# Determining sample size given true proportion.

I'm attempting to solve a problem from a statistics course in regards to finding the sample size I need to take when given the Margin of Error, Confidence interval, and 'true proportion' (probability).

The problem follows:

A researcher is interested in determining the proportion of first year university students who have suffered hearing loss from loud music and other big-city sounds. How many students should she give hearing tests to, in order to determine this proportion with a margin of error (half width of confidence interval) of 0.1 using a 95% confidence interval if she believes from previous studies that the true proportion is 0.25.

I've been having some trouble with this part of the course, and I'm not quite sure which formula to apply here. I know I would need an equation for a sample size that relates the above values, If you can provide a suitable formula I'm certain I can go from there. Any help is appreciated, thank you!.

The (approximated) confidence interval for proportions is $$\large{\left[\hat p-\underbrace{ z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}}_{\textrm{margin error}}, \ \hat p+z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}\right]}$$
We can use the approximated confidence interval if we can apply the central limit theorem. The thumb rule is $$n>30$$. We can only see later if this condition is fulfilled.
The given data are $$\hat p=0.25, \alpha=1-0.95=0.05\Rightarrow 1-\frac{\alpha}{2}=0.975$$. $$z_{0.975}$$ is the $$z$$-value where the cdf of the standard normal distribution is $$0.975$$. $$\Phi^{-1} (0.975)=1.96$$
Next you have to solve the equation $$1.96\cdot \sqrt{\frac{0.25\cdot 0.75}{n}}=0.1$$ for $$n$$.
Finally check if $$n>30$$.