Mixing
p-value of the following test
$begingroup$
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
asked Jan 17 at 20:31
qcc101qcc101
627213
$endgroup$
add a comment |
$begingroup$
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
asked Jan 17 at 20:31
qcc101qcc101
627213
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 17 at 20:31
qcc101qcc101
627213
$endgroup$
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
asked Jan 17 at 20:31
qcc101qcc101
627213
asked Jan 17 at 20:31
qcc101qcc101
627213
asked Jan 17 at 20:31
qcc101qcc101
627213
asked Jan 17 at 20:31
qcc101qcc101
627213
asked Jan 17 at 20:31
qcc101qcc101
627213
627213
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.$$
answered Feb 13 at 23:29
BenBen
1,570215
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:
$$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.$$
answered Feb 13 at 23:29
BenBen
1,570215
$endgroup$
add a comment |
$begingroup$
Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:
$$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.$$
answered Feb 13 at 23:29
BenBen
1,570215
$endgroup$
add a comment |
$begingroup$
Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:
$$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.$$
answered Feb 13 at 23:29
BenBen
1,570215
$endgroup$
Letting $K=sum_i mathbb{I}(X_i = 1)$ be the number of successes, you have $K sim text{Bin}(n,p)$. Thus, your p-value is:
$$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.$$
answered Feb 13 at 23:29
BenBen
1,570215
answered Feb 13 at 23:29
BenBen
1,570215
answered Feb 13 at 23:29
BenBen
1,570215
answered Feb 13 at 23:29
BenBen
1,570215
1,570215
add a comment |
add a comment |
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$begingroup$
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