Question

Assume that you wish to estimate a population proportion, p. For the given margin of error and confidence level, determine the sample size required:

A researcher wants to determine what proportion of adults in one town regularly buy organic food. Obtain a sample size that will ensure a margin of error of at most 0.04 for a 90% confidence interval. In previous years, the proportion has been 0.16 .

Answer 1

The sample size required is 228.

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)}`

90% confidence level

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

In previous years, the proportion has been 0.16.

This means that
`\pi = 0.16`

Obtain a sample size that will ensure a margin of error of at most 0.04 for a 90% confidence interval.

This is n for which M = 0.04. So

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

`0.04 = 1.645\sqrt{(0.16*0.84)/(n)}`

`0.04√(n) = 1.645√(0.16*0.84)`

`√(n) = (1.645√(0.16*0.84))/(0.04)`

`(√(n))^2 = ((1.645√(0.16*0.84))/(0.04))^2`

`n = 227.3`

Rounding up:

The sample size required is 228.