Question

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.05 for the estimation of a population proportion

Answer 1

A sample of 385 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 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:

We need a sample of n.

n is found when M = 0.05.

We dont know the true proportion, so we work with the worst case scenario, which is
`\pi = 0.5`

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

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

`0.05√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.05)`

`(√(n))^(2) = ((1.96*0.5)/(0.05))^(2)`

`n = 384.16`

Rounding up

A sample of 385 is needed.