Question

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 60%. You would like to be 98% confident that your estimate is within 2.5% of the true population proportion. How large of a sample size is required?

Answer 1

A sample size of 2080 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)}`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

Based on previous evidence, you believe the population proportion is approximately 60%.

This means that
`\pi = 0.6`

How large of a sample size is required?

We need a sample of n.

n is found when
`M = 0.025`. So

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

`0.025 = 2.327\sqrt{(0.6*0.4)/(n)}`

`0.025√(n) = 2.327√(0.6*0.4)`

`√(n) = (2.327√(0.6*0.4))/(0.025)`

`(√(n))^(2) = ((2.327√(0.6*0.4))/(0.025))^(2)`

`n = 2079.3`

Rounding up

A sample size of 2080 is needed.