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\def\sol{\bigskip\item{} {\it Solution}\/:\ }
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\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Midterm Exam I -- Solutions}
\centerline{March 1, 2006}
\bigskip
\noindent
I.  Assume that the calorie content
of chocolate Power Bars is normally distributed.  
A random sample of 10 chocolate Power Bars yields
calorie content values of:
$$210,220,234,227,229,240,221,228,225,219$$
\item{A)} (10) Estimate the population mean calorie content of the
chocolate Power Bars.
\sol We estimate $\mu$ with the sample mean $\overline{Y} \doteq 225.3$
\bigskip
\item{B)} (10) The sample variance of the calorie content
values above is $S^2 = 70.68$ cal.   
Find a $99\%$ confidence interval for the population
mean calorie content, $\mu$. 
\sol Since $n = 10$, we use the small sample formula (using percentage
points of the $t$-distribution with $n - 1 = 9$ degrees of freedom):
$$\eqalign{\mu &= 225.3 \pm t_{.005} {\sqrt{70.68}\over \sqrt{8}}\cr
 &= 225.3 \pm (3.25) {8.407\over \sqrt{10}}\cr
 &= 225.3 \pm 8.64\cr}$$
\bigskip 
\item{C)} (10)  Find a 95\% confidence interval for the 
calorie content variance $\sigma^2$.
\bigskip
\sol The 95\% confidence interval is found using the percentage
points of the $\chi^2$ distribution with $n - 1 = 9$ degrees of freedom:
$$\eqalign{
\sigma^2 &\in \left[{9(70.68)\over \chi^2_{.025}}, {9(70.68)\over \chi^2_{.975}}\right]\cr
&= \left[{9(70.68)\over 19.0228}, {9(70.68)\over 2.70039}\right]\cr
&=\left[33.44,235.57\right]\cr}$$
(Note: The fact that this interval is so large comes from the fact
that the number of samples is quite small.  It is difficult to 
pin down the population variance from this little data.)
\bigskip
\noindent
II.  A researcher found that 107 out of 200 rutabaga 
seeds germinated when they were planted in soil maintained 
at $5 {}^\circ C$, while 123 out of 200 seeds germinated when they 
were planted in soil maintained at $15 {}^\circ C$.
\item{A)} (20)  Construct a $95\%$ confidence interval for 
the difference in the proportions of seeds that germinate
at the two temperatures. 
\sol  Since $n = 200$ for the two samples, we use the large-sample
formula for the difference of proportions (with the percentage
points of the standard normal distribution).  The point
estimators for the two proportions are $\hat{p}_{5^\circ} = 107/200 = .535$
and $\hat{p}_{15^\circ} = 123/200 = .615$.  Then
$$p_{5^\circ} - p_{15^\circ} = 
.535-.615 \pm (1.96)\sqrt{{(.535)(.465)\over 200} + {(.615)(.385)/200}}
\doteq -0.08 \pm .0966$$
\bigskip 
\item{B)} (10)  Can we rely on the results of this study to 
conclude that a greater proportion of rutabaga seeds will
germinate at the higher temperature in general?  Why or why not?
\sol No -- the data does not support that conclusion because the
confidence interval contains zero and positive numbers as well
as negative numbers.  The fact that $\hat{p}_{5^\circ} < \hat{p}_{15^\circ}$
might be a result of random variation.  
\bigskip
\noindent
III. (15) Let $Y_1,\ldots,Y_n$ denote a random sample from a population
whose density is given by 
$$f(y) = \cases{3\beta^3 y^{-4} & if $y \ge \beta$\cr
                   0 & otherwise\cr}$$
where $\beta$ is unknown.  Consider the estimator
$\widehat{\beta} = \min(Y_1,\ldots,Y_n)$.  Is $\widehat{\beta}$
biased or unbiased?
