Question

Le Saveu The UL health centre is giving a Pfizer vaccine to the surrounding community members of the 100 people who were vaccinated, 80 developed immunity to the disease. At 90% confidence level which one of the following is incorrect? O A. The sample propotion is 0.80. OB. The population propotion is 0.80. OC. The marginal error is 0.0658 UD. The critical value is 1.645 FE. The standard error is 0.04​

Answer 1

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 $\pi$, and a confidence level of $1-\alpha$, we the confidence interval for the proportion is:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which  z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

80 developed immunity out of 100, thus, the sample proportion is:

$\pi = \frac{80}{100} = 0.8$

Option A is correct.

The standard error is:

$s = \sqrt{\frac{\pi(1-\pi)}{n}}$

Considering $n = 100$

$s = \sqrt{\frac{0.8(0.2)}{100}} = 0.04$

Thus, statement E is correct.

90% confidence level, thus the critical value is the value of z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $z = 1.645$, which means that statement D is correct.

The margin of error is:

$M = zs = 1.645(0.04) = 0.0658$

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 brainly.com/question/16807970