Answer:
416025
Step-by-step explanation:
For confidence interval of 99%, the range is (0.005, 0.995). Using a z-table, the z-score for 0.995 is 2.58.
Margin of error = 0.2% = 0.002.
Proportion is unknown. So, worse case proportion is 50%. p = 50% = 0.5.
\\ [tex]n = \left(\frac{\texttt{z-score}}{\texttt{margin of error}} \right )^2\cdot p\cdot (1-p) \\ = \left(\frac{2.58}{0.002} \right )^2\cdot 0.5\cdot (1-0.5)=416025[/tex]
So, sample size required is 416025.
