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Mar 12, 2023 at 4:41 history edited User1865345 CC BY-SA 4.0
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Dec 9, 2010 at 10:19 vote accept Isaac
Dec 8, 2010 at 20:14 comment added shabbychef The quantity $(N(0,\frac{\sigma^2}{n}))^2$ is a scaled Chi-square. As a check, the variance of a Chi-square with $k$ dof is $2k$, which gives the same result you get.
Dec 8, 2010 at 19:43 comment added shabbychef exactly. The error is in the first line of the question, where the OP defines $\hat{m}$ as the sum of the $X_i$ not the mean, as you do.
Dec 8, 2010 at 19:35 history answered mpiktas CC BY-SA 2.5