Answer:
We are given:
Confidence level = 99%. Therefore, the critical value at 0.01 significance level using the standard normal table is given below:
[tex]z_{\frac{0.01}{2}}=2.576[/tex]
Margin of error is given in the question as:
[tex]E=0.05[/tex]
Since the previous proportion is not given, therefore, we need to assume [tex]\hat{p} = 0.5[/tex]
Therefore, the sample size is:
[tex]n=\hat{p}(1-\hat{p}) \left( \frac{z_{\frac{0.01}{2}} }{E} \right)[/tex]
[tex]=0.5(1-0.5) \left( \frac{2.576}{0.05} \right)[/tex]
[tex]=0.25 \times 51.52^2[/tex]
[tex]=663.6 \approx 664[/tex]
Therefore, 664 sample observations are required.
