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

The margin of error is 0.0421 = 4.21 percentage points

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

In this problem, we have that:

`n = 500, p = 0.36`

So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`M = 1.96\sqrt{(0.36*0.64)/(500)}`

`M = 0.0421`

The margin of error is 0.0421 = 4.21 percentage points