Question

Dr pox plans to conduct a survey to estimate the percentage of adults in 2020 who have had chickenpox he reviewed a previous study that reported 60% of all adults had contracted chickenpox at some time in their life find the number of people who must be surveyed if you want to be 92% confident with a margin of error of 2%​

Answer 1

Answer:

We must sample 2305 people.

Step-by-step 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.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

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

For this problem, we have that:

$\pi = 0.6$

92% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Number of people who must be surveyed:

We must sample n people.

n is found when $M = 0.02$

So

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

$0.02 = 1.96\sqrt{\frac{0.6*0.4}{n}}$

$0.02\sqrt{n} = 1.96\sqrt{0.6*0.4}$

$\sqrt{n} = \frac{1.96\sqrt{0.6*0.4}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.6*0.4}}{0.02})^{2}$

$n = 2304.96$

Rounding up

We must sample 2305 people.