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As title, is there anything like this? I know how to calculate CI for arithmetic mean, but how about geometric mean? Thanks.

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The geometric mean $(\prod_{i=1}^n X_i)^{1/n}$ is an arithmetic mean after taking logs $1/n \sum_{i=1}^n \log X_i$, so if you do know the CI for the arithmetic mean do the same for the logarithms of your data points and take exponents of the upper and lower bounds.

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  • $\begingroup$ When I read the question I wanted to suggest that strategy. But I preferred to wait for other suggestions because something stopped me. What if one of the $X_i$'s is negative? $\endgroup$ – ocram Mar 16 '11 at 7:46
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    $\begingroup$ @Marco, read the footnote in wikipedia for geometric mean. If one goes for the geometric means, he or she assumes that all $X_i$'s are strictly positive (even zero would be not suitable here). Real life data when in levels is mostly positive ^_^ And even if you do some negatives (like gains and loses) split the two and make them positive again ^_^ $\endgroup$ – Dmitrij Celov Mar 16 '11 at 7:57
  • $\begingroup$ i feel that is not appropriate because once taking exponential of the standard deviation doesnt have the meaning. in that time we cant go for confidence interval also $\endgroup$ – user22576 Mar 27 '13 at 7:09
  • $\begingroup$ The answer above is not advocating that. He's saying you calculate $z=\ln x,$ then calculate the arithmetic mean of $z$, call it $\bar z$, along with the corresponding confidence interval $[L,U]$. The geometric mean is then $\exp \{ \bar z \}$, and its CI is $[\exp \{L \},\exp \{U \}].$ You can also do this in a regression setting. $\endgroup$ – Dimitriy V. Masterov Mar 27 '13 at 7:40
  • $\begingroup$ ya i agree that. but is it appropriate?. if you see later that confidence interval. the mean would not come between the confidence interval. According to me, after taking ln and again once we tranformed. then there is no meaningfull interpretation for standard deviation. $\endgroup$ – user22576 Mar 27 '13 at 7:46

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