Question

Fireworks on July4th.A local news outlet reported that 56% of 600 randomly sampled Kansasresidents planned to set off fireworks on July 4th. Determine the margin of error for the 56% point estimateusing a 95% confidence level.1

Answer 1

The margin of error is 3.97 percentage points.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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)`.

For this problem, we have that:

`n = 600, p = 0.56`

95% confidence interval

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`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.56 - 1.96\sqrt{(0.56*0.44)/(600)} = 0.5203`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.56 + 1.96\sqrt{(0.56*0.44)/(600)} = 0.5997`

The margin of error is the upper limit subtracted by the proportion, or the proportion subtracted by the lower limit. They are the same values.

So the margin of error is 0.5997 - 0.56 = 0.56 - 0.5203 = 0.0397 = 3.97 percentage points.