For large $n$, the sampling distribution of $S$ is sometimes approximated with a normal distribution having the mean $\sigma$ and the variance $\frac{\sigma^{2}}{2 n} .$ Show that this approxi- mation leads to the following $(1-\alpha) 100 \%$ large-sample confidence interval for $\sigma$ : $$ \frac{s}{1+\frac{z_{\alpha / 2}}{\sqrt{2 n}}}<\sigma<\frac{s}{1-\frac{z_{\alpha / 2}}{\sqrt{2 n}}} $$
Added by Elizabeth T.
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We are given that the sampling distribution of $S$ is approximated by a normal distribution with mean $\sigma$ and variance $\frac{\sigma^2}{2n}$. Show more…
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