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Question #2 (10 Marks) Let $S_1^2$ denote the sample variance for a random sample of ten (LC50) values for copper and let $S_2^2$ denote the sample variance for a random sample of eight (LC50) values for lead, both using the same species of fish. Now, Assume $S_1^2$ to be independent of $S_2^2$. The population variance for measurements on copper is assumed to be twice the corresponding population variance for measurements on lead. a) Find "b" such that = $P(frac{S_1^2}{S_2^2} le b) = 0.95$ b) Find "a" such that = $P(a le frac{S_1^2}{S_2^2}) = 0.95$ c) If "a" and "b" are as in parts (a) and (b) above, find $P(a le frac{S_1^2}{S_2^2} le b)$

          Question #2 (10 Marks)
Let $S_1^2$ denote the sample variance
for a random sample of ten (LC50)
values for copper and
let $S_2^2$ denote the sample variance
for a random sample of eight (LC50)
values for lead, both using the
same species of fish.
Now, Assume $S_1^2$ to be independent of $S_2^2$.
The population variance for measurements
on copper is assumed to be twice
the corresponding population variance
for measurements on lead.
a) Find "b" such that =
$P(frac{S_1^2}{S_2^2} le b) = 0.95$
b) Find "a" such that =
$P(a le frac{S_1^2}{S_2^2}) = 0.95$
c) If "a" and "b" are as in
parts (a) and (b) above, find
$P(a le frac{S_1^2}{S_2^2} le b)$

$Question #2 (10 Marks) Let S1^2 denote the sample variance for a random sample of ten (LC50) values for copper and let S2^2 denote the sample variance for a random sample of eight (LC50) values for lead, both using the same species of fish. Now, Assume S1^2 to be independent of S2^2. The population variance for measurements on copper is assumed to be twice the corresponding population variance for measurements on lead. a) Find "b" such that = P(fracS1^2S2^2 le b) = 0.95 b) Find "a" such that = P(a le fracS1^2S2^2) = 0.95 c) If "a" and "b" are as in parts (a) and (b) above, find P(a le fracS1^2S2^2 le b)$

Added by Justin S.

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