\magnification=\magstep1
\centerline{MATH 376 -- Mathematical Statistics}
\centerline{Solutions for Exam 1 Sample Problems}
\centerline{March 12, 2012}
\bigskip
\noindent
I.  
\item{A)}  This has a $\chi^2(7)$ distribution (see Theorems 6.4 and 7.2
in the text).
\item{B)}  From the $\chi^2$ table on pages 850-851 of the text, 
we see:
$$1.68987 = \chi^2_{.975}, \quad {\rm and} \quad 14.0671 = \chi^2_{.05}$$
Hence
$$P(1.68987 \le W \le 14.0671) = .975 - .05 = .925$$
(Drawing the picture helps if this is not clear!)
\item{C)}  Leaving out $Z_6^2$ in the sum means $U$ has a $\chi^2(6)$ 
distribution.  Since
$$V = {6 Z_7^2 \over U} = {Z_7^2/1 \over U/6}$$
has the form given in Definition 7.3, $V$ has
an $F$ distribution with 1 d.f. in the numerator
and 6 d.f. in the denominator. 
\item{D)}  By the conventions used in our book, this means
that $y = f_{.05}(1,6)$.  From the $F$-table (on page 852) we find
$y = 5.99$.
\item{E)}  For this, we need the relation derived on Lab 1:
$$f_{1 - \alpha}(n,d) = {1\over f_\alpha(d,n)}$$
Here, 
$$y = f_{.95}(1,6) = {1\over f_{.05}(6,1)} = {1\over 234.0} \doteq .0043$$
\item{F)}  Recalling Definition 7.2, let $X = \sum_{i=1}^5 Z_i^2$.  Then
$$T = {Z_6\over \sqrt{X/5}}$$
has a $t$-distribution with $5$ degrees of freedom.  
\bigskip
\noindent
II.  
\item{A)}  By the formulas from \S 6.7, the cdf for $Y_{(n)}$ is
$$F_{(n)}(y) = (1 - e^{-y/\theta})^n$$
Hence 
$$P(Y_{(n)} \ge 2\theta) = 1 - P(Y_{(n)} < 2\theta) = 1 - (1 - e^{-2\theta/\theta})^n = 1 - (1 - e^{-2})^n$$
Since $0 < 1 - e^{-2} < 1$, 
$$\lim_{n\to\infty} P(Y_{(n)} \ge 2\theta) = \lim_{n\to\infty} 1 - (1 - e^{-2})^n = 1.$$
This should seem intuitively reasonable, since the exponential distributions have very long
right-hand tails and the expected value is $\theta$.  If you take $n\to\infty$, the probability
that the largest sample value exceeds $2\theta$ is tending to 1.  
\item{B)}  The pdf for the sample minimum is 
$$f_{(1)}(y) = n\left(1 - (1 - e^{-y/\theta})\right)^{n-1} {e^{-y/\theta}\over \theta} = 
{e^{-y/(\theta/n)} \over \theta/n}.$$
We recognize the form of an exponential density with mean $\theta/n$.  Therefore 
$E(Y_{(1)}) = \theta/n$.  As a result $n Y_{(1)}$ is an unbiased estimator for $\theta$. 
\bigskip
\noindent
III.  
\item{A)}  The usual point estimator for $\mu$ would be the sample mean:
$$\overline{Y} = {16.0 + 15.2 + 12.0 + 16.9 + 14.4 + 16.3 + 15.6 + 12.9\over 8} \doteq 14.90   $$
The usual point estimator for $\sigma$ would be the sample standard deviation:
$$S = \sqrt{{1\over 7} \sum_{i=1}^8 (Y_i - \overline{Y})^2} \doteq 1.707$$
\item{B)}  Under the assumptions that the data are from a normal population, we can compute 
a $90\%$ confidence interval using the small sample formulas.  The endpoints are
$$\overline{Y} \pm t_{.05} {S\over \sqrt{n}} = 14.9 \pm 1.895 {1.707\over \sqrt{8}} \doteq 14.90 \pm 1.14$$
To see whether the normality assumption is valid, by what we saw in Lab 3,
we could use either a normal quantile-quantile plot and look for approximate 
linearity, or we could use the Shapiro-Wilk normality test.  
\item{C)}  No, $\mu = 16$ is contained in the interval.  It is a ``believable'' value
for $\mu$ based on this data.
\item{D)}  Using the formula from page 435 (also see Lab 3), the ``standard'' way to 
get such an interval is
$$\left({7 S^2\over \chi^2_{.025}(7)}, {7 S^2\over \chi^2_{.975}(7)}\right) \doteq (1.27,12.08).$$
\bigskip
\noindent
IV.  
\item{A)}  $\overline{p_A} = {24\over 642} \doteq .0373$.
\item{B)}  Since $n_A = 642 > 30$, we use the large sample formula and the approximation for
the standard error making use of the value from part A.  We have $z_{.025} = 1.96$, so
$$p_A = .0374 \pm 1.96 \sqrt{{(.0374)(.9626)\over 642}} \doteq .0374 \pm .0147$$
\item{C)}  The interval is 
$$p_A - p_B = .0374 - .0440 \pm 1.96 \sqrt{{(.0374)(.9626)\over 642} + {(.0440)(.9560)\over 500}} \doteq
-.0066 \pm .0232.$$
Since this interval contains both positive and negative numbers, there is not enough evidence
to say that $p_A < p_B$ (i.e. not sufficient evidence to say 
the quality of the output from shop $A$ is higher than the 
quality from shop $B$).  The observed differences in the proportions of defectives could be
explained by random variations in the sampling process.
\item{D)}  In general the best way to answer a question like this is to apply 
the ``conservative'' method to estimate the standard error -- use $p_A = .5$ in the 
standard error formula, since that gives the largest
possible standard error.  Then the estimator is within .01 of the true value
with probability at least when 
$$\eqalign{
{.01 \over \sqrt{(.5)(.5)/n}} &\ge 1.96,\quad {\rm or}\cr
n &\ge \left({(1.96)(.5)\over .01}\right)^2 = 9604\cr}$$
Depending on what the actual value of $p_A$ is, a much smaller $n$ might work too.  For instance, if 
it was true that $p_A = .0374$ exactly, then applying the standard error formula with that value of 
$p_A$, we would require only
$$\eqalign{
{.01 \over \sqrt{(.0374)(.9626)/n}} &\ge 1.96,\quad {\rm or}\cr
n &\ge \left({(1.96)\sqrt{(.0374)(.9626)}\over .01}\right)^2 \doteq 1383.02\cr}$$
So only $n \ge 1384$ is needed.  
\bye
