Question

Determine the sample size required to estimate a population proportion to within 0.038 with 90.4% confidence, assuming that you have no knowledge of the approximate value of the sample proportion

Answer 1

It looks like you'll have to find a sample size in terms of the unknown sample proportion. Call the sample proportion \(\hat{p}\). It also looks like you can assume a binomial distribution, since we're using this estimator.

So, we're interested in an interval such that \(P(|p-\hat{p}|<0.038)=0.904\). This interval will have the form
\[\hat{p}\pm \color{red}{Z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}}\]
The red part gives the error of estimation, which we want to be 0.038. If \(1-\alpha=0.904\) (the confidence level), then \(\dfrac{\alpha}{2}=0.048\).

So, you must now solve for \(n\):
\[Z_{0.048}\sqrt{\frac{p(1-p)}{n}}=0.038\]
Refer to a table to figure out \(Z_{0.048}\). I'd round up to \(Z_{0.05}\), which would be 1.645, giving
\[Z_{\alpha/2}\sigma_{\hat{p}}\approx1.645\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.038~~\Rightarrow~~n\approx1873.979\hat{p}(1-\hat{p})\]