Question

Marge conducted a survey by asking 350 citizens whether they frequent the city public parks. Of the citizens surveyed, 240 responded favorably.

What is the approximate margin of error for each confidence level in this situation?

0.07

0.03

0.04

0.05

0.06

99%

95%

90%

Answer 1

The margin of error for a 99% confidence interval is of 0.0639, that is, approximately 0.06.

The margin of error for a 95% confidence interval is of 0.0486, that is, approximately 0.05.

The margin of error for a 90% confidence interval is of 0.0408, that is, approximately 0.04.

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)}`

350 citizens, 240 responded favorably:

This means that
`n = 350, \pi = (240)/(350) = 0.6857`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

`M = 2.575\sqrt{(0.6857*0.3143)/(350)} = 0.0639`

The margin of error for a 99% confidence interval is of 0.0639, that is, approximately 0.06.

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`.

`M = 1.96\sqrt{(0.6857*0.3143)/(350)} = 0.0486`

The margin of error for a 95% confidence interval is of 0.0486, that is, approximately 0.05.

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

`M = 1.645\sqrt{(0.6857*0.3143)/(350)} = 0.0408`

The margin of error for a 90% confidence interval is of 0.0408, that is, approximately 0.04.