Question

A math teacher was frustrated at the number of students leaving their graphing calculator behind in her classroom at the end of class without a way to locate the student. A random sample of 50 students is selected, and of the students questioned, 32 had their names written on their graphing calculators.

Required:
What would be a 99% confidence interval for the proportion of all students at this school who have their names written on their graphing calculators?

Answer 1

The 99% confidence interval for the proportion of all students at this school who have their names written on their graphing calculators is (0.4652, 0.8148).

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)`.

A random sample of 50 students is selected, and of the students questioned, 32 had their names written on their graphing calculators.

This means that
`n = 50, \pi = (32)/(50) = 0.64`

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)} = 0.64 - 2.575\sqrt{(0.64*0.36)/(50)} = 0.4652`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.64 + 2.575\sqrt{(0.64*0.36)/(50)} = 0.8148`

The 99% confidence interval for the proportion of all students at this school who have their names written on their graphing calculators is (0.4652, 0.8148).