Question

The student body President of a high school claims to know the names of at least 1000 of the 1800 students at the school. To test this claim, the Student Government Advisor randomly selects 100 students from the school and asks the President to name each one. The President is able to correctly name 46 of the students. What would be the 99% confidence interval for p in this case?

Answer 1

The 99% confidence interval for p in this case is (0.3317, 0.5883).

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)`.

Randomly selects 100 students from the school and asks the President to name each one. The President is able to correctly name 46 of the students.

This means that:

`n = 100, \pi = (46)/(100) = 0.46`

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 lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.46 - 2.575\sqrt{(0.46*0.54)/(100)} = 0.3317`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.46 + 2.575\sqrt{(0.46*0.54)/(100)} = 0.5883`

The 99% confidence interval for p in this case is (0.3317, 0.5883).