Solution Found!
Let S2 1 denote the sample variance for a random sample of ten ln(LC50) values for
Chapter 7, Problem 7.36(choose chapter or problem)
Let S2 1 denote the sample variance for a random sample of ten ln(LC50) values for copper and let S2 2 denote the sample variance for a random sample of eight ln(LC50) values for lead, both samples using the same species of fish. The population variance for measurements on copper is assumed to be twice the corresponding population variance for measurements on lead. Assume S2 1 to be independent of S2 2 . a Find a number b such that P S2 1 S2 2 b = .95. b Find a number a such that P a S2 1 S2 2 = .95. [Hint: Use the result of Exercise 7.29 and notice that P(U1/U2 k) = P(U2/U1 1/k).] c If a and b are as in parts (a) and (b), find P a S2 1 S2 2 b .
Questions & Answers
QUESTION:
Let S2 1 denote the sample variance for a random sample of ten ln(LC50) values for copper and let S2 2 denote the sample variance for a random sample of eight ln(LC50) values for lead, both samples using the same species of fish. The population variance for measurements on copper is assumed to be twice the corresponding population variance for measurements on lead. Assume S2 1 to be independent of S2 2 . a Find a number b such that P S2 1 S2 2 b = .95. b Find a number a such that P a S2 1 S2 2 = .95. [Hint: Use the result of Exercise 7.29 and notice that P(U1/U2 k) = P(U2/U1 1/k).] c If a and b are as in parts (a) and (b), find P a S2 1 S2 2 b .
ANSWER:Step 1 of 4
Given data
For n=10 samples of ln(LC50) values of copper
Sample variance is
For n=8 samples of ln(LC50) values of lead
Samples variance is
We are also given that