p-value of the following test












0












$begingroup$


I have the following problem:



I have $X_1 ... X_n$ ~ 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.



I want to find the p-value of the test.



Now this is what I have done:



$p-value = mathbb{P}(sum x_i geq 18)$ = 0.0005240494
However confronting it with the command: 



binom.test(18,20,p=.6,alternative=greater)


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?






probability statistics hypothesis-testing p-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the confidence level and significance level, respectively? Maybe 0.95 and 1-0.95=0.05?
    $endgroup$
    – callculus
    Jan 17 at 21:19








  • 1




    $begingroup$
    I don't have that information. I don't think is needed as the purpose of the p-value is to work without one, from my understanding.
    $endgroup$
    – qcc101
    Jan 17 at 21:56






  • 1




    $begingroup$
    $mathbb{P}(sum X_i geq 18)=0.0036$
    $endgroup$
    – d.k.o.
    Jan 17 at 23:27












  • $begingroup$
    @d.k.o. I see, maybe I messed up my calculation. Still, is this the right way to do it?
    $endgroup$
    – qcc101
    Jan 18 at 9:35












  • $begingroup$
    Yes this is the correct way to do it
    $endgroup$
    – Alex
    Jan 22 at 7:12
















0












$begingroup$


I have the following problem:



I have $X_1 ... X_n$ ~ 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.



I want to find the p-value of the test.



Now this is what I have done:



$p-value = mathbb{P}(sum x_i geq 18)$ = 0.0005240494
However confronting it with the command: 



binom.test(18,20,p=.6,alternative=greater)


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?






probability statistics hypothesis-testing p-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the confidence level and significance level, respectively? Maybe 0.95 and 1-0.95=0.05?
    $endgroup$
    – callculus
    Jan 17 at 21:19








  • 1




    $begingroup$
    I don't have that information. I don't think is needed as the purpose of the p-value is to work without one, from my understanding.
    $endgroup$
    – qcc101
    Jan 17 at 21:56






  • 1




    $begingroup$
    $mathbb{P}(sum X_i geq 18)=0.0036$
    $endgroup$
    – d.k.o.
    Jan 17 at 23:27












  • $begingroup$
    @d.k.o. I see, maybe I messed up my calculation. Still, is this the right way to do it?
    $endgroup$
    – qcc101
    Jan 18 at 9:35












  • $begingroup$
    Yes this is the correct way to do it
    $endgroup$
    – Alex
    Jan 22 at 7:12














0












0








0





$begingroup$


I have the following problem:



I have $X_1 ... X_n$ ~ 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.



I want to find the p-value of the test.



Now this is what I have done:



$p-value = mathbb{P}(sum x_i geq 18)$ = 0.0005240494
However confronting it with the command: 



binom.test(18,20,p=.6,alternative=greater)


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?






probability statistics hypothesis-testing p-value







share|cite|improve this question









$endgroup$




I have the following problem:



I have $X_1 ... X_n$ ~ 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.



I want to find the p-value of the test.



Now this is what I have done:



$p-value = mathbb{P}(sum x_i geq 18)$ = 0.0005240494
However confronting it with the command: 



binom.test(18,20,p=.6,alternative=greater)


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?






probability statistics hypothesis-testing p-value




probability statistics hypothesis-testing p-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 17 at 20:31









qcc101qcc101

629213




629213












  • $begingroup$
    What is the confidence level and significance level, respectively? Maybe 0.95 and 1-0.95=0.05?
    $endgroup$
    – callculus
    Jan 17 at 21:19








  • 1




    $begingroup$
    I don't have that information. I don't think is needed as the purpose of the p-value is to work without one, from my understanding.
    $endgroup$
    – qcc101
    Jan 17 at 21:56






  • 1




    $begingroup$
    $mathbb{P}(sum X_i geq 18)=0.0036$
    $endgroup$
    – d.k.o.
    Jan 17 at 23:27












  • $begingroup$
    @d.k.o. I see, maybe I messed up my calculation. Still, is this the right way to do it?
    $endgroup$
    – qcc101
    Jan 18 at 9:35












  • $begingroup$
    Yes this is the correct way to do it
    $endgroup$
    – Alex
    Jan 22 at 7:12


















  • $begingroup$
    What is the confidence level and significance level, respectively? Maybe 0.95 and 1-0.95=0.05?
    $endgroup$
    – callculus
    Jan 17 at 21:19








  • 1




    $begingroup$
    I don't have that information. I don't think is needed as the purpose of the p-value is to work without one, from my understanding.
    $endgroup$
    – qcc101
    Jan 17 at 21:56






  • 1




    $begingroup$
    $mathbb{P}(sum X_i geq 18)=0.0036$
    $endgroup$
    – d.k.o.
    Jan 17 at 23:27












  • $begingroup$
    @d.k.o. I see, maybe I messed up my calculation. Still, is this the right way to do it?
    $endgroup$
    – qcc101
    Jan 18 at 9:35












