In article <385EBF05.7880A55A at medicine.adelaide.edu.au>,
Robyn Wallace <rwallace at medicine.adelaide.edu.au> wrote:
>I am looking for a simple statistical program for the following problem.
>>Assuming a person normally has 1 in 500 abnormal cells -
>>How many cells need to be counted to reach 95% confidence?
[and a similar question for 1 in 50.]
If the question is, how many cells must one count to have a 95%
chance of seeing at least one abnormal cell, suppose that you count
N cells and have chance p of seeing an abnormal one. The probability
that you miss seeing one is
N
(1-p)
and you want this to get as low as 0.05, the you just need to
solve
0.05 = (1-p)^N
or
ln (0.05) = N ln (1-p)
or N = ln (1-0.05) / ln (1-p)
For the two values you give (1/500 and 1/50) the numbers are
1496 and 148.
If on the other hand the question was "I counted 50 cells and saw
one abnormal one; is that statistically significantly different
from 0?" The answer is always "yes". It's like looking for
Little Green Men on Mars. If your spacecraft expects to see one
but sees none, that is not terribly conclusive. If it expects to
see none but sees one, that is infinitely strongly significant.
--
Joe Felsenstein joe at genetics.washington.edu
Dept. of Genetics, Univ. of Washington, Box 357360, Seattle, WA 98195-7360 USA
