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Estimate of percent of an infected species is accurate to within $\pm 0.04%$ with $95\%$ CI

Between $15\%$ and $35\%$ of the population are infected

Size of the population is between $1100$ and $2300$

Calculate the sample size.

My calculations: For $15\%$ , $n = 306$ For $35\%$, $n = 546$

Since population is between $1100$ and $2300$ , finite population correction is required.

How can I determine the sample size adjusted by finite population correction?

Do I have to take the weighted average of the smaller and greater $n$? Or should I simply use the larger $n$?

Tree
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    I do not think there is a clearly best answer. If we are going to be very conservative, we will use a worst case estimate for variance. So use $35%$ in that part of the calculation. And the small population correction shrinks the variance most when population is small, so worst case uses finite population correction for $N=2300$. – André Nicolas Sep 21 '15 at 17:05
  • Since finite population is a factor, what can I do with it to find the sample size? Should I multiply the square root of the variance by 1.96? – Tree Sep 21 '15 at 22:28
  • I will give a very brief answer that I don't really believe in. – André Nicolas Sep 21 '15 at 23:01

2 Answers2

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n is the sample size, not the population or something else. You do not need to know the true porpotion of the infected people. The equation for the deviation is $$\large{z_{0.975}} \normalsize{\cdot \sqrt{\frac{0.2 \cdot 0.8}{n} }=0.04}$$

$z_{0.975}$ is the value for z, where the cdf of the standard normal distribution is equal to $0.975$

Remark: $0.25$, the mean of $0.35$ and $0.15$, is the best estimation, since you have no other information.

callculus42
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The issue is what we should use for the variance. The variance for any individual trial is $p(1-p)$, where $p$ is the probability of success. What should we use for $p$? The conservative choice is $p=0.35$. There is a very informal case for $0.25$. This part depends to an uncomfortable degree on the preferences of your instructor.

If $N$ is the population size, the small population version of the variance is $$\frac{p(1-p)}{n}\cdot \frac{N-n}{N-1}.$$ So we end up wanting $$1.96\sqrt{\frac{p(1-p)}{n}}\sqrt{\frac{N-n}{N-1}}\approx 0.04.$$ Square both sides and manipulate a little. We end up with a linear equation for $n$. Solve. It is not clear what we should use for $N$. The conservative approach is to use $2300$, since that gives the highest value of variance. But again, your instructor may have different preferences.

André Nicolas
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  • I found the same solution by using n = n0N / (n0)+(N-1) – Tree Sep 22 '15 at 01:13
  • You were given a "canned" formula for $n$ that comes from the analysis I made. That uses the somewhat more basic adjusted formula for variance that I quoted in the answer. – André Nicolas Sep 22 '15 at 01:22