# Smallest sample size for true proportion estimation (Binomial distribution with 1 found defective unit)

Suppose that we allow only 1 defective unit in the control sample. What is the smallest sample it should be in order to conclude that my true defective proportion of the population is smaller than 0.5% (with 95% confidence level).

I have tried to solve this problem using the binom.confint function in R. Since my defective rate is too small (near 0), I have calculate the confidence interval of the true proportion with the Wilson interval. Then I look for the sample size who gives the nearest upper limit from the 0.5% value. The sample size that I found is 1130 since the upper limit of the IC is 0.4999%. So I can conclude that if I found 1 defective unit in a sample of 1130 units, my true defective proportion should be lower than 0.5% (with 5% error)

binom.confint(x = 1, n = 1130)


Is this a good approach? Do you have other suggestion?

Thank you

This is a very good approximate approach. For exact inference you can reference the CDF of a binomial distribution with $$p=0.005$$ and identify the $$n$$ such that the probability of observing $$1$$ or fewer events is $$2.5\%$$. This approach yields an $$n$$ of 1,113. The resulting two-sided equal-tailed $$95\%$$ confidence interval from inverting the binomial CDF would be would be $$(0.00002275, 0.00499)$$.
It sounds like you will want a one-sided confidence interval since you are interested in testing whether the proportion is less than 0.5%. The binom.confint function uses a two-sided interval.
The upper limit for an exact one-sided 95% interval is equal to the 95% quantile of a $$\text{Beta}(x + 1, n - x)$$ random variable, so you need to find $$n$$ such that qbeta(0.95, 2, n-1) $$\leq 0.005$$. This is the case when $$n \geq 947$$.