Question

Ian is a phycologist interested in determining the proportion of algae samples from a local rivulet that belonged to a particular phylum. A random sample of 50 alga were obtained and each was categorized as either being cyanobacteria or not. It was found that 38 were, in fact, cyanobacteria.Without relying on any previous knowledge Ian wanted to estimate the proportion that were cyanobacteria with a margin of error of at most 0.01 in a 99% confidence interval. How large a sample size would be required?

Answer 1

Answer: 16590

Step-by-step explanation:

The formula we use to find the sample size is given by :-

, where p = prior estimate of population proportion.

$z^*$= Critical z-value (Two- tailed)

E= Margin of error .

Let p be the population proportion of cyanobacteria.

Given : E = 0.01

Critical value for 99% confidence interval : $z^*=2.576$

If Ian does not want to rely on previous knowledge , then we assume p=0.5 because the largest standard error is at p=0.5.

Now, the required sample size = $n=0.5(1-0.5)(\frac{2.576}{0.01})^2$

$n=0.25(257.6)^2=0.25(66357.76)=16589.44\approx16590$

Required sample size = 16590