Question

The National Health Interview Survey, which included a questionnaire administered during in-person interviews with 21,781 adults, found that 20.6 percent of them were smokers in 2008. (New York Times, Nov 18, 2009). Round your numbers to 3 decimal places. Find a 95% confidence interval for the proportion of American adults who smoked in 2008

Answer 1

The 95% confidence interval for the proportion of American adults who smoked in 2008 is (20.1%, 21.1%).

Explanation:

The critical value of z for 95% confidence level is:

z = 1.96

Compute the 95% confidence interval for the proportion of American adults who smoked in 2008 as follows:

`CI=\hat p\pm z_(\alpha/2)\cdot\sqrt{(\hat p(1-\hat p))/(n)}`

`=0.206\pm 1.96*\sqrt{(0.206(1-0.206))/(21781)}\\\\=0.206\pm 0.0054\\\\=(0.2006, 0.2114)\\\\\approx (0.201, 0.211)`

Thus, the 95% confidence interval for the proportion of American adults who smoked in 2008 is (20.1%, 21.1%).