 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2011/o/1/c/solution

Since the true mean is known to be zero, the chi-squared distribution gives

Q / θ \sim χ_{n}^{2}

. Let

q_{β}

be the

β

quantile of this law. The probability

P (q_{α /2} \leq Q / θ \leq q_{1 - α /2}) = 1 - α

rearranges, using positivity, to the exact confidence interval

[\frac{\sum _{i} X _{i}^{2}}{q _{1 - α /2}}, \frac{\sum _{i} X _{i}^{2}}{q _{α /2}}] .

There are

n

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