X 1 . . . X n ~ Bernoulli(p) independent. I also have the following test hypothesis: H 0 : p = p 0 and H 1 : p &gt; 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?

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X    1    .  .  .      X    n   ~ Bernoulli(p) independent. I also have the following test hypothesis:       H    0    :  p  =      p    0   and       H    1    :  p  &amp;gt;      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 &quot;number of succession is very high&quot;. How do I work around this?

aqueritztv · Accepted Answer

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.

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

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