CLT confidence intervals I have come across the following statement:

$(*)$  The width of CLT-based 99% confidence intervals is $6\sigma n^{-1/2}$.      

How does one derive this? Is there a general formula?
I tried to check it myself:
The CLT states that a sum $S_n$ of iid rv's approaches the following distribution.
$$p(S_n = s) = \frac 1{2\pi n\sigma^2}\mbox{exp}\left(-\frac{(s-n\mu)^2}{2n\sigma^2}\right)$$
Now as I understand it, $(*$) is basically saying that
$$p(S_n\leq -3\sigma n^{-1/2}) + p(S_n \geq 3\sigma n^{-1/2}) = 0.01$$
where I have assumed (wlog?) $\mu = 0$.
Thus I was hoping that I might find 
$$\int_{-\infty}^{-3\sigma n^{-1/2}}p(s)ds = 0.005$$
where $n$ and $\sigma$ have magically cancelled out. However, this doesn't seem to be the case.
Can someone explain? 
 A: Presumably, a "CLT-based confidence interval" of confidence $1-\alpha$ for a sample $x_1,\ldots, x_n$ drawn randomly from a distribution with mean $\mu$ and variance $\sigma^2$ means the interval for $\mu$ having endpoints $\bar x \pm Z_{\alpha/2} s/\sqrt{n}$ where $\bar x$ is the sample mean, $s^2$ is the unbiased estimator of $\sigma^2,$ and $Z_{\alpha/2}$ is the $\alpha/2$ quantile of the standard Normal distribution.
This interval, as a function of the sample, also is random; but we may evaluate its expected width,
$$E\left[|(\bar x - Z_{\alpha/2}s/\sqrt{n}) - (\bar x + Z_{\alpha/2}s/\sqrt{n})|\right] = 2 Z_{\alpha/2}\,E[s]/\sqrt{n}.$$
We know $E[s]$ is finite (because $E[s^2]=\sigma^2$ is assumed finite for the CLT) and underestimates $|\sigma|$ (by virtue of Jensen's Inequality).  For guidance, note that when the distribution truly is Normal, the bias in this estimate is of the order $1/n.$  For anything but the smallest sample sizes, then, we may take $E[s]\approx \sigma$ in this analysis.

Consequently, the expected width of the confidence interval is $2 |Z_{\alpha/2}\,\sigma|/\sqrt{n}.$

When $1-\alpha=0.99,$ $\alpha/2 = 0.005$ and $Z_{\alpha/2} \approx -2.58,$ giving an expected width of $5.16\,\sigma.$  Anyone wishing to state a conservative rule of thumb will wish to overestimate this result and might want to add another fudge factor for all the approximations implied by use of the CLT and replacing $E[s]$ by $\sigma.$ They will also want to introduce simple, memorable coefficients.  These considerations lead to replacing $5.16$ by the next largest integer, yielding the statement in the question.
