I am trying to learn some statistics using the book, Biometry by Sokal and Rohlf (3e). This is an exercise in the 5th chapter which covers probability, the binomial distribution, and Poisson distribution. enter image description here

I realize there is a formula to produce an answer to this question: $$ n = \frac 4 {( \sqrt{p} - \sqrt{q} )^2} $$ However, this equation is not in this text. I'd like to know how to calculate sample size knowing only the probability, the desired level of confidence, and the binomial distribution. Are there any resources covering this topic that I can be pointed to? I've tried Google, but what I've seen so far requires information I don't have access to in this problem.

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    $\begingroup$ Do you want to be guided on a journey to figure out the answer or would you prefer to just be given the answer, along with an explanation of why it's the answer? $\endgroup$
    – jbowman
    Oct 16, 2013 at 20:05
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    $\begingroup$ A journey sounds nice. This isn't for a class and the answer is given at the end of the question. I don't care to just know the answer - I already know it! I've taken a stats course many years ago, but I didn't appreciate it enough then. I'm trying to remedy that now and really begin to understand the underlying patterns. I'd appreciate the help. This particular problem doesn't seem to fit with the others from this section and a proper approach isn't clearly demonstrated (to me) from the text's information on the binomial distribution nor its examples given. $\endgroup$
    – baffled
    Oct 16, 2013 at 21:11
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    $\begingroup$ I would be very interested in reading a detailed answer (with pointers to further reading where necessary) to this question. $\endgroup$
    – Zhubarb
    Oct 17, 2013 at 12:00
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    $\begingroup$ Let's consider a concrete, simple example; you have 5 slides from a person who has the pathogen. What is the probability that you fail to correctly identify this person as having the pathogen? A hidden assumption is that the presence / absence of the pathogen on a slide is independent of the presence / absence of the pathogen on other slides taken from the same specimen. $\endgroup$
    – jbowman
    Oct 17, 2013 at 15:28
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    $\begingroup$ That would be the probability of obtaining 5 false negatives in a row: $\endgroup$
    – baffled
    Oct 17, 2013 at 16:35

1 Answer 1


That would be the probability of obtaining a false negative in 5 slides:

(0.80)^5 = 0.32768

Ahhh, so in order to decrease the probability of false negatives below 1% you can do:

> x <- matrix(c(0), nrow=25)
> for(i in 1:25) x[i] = (0.8)^i
> x
 [1,] 0.800000000
 [2,] 0.640000000
 [3,] 0.512000000
 [4,] 0.409600000
 [5,] 0.327680000
 [6,] 0.262144000
 [7,] 0.209715200
 [8,] 0.167772160
 [9,] 0.134217728
 [10,] 0.107374182
 [11,] 0.085899346
 [12,] 0.068719477
 [13,] 0.054975581
 [14,] 0.043980465
 [15,] 0.035184372
 [16,] 0.028147498
 [17,] 0.022517998
 [18,] 0.018014399
 [19,] 0.014411519
 [20,] 0.011529215
 [21,] 0.009223372
 [22,] 0.007378698
 [23,] 0.005902958
 [24,] 0.004722366
 [25,] 0.003777893

And find that the false positive rate is less than 1% at i = 21.

Great! Thanks. I can't believe I didn't see that. I was trying all kinds of conditional probabilities and such for some reason. Keep it simple, stupid...

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    $\begingroup$ Yes, sometimes the easiest problems are the hardest! $\endgroup$
    – jbowman
    Oct 17, 2013 at 17:45

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