Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 87, x = 26; 98 percent (0.185, 0.413) (0.202, 0.396) (0.184, 0.414) (0.203, 0.395)

Answer 1

(0.185, 0.413)

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:

`n = 87, x = 26, p = (x)/(n) = (26)/(87) = 0.2989`

98% confidence interval

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.325`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.2989 - 2.325\sqrt{(0.2989*0.7011)/(87)} = 0.185`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.2989 + 2.325\sqrt{(0.2989*0.7011)/(87)} = 0.413`

So the correct answer is:

(0.185, 0.413)