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Options I course of the University of Oxford

Past exam of the options I course of the University of Oxford

2011

{c}

Since the true mean is known to be zero, the <chi-squared distribution> gives $Q/\theta\sim\chi_n^2$. Let $q_\beta$ be the $\beta$ quantile of this law. The probability
$$
\mathbb P\left(q_{\alpha/2}\leq Q/\theta\leq q_{1-\alpha/2}\right)=1-\alpha
$$
rearranges, using positivity, to the exact <confidence interval>
$$
\boxed{\left[\frac{\sum_iX_i^2}{q_{1-\alpha/2}},\ \frac{\sum_iX_i^2}{q_{\alpha/2}}\right].}
$$
There are $n$ degrees of freedom, because no unknown mean was fitted.

Solution

Past exam of the options I course of the University of Oxford / 2011 / O / 1 / c / Solution

Past exam of the options I course of the University of Oxford / 2011 / O / 1 / c / Solution

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