Answer:
D. As the sample size is appropriately large, the margin of error is ±0.15
Step-by-step explanation:
The number of students in the sample, n = 32 students
The percentage of the students that preferred studying abroad, [tex]\hat p[/tex] = 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;
[tex]M.O.E. = z^*\times \sqrt{\dfrac{\hat{p} \cdot(1-\hat{p})}{n}}[/tex]
Therefore, we get;
[tex]M.O.E = 1.96\times \sqrt{\dfrac{0.25\times(1-0.25)}{32}} \approx \pm0.15[/tex]
Therefore, the correct option is that as the sample size is appropriately large, the margin of error is ±0.15.