Question

Use the given data to construct a confidence interval for the population proportion of the requested level. x=52, n=72, confidence level 99.9% Round the answer to at least three decimal places.

Answer 1

The 99.9% confidence interval for the population proportion is (0.548, 0.896).

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 = 72, \pi = (x)/(n) = (52)/(72) = 0.722`

99.9% confidence level

So
`\alpha = 0.001`, z is the value of Z that has a pvalue of
`1 - (0.001)/(2) = 0.9995`, so
`Z = 3.29`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.722 - 3.29\sqrt{(0.722*0.278)/(72)} = 0.548`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.722 + 3.29\sqrt{(0.722*0.278)/(72)} = 0.896`

The 99.9% confidence interval for the population proportion is (0.548, 0.896).