Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 133 with 61.7% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Answer 1

The 99% confidence interval is (0.508, 0.726).

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 = 133, p = 0.617`

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)} = 0.617 - 2.575\sqrt{(0.617*0.383)/(133)} = 0.508`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.617 + 2.575\sqrt{(0.617*0.383)/(133)} = 0.726`

The 99% confidence interval is (0.508, 0.726).