We are given
Proportion (p)
[tex]p=19\%=0.19[/tex]Level of significance = 99%
margin of error = 1.5% = 0.015
We want to find n
Solution
Note: The margin of error formula is given as
[tex]MarginOfError=z_{\frac{\alpha}{2}}\times\sqrt[]{\frac{p(1-p)}{n}}[/tex]The image of the formula to use is
In the question, we have
[tex]\begin{gathered} \alpha=0.01 \\ \text{from the statistics table} \\ z_{\frac{\alpha}{2}}=z_{\frac{0.01}{2}}=z_{0.005}=2.576 \end{gathered}[/tex][tex]\begin{gathered} p=0.19 \\ MarginOfError=0.015 \end{gathered}[/tex]Substituting the parameters
[tex]\begin{gathered} MarginOfError=z_{\frac{\alpha}{2}}\times\sqrt[]{\frac{p(1-p)}{n}} \\ 0.015=2.576\times\sqrt[]{\frac{0.19(1-0.19)}{n}} \\ \sqrt[]{\frac{0.19(1-0.19)}{n}}=\frac{0.015}{2.576} \\ \frac{0.19(0.81)}{n}=(\frac{15}{2576})^2 \\ \frac{n}{0.19(0.81)}=(\frac{2576}{15})^2 \\ n=0.19(0.81)(\frac{2576}{15})^2 \\ n=4538.870784 \\ n=4539\text{ (to the nearest whole number)} \end{gathered}[/tex]Therefore, the sample size is as large as
[tex]n=4539[/tex]
