Diagnostic Placement Quiz

$\theta$

$1/\theta$

$\theta^2$

$1/\theta^2$

$45$

$51$

$39$

$55$

$\bar X_n \sim N(\mu, \sigma^2/n)$ exactly for any $n \geq 1$.

$\sqrt{n}(\bar X_n - \mu)/\sigma \xrightarrow{d} N(0,1)$ as $n \to \infty$.

$\bar X_n \xrightarrow{d} \mu$ implies $\bar X_n \xrightarrow{P} \mu$ only if the $X_i$ are normal.

$\bar X_n \xrightarrow{P} \mu$ requires $E[X_1^4] < \infty$.

$n \geq 80$

$n \geq 800$

$n \geq 8000$

$n \geq 80000$

$N(0, \sigma^2)$

$N(0, \sigma^2 \cdot [g'(\theta)]^2)$

$N(g(\theta), \sigma^2 [g'(\theta)]^2)$

$N(0, \sigma^2 [g''(\theta)]^2)$

$\hat\lambda = \bar X_n$

$\hat\lambda = 1/\bar X_n$

$\hat\lambda = \frac{1}{n}\sum_i (X_i - \bar X_n)^2$

$\hat\lambda = \max_i X_i$

$\hat\theta = \bar X_n$

$\hat\theta = 2\bar X_n$

$\hat\theta = X_{(n)} = \max_i X_i$

$\hat\theta = X_{(1)} = \min_i X_i$

$\hat\sigma^2_{\text{MLE}} = \frac{1}{n-1}\sum_i (X_i - \bar X_n)^2$ and is unbiased.

$\hat\sigma^2_{\text{MLE}} = \frac{1}{n}\sum_i (X_i - \bar X_n)^2$ and is biased downward.

$\hat\sigma^2_{\text{MLE}} = \frac{1}{n}\sum_i X_i^2$ and is unbiased.

The MLE of $\sigma^2$ does not exist when $\mu$ is unknown.

$\mathrm{MSE}(T_n) = \mathrm{Var}(T_n) - [\mathrm{bias}(T_n)]^2$

$\mathrm{MSE}(T_n) = \mathrm{Var}(T_n) + [\mathrm{bias}(T_n)]^2$

$\mathrm{MSE}(T_n) = [\mathrm{bias}(T_n)]^2$ only when $T_n$ is unbiased.

$\mathrm{MSE}(T_n) = E[T_n]^2 - \mathrm{Var}(T_n)$

$T = X_{(1)} = \min_i X_i$

$T = X_{(n)} = \max_i X_i$

$T = \sum_{i=1}^n X_i$

$T = (\sum X_i, \sum X_i^2)$

Gamma$(\alpha + \sum X_i, \beta + n)$

Gamma$(\alpha + n, \beta + \sum X_i)$

Gamma$(\alpha \cdot \sum X_i, \beta \cdot n)$

Beta$(\alpha + \sum X_i, \beta + n - \sum X_i)$

$\chi^2_n$

$\chi^2_{n-1}$

$N(0, 2(n-1)\sigma^4)$

$t_{n-1}$

$12.5 \pm 1.96 \cdot \frac{2}{\sqrt{16}} = (11.52,\ 13.48)$

$12.5 \pm 2.131 \cdot \frac{2}{\sqrt{16}} = (11.43,\ 13.57)$

$12.5 \pm 2.131 \cdot \frac{4}{\sqrt{16}} = (10.37,\ 14.63)$

$12.5 \pm 2.131 \cdot \frac{2}{16}$

$\theta_0$ falls OUTSIDE the corresponding $(1-\alpha)$ confidence interval.

$\theta_0$ falls INSIDE the corresponding $(1-\alpha)$ confidence interval.

$\theta_0$ falls outside the $\alpha$ confidence interval (not $(1-\alpha)$).

Test/CI duality only holds for one-sided alternatives.

$\theta(1-\theta)$

$\frac{1}{\theta(1-\theta)}$

$\frac{1}{\theta^2}$

$\frac{n}{\theta(1-\theta)}$

Bernoulli$(\theta)$

Poisson$(\lambda)$

$N(\mu, 1)$

Uniform$([0, \theta])$

No, because $\bar X_n$ is biased.

Yes — $\bar X_n$ is unbiased and $\mathrm{Var}(\bar X_n) = \theta/n = 1/(n I(\theta))$, achieving the CRLB.

No — efficiency only applies to normal data.

No — Poisson does not satisfy the CRLB regularity conditions.

$\sup_{\theta \in \Omega_0} P_\theta(\text{reject } H_0)$

$\sup_{\theta \in \Omega_1} P_\theta(\text{reject } H_0)$

$\sup_{\theta \in \Omega_1} P_\theta(\text{accept } H_0)$

$\inf_{\theta \in \Omega_0} P_\theta(\text{reject } H_0)$

$0.0359$

$0.0718$

$0.9641$

$0.5 - 0.0359 = 0.4641$

The likelihood ratio $L(\theta_1)/L(\theta_0) > k$ for some constant $k$ chosen so the test has size $\alpha$.

The likelihood ratio $L(\theta_0)/L(\theta_1) > k$.

The MLE $\hat\theta$ is closer to $\theta_1$ than to $\theta_0$.

The Fisher information evaluated at $\theta_0$ exceeds the threshold.

No UMP test exists because the parameter space is composite.

The UMP level-$\alpha$ test rejects when $\sqrt{n}(\bar X - \mu_0)/\sigma > z_{\alpha}$.

The UMP test rejects when $|\sqrt{n}(\bar X - \mu_0)/\sigma| > z_{\alpha/2}$.

A UMP test exists only if $\sigma^2$ is unknown.

$Z = \sqrt{n}(\bar X - \mu_0)/\sigma \sim N(0,1)$

$T = \sqrt{n}(\bar X - \mu_0)/s \sim t_{n-1}$

$T = \sqrt{n-1}(\bar X - \mu_0)/s \sim t_n$

$F = (\bar X - \mu_0)^2 / s^2 \sim F_{1, n-1}$

$T = (\bar X - \bar Y) / s_p \sqrt{1/m + 1/n}$, with $T \sim t_{m+n-2}$

$F = s_X^2 / s_Y^2$ with $F \sim F_{m-1, n-1}$ under $H_0$

$\chi^2 = (m+n-2) s_p^2 / \sigma_0^2$ with $\chi^2 \sim \chi^2_{m+n-2}$

$Z = (s_X^2 - s_Y^2)/\sqrt{2(s_X^4/m + s_Y^4/n)}$ with $Z \sim N(0,1)$

$T = 2.0$, fail to reject $H_0$.

$T = 2.0$, reject $H_0$.

$T = 4.0$, reject $H_0$.

$T = 1.0$, fail to reject $H_0$.

$N(0, 1)$ regardless of dimension.

$\chi^2_d$ where $d = \dim(\Omega) - \dim(\Omega_0)$.

$\chi^2_n$ where $n$ is the sample size.

$F_{d, n-d}$ where $d = \dim(\Omega) - \dim(\Omega_0)$.