Question

I want to take a survey of students at my university to find out what proportion like the new bus service on campus. I believe that 75% of students like the service. How many students will I need to survey if I want to estimate, with 99% confidence, the true proportion to within 2%?

Answer 1

Answer:

You will need to sample at least 3108 students.

Step-by-step explanation:

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

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

In this problem, we have that:

$\pi = 0.75$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

How many students will I need to survey if I want to estimate, with 99% confidence, the true proportion to within 2%?

You need a sample size of at least n.

n is found when M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.575\sqrt{\frac{0.75*0.25}{n}}$

$0.02\sqrt{n} = 1.115$

$\sqrt{n} = \frac{1.115}{0.02}$

$\sqrt{n} = 55.75$

$\sqrt{n}^{2} = (55.75)^{2}$

$n = 3108$

You will need to sample at least 3108 students.