# Sample Size Calculation for One Sided Hypothesis Testing

I have 1100 objects to inspect whether they meet the standard or not (yes or no question), of which 100 have been inspected already and 99% of them passed the test.

Due to resource constraints, we can't inspect all the remaining 1000 objects and hence need to take a random sample of them. How can I calculate the minimum sample size required to test the hypothesis that at least 98% of them will pass the test.

Note: 95% Significance Level & 99% Confidence Level

The minimum number of additional samples you need to test is $$135$$.

In your problem you have a Bernoulli random variable $$X$$ with an unknown parameter $$p$$. Your current estimate for $$p$$ is that it is $$.99$$, of course, because of uncertainty we are not $$95\%$$ sure that this estimate is correct.

If we assume a uniform proir on $$p$$, then the posterior distribution for $$p$$, after gathering data of $$99$$ successes and $$1$$ failure is given by the function, $$f(p) = \frac{101!}{(99!\times 1!)}p^{99}(1-p)^1$$

The minimum number of additional samples $$n$$ will occur when they are all successes (otherwise you will need to keep on gathering more data). If we gather $$n$$ more samples, and if they are all successes, then the posterior distribution changes to, $$f(p) = \frac{(101+n)!}{(99+n)!\times 1!}p^{99+n}(1-p)^1 = (101+n)(100+n)p^{99+n}(1-p)$$

You want to be $$95\%$$ sure that the true value of $$p$$ is at least $$.98$$.

Therefore, you need to find the minimum $$n$$ such that, $$\int_{.98}^1 (101+n)(100+n)p^{99+n}(1-p) ~ dp \geq .95$$

Using WolframAlpha,

integrate (101+n)(100+n) x^(99+n) (1-x) dx from .98 to 1 where n = 134


We find this integral computes to $$.949$$, however at $$n=135$$ it will exceed $$.95$$.

• Note the Bernoulli variables are not independent though - you in fact have a hypergeometric rather than a binomial sampling scheme. Mar 13 at 17:16
• @Scortchi-ReinstateMonica You are correct. My answer is wrong, however, I still think it is helpful to keep it posted since it does answer what happens when the samples are chosen with replacement. Mar 14 at 1:49
• Not precisely, as $p$ can only take values in $0, \frac{1}{1100}, \ldots, \frac{1099}{1100}, 1$. Your method might be best billed as an approximation for when the sample is a small fraction of a large population. Mar 14 at 7:52