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 99% 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  $\pm 5\%$ with 99 %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:

The value is  $n = 666$

Step-by-step explanation:

From the question we are told that

   The  margin of error is  $E = 0.05$

From the question we are told the confidence level is  99% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.01$

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

   $Z_{\frac{\alpha }{2} } = 2.58$

Here we will assume that the sample proportion of those who support a proposed gun control law to be  $\^ p = 0.5$ because  from the question they do not have any prior knowledge about the proportion who might support the law

Generally the sample size is mathematically represented as

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

=>   $n = 666$