How could calculate sampling size for 95% confidence in a unique distribution? I made a big list of words that I claimed more than x% of them have dictation error. For showing my confidence of claim I need to show a sample(random) that prove it(because I can survey a sample list by observation). I want a 95% confidence, so my sampling size should be how big?
Problem is where my population is not normal distribution. I saw each word one time, so my distribution is unique.
 A: The formula to calculate sampling size is as follow :
$$n = \frac{t^{2}*p*(1-p)}{m^{2}}$$
with $t$ the level of confidence and a level of risk set
$p$ the probability of realisation of the event (ie the word being wrong)
and $m$ the margin of error.
$t$ can be found in a gaussian table if the number is high enough (by virtue of the Central Limit Theorem, and Law of Large Numbers), ie if that number is $>30$.
A: You can go in few directions, answering close but not similar questions


*

*Take your x% as the null hypothesis and see if it hold on your data set. The number of samples needed depends on the difference between your assumption and the behaviour on the data set. If you are way off, a small number of samples will be enough.

*Compute the confidence interval of the dictation error with confidence x%. After that check in your assumed x% falls in this interval.

*You can compute the minimal probability of dictation errors that will lead to the rate at least as high as you see in the data set with probability at least 95%.  

