Question

You intend to estimate a population proportion with a confidence interval. The data suggests that the normal distribution is a reasonable approximation for the binomial distribution in this case. While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 97.1%.

Answer 1

The critical value that corresponds to a confidence level of 97.1% is
`Z = 2.18`.

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

97.1% confidence level

So
`\alpha = 0.029`, z is the value of Z that has a p-value of
`1 - (0.029)/(2) = 0.9855`, so
`Z = 2.18`.

The critical value that corresponds to a confidence level of 97.1% is
`Z = 2.18`.