Part 1 In 2013, the Pew Research Foundation reported that “45% of U.S. adults re
ID: 3300657 • Letter: P
Question
Part 1 In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”. However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also interpret the confidence interval in the context of the study.
Part 2
(a) We can say with certainty that the confidence interval from Exerise 4.7 contains the true percentage of U.S. adults who suer from a chronic illness.
(b) If we repeated this study 1,000 times and constructed a 95% confidence interval for each study, then approximately 950 of those confidence intervals would contain the true fraction of U.S. adults who suer from chronic illnesses.
(c) The poll provides statistically significant evidence (at the = 0.05 level) that the percentage of U.S. adults who suer from chronic illnesses is below 50%.
(d) Since the standard error is 1.2%, only 1.2% of people in the study communicated uncertainty about their answer.
Explanation / Answer
part 1) here std error -1.2%
and for 95% confidence interval; critical value of z =1.96
therefore 95% confidence interval =sample proportion -/+ z*std error =45 -/+ 1.96*1.2 =42.648% ; 47.352%
part 2)
option B is correct.
If we repeated this study 1,000 times and constructed a 95% confidence interval for each study, then approximately 950 of those confidence intervals would contain the true fraction of U.S. adults who suer from chronic illnesses.