Respuesta :

In an interval estimation for a proportion of a population, the value of z at 99.2% confidence is 2.65.

Answer:

[tex]z = 2.65[/tex]

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

99.2% confidence level

So [tex]\alpha = 0.008[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.008}{2} = 0.996[/tex], so [tex]z = 2.65[/tex].