Question

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 73%. You would like to be 95% confident that your estimate is within 4% of the true population proportion. How large of a sample size is required

Answer 1

A sample size of 474 is required.

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 of:

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

Based on previous evidence, you believe the population proportion is approximately 73%.

This means that
`\pi = 0.73`

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`.

How large of a sample size is required?

A sample size of n is required, and n is found when M = 0.04. So

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

`0.04 = 1.96\sqrt{(0.73*0.27)/(n)}`

`0.04√(n) = 1.96√(0.73*0.27)`

`√(n) = (1.96√(0.73*0.27))/(0.04)`

`(√(n))^2 = ((1.96√(0.73*0.27))/(0.04))^2`

`n = 473.24`

Rounding up:

A sample size of 474 is required.