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].