data is log-normal?
> do you think a taylor approximation is a really fun way to get an
estimate?
>
> a few things you might need to think about.
>
> i'm sure i got a few things wrong in here but the list is good at
correcting
> me and, as i said, you should really ask someone about this.
>
> good luck with your problem.
> bob
>
> -----Original Message-----
> From: Jaume [mailto:jaguado@medicina.ub.es]
> Sent: Friday, May 24, 2002 5:53 AM
> To: S-News (E-mail)
> Subject: [S] geometric mean
>
>
> I know that this question is not new but I haven't seen the answer yet.
How
> could I compute the standard deviation of the geometric mean with S-Plus?.
> Is there a function to do it? Is it correct to work with the log-data and
> then make the antilog of the confidence interval limits?.
>
> Thank you.
>
> PD: this is my first post. Spanish undergraduate student. So sorry about
my
> english.
>
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