Answer:
D. As the sample size is appropriately large, the margin of error is ±0.15
Explanation:
The number of students in the sample, n = 32 students
The percentage of the students that preferred studying abroad,
= 25%
The confidence level for the study = 95%
As a general rule, a sample size of 30 and above are taken as sufficient
The z-value at 95% confidence level, z = 1.96
The margin of error of a proportion formula is given as follows;
![M.O.E. = z^** \sqrt{\frac{\hat{p} \cdot(1-\hat{p})}{n}}](https://img.qammunity.org/2022/formulas/mathematics/high-school/egg0x16eyjgog4rq9pnqw3sd6sz6m3zrj1.png)
Therefore, we get;
![M.O.E = 1.96* \sqrt{(0.25*(1-0.25))/(32)} \approx \pm0.15](https://img.qammunity.org/2022/formulas/mathematics/high-school/8vf421v49284gqay2756p6hjcqg3bkw8py.png)
Therefore, the correct option is that as the sample size is appropriately large, the margin of error is ±0.15.