Question

A poll asked 1057 American adults if they believe there was a conspiracy in the assassination of President Kennedy, and found that 614 believe there was a conspiracy. Estimate the proportion of Americans who believe in this conspiracy using a 95% confidence interval. Round to three decimal places.

Answer 1

Answer:

The 95% confidence interval of the proportion of Americans who believe in the conspiracy is  $0.551< p < 0.610$

Step-by-step explanation:

From the question we are told that

   The sample size is  n =1057

   The number that believe there was a conspiracy is  k = 614

Generally the sample proportion is mathematically represented as

      $\^ p = \frac{614}{1057 }$

=>   $\^ p = 0.5809$

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 margin of error is mathematically represented as  

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

=>   $E = 1.96 * \sqrt{\frac{0.5809 (1- 0.5809)}{1057} }$    

=>   $E = 0.02975$  

Generally 95% confidence interval is mathematically represented as  

      $\^ p -E < p < \^ p +E$

=>    $0.5809 - 0.02975< p < 0.5809 + 0.02975$

=>    $0.551< p < 0.610$