There's a question I'm not really sure if I did it right or even understand what its trying to say.
There is a coin which produces heads with an unknown probability p. How many times should we throw this coin if the proportion of heads is to lie within 0.05 of p with probability at least 0.9? Hint: Answer is not complete if it relies on p and do not worry about continuity correction.
This question comes from a chapter about binomial normal approximation.
So far I know I have a B~(n,p) and I need to find the number of n so that P ( c/n <= 0.05*p) >=0.9 with c as the number of heads? I'm not sure if this is right.
In turn P( c <= 0.05pn ) >=0.9
So I'll normalize the binomal such that
P( Z <= [(0.05pn - n*p)/(np(1-p)] +0.5 ) >=0.9
with 0.5 as the continuity correction.
Am I on the right track? I thought about using confidence intervals but we haven't discussed about that in the chapter or lecture.
I believe you are expected to use the normal distribution approximation. From the normal distribution table, $1.645$ standard deviations around the mean include (just over) $0.9$ of the area. For $N$ flips the standard deviation in the number of heads is $\sqrt{Np(1-p)}$, which is largest at $p=\frac 12$, so $\sigma \leq\frac 12 \sqrt N$. We want $1.645\sigma \le\frac 121.645\sqrt N\lt 0.05N$, so $N \gt 16.45^2\approx 270.6$ Take $N=271$