Question

In an interval estimation for a proportion of a population, the value of z at 99.2% confidence is

Answer 1

`z = 2.65`

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)`.

99.2% confidence level

So
`\alpha = 0.008`, z is the value of Z that has a pvalue of
`1 - (0.008)/(2) = 0.996`, so
`z = 2.65`.