Asymmetric confidence intervals Suppose we have iid data $X_i$ with known variance $\sigma^2$, and wish to write an asymptotic $1-\alpha $ coverage CI for the population mean $\mu$. CLT implies that if $z_q$ represents the $q$ quantile of a standard normal,
$$z_{\alpha/2}=-z_{1-\alpha/2}\leq \frac{\bar X-\mu}{\sigma/\sqrt  n}\leq z_{1-\alpha/2}$$
occurs (asymptotically) with probability $1-\alpha$ and thus implies a CI for $\mu$ of $\bar X\pm z_{1-\alpha/2}\frac{\sigma}{\sqrt n}.$
Any particular reason we take symmetric bounds, or is this just a matter of simplicity? For instance, it seems to me we could have also used
$$ z_{q_1}\leq \frac{\bar X-\mu}{\sigma/\sqrt  n}\leq z_{q_2}$$
for any $q_2-q_1=1-\alpha.$

Update: By "symmetric," I mean using $q_2=1-q_1.$

There is no single best way of constructing a confidence interval.
In the below graph you see some different possibilities of a 95% confidence interval for a z-test.
In this graph the confidence interval is expressed as the boundaries in term of the amount of $\pm \sigma$ deviation from the observed $X$ and taken as a constant independent of $X$ (you could make confidence intervals with different shapes/sizes depending on $X$).
For instance for the percentage on the left tail, $q_1 = 0.025$, you get the middle interval which is $[\bar{x} -1.96\sigma, \bar{x}+1.96 \sigma]$.

The symmetric confidence intervals (in the middle on the graph, wich is based on a two-tailed hypothesis test), divides the attention equally to both sides. It is advantageous because it doesn't favor any side but also because it is a smaller confidence interval compared to the other choices.
On the extreme of the graph, we see confidence intervals based on one-tailed hypothesis tests. These confidence intervals are larger for the z-test, but the boundary on one side is less more close to the mean. So there are advantages and disadvantages and which is chosen depends on the application.
