Kentucky has a population of approximately 12, 501 U.S. Indians. In a 2012 survey of Asian Americans, Pew Research Center found that \( 70 \% \) of U.S. Indians celebrate Diwali from a sample size of 580. (a) What is the estimated mean of the sampling distribution of sample proportions of this same sample size? \( p=12501 \) \( \hat{p}=0.70 \) \( \mu_{\hat{p}}=0.70 \) \( p=0.70 \) (b) What is the standard error for the sampling distribution of sample proportions of this same sample size? \( \sqrt{\frac{0.70(1-0.70)}{580}} \) \( \sqrt{\frac{0.70(1-0.70)}{12501}} \) \( \sqrt{\frac{12501(1-12501)}{580}} \) \( \sqrt{\frac{580(1-580)}{12501}} \)

When we look at the sampling distribution of sample proportions, the estimated mean of the sampling distribution is simply the population proportion, \( \hat{p} \). So, in this case, \( \mu_{\hat{p}} \) would indeed be \( 0.70 \) based on the data from the survey. For the standard error, you’re on the right track with the formula for standard error of a sample proportion! The correct formula should be calculated using the sample size, which in this case is 580. Hence, it would be \( \sqrt{\frac{0.70(1-0.70)}{580}} \), giving you the measure of how much the proportion would vary from sample to sample. Keep your calculators handy; math is more fun with numbers!