# Thread: Confidence interval question

1. ## Confidence interval question

Hi
The following question i am having difficulties solving.

In certain water-quality studies, it is important to check for the presence or absence of various types of micro-organisms. Suppose that 20 out of 100 randomly selected
samples of a fixed volume show the presence of a particular micro-organism.
(a) Estimate the true probability of finding this micro-organism in a sample of the
same volume, using a 95% confidence interval.

i know that 20/100 = .20 shows the presence of a particular micro-organism and .80 does not.

the confident interval is .95%

so i get something like this

$\displaystyle .20 + 1.95 * \sqrt(.20*.80/100)$

however this is incorrect

(b) Can it reasonably be said that there is only a 10% chance that a sample of the
given volume will contain the micro-organism?

P.S

2. ## Re: Confidence interval question

would 1.96 instead of the 1.95 give you the correct answer?
1.96 is right, but your error is minimal.

3. ## Re: Confidence interval question

the answer is for 8a) (0.1216,0.2784)

edit: got the correct answer

4. ## Re: Confidence interval question

that is correct with 1.96 used as the approximate percentile point
this is a normal approximation to the binomial distribution