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
 A: 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 $.95$.
