X 1 </msub> . . . X n </msub> ~ Bernoulli(p) independe

Case Nixon

Case Nixon

Answered question

2022-05-22

X 1 . . . X n ~ Bernoulli(p) independent. I also have the following test hypothesis: H 0 : p = p 0 and H 1 : p > p 0 . I define my test as follows:
T : I refuse H 0 in favour of P 1 if the number of successes is very high.
Now suppose p 0 = 0.6 and n = 20, we find that we have 18 successes.
Find the p-value of the test.
Now this is what I have done:
p v a l u e = P ( x i 18 ) = 0.0005240494 However confronting it with the command:
b i n o m . t e s t ( 18 , 20 , p = .6 , a l t e r n a t i v e = g r e a t e r )
It is not right (by a factor of 10). My problem is that I am not able to correctly asses what the "number of succession is very high". How do I work around this?

Answer & Explanation

aqueritztv

aqueritztv

Beginner2022-05-23Added 10 answers

Letting K = i I ( X i = 1 ) be the number of successes, you have K Bin ( n , p ). Thus, your p-value is:
p p ( k ) P ( K k | H 0 ) = P ( K k | K Bin ( n , p 0 ) ) = r = k n Bin ( r | n , p 0 ) .
With p 0 = 0.6, n = 20 and k = 18 you get the p-value:
p = r = 18 20 Bin ( r | 20 , 0.6 ) = 0.003611472.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?