# Confidence interval for geometric mean

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

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.
• 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? – ocram Mar 16 '11 at 7:46
• @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 ^_^ – Dmitrij Celov Mar 16 '11 at 7:57
• 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. – Dimitriy V. Masterov Mar 27 '13 at 7:40