When designing a study to determine this population proportion, what is the minimum number of drivers you would need to survey to be 95% confident that the population proportion is estimated to within 0.02

Respuesta :

Answer:

The sample size is [tex]n =2401 [/tex]

Step-by-step explanation:

From the question we are told that

    The margin of error is  [tex]E = 0.02[/tex]

From the question we are told the confidence level is  95% , hence the level of significance is    

      [tex]\alpha = (100 - 95 ) \%[/tex]

=>   [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]

Generally we will assume the sample proportion to be  [tex]\^ p = 0.5[/tex]

Generally the sample size is mathematically represented as

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

=> [tex]n = [\frac{1.96 }{0.02} ]^2 *  0.5 (1 - 0.5) [/tex]

=> [tex]n =2401 [/tex]