Aleah Choi

2022-03-27

Constructing confidence Interval (standard normal)
I am asked to construct 95 percent confidence interval of the variance,
sigma squared using this: $\frac{\sum _{j=1}^{n}\left(\frac{|{Y}_{j}|}{\sigma }|-\mathbb{E}|Z\mid \right)}{\sqrt{nVar\left(|Z|\right)}}$?
I know the expression given above is standard normal with mean, 0 and variance 1. Note I have I have calculated E|z| (the expectation of a standard normal) to be $\sqrt{\frac{2}{\pi }}$ and Var|z| to be $1-\frac{2}{\pi }$. Note Y1...Yn is a sample from N(0,sigma squared).
Note: $\sum _{j=1}^{n}|\frac{{T}_{j}}{\sigma }|$ are pivots, meaning is a pivot for variance, distribution does not depend on variance.
PROBLEM: The problem I'm having in constructing the confidence interval is isolating sigma squared (variance) from the expression given at the top to get an interval for variance. Please provide help, suggestions. Thank you

Dixie Reed

Explanation:
You have $d\frac{\frac{\sum _{j=1}^{n}|{Y}_{j}|}{\sigma }-n\sqrt{\frac{2}{\pi }}}{\sqrt{n\left(1-\frac{2}{\pi }\right)}}\sim N\left(0,1\right).$
Put this fraction between critical values $±1.96$ for standard normal distribution and solve with respect to $\sigma$ the inequality:
$-1.96\le d\frac{\frac{\sum _{j=1}^{n}|{Y}_{j}|}{\sigma }-n\sqrt{\frac{2}{\pi }}}{\sqrt{n\left(1-\frac{2}{\pi }\right)}}\le 1.96.$.
For n sufficently large this inequality leads to two-sided inequality for $\sigma$. Then square all parts.

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