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Hitman42 [59]
3 years ago
10

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
Mathematics
1 answer:
Flauer [41]3 years ago
3 0

Answer:

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

Step-by-step explanation:

The critical value of <em>z</em> for 95% confidence level is:

<em>z</em> = 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{\frac{\hat p(1-\hat p)}{n}}

     =0.206\pm 1.96\times\sqrt{\frac{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%).

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