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You might start this problem as follows:
P(s^2/true variance <= 2) = .95 is equivalent to
P(s^2 * (n-1) / true variance <= (n-1)*2) = .95
We know that the left hand side of the inequality is a chi-squared
random variable with n-1 degrees of freedom. So, looking at
Table A.2, we would choose df=1 and find that (n-1)*2 = 3.841,
which implies that n=3.
However, note that this solution doesn't work. That is,
P(X <= (3-1)*2) < .95
if X is a chi-squared rv with 2 df.
A trial and error method shows you that n=9 is the lowest n that
"works".
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