\line{\hfil Michael W. Daniels}
\line{\hfil MATH 363}
\line{\hfil July 27, 1998}

\parindent = 0 pt
\parskip = 10 pt
\def\sonen{\sum_{i=1}^n}
\def\ponen{\prod_{i=1}^n}
\def\plglt{{\partial \ln L(x;\theta) \over \partial\theta}}
\def\izinf{\int\limits_0^\infty}
\def\extt{e^{-{x^2 \over 2\theta}}}

6.1.5)

a) We are given $f(x;\theta) = {xe^{-{x \over \theta}} \over
\theta^2}$, so $$\eqalign{
L(x;\theta) &= \ponen f(x_i;\theta) \cr
&= \theta^{-2n} \left(\ponen x_i\right)\left(e^{-{\sonen x_i \over
\theta}}\right) \cr
\ln L(x;\theta) &= -2n \ln \theta + \sonen \ln x_i - {\sonen x_i \over
\theta} \cr
\plglt &= -{2n \over \theta} + {\sonen x_i \over \theta^2}}$$

Solving this for zero yields $$\eqalign{
{2n \over \hat \theta} &= {\sonen x_i \over \hat\theta^2} \cr
2n\hat\theta &= \sonen x_i \cr
\hat\theta = {\sonen x_i \over 2n} = {\bar x \over 2}}$$

b) We are given $f(x; \theta) = {x^2e^{-{x \over \theta}} \over
2\theta^3}$, so $$\eqalign{
L(x;\theta) &= \ponen f(x_i; \theta) \cr
&= {\theta^{-3n} \over 2}\left(\ponen x_i^2\right)\left(e^{-{\sonen
x_i \over \theta}}\right) \cr
\ln L(x;\theta) &= -3n \ln 2\theta + \sonen 2 \ln x_i - {\sonen x_i
\over \theta} \cr
\plglt &= -{6n \over 2\theta} + {\sonen x_i \over \theta^2}}$$

Solving this for zero yields $$\eqalign{
{3n \over \hat\theta} = {\sonen x_i \over \hat\theta^2} \cr
3n\hat\theta = \sonen x_i \cr
\hat\theta = {\sonen x_i \over 3n} = {\bar x \over 3}}$$

c) We are given $f(x; \theta) = {e^{-|x - \theta|} \over 2}$, so
$$\eqalign{
L(x;\theta) &= \ponen f(x_i; \theta) \cr
&={e^{-\sonen |x_i - \theta|} \over 2^n} \cr
\ln L(x;\theta) &= -n \ln 2 - \sonen |x_i - \theta| \cr
\plglt &= \sonen {|x_i - \theta| \over x_i - \theta}}$$

Since this is the sum of a function that can only take the values $1$
and $-1$, it will be zero when exactly half of the $x_i - \hat\theta$
terms are $1$ and half are $-1$, or when half of the $x_i$ terms are
greater than $\hat\theta$ and half less than $\hat\theta$.  This
occurs when $\hat\theta$ is the sample median.

6.1.7)

a) \vskip 2 in

b) We are given $f(x;\theta) = \theta x^{\theta - 1}$, so $$\eqalign{
L(x;\theta) &= \ponen f(x_i;\theta) \cr
&= \theta^n \ponen x_i^{\theta - 1} \cr
\ln L(x;\theta) &= n \ln \theta + \sonen (\theta - 1)\ln x_i \cr
\plglt &= {n \over \theta} + \sonen \ln x}$$

Solving this for zero yields $$\eqalign{
-{n \over \hat\theta} = \sonen \ln x_i \cr
\hat\theta = {-n \over \ln \ponen x_i}}$$

c) From the example on page~289, we know that the method-of-moments
estimator $\tilde\theta$ is ${\bar x \over 1 - \bar x}$.  For the
first set, $\hat\theta = 0.5493$ and $\tilde\theta = 0.5975$; for the
second set, $\hat\theta = 2.2101$ and $\tilde\theta = 2.4004$; for the
third set, $\hat\theta = 0.9586$ and $\tilde\theta = 0.8646$.

