\magnification=\magstep1
\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Solutions for Midterm Exam 1 -- March 11, 2010}
\bigskip
\noindent
I. In a random poll of female students at BigState University, 
390 out of 500 sorority women were in favor of lengthening the 
spring break, while 64 out of 100 non-sorority women were 
in favor.  
\item{A)}  (15)  Give a $99\%$ confidence interval for
the difference of the proportions favoring lengthening spring break.
\bigskip
\item{} {\it Solution -- } The point estimators for the two
population proportions are ${390\over 500} = .78$ and ${64\over 100} = .64$.
By the large sample formulas, $z_{.005} \doteq 2.575$, and the confidence interval is
$$(.78 - .64) \pm 2.575 \sqrt{{(.78)(.22)\over 500} + {(.64)(.36)\over 100}} \doteq
.14 \pm .1325.$$
\item{B)}  (5)  Now suppose a new poll of equal numbers of women
from both groups is being planned.  How many women should be 
polled so that error of estimation is no larger than $.08$ according
to a two standard error bound?  (Be ``conservative.'')      
\bigskip
\item{} {\it Solution -- }  We want to choose the number $n$ of 
women to be polled so that 
$$2 \sqrt{{(.5)(.5)\over n} + {(.5)(.5)\over n}} \le .08.$$
This says $n \ge 312.5$, so $n \ge 313$.    
\bigskip
\noindent
II.  Two methods for teaching reading were applied to two randomly
selected groups of school children, and compared by means of 
a common test.  The sample means ($\overline{y}$) and variances
($s^2$) of the test scores for the two groups are as follows:
$$\matrix{{\rm Statistic} & {\rm Method\ 1} & {\rm Method\ 2}\cr
          {\rm Number\ of\ Children} & 11 & 14 \cr
          \overline{y} & 64 & 69 \cr
          s^2 & 20 & 15 \cr}$$
\item{A)}  (10)  Find a $95\%$ confidence interval for
$\sigma_1^2$ (the variance of the test scores for students
taught with Method 1.  What assumptions are you making?
\bigskip
\item{}{\it Solution -- }  We need to assume that the test scores
are from a normally distributed population.  Then using the $\chi^2$ 
table, the confidence interval is
$${(n_1 - 1) S_1^2\over \chi^2_{.025}} = {{10 \cdot 20} \over 20.4831} \le \sigma_1^2 \le {{10 \cdot 20}\over 3.24697} = {(n_1 - 1) S_1^2\over \chi^2_{.975}}$$
or 
$$9.764 \le \sigma_1^2 \le 61.596.$$
\item{B)}  (20)  Find a $95\%$ confidence interval for 
$\mu_1 - \mu_2$ (the difference in the population mean test 
scores with the two methods).  What assumptions are you making?
\bigskip
\item{} {\it Solution -- }  We need to assume the samples are 
independent from normal populations with equal variance.  
(Note that the sample variance from the students taught with
Method 2 is in the interval from part A, which is a limited
indication that the equality of variances is reasonable.)
From the small sample formulas, we have that the pooled
estimator for the variance is
$$S_p^2 = {10\cdot 20 + 13 \cdot 15\over 23} \doteq 17.1739.$$
Then since $t_{.025}(23) = 2.069$, 
$$\mu_1 - \mu_2 = (64-69) \pm 2.069 \cdot \sqrt{17.1739} \cdot \sqrt{1/11 + 1/14} \doteq 
-5 \pm 3.4547.$$
\item{C)}  (5)  Do your results support the conclusion that
Method 2 is a superior method for teaching reading?   Explain.  
\bigskip
\item{} {\it Solution -- } Yes, since the interval in part B consists
of only negative numbers, there is evidence (at the $95\%$ confidence
level) to say that Method 2 yields superior results.  
\bigskip
\noindent
III.  (15)  A random sample of size $100$ was taken from a 
normal population with $\sigma^2 = 8$.  A confidence interval
for the population mean was given as $6 \pm .047$.  What
was the confidence coefficient associated with this interval
(that is the $\alpha$ value in the $(1 - \alpha)\times 100\%$)?
\bigskip
\item{} {\it Solution -- }  Since the number of samples
was $100 > 30$, the large sample formulas would be used:
$$.047 = z_{\alpha/2} {\sqrt{8}\over \sqrt{100}},$$
so $z_{\alpha/2} = .47/\sqrt{8} \doteq .166$.  Using the closest
$z$ value 
from the standard normal table, this says $\alpha/2 \doteq .4325$.
Hence $\alpha \doteq .865$.  This was a $13.5\%$ confidence interval.
(Not one of the standard values!  And actually, this is a somewhat
nonsensical value because this problem contained
an unfortunate typo -- I had meant to make the reported interval
$6\pm 0.47$ but I switched the 0 and the decimal point 
and I did not catch this(!))
\bigskip
\noindent
IV.  from a population described by the pdf:
$$f(y) = \cases{ {2y\over \theta^2} & if $0 \le y \le \theta$\cr
                     0 & otherwise. \cr}$$
Let $Y_{(n)}$ be the sample maximum and let $U = {1\over \theta} Y_{(n)}$.
