Answer:
2,655 students
Step-by-step explanation:
The z-score for a 99% confidence interval is z = 2.576
The standard error for a proportion p is:
[tex]SE = z*\sqrt{\frac{p*(1-p)}{n} }[/tex]
For a proportion of p =0.20, in order to ensure a standard error of 0.02, the sample size 'n' must be:
[tex]0.02 = 2.576*\sqrt{\frac{0.2*(1-0.2)}{n}}\\n=2,654.31[/tex]
Rounding up to the next whole student, the sample size needed is 2,655 students.
