Question

On the basis of sample of 25 students in a large campus laboratory, it appears that 8 students smoke. A 95% CI estimate of the proportion of smokers in the lab is: a. [0.137, 0.503] b. [0.037, 0.050] c. [1.137, 1.503]

Answer 1

a. [0.137, 0.503]

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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:

A sample of 25 students in the lab, of whom 8 smoke. So
`n = 25, \pi = (8)/(25) = 0.32`

95% confidence interval

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 lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.32 - 1.96\sqrt{(0.32*0.68)/(25)} = 0.137`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.32 + 1.96\sqrt{(0.32*0.68)/(25)}{119}} = 0.503`

The correct answer is:

a. [0.137, 0.503]