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

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