\item{A)}  (10)  Find a constant $c$ so that 
$\widehat{\theta} = cY_{(n)}$ is an unbiased estimator for $\theta$.
\bigskip
\item{} {\it Solution -- } We have 
$$F(y) = \cases{ 0 & if $y < 0$\cr
{y^2\over \theta^2} & if $0 \le y\le \theta$\cr
1 & if $y > \theta$.\cr}$$
Therefore from the formulas for order statistics, the PDF for
the sample maximum is 
$$f_{(n)}(y) = n \left({y^2\over \theta^2}\right)^{n-1} {2y\over \theta^2}.$$
This implies
$$\eqalign{E(Y_{(n)}) &= \int_0^\theta y \cdot n \left({y^2\over \theta^2}\right)^{n-1} {2y\over \theta^2}\ dy\cr
&= \int_0^\theta 2n {y^{2n}\over \theta^{2n}}\ dy\cr
&= \left.{2n\over (2n+1)}{y^{2n+1}\over\theta^{2n}} \right|_0^\theta\cr
&= {2n\theta\over 2n+1}.\cr}$$
This shows that $Y_{(n)}$ is a biased estimator for
$\theta$, but $\widehat{\theta} = {2n+1\over 2n} Y_{(n)}$ is unbiased.
\bigskip
\item{B)}  (10)  Show that $U$ has CDF given by 
$$F_U(u) = \cases{0 & if $u < 0$\cr
                  u^{2n} & if $0 \le u \le 1$\cr
                  1 & if $u > 1$\cr}$$
\item{} {\it Solution -- } By definition the CDF of $U$ is
$$\eqalign{
F_U(u) &= P(U \le u)\cr
       &= P(Y_{(n)} \le \theta u)\cr
       &= \cases{0 & if $u < 0$\cr
     \int_0^{\theta u}  n \left({y^2\over \theta^2}\right)^{n-1} {2y\over \theta^2}\ dy & if $0 \le u \le 1$\cr
         1 & if $u > 1$.}\cr
&=\cases{0 & if $u < 0$\cr
                  u^{2n} & if $0 \le u \le 1$\cr
                  1 & if $u > 1$\cr}
}$$
as we wanted to show.
\bigskip       
\item{C)}  (10)  Note that the distribution of $U$ does 
not depend on $\theta$.  Use this information to derive
a $95\%$ {\it upper} confidence bound on $\theta$ from the 
samples.
\bigskip
\item{} {\it Solution -- }  $P(U \ge u) = 1 - P(U \le u) = .95$
when $P(U \le u) = .05$.  By part B, this says $u = (.05)^{1/(2n)}$.   Hence
$$P\left({1\over \theta}Y_{(n)} \ge (.05)^{1/(2n)}\right) = .95,$$
which is equivalent to 
$$P\left(\theta \le {Y_{(n)} \over (.05)^{1/(2n)}}\right) = .95.$$
So the $95\%$ upper confidence bound is 
$$\widehat{\theta_U} = {Y_{(n)} \over (.05)^{1/(2n)}}.$$
 \bigskip
\noindent
{\it Extra Credit.}  
\item{A)}  (5)  (Multiple Choice, and explain.)
A survey was conducted to determine what features
users want included in cell phone service.  The results 
said a $95\%$ confidence interval showed that
$73\% \pm 4\%$ of users wanted email services.
What is meant by the $\pm 4\%$?
\itemitem{1)}  As many as $4\%$ of those polled may change 
their minds after the time the survey was conducted.
\itemitem{2)}  The probability is $.95$ that the true 
population percentage favoring email service is
in the interval $(69\%,77\%)$.
\itemitem{3)}  It would be unlikely to get the observed
sample proportion if the true population percentage
was outside the interval $(69,77)$.  
\itemitem{4)}  Only $4\%$ of cellphone users were polled.
\bigskip
\item{} {\it Solution -- } Only 3 is correct.  1 and 4 are
clearly nonsense.  2 is the incorrect interpretation of
confidence intervals that we discussed in Lab 1 (recall
that there is no chance involved in the location of the
population percentage relative to the interval -- the 
chance is in the sampling process that produces the confidence 
interval).
\bigskip  
\item{B)}   (5) Let $Z_1,\ldots,Z_5$ be independent samples
from a $N(0,1)$ population, and define:
$$W = {2 Z_1\over \sqrt{Z_2^2 + \cdots + Z_5^2}}.$$
What is the distribution of $W^2$?  Explain.
\bigskip 
\item{} {\it Solution -- }  By definition, $W$ has a
$t$-distribution with $4$ degrees of freedom.  Therefore
$$W^2 = {(Z_1^2)/1 \over (Z_2^2 + \cdots + Z_5^2)/4}$$
is a ratio of $\chi^2$'s divided by their numbers
of degrees of freedom.  Thus $W$ has an $F$-distribution
with $1$ numerator degree of freedom and 4 denominator
degrees of freedom.  
\bye
