Question

A phycologist is interested in determining the proportion of algae samples from a local rivulet that belong to a particular phylum. She wants to estimate the proportion using a 99% confidence interval with a margin of error of at most 0.01. How large a sample size would be required

Answer 1

A sample of 16577 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)}`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a p-value of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

She wants to estimate the proportion using a 99% confidence interval with a margin of error of at most 0.01. How large a sample size would be required?

We have no estimation for the proportion, and thus we use
`\pi = 0.5`, which is when the largest sample size will be needed.

The sample size is n for which M = 0.01. So

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

`0.01 = 2.575\sqrt{(0.5*0.5)/(n)}`

`0.01√(n) = 2.575*0.5`

`√(n) = (2.575*0.5)/(0.01)`

`(√(n))^2 = ((2.575*0.5)/(0.01))^2`

`n = 16576.6`

Rounding up:

A sample of 16577 is required.