How to estimate $SE$ from $95\%$CI, $\alpha $ and $n$? Background
I am conducting a quantitative synthesis (meta-analysis) that includes previously published data. For the model that I am using, an exact or inflated estimate of SE is acceptable.
Question
Given a 95% Confidence Interval (measured from mean to upper or lower confidence limit), $\alpha$, and $n$, is it correct to estimate the standard error thus:
$$SE = \frac{CI}{t_{(1-\alpha/2,n)}}$$
?
This is based on the calculation by Saville, 2003, but most calculations of CI using the Z score. Unfortunately, $Z=f(SE)$, and I don't think it can be otherwise estimated.
note: this question is a subset of one that has remained unanswered on stats.stackexchange
 A: I would say yes.  
If your sample isn't large enough you should be using $t$ scores to find the confidence interval rather than $z$ scores, anyway.  (Some folks argue that you should always use $t$-scores, since the assumption required for using $z$-scores - that you know the population standard deviation $\sigma$ - is rarely satisfied.)  With $t$ scores, and using your notation, the formula for $CI$, half of the width of the confidence interval for a population mean, is known to be
$CI = t_{(1 - \alpha/2,n)} SE$, 
so solving that for $SE$ yields the expression you have in your question.
Update: Now that Henry has fixed the link, I see that the expression I give is exactly that in Saville's paper.  It's a well-known expression for the confidence interval using $t$-scores, and so, again, I don't see why it would be invalid to use it.  For example, see just about any introductory statistics text, such as Moore, McCabe, and Craig, Introduction to the Practice of Statistics, 6th edition, p. 420.
