Answer:
The minimum sample size is [tex]n = 2123[/tex]
Step-by-step explanation:
From the question we are told that
The margin of error is [tex]E = 0.028[/tex]
Given that the confidence level is 99% then the level of significance is evaluated as
[tex]\alpha = 100 - 99[/tex]
[tex]\alpha = 1 \%[/tex]
[tex]\alpha =0.01[/tex]
Next we obtain the critical value of [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table
The value is [tex]Z_{\frac{ \alpha }{2} } = 2.58[/tex]
Now let assume that the sample proportion is [tex]\r p = 0.5[/tex]
hence [tex]\r q = 1 - \r p[/tex]
=> [tex]\r q = 0.50[/tex]
Generally the sample size is mathematically represented as
[tex]n =[ \frac{Z_{\frac{ \alpha }{2} }}{ E} ]^2 * \r p * \r q[/tex]
[tex]n =[ \frac{2.58}{ 0.028} ]^2 * 0.5 * 0.5[/tex]
[tex]n = 2123[/tex]
