How could you find the probability that the estimator is within 0.03 of the mean?

94 Views Asked by At

p = fraction of large population that smokes

n = sample size

y = # in sample that smoke

The maximum likelihood estimate of p is p-hat = y/n

Consider the random variable Y and estimator F = Y/n

How could you work out P(-0.03 < F - p < 0.03) knowing n and p?

This is what I have so far, I'm just not sure how to get rid of the st.dev variable:

let x be the standard deviation

Y ~ G(p,x)

Y-bar ~ G(p, x/ sqrt n)

P (-0.03 < F - p < 0.03) = P(p - 0.03 < Y-bar < p + 0.03)

= P[ ((p - 0.03) -p)/(x/sqrtn) < (Y-bar - p) / (x/sqrt n) < ((p + 0.03)-p)/(x/sqrt n)]

= P (-0.03*sqrtn/x < Z < 0.03*sqrtn/x)

So how would I figure this out without knowing x (standard deviation)?