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Aug 28 at 22:12 vote accept Adrian Keister
Aug 28 at 22:12 comment added Adrian Keister Yep, I think you're right. On page 80-1, the author mentions this approximation, and in context, the $\sigma^2/N$ for independent $X$'s is compared to this expression. So it has to be $\sigma_X^2.$
Aug 28 at 22:05 comment added Adrian Keister Oh, interesting! The book's problem statement is exactly how I have it: there's no subscript on the $\sigma^2,$ nor any clarification as to which variance the $\sigma^2$ is referring. Your explanation would certainly fit what's going on, here. Many thanks!
Aug 28 at 21:46 history answered Thomas Lumley CC BY-SA 4.0