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

Answer:

A sample of 16577 is required.

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, 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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.01\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.01}$

$(\sqrt{n})^2 = (\frac{2.575*0.5}{0.01})^2$

$n = 16576.6$

Rounding up:

A sample of 16577 is required.