Minimum Sample Size for Confidence Interval

Date: 09/21/2000 at 11:31:54
From: Steve
Subject: Sample size for estimate for population proportion

Let's say I have a population of 400 people. I need to find out if a 
value for each person is 1 or 0. Sampling each person is very time-
consuming, however, so I don't want to have to sample any more than 
necessary.

I need to find out how many people I need to sample to say that the 
proportion correct (1 = correct) in the sample is representative of 
the proportion correct in the population. I am trying to make a case 
that if the randomly drawn sample I choose is done right, I am 
"whatever" amount confident that the entire population is correct at 
the same rate.

Thanks,
Steve

Date: 09/21/2000 at 15:55:56
From: Doctor Anthony
Subject: Re: Sample size for estimate for population proportion

Since we are talking about the means of samples we can by the Central 
Limit theorem use a normal distribution.

If p = proportion of 1's, then if we take a random sample of size n, 
and ps is the unbiased estimator of p we get (using the normal 
approximation to the binomial),

            ps - p
     z = -------------
         sqrt(ps.qs/n)

where qs = 1 - ps.

With 90% confidence limits we have:

                       ps-p
     Prob[-1.645 < ------------- < 1.645] = 0.90
                   sqrt(ps.qs/n)


     Prob[ps-1.645*sqrt(ps.qs/n) < p < ps+1.645*sqrt(ps.qs/n)] = 0.90

Now if we are to have a margin of error of not more than say 10% we 
require:

     ps + 1.645 sqrt(ps.qs/n) = ps + 0.1
so
          1.645 sqrt(ps.qs/n) = 0.1

since we don't know the value of ps, we must assume the worst case 
with ps.qs a maximum. That is:

     ps(1 - ps) = maximum

      ps - ps^2 = maximum

Differentiating:

     1 - 2.ps = 0
so
     ps = 1/2   and   qs = 1 - ps = 1/2

So we can put

     ps.qs = 0.5 x 0.5 = 0.25

In the worst case

     1.645*sqrt(0.25/n) = 0.1

squaring

     2.706 (0.25/n) = 0.01

                  n =  2.706 x 0.25/0.01

                    =  67.65

So take a sample of size 67.65 to get a 90% confidence limit with a 
maximum of 10% margin of error.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/