Question

In 2013 , the Pew Research Foundation reported that " 35% 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 1.9\%, 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. Report your answer as a percentage rounded to 3 decimal places. We are 95% confident that the true proportion of U.S. adults who live with one or more chronic conditions is between % and

Answer 1

To create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions, we can use the formula: CI = p-hat ± z * sqrt((p-hat)(1-p-hat)/n), where p-hat is the sample proportion, z is the z-score corresponding to the desired confidence level, and n is the sample size.

Step-by-step explanation:

To create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions, we can use the formula:

CI = p-hat ± z * sqrt((p-hat)(1-p-hat)/n)

Where p-hat is the sample proportion, z is the z-score corresponding to the desired confidence level (in this case, 95% confidence level), and n is the sample size.

Given that the sample proportion is 35% (or 0.35), the standard error is 1.9%, and we have a normal model, we can calculate the confidence interval as follows:

CI = 0.35 ± 1.96 * sqrt((0.35)(0.65)/n)

Since we don't have the sample size (n), we cannot calculate the confidence interval.