\sol We need to compute the expected value of $\widehat{\beta} = Y_{(1)}$, 
so we need the pdf for the sample minimum in this case.  The 
general formula is $f_{(1)}(y) = n(1 - F(y))^{n-1}f(y)$, where
$F(y)$ is the cdf for the distribution.
$$F(y) = \int 3\beta^3 y^{-4}\ dy = -\beta^3 y^{-3} + c$$
(for $y \ge \beta$, 0 otherwise).
The constant $c$ must be chosen to make this a valid
cdf.  Since $\lim_{y\to +\infty}F(y) = 1$, we see $c = 1$.
Then 
$$f_{(1)}(y) = n(\beta^3 y^{-3})^{n-1}\cdot 3\beta^3 y^{-4}$$
and 
$$\eqalign{
E(Y_{(1)}) &= \int_\beta^\infty y\cdot n(\beta^3 y^{-3})^{n-1}\cdot 3\beta^3 y^{-4}\ dy\cr
&=3n\beta^{3n} \int_\beta^\infty y^{-3n}\ dy\cr
&={3n\over 3n - 1}\beta\cr}$$
Since we do not get exactly $\beta$, this is a {\it biased} estimator.
\bigskip
\noindent
IV.  Let $X_1,\ldots,X_n$ and $Y_1, \ldots, Y_n$ be two 
independent random samples of equal sizes from a normal distribution with
{\it known} mean $\mu$, but {\it unknown}
variance $\sigma^2$.  Let $\Sigma_X^2$ and $\Sigma_Y^2$
be the statistics:
$$
\Sigma_X^2 = {1\over n}\sum_{i=1}^n (X_i - \mu)^2 \qquad
\Sigma_Y^2 = {1\over n}\sum_{i=1}^n (Y_i - \mu)^2$$
\item{A)} (10) What is the distribution of $n\Sigma^2_X/\sigma^2$?
Explain.
\sol We see
$$n\Sigma^2_X/\sigma^2 = \sum_{i=1}^n \left({X_i - \mu\over\sigma}\right)^2$$
is the sum of squares of $n$ independent standard normals.  Hence
this has a $\chi^2$ distribution with $n$ degrees of freedom. 
(It's {\it not} $n-1$ degrees of freedom here because we know the exact 
population $\mu$ and we use that rather than the sample mean $\overline{X}$.)
\item{B)} (5) Use your answer from part A to show that 
$\Sigma_X^2$ is an unbiased estimator for $\sigma^2$.
\sol Recall that $E(Z^2) = V(Z) + (E(Z))^2 = 1$.  So 
a $\chi^2$ random variable with $n$ degrees of freedom has 
expected value $n$.  Hence
$${n\over\sigma^2}E(\Sigma^2_X) = E\left(n\Sigma^2_X/\sigma^2\right) = n \Rightarrow
E(\Sigma_X^2) = \sigma^2,$$
so this is an unbiased estimator for $\sigma^2$.  (Once again,
this is different from the case we discussed in class because
we have $\mu$ rather than $\overline{X}$.)
\item{C)}  (10) What is the distribution of $\Sigma_X^2/\Sigma_Y^2$? Explain.
\sol Rewriting the quotient as 
$$\Sigma_X^2/\Sigma_Y^2 = {{n\Sigma_X^2\over \sigma^2}/n \over {n\Sigma_Y^2\over \sigma^2}/n}$$
we see that $\Sigma_X^2/\Sigma_Y^2$ has an $F(n,n)$ distribution.
\bigskip
\noindent
{\it Extra Credit} (10)  The {\it median} of a sample is 
the ``middle'' value, that is the value with the property that
half the sample values are greater than or equal to the median
and half are less than or equal to the median.  Give formulas
for computing the sample median in terms of the {\it order statistics}
of the sample.  (Note: the cases $n$ even and $n$ odd are different.)
\sol  When $n$ is odd, the middle value is the order statistic
$Y_{((n+1)/2)}$.  When $n$ is even, we take the average of the 
two order statistics on either side of the middle point:
$${1\over 2}(Y_{(n/2)} + Y_{(n/2 + 1)})$$
\bye
