Question

checkpoint question (1/3): we are now going to perform the calculations for a 99% confidence interval for ( p ) = the true proportion of u.s. adults who \often or always\ got enough sleep during the past 7 nights. given ( n = 1029 ) and ( hat{p} = 0.48 ), what is the mean and standard error of the sampling distribution of ( hat{p} )? ( mu_{hat{p}} = 0.48 ) ( s_{hat{p}} = sqrt{\frac{(0.48)(0.52)}{1029}} ) ( mu_{hat{p}} = 0.48 ) ( s_{hat{p}} = sqrt{1029(0.48)(0.52)} ) ( mu_{hat{p}} = 1029(0.48) ) ( s_{hat{p}} = sqrt{\frac{(0.48)(0.52)}{1029}} )

Explanation:

Step1: Recall the formula for the mean of the sampling distribution of \(\hat{p}\)

The mean of the sampling distribution of \(\hat{p}\), denoted as \(\mu_{\hat{p}}\), is equal to the population proportion \(p\). When we are given a sample proportion \(\hat{p}\), and assuming the sampling is done correctly (random sampling, etc.), \(\mu_{\hat{p}}=\hat{p}\). Given \(\hat{p} = 0.48\), so \(\mu_{\hat{p}}=0.48\)

Step2: Recall the formula for the standard error of the sampling distribution of \(\hat{p}\)

The formula for the standard error of the sampling distribution of \(\hat{p}\), denoted as \(s_{\hat{p}}\), is \(s_{\hat{p}}=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\). Substitute \(\hat{p}=0.48\) and \(n = 1029\) into the formula. Since \(1-\hat{p}=1 - 0.48=0.52\), we get \(s_{\hat{p}}=\sqrt{\frac{(0.48)(0.52)}{1029}}\)

Answer:

The first option (where \(\mu_{\hat{p}} = 0.48\) and \(s_{\hat{p}}=\sqrt{\frac{(0.48)(0.52)}{1029}}\)) is correct.