Question

Suppose that the epidemiologist wants to re-estimate the population proportion and wishes for her 95% confidence interval to have a margin of error no larger than 0.04. How large a sample should she take to achieve this? Please carry answers to at least six decimal places in intermediate steps.

Answer 1

She needs a sample of at least 385.

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)}`

95% 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`.

How large a sample should she take to achieve this?

She needs a sample of size at least n.

n is found when M = 0.04.

We do not know the true population proportion, so we use
`\pi = 0.5`, which is the case for which we are going to need the largest sample size.

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

`0.05 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.05√(n) = 1.96*0.5`

Dividing both sides by 0.05

`√(n) = 19.6`

`(√(n))^(2) = 19.6^(2)`

`n = 384.2`

Rounding up

She needs a sample of at least 385.