Using confidence interval concepts, it is found that the incorrect statement is given by:
B. The population proportion is 0.80.
In a sample of n people with a proportion of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we the confidence interval for the proportion is:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
80 developed immunity out of 100, thus, the sample proportion is:
[tex]\pi = \frac{80}{100} = 0.8[/tex]
Option A is correct.
The standard error is:
[tex]s = \sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Considering [tex]n = 100[/tex]
[tex]s = \sqrt{\frac{0.8(0.2)}{100}} = 0.04[/tex]
Thus, statement E is correct.
90% confidence level, thus the critical value is the value of z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]z = 1.645[/tex], which means that statement D is correct.
The margin of error is:
[tex]M = zs = 1.645(0.04) = 0.0658[/tex]
Thus, statement C is correct, which leaves B as the false statement, as we have the sample proportion, and we can only estimate, not guarantee the population proportion.
A similar problem is gien at https://brainly.com/question/16807970
