Constructing an upper confidence limit for $σ^2$
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Suppose that $X_1, ..., X_n$ form a random sample from the normal distribution with unknown mean µ and unknown standard deviation σ. Construct an upper confidence limit U(X1, ..., Xn) for $σ^2$ such that $P[σ^2 < U(X_1, ..., X_n)]= 0.99$ . What I've done is so far Since $dfrac {(n-1)s^2}{σ^2}$ ~ $X^2(n-1)$ , $P(X^2_.99 <dfrac {(n-1)s^2}{σ^2})$ = $P(σ^2<dfrac {(n-1)s^2}{X^2_.99})$ Therefore $U(X_1, ..., X_n)=dfrac {(n-1)s^2}{X^2_.99}$ Now is it correctly done?
sampling estimation-theory confidence-interval interval-arithmetic
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edited Dec 2 '18 at 6:38
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