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\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Problem Set 5}
\centerline{March 3, 2004}
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A) (A follow-up problem to C from Lab Project 2.)  Recall that 
in Lab Project 2, to determine the endpoints of the 
confidence interval for the ratio of variances $\sigma_1^2/\sigma_2^2$,
you needed to work with a formula like this 
$$P\left({S_1^2/S_2^2\over f_{\alpha/2}} \le \sigma_1^2/\sigma_2^2
\le {S_1^2/S_2^2\over f_{1-\alpha/2}}\right) = 1 - \alpha.$$
In the problem from the Lab Project, you were looking for a 
$95\%$ confidence interval, so you needed $f_{.025}, f_{.975}$.  The $F$-table
in the back of the book gives the values $f_{\alpha}$ for
$\alpha = .1, .05, .025, .01, .005$, but not the complementary 
values $\alpha = .9, .95, .975, .99, .995$.  As you might have
wondered, {\it why aren't those other entries there??}
Fortunately, with the {\tt FCDF} function from our Maple 
package, you could compute $f_{.975}$, etc., by the same method
we used for $\chi^2$ and $t$ distributions in the lab, 
so this wasn't a problem.
In this question you will show that the other values could be
derived from the table entries given with a little cleverness, too!
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\item{1)} Let the numbers $f_\alpha(n,m)$ be defined by 
$$P(Y \ge f_\alpha(n,m)) = \alpha$$
if $Y$ has an $F$-distribution with $n$ numerator degrees of freedom
and $m$ denominator degrees of freedom.
Show that for all $0 < \alpha < 1$, these numbers satisfy:
$$f_{1-\alpha}(n,m) = {1\over f_{\alpha}(m,n)}$$
(note the reversal in the numbers of degrees of freedom on the 
right.)  {\bf Hint:}  What is the standard way to get a random variable
that has an $F$-distribution?
\item{2)}  Using the table in the text and part 1, determine the 
number $f_{.99}(7,10)$.
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{\it Additional problems from the text:}  
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\item{$\bullet$} Chapter 8/73, 76, 81, 83, 84;
\item{$\bullet$} Chapter 9/13, 19, 61, 67, 75, 80.
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{\it Due:}  In class on Friday, March 19.
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