Final answer:
The sample proportion (\(\hat{p}\)) follows a normal distribution with mean \(\mu = p\) and standard deviation \(\sigma = \sqrt{\frac {p \cdot q}{n}}\), given \(np > 5\) and \(nq > 5\). The standard error for the proportion is \(SE = \sqrt{\frac {\hat{p} \cdot \hat{q}}{n}}\).
Step-by-step explanation:
The formulas for the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the exact distribution of the sample proportion (\(\hat{p}\)) are based on the binomial distribution, under conditions where it can be approximated using a normal distribution. This is relevant when you're dealing with a proportion problem involving categorical data with two outcomes (success or failure). For a given population with a proportion of success (\(p\)), and proportion of failure (\(q = 1 - p\)), for a large enough sample size (\(n\)), the sample proportion (\(\hat{p}\)) can be modeled with a normal distribution with a mean of \(\mu = p\) and a standard deviation of \(\sigma = \sqrt{\frac {p \cdot q}{n}}\).
When conducting a hypothesis test of a single population proportion, if the sample size is sufficient so that both \(np\) and \(nq\) are greater than 5, the distribution of the sample proportions is approximately normal. In this scenario, the standard error (SE) is calculated as \(SE = \sqrt{\frac {\hat{p} \cdot \hat{q}}{n}}\), where \(\hat{p}\) and \(\hat{q} = 1 - \hat{p}\) are the sample estimates of population proportions. This allows us to construct confidence intervals or conduct hypothesis tests regarding the population proportion.