To determine the margin of error for the population proportion based on the survey results, follow these steps:
1. Calculate the sample proportion ([tex]\( \hat{p} \)[/tex]):
- Given that 90% of the students in the sample have more than one active email account, [tex]\( \hat{p} = 0.90 \)[/tex].
2. Find the sample size ([tex]\( n \)[/tex]):
- The sample size is 400 students.
3. Calculate the standard error ([tex]\( \text{SE} \)[/tex]) for the sample proportion:
- [tex]\(\text{SE} = \sqrt{ \frac{\hat{p} (1 - \hat{p})}{n} }\)[/tex]
- With [tex]\( \hat{p} = 0.90 \)[/tex] and [tex]\( n = 400 \)[/tex], the standard error is approximately 0.015.
4. Determine the z-score for the desired confidence level:
- For a 95% confidence level, the z-score is 1.96.
5. Calculate the margin of error (ME):
- [tex]\( \text{ME} = z \cdot \text{SE} \)[/tex]
- Using a z-score of 1.96 and a standard error of 0.015, the margin of error is approximately 0.0294.
6. Convert the margin of error to a percentage:
- The margin of error as a percentage is roughly 2.94%.
Therefore, the correct margin of error for the population proportion is approximately ±2.94%. Thus, the correct answer is not directly listed in the given options. None of the provided options (not OA, OB, OC, or OD) are correct since the correct margin of error is ±2.94%.
