Smallest sample size $n$ to have $90\%$ confidence

Question

Task:

a) is quite easy but I'm looking for tips or a solution for b)

$0.9=P(|X_n-p|\le0.01)=P(X_n\le0.01+p)+P(X_n\ge p-0.01)=P(X_n\le 0.01+p)+1-P(X_n\le p-0.01)$.

I guess I need to somehow use CLT and then approximate the proability using a $Z \sim N(0,1)$ random variable but I don't know how to continue from up there.

Answer 1

So there is a few things. First note the sampling distribution for X_n is going to be, for large n,

N(p, p*(1-p)/n)

So worst case scenario for our variance is if p = .5.

To be within .01 of p with 90% confidence means our margin of error, at 90% confidence, is .01. The formula for margin of error from a normal distribution is 1.645*(st.dev of sampling distribution). Here though we kind of know of the sampling distribution by assuming worst case for p. So....all together

.01 = 1.645*sqrt(p * (1-p)/n) with p = .5. Now solve for n

(should all work but I might be mis understanding something)