Question

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 32%. You would like to be 90% confident that your estimate is within 4% of the true population proportion. How large of a sample size is required?

Answer 1

We need a sample size of at least 369.

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 = 0.32`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

How large of a sample size is required?

We need a sample size of at least n.

n is found when
`M = 0.04`. So

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

`0.04 = 1.645\sqrt{(0.32*0.68)/(n)}`

`0.04√(n) = 1.645√(0.32*0.68)`

`√(n) = (1.645√(0.32*0.68))/(0.04)`

`(√(n))^(2) = ((1.645√(0.32*0.68))/(0.04))^(2)`

`n = 368.02`

Rounding up

We need a sample size of at least 369.