1. How many times do we need to toss the coin, keeping track of the number of tosses and the number of success, to ensure that P(0.95p ≤ pˆ ≤ 1.05p) ≥ 0.95? Please justify your answer as fully as you can. Note: That is, find an n that ensures that the probability that ˆp is within 5% of p is at least 0.95.

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Chapter1: Combinatorial Analysis
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1. How many times do we need to toss the coin, keeping track of the number of tosses and the number of success, to ensure that P(0.95p ≤ pˆ ≤ 1.05p) ≥ 0.95? Please justify your answer as fully as you can.

Note: That is, find an n that ensures that the probability that ˆp is within 5% of p is at least 0.95.

p of
Suppose we have a coin with a probability p of coming up heads and q 1
coming up tails on any given toss (so the coin is biased unless p = 0.5), but we are not
given what p or q are. We know, from class or the textbook, that the expected value of
the number of heads in n tosses is E(X) = np. If we repeatedly flip the coin and record
the results, the number of heads that actually turn up, call it Ê, can be divided by the
number of tosses, n, to give an estimate, ô = Ê/n, of p. The problem is that we're likely
to have to be pretty lucky for p = p, so the real question is how likely it is that p is close
to p. (This sort of thing is what statistics is all about: try infer from actual data what is
really true, and also provide an estimate of the likelihood that what is inferred is close to
reality.)
Let us suppose,
for the sake of argument, that we know [somehow!] that 0.4 ≤p ≤ 0.6
(so 0.4 < q = 1- p ≤ 0.6) for our possibly biased coin and that we have the time and
patience to toss it any finite number of times n that we need to. Further, let us suppose that
we desire to get an estimate of p that is within 5% of the real value, i.e. 0.95p ≤ p ≤ 1.05p.
Transcribed Image Text:p of Suppose we have a coin with a probability p of coming up heads and q 1 coming up tails on any given toss (so the coin is biased unless p = 0.5), but we are not given what p or q are. We know, from class or the textbook, that the expected value of the number of heads in n tosses is E(X) = np. If we repeatedly flip the coin and record the results, the number of heads that actually turn up, call it Ê, can be divided by the number of tosses, n, to give an estimate, ô = Ê/n, of p. The problem is that we're likely to have to be pretty lucky for p = p, so the real question is how likely it is that p is close to p. (This sort of thing is what statistics is all about: try infer from actual data what is really true, and also provide an estimate of the likelihood that what is inferred is close to reality.) Let us suppose, for the sake of argument, that we know [somehow!] that 0.4 ≤p ≤ 0.6 (so 0.4 < q = 1- p ≤ 0.6) for our possibly biased coin and that we have the time and patience to toss it any finite number of times n that we need to. Further, let us suppose that we desire to get an estimate of p that is within 5% of the real value, i.e. 0.95p ≤ p ≤ 1.05p.
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