If I have a sample of a lognormal random variable given by some values $x_1$, $x_2$, $x_3$, $x_4$, $x_5$,$x_6$, $x_7$, $x_8$, $x_9$, $x_{10}$, and this sample was truncated in some fixed value $x_0$; how can I calculate the 95% confidence interval for $\mu$?
I know that for a non-truncated sample I could find the maximum likelihood estimator of $\mu$ and $\sigma$, which are:
$\hat{\mu}_{MLE}= \frac{\sum_{i=1}^{10} ln(x_i)}{10}$
$\hat{\sigma}_{MLE}=\sqrt{\frac{\sum_{i=1}^{10} (ln(x_i)-\hat{\mu}_{MLE})^2}{10}}$
And then I would have:
Confidence Interval $\sim \hat{\mu}_{MLE} \pm 1.96 \cdot \frac{\hat{\sigma}_{MLE}}{\sqrt{10}}$
I have tried modifying my distribution function for another one that takes my left truncation into account and computing its maximum likelihood estimator to follow a similar procedure as above but I find it difficult to obtain the estimator of the function.