In our sample of 1000 people, 861 have health insurance and 139 do not.
(a) Use bootstrapping to generate a 90% confidence interval for the proportion of US residents who do not have health insurance.
(b) Use the appropriate formula and distribution to calculate a 90% confidence interval for the proportion of US residents who do not have health insurance. How does this answer compare to your answer from part (a)?
(c) Interpret your answer to either (a) or (b) in context.
(d) What is the sample statistic and corresponding margin of error in part (b)?
(e) We can also use the website http://factfinder2 .census.gov to find the sample statistic and corresponding margin of error based on the entire American Community Survey sample. There we find p̂ = 0.155 with a margin of error of 0.001 (for 90% confidence). Why is this margin of error so different from the one you computed in (d)?
(f) Use the information given in (e) to generate and interpret a 90% confidence interval for the proportion of US residents who do not have health insurance, based on the entire American Community Survey sample.

The American Community Survey, administered by the US Census Bureau, is given every year to a random sample of about 3.5 million households (about 3% of all US households). It has been crucial for government and policy decisions, and helps to determine how over 400 billion dollars of government funds are distributed each year. Unfortunately, the House voted in May 2012 to eliminate this valuable source of information. Data on a random sample of 1% of all US residents are made public (after ensuring anonymity), and we have selected a random sub-sample of n = 1000 from the 2010 data, stored in ACS.