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?