Question

10.

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that the study select 2017 people using a simple random sample and finds 520 smokers. At 95% confidence, what is the margin of error? Round your answers to four decimal places.

Answer 1

Solution :

Given that,

n = 2017

x = 520

Point estimate = sample proportion = = x / n = 520 / 2017 = 0.2578

1 - = 1 - 0.2578 = 0.7422

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

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

E = 1.96 (((0.2578 * 0.7422) / 2017)

E = 0.0191