Let's say what you're asking for is the standard error of the estimation of the proportion defective, and that these can be represented as independent draws from a binomial distribution.

This standard error is $\sqrt{\frac{p(1-p)}{n}}$ where $p$ is the measured proportion defective. For 95 % one-sided confidence, we want to see a measurement at least 1.645 standard errors away from 2 %. Thus, the full equation we want to solve is

$$p + 1.645\sqrt{\frac{p(1-p)}{n}} = 0.02.$$

As you can see, the sample size required actually depends on what proportion defective you expect to find. Here are some examples:

- p = 0.5 %, n = 60
- p = 1.0 %, n = 268
- p = 1.5 %, n = 1600
- p = 1.8 %, n = 11958

As you can probably intuit, the closer the observed proportion is to 2 %, the more data you need to be really sure it's not greater than 2 %.

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Note that this assumes the total population is practically infinite. You haven't specified the size exactly, but once you're sampling a significant proportion of the entire population (If you're doing non-destructive testing, anyway) you get a bonus to the standard error that's roughly the equivalent of multiplying by (N-n)/N.

However, this would only kick in once you have a really huge sample anyway, and the asymptote is still inspecting the entire population, so I'm not sure how much that helps you.

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For more on this type of sampling for proportion defective, Deming wrote a lot about it in his books on sampling and statistics, as that was a big part of what he worked on.

(In particular, when doing destructive testing, depending on how lots are drawn, etc, you can have funny effects where discovering a large number of defective in the sample means there are *fewer* defective going to the customer.)