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

Solution :

Given that,

Point estimate = sample proportion = = 0.206

1 - = 1 - 0.206 = 0.794

Z/2 = Z0.025 = 1.96  

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 (((0.206 * 0.794) / 21781 )

= 0.005

A 95% confidence interval for population proportion p is ,

± E

= 0.206 ± 0.005

= ( 0.201, 0.211 )