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\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Midterm Exam 1 -- March 11, 2010}
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{\it Directions:}  
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Do all work in the blue exam booklet.  There
are 100 possible points and 10 Extra Credit points (Please turn over for 
part C of question IV and the Extra Credit problem.)
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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.
\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.'')      
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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?
\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?
\item{C)}  (5)  Do your results support the conclusion that
Method 2 is a superior method for teaching reading?  Explain.
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III.  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 $(1 - \alpha)\times 100\%$)?
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IV.  Let $Y_1,\ldots,Y_n$ denote a random sample of size $n$
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 $\theta$.
\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{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.  
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{\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.
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\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.
\bye
