Question

In a poll conducted by the Gallup organization in April 2013, 48% of a random sample of 1022 adults in the U.S. responded that they felt that economic growth is more important than protecting the environment. We can use this information to calculate a 95% confidence interval for the proportion of all U.S. adults in April 2013 who felt that economic growth is more important than protecting the environment. Make sure to include all steps.

Answer 1

Answer:

The  95% confidence interval is  $0.449 < p < 0.48 + 0.511$

Step-by-step explanation:

From the question we are told that  

     The sample proportion is $\r p = 0.48$

      The sample size is $n = 1022$

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

       $\alpha = 100 - 95$

       $\alpha = 5 \%$

       $\alpha = 0.05$

Next we obtain the critical value of  $\frac{\alpha }{2}$ from the z-table , the value is

     $Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96$

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

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

    Generally the margin of error is mathematically represented as

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

substituting values

          $E = 1.96* \sqrt{\frac{0.48 (1- 0.48 )}{1022} }$

          $E = 0.03063$

The 95% confidence interval is mathematically represented as

      $\r p - E < p < \r p + E$

substituting values

       $0.48 - 0.03063 < p < 0.48 + 0.03063$

       $0.449 < p < 0.48 + 0.511$