probability of a biased coin

Suppose we have a biased coin, with probabilities p of coming up heads and q = 1-p
of coming up tails on any given toss, but we are not given what p or q are. We know
from class 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 E^, can be divided by the number of tosses, n, to give an estimate, ^p = E^/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:
trying to infer from actual data what is really true and then estimate the likelihood that
what is inferred is close to reality.)


Let us suppose, for the sake of argument, that we know that 0:5 < p < 0:6 (so
0:4 < q = 1 -p < 0:5) for our biased coin and that we have the time and patience to toss
it any nite number of times n that we want. 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.


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<  p^  <1:05p)  0:95? Please justify
your answer as fully as you can.


Note: That is, what does n have to be to ensure that the probability that ^p is within 5%
of p is at least 0:95?

danicathea said:

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:
trying to infer from actual data what is really true and then estimate the likelihood that
what is inferred is close to reality.)

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< p^ <1:05p) 0:95? Please justify
your answer as fully as you can.


Note: That is, what does n have to be to ensure that the probability that ^p is within 5%
of p is at least 0:95?

Click to expand...

After N trials, your sample \(\displaystyle \hat p\) is your best estimator of the population \(\displaystyle p\). Do you know the properties of the Binomial Distribution? The mean and standard deviations for the distribution of number of "heads" are

\(\displaystyle \mu = N\ p\)

\(\displaystyle \sigma = \sqrt{N\ p\ (1-p)}\)

Dividing by \(\displaystyle N\) to solve for the distribution of \(\displaystyle p\) in terms of the observed \(\displaystyle \hat\mu\),

\(\displaystyle \bar p = \hat\mu\ /N\)

\(\displaystyle \sigma(p) = \sigma/N = \sqrt{p\ (1-p)/N} \approx \dfrac{1}{2\ \sqrt{N}}\)

When \(\displaystyle N\) and \(\displaystyle Np\) are large enough, the binomial distribution is equal to a normal distribution with same mean and standard deviation. Thus you can use confidence limits of the normal distribution to answer all your questions. Note especially that \(\displaystyle \sigma(p)\) decreases with \(\displaystyle \sqrt{N}\).