d) \vskip 2 in

6.1.8)

a) We are given $f(x;\theta) = {x^{{1 - \theta \over \theta}} \over
\theta}$, so $$\eqalign{
L(x;\theta) &= \ponen f(x_i;\theta) \cr
&= {\ponen x_i^{{1 - \theta \over \theta}} \over \theta^n} \cr
\ln L(x;\theta) &= -n \ln \theta + \sonen \left({\ln x_i \over \theta}
- \ln x_i\right) \cr
\plglt &= -{n \over \theta} + \sonen -{\ln x_i \over \theta^2}}$$

Solving this for zero yields $$\eqalign{
{n \over \hat\theta} &= -{\sonen \ln x_i \over \theta^2} \cr
\hat\theta &= -{\sonen \ln x_i \over n}}$$

b) We have $E(\hat\theta) = -{E\left(\sonen \ln x_i\right) \over n}$.
Since $E\left(\sonen \ln x_i\right) = -n\theta$, this expression
equals ${n\theta \over n} = \theta$.  Since $E(\hat\theta) = \theta$,
the estimator is unbiased.

6.1.20)

This is a hypergeometric distribution with $n = 8$, $N = 64$, $N_2 =
64 - N_1$, and parameter $\theta = N_1$.  For such a distribution,
$\mu = n{N_1 \over N} = {\theta \over 8}$.  By the method of moments,
then, we can use $\tilde\theta = 8\bar x$ as an estimator for
$\theta$.  Then $\bar x = 1.467$ and $\tilde\theta = 11.733$.  We
therefore guess that there are 12 orange balls and 52 blue balls in
the urn.

\vfill\eject

6.2.6)

The confidence interval is given by $$
\bar x \pm t_{\alpha /2}(n-1) {s \over \sqrt{n}} = 11.95 \pm
{(1.96)(11.8) \over \sqrt{37}} = (8.148, 15.752)$$

6.2.14)

The confidence interval is given by $$
\bar x \pm z_{\alpha /2} {s \over \sqrt{n}} = 6.05 \pm
{(2.576)(0.02) \over \sqrt{1219}} = (6.0485, 6.0515)$$

6.3.2) 

Here, the variances of the populations are equal.  We have $n=5$,
$\bar x = 539.2$, $s_x = 62.92$, $m=8$, $\bar y = 544.625$, and $s_y =
61.538$.  The confidence interval is given by $$(\bar x - \bar y) \pm
t_{\alpha /2}(m+n-2)(s_p)\sqrt{{1 \over n} + {1 \over m}}$$  Since
$$s_p = \sqrt{{(n -1)s_x^2 + (m-1)s_y^2 \over m+n-2}} = 62.044$$ the
interval is $$-5.425 \pm (1.796)(62.044)(0.5701) = (-68.95, 58.100)$$

6.3.4)

Here, the variances of the populations are not equal.  We have $n=7$,
$\bar x = 1511.714$, $s_x = 206.335$, $m=10$, $\bar y = 1118.4$, and
$s_y = 117.336$.  

a) The point estimate is $\bar x - \bar y = 393.314$.

b) The confidence interval is given by $$(\bar x - \bar y) \pm
t_{\alpha /2}(m+n-2)\sqrt{{s_x^2 \over n} + {s_y^2 \over m}} = 393.314
\pm (2.131)(78.062) = (226.963, 559.665)$$

6.3.8)

a) The point estimate is $\bar x = 0.07875$.

b) The confidence interval is given by $$
\bar x \pm t_{\alpha /2}(n-1) {s \over \sqrt{n}} = 0.07875 \pm
{(2.069)(0.2496) \over \sqrt{24}} = (-0.02665, 0.1842)$$

c) Since the confidence interval is centered around a positive number,
we can say that the program was probably effective.

