Confidence interval for mean of lognormal distributed data

Question

I have a variable X that is distributed log-normally.

I let Y = lnX ~ N($\mu$, $\sigma^2$) and I've been given that $\sigma$=0.3, $\bar{y}$ = 0.12 and n = 40.

So I find a confidence interval for the mean of the log-transformed data like this:

$(\bar{y}-z_{1-\alpha/2}\times\frac{\sigma}{\sqrt n}, \bar{y}+z_{1-\alpha/2}\times\frac{\sigma}{\sqrt n})\\ (0.12-1.96\times\frac{0.3}{\sqrt 40}, 0.12+1.96\times\frac{0.3}{\sqrt 40})\\ (0.027, 0.213)$

To get the 95% confidence interval for E(X) (the original variable) I just raise e to the power of the endpoints of the interval I just calculated.

so the interval would be $(e^{0.027}, e^{0.213})=\\$ $(1.03, 1.24)$

Is this correct?

Thanks any help appreciated

Answer 1

Quoting https://ww2.amstat.org/publications/jse/v13n1/olsson.html using $\theta=EX$

It would seem natural to use the following "naïve" approach for calculating a confidence interval for $\theta$. A confidence interval for $\mu$ is calculated using standard methods. The limits of the confidence interval are back-transformed to give the limits in a confidence interval for $\theta$. [...an example illustrated...] This illustrates the fact that the naïve method gives a biased estimator of $\theta$

The method outlined by OP is this biased method.

The paper offers a few alternatives including Cox Method. If, as in OP's case $\sigma^2$ is known, a modification should be made to the Cox method. This modification would be to add $(1/2)\sigma^2$ to both ends of the confidence interval for $\mu$ and then take anti-log's.

The end of the paper also provides simulation studies to assess the coverage of the confidence intervals under each method.

Answer 2

Quantiles and percentiles are preserved under increasing transformations like $\exp(x)$ and $\ln(x)$ so the 5th and 95th percentile of $X$ will still be the 5th and 95th percentile when transformed through $\exp$. (the preservation property is mentioned on Wikipedia).