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

We must sample 2305 people.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\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 - (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{(\pi(1-\pi))/(n)}`

`0.02 = 1.96\sqrt{(0.6*0.4)/(n)}`

`0.02√(n) = 1.96√(0.6*0.4)`

`√(n) = (1.96√(0.6*0.4))/(0.02)`

`(√(n))^(2) = ((1.96√(0.6*0.4))/(0.02))^(2)`

`n = 2304.96`

Rounding up

We must sample 2305 people.