\vfill\eject

6.5.4)

Let $\hat p = {y \over n} = 0.700$.  Then the desired interval is
given by $$\hat p \pm z_{\alpha /2} \sqrt{{\hat p(1 - \hat p) \over
n}} = 0.7 \pm 1.96\sqrt{{(0.7)(0.3) \over 1234}} = (0.6744, 0.7256)$$

6.5.18)

a) Let $\hat p_A = {y_A \over n_A} = 0.369$ and $\hat p_B = {y_B \over
n_B} = 0.320$.  Then the desired interval is given by $$\hat p_A -
\hat p_B \pm z_{\alpha /2} \sqrt{{\hat p_A (1 - \hat p_A) \over n_A} +
{\hat p_B (1 - \hat p_B) \over n_B}} = 0.04911 \pm 1.96\sqrt{0.0005062
+ 0.0004945} = (-0.01289, 0.1111)$$

b) If the two versions were consistent, we would expect the interval
to be centered around zero, as an ideal difference would equal zero.
Since this interval is skewed toward the right, the two versions are
not consistent.

Handout \#7)

We are given $f(x) = {x \over \theta} \extt$, so $$\eqalign{
E(Y) &= \izinf x^2 {x \over \theta} \extt\,dx \cr
&= -x^2 \extt + \izinf 2x\extt\,dx \cr
&= \left[-x^2 \extt - 2\theta\extt\right]_0^\infty \cr
&= \left[-\extt (2\theta + x^2)\right]_0^\infty \cr
&= 2\theta}$$

$Var(Y) = E(Y^2) - (E(Y))^2 = E(X^4) - 4\theta^2$.  We note that
$$\eqalign{
E(x^4) &= \izinf x^4 \extt\,dx \cr
&= -x^4 \extt + \izinf 4x^3 \extt\,dx \cr
&= -x^4 \extt - 4\theta x^2\extt + \izinf 8\theta x\extt\,dx \cr
&= \left[-x^4 \extt - 4\theta x^2\extt - 8\theta^2
\extt\right]_0^\infty \cr
&= \left[-\extt (x^4 + 4\theta x^2 + 8\theta^2)\right]_0^\infty \cr
&= 8\theta^2}$$

and therefore $Var(Y) = 8\theta^2 - 4\theta^2 = 4\theta^2$.

\vfill\eject

Handout \#8)

a) We are given $f(x) = {x \over \theta} \extt$, so $$\eqalign{
L(x;\theta) &= \ponen f(x_i;\theta) \cr
&= {\ponen x_i \over \theta^n} e^{-{\sonen x_i^2 \over 2\theta}}}$$

b) $$\eqalign{
\ln L(x;\theta) &= \ln \ponen x_i - \ln (\theta^n) + \sonen -{x_i^2
\over 2\theta} \cr
&= \sonen \ln x_i - n \ln \theta - {\sonen x_i^2 \over 2\theta} \cr
\plglt &= -{n \over \theta} + {\sonen x_i^2 \over 2\theta^2}}$$

Solving this for zero yields $$\eqalign{
{\sonen x_i^2 \over 2\hat\theta^2} &= {n \over \hat\theta} \cr
\hat\theta &= {\sonen x_i^2 \over 2n}}$$

c) $$\eqalign{
E(\hat\theta) &= {E(\sonen x_i^2) \over 2n} \cr
&= {nE(X^2) \over 2n} \cr
&= {2\theta \over 2} = \theta}$$

Since $E(\hat\theta) = \theta$, $\hat\theta$ is an unbiased estimator
of $\theta$.

d) $$\eqalign{
Var(\hat\theta) &= {Var(\sonen x_i^2) \over (2n)^2} \cr
&= {n(Var(X^2)) \over 4n^2} \cr
&= {4\theta^2 \over 4n} \cr
&= {\theta^2 \over n}}$$

$$SE(\hat\theta) = {SD(\hat\theta) \over \sqrt{n}} = {\theta
\over \sqrt{n}}\sqrt{n} = \theta$$

e) Since $\hat\theta$ is unbiased, $b(\hat\theta) = 0$ for all n.  We
note that $Var(\hat\theta) = {\hat\theta^2 \over n} \to 0$ as $n \to
\infty$.  Therefore $\hat\theta$ is a consistent estimate of $\theta$.

f) The 95\% confidence interval is given by $\bar x \pm
{(1.96)\sqrt{Var(\theta)} \over \sqrt{n}} = \bar x \pm 1.96\theta$.

\bye