Question

Suppose you had been in charge of designing the study. What sample size would be needed to construct a margin of error of 2% with 95% confidence

Answer 1

A sample size of 2401 is needed.

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is given by:

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

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

What sample size would be needed to construct a margin of error of 2% with 95% confidence?

This is M for which n = 0.02.

Supposing we have no estimate for the true proportion, we use
`\pi = 0.5`.

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

`0.02 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.02√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.02)`

`(√(n))^2 = ((1.96*0.5)/(0.02))^2`

`n = 2401`

A sample size of 2401 is needed.