Final

Problem 1

The test statistic is
$$\begin{align*} Z & = \frac{\overline{X} - \mu}{\sigma / \sqrt{n}} \ & = \frac{23.6 - 20}{10 / \sqrt{20}} \ & = 1.61\end{align*}$$
Under the null hypothesis Z is standard normal.

From Table I in Lindgren, the one-tailed P-value is 0.0537.

Problem 2

Welch's procedure uses the test statistic
$$\begin{align*} T & = \frac{\overline{X}_1 - \overline{X}_2}{\sqrt{\frac{S_1^2}... ...538}{\sqrt{\frac{0.241^2}{10} + \frac{0.313^2}{8}}} \ & = 1.8904\end{align*}$$
Under the null hypothesis T has an approximate Student t-distribution with noninteger d. f. given by
$$\begin{align*} \hat{\nu} & = \frac{\left(\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}... ...)^2 + \frac{1}{7} \left(\frac{0.313^2}{8}\right)^2} \ & = 12.95\end{align*}$$

From Table IIIa in Lindgren, the one-tailed P-value is 0.041 and the two-tailed is (double that) 0.082. (The exact P-value using a computer is 0.081.)

Problem 3

The test statistic

$$Z = \frac{\overline{X} - 10}{S / \sqrt{n}}$$

has an asymptotic standard normal distribution under the null hypothesis $\mu = 10$.For an $\alpha = 0.05$ level upper-tail test, we reject when Z > 1.6449 (Table 1a in Lindgren).

Under the alternative hypothesis $\mu = 11$

$$Z_A = \frac{\overline{X} - 11}{S / \sqrt{n}} = Z + \frac{10 - 11}{S / \sqrt{n}}$$

is approximately standard normal. We don't know what S will be. In order to do a calculation we plug in the estimate from preliminary data (S = 4).

The power condition is

$$0.9 = P_A(Z \gt 1.6449) = P_A\left(Z_A \gt 1.6449 + \frac{-1}{4 / \sqrt{n}} \right)$$

where P_A here means probability under the alternative $\mu = 11$(under which Z_A is standard normal). From Table 1a in Lindgren we see that P_A(Z_A > -1.2816) = 0.9 so we find n by solving the equation

$$- 1.2816 = 1.6449 - \frac{\sqrt{n}}{4}$$

or

$$\frac{\sqrt{n}}{4} = 1.6449 + 1.2816 = 2.9265$$

or

$$n = \left(4 \times 2.9265\right)^2 = 137.03$$

Round to 137 (or 138 if you want to be conservative).

Of course, since the actual population standard deviation is not 4, we won't get exactly the desired power, but this is the best we can do with the data at hand.

Problem 4

(a)

The likelihood for $\lambda$ is
$$\begin{align*} L_n(\lambda) & = \prod_{i=1}^n \left(1 - e^{- \lambda}\right) e... ... \ & = \left(1 - e^{- \lambda}\right)^n e^{- \lambda \sum_i X_i}\end{align*}$$
The log likelihood is

$$\begin{align*} l_n(\lambda) & = - \lambda {\textstyle \sum_i X_i}+ n \log\left(1 - e^{- \lambda}\right)\end{align*}$$
and

$$l_n'(\lambda) = - {\textstyle \sum_i X_i}+ n \frac{e^{- \lambda}}{1 - e^{- \lambda}}$$

which is equal to zero when

$$\overline{X}_n = \frac{e^{- \lambda}}{1 - e^{- \lambda}} = \frac{1}{e^{\lambda} - 1}$$

Solving for $\lambda$ gives

$$\hat{\lambda}_n = \log\left(1 + \frac{1}{\overline{X}_n}\right)$$

(b)

$$\begin{align*} l_1''(\lambda) & = - \frac{e^{- \lambda} (1 - e^{- \lambda}) + ... ...\lambda})^2} \ & = - \frac{e^{- \lambda}}{(1 - e^{- \lambda})^2}\end{align*}$$
Since this does not contain X_i, it is nonrandom, and hence its negative is the Fisher information

$$I_1(\lambda) = \frac{e^{- \lambda}}{(1 - e^{- \lambda})^2}$$

and the Fisher information for a sample of size n is $I_n(\lambda) = n I_1(\lambda)$.

(c)

A 95% C. I. for $\theta$ is

$$\hat{\lambda}_n \pm 1.96 \frac{1}{\sqrt{I_n(\hat{\lambda}_n)}}$$

Problem 5

The mean of a $\text{Beta}(s, t)$ distribution is s / (s + t), equation (17), p. 176 in Lindgren, so the mean of a $\text{Beta}(\alpha, 2)$ is

$$\mu = \frac{\alpha}{\alpha + 2}$$

A method of moments estimator is found by solving this equation for $\alpha$giving

$$\alpha = \frac{2 \mu}{1 - \mu}$$

and then plugging the corresponding empirical moment $\overline{X}_n$ for the theoretical moment $\mu$, giving

$$\hat{\alpha}_n = \frac{2 \overline{X}_n}{1 - \overline{X}_n}$$

Problem 6

(a)

The likelihood is

$$L(\lambda) \propto \lambda^\alpha e^{- \lambda x}$$

The terms $x^{\alpha - 1}$ and $\Gamma(\alpha)$ don't contain the parameter ($\lambda$) and can be omitted. The prior is

$$g(\lambda) \propto \lambda^{\alpha_0 - 1} e^{- \lambda_0 \lambda}$$

The terms $\lambda_0^{\alpha_0}$ and $\Gamma(\alpha_0)$ don't contain $\lambda$and can also be omitted.

Thus the unnormalized posterior is

$$L(\lambda) g(\lambda) \propto \lambda^{\alpha + \alpha_0 - 1} e^{- (x + \lambda_0) \lambda}$$

which we recognize as an unnormalized $\text{Gam}(\alpha + \alpha_0, x + \lambda_0)$density. So that's the posterior.

(b)

The mean of a $\text{Gam}(\alpha + \alpha_0, x + \lambda_0)$ is the thing in the first slot divided by the thing in the second slot, hence

$$E(\lambda \vert X = x) = \frac{\alpha + \alpha_0}{x + \lambda_0}$$

Problem 7

The variance $\overline{X}_n$ is $\sigma^2 / n$. The variance of $\widetilde{X}_n$ is

$$\frac{1}{4 n f(\mu)^2}$$

where $\mu$ is the population median. In this problem, the densities are symmetric about $\mu$, hence $\mu$ is the population mean and median both. (X has moments of all orders because it is bounded).

Calculation of $\sigma^2$.

This one we have to do by brute force because this distribution isn't one of the usual suspects.
$$\begin{align*} \mathop{\rm var}\nolimits(X) & = \int_{\mu - \lambda}^{\mu + \l... ... & = 2 \lambda^2 \frac{2 ! 1 !}{4 !} \ & = \frac{\lambda^2}{6}\end{align*}$$
Thus

$$\mathop{\rm var}\nolimits(\overline{X}_n) = \frac{\lambda^2}{6 n}$$

Asymptotic variance for $\widetilde{X}_n$.

$$\begin{align*} \mathop{\rm var}\nolimits{\widetilde{X}_n} & \approx \frac{1}{4 n f(\mu)^2} \ & = \frac{\lambda^2}{4 n}\end{align*}$$

The one with the smaller asymptotic variance ($\overline{X}_n$) is better. The ARE (not asked, but just in case anyone was wondering) is 4 / 6 = 2 / 3.