Question

A sample of 1000 college students at NC State University were randomly selected for a survey. Among the survey participants, 102 students suggested that classes begin at 8 AM instead of 8:30 AM. The sample proportion is 0.102.What is the margin of error for a 99% confidence interval for this sample? Give your answer to three decimal places.

Answer 1

The margin of error is 0.025 = 2.5 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)}`

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`.

The sample proportion is 0.102.

`p = 0.102`

1000 college students at NC State University were randomly selected for a survey.

So
`n = 1000`

What is the margin of error for a 99% confidence interval for this sample?

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

`M = 2.575\sqrt{(0.102*0.898)/(1000)} = 0.025`