Question

The 2003 Statistical Abstract of the United States reported the percentage of people years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of . a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of

Answer 1

Complete Question

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and non smokers uses a preliminary estimate of the proportion who smoke of 0.30.

a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of 0.02? use 95% confidence.

b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population?

c. What is the 95% confidence interval for the proportion of smokers in the population?

Answer:

a

$n = 2017$

b

$\^ p = 0.2578$

c

$0.238 < 0.278$

Step-by-step explanation:

Considering question a

From the question we are told that

    The margin of error is  $E = 0.02$

     The preliminary estimate of the proportion who smoke is  $p = 0.30$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the sample size is mathematically  represented as  

      $n = [ \frac{Z_{\frac{\alpha }{2} } }{E} ] ^2 * p(1 - p )$

=>   $n = [ \frac{1.96 }{0.02} ] ^2 * 0.30(1 - 0.30 )$

=>   $n = 2017$

Considering question b

 From the question we are told that

   The number of smokers  is  k =  520  

Generally the  point estimate of the proportion of smokers in the population is mathematically represented as

        $\^ p = \frac{520 }{2017 }$

=>    $\^ p = 0.2578$

Considering  question c

Generally 95% confidence interval is mathematically represented as  

      $\^ p -E < p < \^ p +E$

=>   $0.2578 - 0.02 < 0.2578 + 0.02$

=>   $0.238 < 0.278$