# problem-7.11

problem-7.11 We have z* \SE(\phat) £ 0.01. We know z* as a is given, so we have

| Ö | \frac\phat(1-\phat)n | £ \frac0.01z*.

Solving for n we get

(\fracz*0.01)2 \phat(1-\phat) £ n.

No matter what \phat is, the expression \phat(1-\phat) is largest when \phat = 1/2, so we have n is large enough if

(\fracz*0.01)2 \frac12(1 - \frac12) £ n.

For our example, we have

    ## for 90% confidence
    > alpha = 0.1; zstar = qnorm(1 - alpha/2)
    > (zstar/0.01)^2/4
    [1] 6764
    ## for 80% confidence
    > alpha = 0.2; zstar = qnorm(1 - alpha/2)
    > (zstar/0.01)^2/4
    [1] 4106