  • $begingroup$
    Yes this is the correct way to do it
    $endgroup$
    – Alex
    Jan 22 at 7:12
















$begingroup$
What is the confidence level and significance level, respectively? Maybe 0.95 and 1-0.95=0.05?
$endgroup$
– callculus
Jan 17 at 21:19






$begingroup$
What is the confidence level and significance level, respectively? Maybe 0.95 and 1-0.95=0.05?
$endgroup$
– callculus
Jan 17 at 21:19






1




1




$begingroup$
I don't have that information. I don't think is needed as the purpose of the p-value is to work without one, from my understanding.
$endgroup$
– qcc101
Jan 17 at 21:56




$begingroup$
I don't have that information. I don't think is needed as the purpose of the p-value is to work without one, from my understanding.
$endgroup$
– qcc101
Jan 17 at 21:56




1




1




$begingroup$
$mathbb{P}(sum X_i geq 18)=0.0036$
$endgroup$
– d.k.o.
Jan 17 at 23:27






$begingroup$
$mathbb{P}(sum X_i geq 18)=0.0036$
$endgroup$
– d.k.o.
Jan 17 at 23:27














$begingroup$
@d.k.o. I see, maybe I messed up my calculation. Still, is this the right way to do it?
$endgroup$
– qcc101
Jan 18 at 9:35






$begingroup$
@d.k.o. I see, maybe I messed up my calculation. Still, is this the right way to do it?
$endgroup$
– qcc101
Jan 18 at 9:35














$begingroup$
Yes this is the correct way to do it
$endgroup$
– Alex
Jan 22 at 7:12




$begingroup$
Yes this is the correct way to do it
$endgroup$
– Alex
Jan 22 at 7:12










1 Answer
1






active

oldest

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0












$begingroup$

Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:



$$begin{equation} begin{aligned}
p equiv p(k)
&equiv mathbb{P}( K geqslant k | H_0) \[6pt]
&= mathbb{P}( K geqslant k | K sim text{Bin}(n,p_0)) \[6pt]
&= sum_{r=k}^n text{Bin}(r|n,p_0). \[6pt]
end{aligned} end{equation}$$



With $p_0=0.6$, $n=20$ and $k=18$ you get the p-value:



$$p = sum_{r=18}^{20} text{Bin}(r|20, 0.6)
= 0.003611472.$$






share|cite|improve this answer









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    oldest

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    0












    $begingroup$

    Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:



    $$begin{equation} begin{aligned}
    p equiv p(k)
    &equiv mathbb{P}( K geqslant k | H_0) \[6pt]
    &= mathbb{P}( K geqslant k | K sim text{Bin}(n,p_0)) \[6pt]
    &= sum_{r=k}^n text{Bin}(r|n,p_0). \[6pt]
    end{aligned} end{equation}$$



    With $p_0=0.6$, $n=20$ and $k=18$ you get the p-value:



    $$p = sum_{r=18}^{20} text{Bin}(r|20, 0.6)
    = 0.003611472.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:



      $$begin{equation} begin{aligned}
      p equiv p(k)
      &equiv mathbb{P}( K geqslant k | H_0) \[6pt]
      &= mathbb{P}( K geqslant k | K sim text{Bin}(n,p_0)) \[6pt]
      &= sum_{r=k}^n text{Bin}(r|n,p_0). \[6pt]
      end{aligned} end{equation}$$



      With $p_0=0.6$, $n=20$ and $k=18$ you get the p-value:



      $$p = sum_{r=18}^{20} text{Bin}(r|20, 0.6)
      = 0.003611472.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:



        $$begin{equation} begin{aligned}
        p equiv p(k)
        &equiv mathbb{P}( K geqslant k | H_0) \[6pt]
        &= mathbb{P}( K geqslant k | K sim text{Bin}(n,p_0)) \[6pt]
        &= sum_{r=k}^n text{Bin}(r|n,p_0). \[6pt]
        end{aligned} end{equation}$$



        With $p_0=0.6$, $n=20$ and $k=18$ you get the p-value:



        $$p = sum_{r=18}^{20} text{Bin}(r|20, 0.6)
        = 0.003611472.$$






        share|cite|improve this answer









        $endgroup$



        Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:



        $$begin{equation} begin{aligned}
        p equiv p(k)
        &equiv mathbb{P}( K geqslant k | H_0) \[6pt]
        &= mathbb{P}( K geqslant k | K sim text{Bin}(n,p_0)) \[6pt]
        &= sum_{r=k}^n text{Bin}(r|n,p_0). \[6pt]
        end{aligned} end{equation}$$



        With $p_0=0.6$, $n=20$ and $k=18$ you get the p-value:



        $$p = sum_{r=18}^{20} text{Bin}(r|20, 0.6)
        = 0.003611472.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 13 at 23:29









        BenBen

        1,900215




        1,900215






























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