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
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.01[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[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]
=> [tex]n = 666 [/tex]
