Answer:
The minimum sample required = 1296.65
Explanation:
Given that:
Variance
![\sigma^2 = 2.89](https://img.qammunity.org/2021/formulas/mathematics/college/fkkt9f7j0bj9dlwgf3x2u3fxruqlyt4hcy.png)
Standard deviation
![\sigma = √(2.89)](https://img.qammunity.org/2021/formulas/mathematics/college/va048g1v56mx24b1klnnvryapso17pvg05.png)
Standard deviation
![\sigma = 1.7](https://img.qammunity.org/2021/formulas/mathematics/college/o8g0xusg1rv80zqthzcepxtcrxpwz5zlxt.png)
Margin of error = 0.11
Confidence Interval = 98%
Level of significance = 1 - 0.98 = 0.02
The critical value =
![Z _(\alpha//2) = Z_(0.02/2) = Z_(0.01)](https://img.qammunity.org/2021/formulas/mathematics/college/lhzas4of9l5uki5jyq08c5ob04qgqajvr9.png)
= 2.33
Thus, the minimum sample size is given by the formula:
![n = \bigg ( (Z_(\alpha/2) * \sigma )/(E) \bigg)^2](https://img.qammunity.org/2021/formulas/mathematics/college/3jjc7rvvwta9b5mty0qizjki3rpkqzomhx.png)
![n = \bigg ((2.33 * 1.7 )/(0.11) \bigg)^2](https://img.qammunity.org/2021/formulas/mathematics/college/l7kdbzxxdhgvdzxz68dbe63ud2w3n1hwkm.png)
n = 1296.65