Question

A recent poll of 500 residents in a large town found that only​ 36% were in favor of a proposed referendum to build a new high school. Find the margin of error for this poll if we want​ 95% confidence in our estimate of the percentage of residents in favor of this proposed referendum.

Answer 1

Answer:

The margin of error is 0.0421 = 4.21 percentage points

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 $1 - \frac{\alpha}{2}$.

The margin of error is:

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

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

In this problem, we have that:

$n = 500, p = 0.36$

So

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

$M = 1.96\sqrt{\frac{0.36*0.64}{500}}$

$M = 0.0421$

The margin of error is 0.0421 = 4.21 percentage points