Question

A random sample of size 300 is to be selected from a population. Determine the mean and standard deviation of the sampling distribution of \(\hat{p}\) for each of the following population proportions. a. \(p=0.20\) b. \(p=0.45\) c. \(p=0.70\) d. \(p=0.90\)

Short answer

The mean and standard deviation for each case are: a. \(\mu_{\hat{p}} = 0.20, \sigma_{\hat{p}} = \sqrt{\frac{0.20 * (1-0.20)}{300}}\)b. \(\mu_{\hat{p}} = 0.45, \sigma_{\hat{p}} = \sqrt{\frac{0.45 * (1-0.45)}{300}}\)c. \(\mu_{\hat{p}} = 0.70, \sigma_{\hat{p}} = \sqrt{\frac{0.70 * (1-0.70)}{300}}\)d. \(\mu_{\hat{p}} = 0.90, \sigma_{\hat{p}} = \sqrt{\frac{0.90 * (1-0.90)}{300}}\)

Step-by-step solution

Calculate mean for each population proportion

The mean \(\mu_{\hat{p}}\) of the sampling distribution of the sample proportion is equal to the population proportion. So, for each case: a. \(\mu_{\hat{p}} = p = 0.20\)b. \(\mu_{\hat{p}} = p = 0.45\)c. \(\mu_{\hat{p}} = p = 0.70\)d. \(\mu_{\hat{p}} = p = 0.90\)

Calculate standard deviation for each population proportion

The standard deviation \(\sigma_{\hat{p}}\) of the sampling distribution of the sample proportion is calculated using the formula \(\sqrt{\frac{p(1-p)}{n}}\), where \(n\) is the sample size and \(p\) is the population proportion. For each case:a. \(\sigma_{\hat{p}} = \sqrt{\frac{0.20 * (1-0.20)}{300}}\)b. \(\sigma_{\hat{p}} = \sqrt{\frac{0.45 * (1-0.45)}{300}}\)c. \(\sigma_{\hat{p}} = \sqrt{\frac{0.70 * (1-0.70)}{300}}\)d. \(\sigma_{\hat{p}} = \sqrt{\frac{0.90 * (1-0.90)}{300}}\)