hypothesis testing to check for occurrence I am struggling to understand if I can make such a hypothesis or not and if so, which test is best to validate it.
I have a data with a size of 50 (quality of wine) I see that a sensory quality happening there (lets say a number is repeated few times). To make it more understandable, I have 50 signs , taste, color .... I see that taste is repeated few times in that 50 population. SO my hypothesis is if this number is by chance or just correct
so I want to test this hypothesis if I pick up the same size of the data for as many times as I want (100000) from a database with so many wine samples what is the chance to get to the number of taste that I got for my set of data . 
Which test is suitable to check for this hypothesis ?
 A: For the first question testing whether the counts for quality in your sample are all roughly equally frequent you can use a Chi-square test. The null hypothesis is independence between the categories.
For the second question testing whether these proportions will be similar if you repeat this procedure many times a permutation / bootstrap test can be used. You would bootstrap your sample many times, record the counts / frequencies, and then compare these with the values from your original sample.
Edit: here is an example in R using mtcars dataset. We will look at the gear column and treat it as a count.
> table(mtcars$gear)

 3  4  5 
15 12  5 

Chi-square test
> chisq.test(table(mtcars$gear))

    Chi-squared test for given probabilities

data:  table(mtcars$gear)
X-squared = 4.9375, df = 2, p-value = 0.08469

So we keep the null hypothesis. Now let's use bootstrapping on the X-squared statistic, repeating 1000 times.
> res=replicate(1000,{
>   tmp=sample(mtcars$gear,replace=T)
>   chisq.test(table(tmp))$statistic
> })

> summary(res)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.0625  3.8125  6.4375  6.7245  8.6875 25.0000

Now we compare the original X-square statistic of the original sample with the resulting vector of X-square values obtained through bootstrapping.
> mean(res>=4.9375)
[1] 0.649

So based on the bootstrap our results are quite likely under the null hypothesis.
