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  [tex]\pm 5\%[/tex] 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  [tex]n = 666 [/tex]
Step-by-step explanation:
From the question we are told that
  The  margin of error is  [tex]E = 0.05[/tex]
From the question we are told the confidence level is  99% , hence the level of significance is  Â
   [tex]\alpha = (100 - 95 ) \%[/tex]
=> Â [tex]\alpha = 0.01[/tex]
Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is Â
  [tex]Z_{\frac{\alpha }{2} } = 2.58[/tex]
Here we will assume that the sample proportion of those who support a proposed gun control law to be  [tex]\^ p = 0.5[/tex] 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
   [tex]n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p ) [/tex]
=> Â [tex]n = 666 [/tex]
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