Kassandra Ross

2022-06-20

From a collection of objects numbered $\{1,2,....,K\}$ objects are picked and replaced. We want to test ${H}_{0}:K=100000$ against ${H}_{1}<100000$, with the highest ranking number $M$ of our sample as test statistic. We find for our realisation for $M$ the value $81115$.

What is the $P$ value?

The correct answer is: $0.015$

I know that the definition of the $p$-value is:

The p-value is the probability of getting the observed value of the test static or a value with even greater evidence against ${H}_{0}$, if the hypothesis is actually true

or in formula form $P(T\ge t)$

I have the following questions:

What are $T$ and $t$?

I think that distribution is uniform, but how do I calculate the $p$-value?

What is the $P$ value?

The correct answer is: $0.015$

I know that the definition of the $p$-value is:

The p-value is the probability of getting the observed value of the test static or a value with even greater evidence against ${H}_{0}$, if the hypothesis is actually true

or in formula form $P(T\ge t)$

I have the following questions:

What are $T$ and $t$?

I think that distribution is uniform, but how do I calculate the $p$-value?

mar1nerne

Beginner2022-06-21Added 20 answers

In a uniform distribution defined between the values $a$ and $b$, the cdf for $k\phantom{\rule{thinmathspace}{0ex}}\u03f5\phantom{\rule{thinmathspace}{0ex}}[a,b]$ is $(k-a)/(b-a)$.

Dale Tate

Beginner2022-06-22Added 5 answers

The probability that one randomly chosen object is at most $81115$ is $q=81115/100000$. Since the draws are independent (because they are replaced after each pick), the probability that all $20$ choices are less than or equal to 81115 is $p={q}^{20}=\mathrm{0.015.}$.