Question

The 2010 General Social Survey reported a sample where about 48% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% confidence interval to 4%, about how many Americans would we need to survey

Answer 1

Answer:

The sample size is  $n = 600$

Step-by-step explanation:

From the question we are told that

    The  sample proportion is  $\r p = 0.48$

     The  margin of error is  $MOE = 0.04$

Given that the confidence level is 95%  the level of significance is mathematically represented as

        $\alpha = 100 - 95$

        $\alpha = 5 \%$

        $\alpha = 0.05$

Next  we obtain the critical value of  $\frac{\alpha }{2}$ from the normal distribution table , the values is

                   $Z_{\frac{\alpha }{2} } = 1.96$

The reason we are obtaining critical value of    $\frac{\alpha }{2}$ instead of    $\alpha$ is because  

$\alpha$ represents the area under the normal curve where the confidence level interval (  $1-\alpha$) did not cover which include both the left and right tail while  

$\frac{\alpha }{2}$ is just the area of one tail which what we required to calculate the margin of error

Generally the margin of error is mathematically represented as

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

substituting values

          $0.04= 1.96* \sqrt{ \frac{0.48(1- 0.48 )}{n} }$

         $0.02041 = \sqrt{ \frac{0.48(52 )}{n} }$

         $0.02041 = \sqrt{ \frac{ 0.2496}{n} }$

          $0.02041^2 = \frac{ 0.2496}{n}$

           $0.0004166 = \frac{ 0.2496}{n}$

=>       $n = 600$