Question

The student union wants to negotiate a subsidized bus pass with your school's administration. They have polled 270 students as to whether they support this proposal, which would add a small fee to their tuition since both the college and the students would share the cost. The union found that opinions were evenly split: 50% of students were in favor of the proposal, and the rest were opposed. The margin of error for their finding at the 90% level of confidence is ± _____ %.

Answer 1

The margin of error for the student union poll at the 90% confidence level is ± 5%, calculated using the Z-score for 90% confidence and the formula for margin of error of a proportion.

Step-by-step explanation:

To calculate the margin of error at the 90% level of confidence for a proportion when the sample size is 270 and the proportion is 0.5 (50%), we use the formula for the margin of error of a proportion, which is:

Margin of Error = Z * sqrt(p(1-p)/n)

Where:

Z is the Z-score corresponding to the desired level of confidence

p is the sample proportion

n is the sample size

For a 90% confidence level, the Z-score is approximately 1.645. Given that p=0.5 and n=270:

Margin of Error = 1.645 * sqrt(0.5(1-0.5)/270)

Margin of Error = 1.645 * sqrt(0.25/270)

Margin of Error = 1.645 * sqrt(0.0009259)

Margin of Error = 1.645 * 0.0304282

Margin of Error = approximately 0.05, or 5%

Therefore, the estimated margin of error for the student union poll at the 90% confidence level is ± 5%.