# Standard error for calculating non-symmetric large-sample confidence intervals

As an illustrative example, given observations of a random variable $X$ of interest, distributed as a $N(\mu, \sigma^2)$ and taken from a large sample of size $n$, based on knowing a large-sample (asymptotic) 95% confidence interval for $\mu$, one can go about finding $SE[X]$ using the equation

$SE[X] = \frac{U - L}{2z_{\alpha/2}}$,

where $U$ and $L$ are the upper and lower confidence limits respectively. The above equation can be easily derived for a symmetric confidence interval around $\mu$ for any confidence level.

Is there a known general expression that takes into account $Skew[X]$ (or some other measure of skewness) for a non-symmetric interval for a skewed distribution (such as the exponential)?

• I'm sorry, but a normal distribution is not skewed. – Roland Sep 19 '17 at 7:02
• I have made it more clear. Just using the Normal as an illustrative example. – compbiostats Sep 19 '17 at 13:24
• As you are still with a large $n$, you still have the CLT and can compute confidence intervals for $\mu$ the usual way for any distribution. Any particular reason that is not sufficient? – David Ernst Sep 19 '17 at 13:59
• @DavidErnst My statistic of interest is actually the inverse estimate $x_0$, the value of the predictor that corresponds to a particular value of the response. Bootstrapping shows that the distribution of resamples is left-skewed, but approaching normality nonetheless. I would just like to know if perhaps there's a closed form formula to compute the standard error. Perhaps though, given an interval ($U$, $L$), it is sufficient just to use the equation I have given in my question. – compbiostats Sep 19 '17 at 14:18