Question

estimate the number of first-year students at an assembly meeting. He surveys 60 students and finds that 24 of them are first-year students. Identify the values needed to calculate a confidence interval at the 95% confidence level. Then find the confidence interval.

Answer 1

The values needed to calculate a confidence interval at the 95% confidence level are
`z = 1.96, n = 60, \pi = (24)/(60) = 0.4`

The 95% confidence interval for the proportion of first-year students at an assembly meeting is (0.276, 0.524).

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)`.

For this problem, we have that:

`n = 60, \pi = (24)/(60) = 0.4`

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`.

The values needed to calculate a confidence interval at the 95% confidence level are
`z = 1.96, n = 60, \pi = (24)/(60) = 0.4`

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.4 - 1.96\sqrt{(0.4*0.6)/(60)} = 0.276`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4 + 1.96\sqrt{(0.4*0.6)/(60)} = 0.524`

The 95% confidence interval for the proportion of first-year students at an assembly meeting is (0.276, 0.524).