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 ± _____ %.

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

A sample of 270 students, so [tex]n = 270[/tex].

50% of students were in favor of the proposal, so [tex]\pi = 0.50[/tex].

We are working with a 90% confidence interval:

So [tex]\alpha = 0.10[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.10}{2} = 0.95[/tex], so Z = 1.645.

The margin of error for their finding at the 90% level of confidence is

The margin of error is [tex]\pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex][/tex]

So

[tex]M = \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]M = \pm 1.645*\sqrt{\frac{0.50(0.50)}{270}} = \pm 0.05[/tex]

The margin of error for their finding at the 90% level of confidence is [tex]\pm 0.05[/tex].