1 The random variable \(X\) has the Normal distribution with mean 0 and variance \(\theta\), so that its probability density function is $$\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 \pi \theta } } \mathrm { e } ^ { - x ^ { 2 } / 2 \theta } , \quad - \infty < x < \infty$$ where \(\theta ( \theta > 0 )\) is unknown. A random sample of \(n\) observations from \(X\) is denoted by \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\).

1. Find \(\hat { \theta }\), the maximum likelihood estimator of \(\theta\).
2. Show that \(\hat { \theta }\) is an unbiased estimator of \(\theta\).
3. In large samples, the variance of \(\hat { \theta }\) may be estimated by \(\frac { 2 \hat { \theta } ^ { 2 } } { n }\). Use this and the results of parts (i) and (ii) to find an approximate \(95 \%\) confidence interval for \(\theta\) in the case when \(n = 100\) and \(\Sigma X _ { i } ^ { 2 } = 1000\).