Question

On September 16, 2011, the New York Times reported that the American public's approval rating of Congress had sunk to an all-time low of 12%. Suppose we wanted to carry out another poll the following week. How many individuals must we survey to estimate the proportion who approve of Congress at that point to within 3% at 95% confidence?

Answer 1

Answer:

The sample size is  $n = 451$

Step-by-step explanation:

From the question we are told that

   The margin of error is  E = 0.03

    The sample proportion is  $\^ p = 0.12$

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  

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

=>   $n = [\frac{1.96}{0.03} ]^2 *0.12 (1 - 0.12 )$

=>   $n = 451$