Answer:
We are given:
![\hat{p}=0.65](https://img.qammunity.org/2020/formulas/mathematics/middle-school/sr6kejpnnbrhfijlpo8bqlsuae1a13zqw2.png)
Margin of error
![E=0.02](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1pjkec3kol8kwr9c2h6exynx7s2eh3brec.png)
Confidence level = 0.99
We need to find the sample size to estimate the proportion of people who carry the TAS2R388 gene.
The formula for finding the sample size is:
![n=\hat{p} (1-\hat{p})\left (\frac{z_{(0.01)/(2)}}{E} \right )^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rpcp6po5ut2rx82o530llg5cgyz9del0l6.png)
Where:
is the critical value at 0.01 significance level.
Therefore, the required sample size is:
![n=0.65(1-0.65) \left((2.576)/(0.02) \right)^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lf9t7bc7ayix4vidc80d7ilxmqzfj8idbi.png)
![=0.2275 * 16589.44](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4dobu7myag3qnjwvk4exlnczrfaqyyeiq7.png)
![=3774.1 \approx 3774](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nxv7ifxwmui8jhvbwechvalauchxsuet44.png)
Therefore, the required sample size is 3774