der a probler stimating the difference of proporti рopulatior sample 1, out of n1 subjects, S1 of them are "successes" and the rest are "failures". In sample 2, out of n2 bjects, S2 of them are "successes" and the rest are "failures". is known that S1 · B(n1,P1) and S2 ~ B(n2, p2). e are interested in estimating pi – P2- Denote îi = 1 and p2 = 2. Show that Pi – P2 is an unbiased estimator for p1 – P2. And what is the mean squared error of it? n2 Explain how CLT would be applied to approximate the distribution of pi – P2 so that pi(1 — р1) , Р2(1 — рә) Pi – P2 ~ N ( Pi – P2, + approximately n1 n2

College Algebra
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ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Counting And Probability
Section9.3: Binomial Probability
Problem 2E: If a binomial experiment has probability p success, then the probability of failure is...
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der a probler
stimating the difference of proporti
рopulatior
sample 1, out of n1 subjects, S1 of them are "successes" and the rest are "failures". In sample 2, out of n2
bjects, S2 of them are "successes" and the rest are "failures".
is known that S1
· B(n1,P1) and S2 ~ B(n2, p2).
e are interested in estimating pi – P2-
Denote îi = 1 and p2 = 2. Show that Pi – P2 is an unbiased estimator for p1 – P2. And what is the
mean squared error of it?
n2
Explain how CLT would be applied to approximate the distribution of pi – P2 so that
pi(1 — р1) , Р2(1 — рә)
Pi – P2 ~ N ( Pi – P2,
+
approximately
n1
n2
Transcribed Image Text:der a probler stimating the difference of proporti рopulatior sample 1, out of n1 subjects, S1 of them are "successes" and the rest are "failures". In sample 2, out of n2 bjects, S2 of them are "successes" and the rest are "failures". is known that S1 · B(n1,P1) and S2 ~ B(n2, p2). e are interested in estimating pi – P2- Denote îi = 1 and p2 = 2. Show that Pi – P2 is an unbiased estimator for p1 – P2. And what is the mean squared error of it? n2 Explain how CLT would be applied to approximate the distribution of pi – P2 so that pi(1 — р1) , Р2(1 — рә) Pi – P2 ~ N ( Pi – P2, + approximately n1 n2
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