Question

A survey is planned to estimate the proportion of voters who support a proposed gun control law. The estimate should be within a margin of error of with 90% confidence, and we do not have any prior knowledge about the proportion who might support the law. How many people need to be included in the sample

Answer 1

Complete Question

A survey is planned to estimate the proportion of voters who support a proposed gun control law. The estimate should be within a margin of error of ±5% with 90% confidence, and we do not have any prior knowledge about the proportion who might support the law. How many people need to be included in the sample?

Round your answer up to the nearest integer.

sample size = _____

Answer:

The sample size is $n = 271$    

Step-by-step explanation:

From the question we are told that  

   The margin of error is  E = 5% = 0.05

Generally given that there was no prior knowledge about the proportion of who might support the law, we will assume the sample proportion to be  

     $\^ p = 0.5$

From the question we are told the confidence level is  90% , hence the level of significance is    

      $\alpha = (100 - 90 ) \%$

=>   $\alpha = 0.05$

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

   $Z_{\frac{\alpha }{2} } = 1.645$

Generally the sample size is mathematically represented as  

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

=>   $n = [\frac{1.645 }{0.05} ]^2 * 0.5 (1 - 0.5 )$

=>   $n = 271$