Point and Interval Estimation Questions and Answers

Answer: A

The answer is both \(T_{1}\) and \(T_{2}\) are equally efficient when \(n = 2\).

With two observations, \(X_{(1)} + X_{(2)} = X_{1} + X_{2}\), so \(T_{2} = (X_{(1)} + X_{(2)})/2 = \bar{X} = T_{1}\). They are the same statistic and have the same variance.

For larger \(n\) the midrange is far better: \(\mathrm{Var}(T_{1}) = \theta^{2}/(12n)\), while \(\mathrm{Var}(T_{2}) = \theta^{2}/(2(n + 1)(n + 2))\), which falls like \(1/n^{2}\). So the efficiency of \(T_{1}\) relative to \(T_{2}\) is below 1 for \(n \gt 2\) and tends to 0, which rules out (B) and (D), and (C) fails at \(n = 2\), where the ratio is exactly 1.

Answer: D

The confidence coefficient is \(e^{-1/2} - e^{-1}\).

The interval covers \(1/\theta\) when \(X \le 1/\theta \le 2X\), that is, when \(\frac{1}{2\theta} \le X \le \frac{1}{\theta}\). With \(P(X \gt x) = e^{-\theta x}\):

\[ \begin{aligned} P\left(\tfrac{1}{2\theta} \le X \le \tfrac{1}{\theta}\right) &= e^{-\theta \cdot \frac{1}{2\theta}} - e^{-\theta \cdot \frac{1}{\theta}} \\ &= e^{-1/2} - e^{-1} \approx 0.239 \end{aligned} \]

It does not depend on \(\theta\), as a confidence coefficient must not. \(e^{-1/2}\) and \(e^{-1}\) are the two tail probabilities separately, and 1/2 does not arise.

Answer: C

The power and size are (1, 0).

Under \(H_{0}\) (\(\theta = -\frac{1}{2}\)) the observations lie in \((-1, 0)\), so \(X_{(1)} \gt 0\) is impossible: the size, the probability of rejecting a true \(H_{0}\), is 0.

Under \(H_{1}\) (\(\theta = \frac{1}{2}\)) the observations lie in \((0, 1)\), so \(X_{(1)} \gt 0\) with probability 1: the power is 1.

Because the two supports do not overlap, this test makes no errors at all. (0, 0), (0, 1) and (1, 1) mix up which probability belongs to which hypothesis.

Answer: B

The theorem used to obtain sufficient statistics is the Neyman factorization theorem.

It states that T(X) is sufficient for \(\theta\) if and only if the joint density factorises as \(f(x; \theta) = g(T(x); \theta)\, h(x)\), with \(h\) free of \(\theta\). Reading off such a factorisation is the standard way to find a sufficient statistic.

The Neyman-Pearson lemma gives most powerful tests. Basu's theorem uses a complete sufficient statistic to prove independence from ancillary statistics, and the Rao-Blackwell theorem improves an estimator by conditioning on a sufficient statistic; both take sufficiency as given rather than finding it.

Answer: C

The statement that is not true is that T(X) is unbiased for \(\theta\).

The density is a one-parameter exponential family, so \(T = \sum d(X_{i})\) is sufficient (factorisation) and, when the natural parameter space contains an interval, complete (A). By the Lehmann-Scheffe theorem a complete sufficient statistic is the UMVUE of its own expectation (B). Its distribution depends on \(\theta\) (D).

But \(E[T] = nE[d(X)]\), a function of \(\theta\) that is generally not \(\theta\) itself. For example, for the Poisson family \(d(x) = x\) and \(E[T] = n\theta\). So T is not, in general, unbiased for \(\theta\).

Answer: D

The statement that fits a confidence interval is option D: the interval covers the true value of \(\mu\) with probability 0.95.

Here \(\mu\) is a fixed, unknown number and the endpoints are computed from the random sample. The 95% belongs to the method: over many repeated samples, about 95% of the intervals built this way cover \(\mu\). Option D says this, with the interval as the thing that covers. Strictly, once the numbers (15, 25] have been computed, that one interval either contains \(\mu\) or does not; D is the usual textbook reading of the coverage property, and the key follows it.

Option B turns \(\mu\) into a random variable that falls in a fixed interval with probability 0.95, which is the Bayesian credible-interval reading and not what a confidence interval says. Option C states certainty, and option A picks the midpoint, which the interval does not identify.

Answer: C

The minimal sufficient statistic is \(\max\{-X_{(1)}, X_{(n)}\}\).

The joint density is \((2\theta)^{-n}\) when every \(|x_{i}| \le \theta\), i.e. when \(\max_{i}|x_{i}| \le \theta\), and 0 otherwise. By the factorisation theorem, \(\max|X_{i}| = \max\{-X_{(1)}, X_{(n)}\}\) is sufficient, and the likelihood ratio of two samples is free of \(\theta\) exactly when this maximum is the same, so it is minimal.

\(-X_{(1)}\) or \(X_{(n)}\) alone captures only one end of the sample, and the likelihood depends on whichever of them is larger. The minimum of the two is not sufficient.

Answer: D

The MLE of \(\theta\) is \(\max\{|X_{1}|, \ldots, |X_{n}|\}\).

The likelihood \(L(\theta) = (2\theta)^{-n}\) for \(\theta \ge \max|x_{i}|\), and 0 otherwise. It decreases in \(\theta\), so it is maximised at the smallest admissible value, \(\hat{\theta} = \max|X_{i}|\).

\(X_{(n)}\) or \(-X_{(1)}\) alone can be smaller than \(\max|X_{i}|\), and then the likelihood at that value is 0 because some observation lies outside \((-\theta, \theta)\). \((X_{(1)} + X_{(n)})/4\) is not even guaranteed to be positive.

Answer: A

The statement that is not true is that \(X_{(n)}\) is consistent for \(\theta\).

The upper end of the support is \(\theta + 1\), and the sample maximum converges to it: \(X_{(n)} \to \theta + 1\) in probability. So \(X_{(n)}\) is consistent for \(\theta + 1\), not for \(\theta\), and \(X_{(n)} - 1\) is consistent for \(\theta\) (D).

The distribution is symmetric about \(\theta\), so the sample mean (by the law of large numbers) and the sample median (which converges to the population median \(\theta\)) are both consistent, making B and C true.

Answer: C

The MLE is \(\frac{X + 1}{3}\).

The likelihood of one Bernoulli observation is \(p^{X}(1 - p)^{1 - X}\). If \(X = 1\), it is \(p\), increasing in \(p\), so on \([1/3, 2/3]\) it is maximised at the right end \(p = 2/3\). If \(X = 0\), it is \(1 - p\), decreasing, so it is maximised at \(p = 1/3\).

The formula \((X + 1)/3\) gives exactly 1/3 for \(X = 0\) and 2/3 for \(X = 1\).

The unrestricted MLE X takes the values 0 and 1, which lie outside the parameter space. \((2X + 1)/3\) gives 1 when \(X = 1\), also outside, and \((X - 1)/3\) is negative or zero, never in \([1/3, 2/3]\).