p-value of a test statistic on a two-sided test
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For coursework, I am doing a two-sided test ($H_0 beta = 0, H_a beta neq 0$). The test itself is a generalized likelihood ratio. The test statistic is the ratio LR:
$$LR=frac{L(beta=0)}{argmax_{beta in R}L(beta)}$$
Where L is the likelihood function.Then $-2ln(LR)$ follows a $chi_1^2$.
I am trying to calculate the p-value for a specific value of LR. Say I look for at a $chi_1^2$ table (or online calc) and find out that
$$ P(chi_1^2 geq -2ln(LR)) = alpha $$
Is $alpha$ my final p-value?
Or do I need to account for the fact that the test is two-sided and set the p-value $= 2*alpha$?
I think I should do the later but I am a bit unsure. Some extra intuition would help.
Edit: added the LR function.
statistics statistical-inference hypothesis-testing p-value
asked Nov 15 at 16:08
LucasMation
1012
add a comment |
up vote
0
down vote
favorite
For coursework, I am doing a two-sided test ($H_0 beta = 0, H_a beta neq 0$). The test itself is a generalized likelihood ratio. The test statistic is the ratio LR:
$$LR=frac{L(beta=0)}{argmax_{beta in R}L(beta)}$$
Where L is the likelihood function.Then $-2ln(LR)$ follows a $chi_1^2$.
I am trying to calculate the p-value for a specific value of LR. Say I look for at a $chi_1^2$ table (or online calc) and find out that
$$ P(chi_1^2 geq -2ln(LR)) = alpha $$
Is $alpha$ my final p-value?
Or do I need to account for the fact that the test is two-sided and set the p-value $= 2*alpha$?
I think I should do the later but I am a bit unsure. Some extra intuition would help.
Edit: added the LR function.
statistics statistical-inference hypothesis-testing p-value
asked Nov 15 at 16:08
LucasMation
1012
@LinAlg, tks. I realized that in general tests (of a transformation, standardization of X) we need to look at both sides. But in this case, we have a likelihood ratio of a constrained likelihood (in the numerator) divided by an unconstrained likelihood (in the denominator). Thus the smaller the value the greatest the evidence against H0.
– LucasMation
Nov 17 at 20:39
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
Nov 17 at 20:40
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
Nov 17 at 20:41
could you explicitly add what likelihood was in the denominator?
– LinAlg
Nov 17 at 21:05
@LinAlg see above
– LucasMation
Nov 19 at 2:48
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For coursework, I am doing a two-sided test ($H_0 beta = 0, H_a beta neq 0$). The test itself is a generalized likelihood ratio. The test statistic is the ratio LR:
$$LR=frac{L(beta=0)}{argmax_{beta in R}L(beta)}$$
Where L is the likelihood function.Then $-2ln(LR)$ follows a $chi_1^2$.
I am trying to calculate the p-value for a specific value of LR. Say I look for at a $chi_1^2$ table (or online calc) and find out that
$$ P(chi_1^2 geq -2ln(LR)) = alpha $$
Is $alpha$ my final p-value?
Or do I need to account for the fact that the test is two-sided and set the p-value $= 2*alpha$?
I think I should do the later but I am a bit unsure. Some extra intuition would help.
Edit: added the LR function.
statistics statistical-inference hypothesis-testing p-value
asked Nov 15 at 16:08
LucasMation
1012
For coursework, I am doing a two-sided test ($H_0 beta = 0, H_a beta neq 0$). The test itself is a generalized likelihood ratio. The test statistic is the ratio LR:
$$LR=frac{L(beta=0)}{argmax_{beta in R}L(beta)}$$
Where L is the likelihood function.Then $-2ln(LR)$ follows a $chi_1^2$.
I am trying to calculate the p-value for a specific value of LR. Say I look for at a $chi_1^2$ table (or online calc) and find out that
$$ P(chi_1^2 geq -2ln(LR)) = alpha $$
Is $alpha$ my final p-value?
Or do I need to account for the fact that the test is two-sided and set the p-value $= 2*alpha$?
I think I should do the later but I am a bit unsure. Some extra intuition would help.
Edit: added the LR function.
statistics statistical-inference hypothesis-testing p-value
statistics statistical-inference hypothesis-testing p-value
asked Nov 15 at 16:08
LucasMation
1012
asked Nov 15 at 16:08
LucasMation
1012
edited Nov 19 at 2:43
asked Nov 15 at 16:08
LucasMation
1012
asked Nov 15 at 16:08
LucasMation
1012
asked Nov 15 at 16:08
LucasMation
1012
1012
@LinAlg, tks. I realized that in general tests (of a transformation, standardization of X) we need to look at both sides. But in this case, we have a likelihood ratio of a constrained likelihood (in the numerator) divided by an unconstrained likelihood (in the denominator). Thus the smaller the value the greatest the evidence against H0.
– LucasMation
Nov 17 at 20:39
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
Nov 17 at 20:40
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
Nov 17 at 20:41
could you explicitly add what likelihood was in the denominator?
– LinAlg
Nov 17 at 21:05
@LinAlg see above
– LucasMation
Nov 19 at 2:48
add a comment |
@LinAlg, tks. I realized that in general tests (of a transformation, standardization of X) we need to look at both sides. But in this case, we have a likelihood ratio of a constrained likelihood (in the numerator) divided by an unconstrained likelihood (in the denominator). Thus the smaller the value the greatest the evidence against H0.
– LucasMation
Nov 17 at 20:39
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
Nov 17 at 20:40
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
Nov 17 at 20:41
could you explicitly add what likelihood was in the denominator?
– LinAlg
Nov 17 at 21:05
@LinAlg see above
– LucasMation
Nov 19 at 2:48
@LinAlg, tks. I realized that in general tests (of a transformation, standardization of X) we need to look at both sides. But in this case, we have a likelihood ratio of a constrained likelihood (in the numerator) divided by an unconstrained likelihood (in the denominator). Thus the smaller the value the greatest the evidence against H0.
– LucasMation
Nov 17 at 20:39
@LinAlg, tks. I realized that in general tests (of a transformation, standardization of X) we need to look at both sides. But in this case, we have a likelihood ratio of a constrained likelihood (in the numerator) divided by an unconstrained likelihood (in the denominator). Thus the smaller the value the greatest the evidence against H0.
– LucasMation
Nov 17 at 20:39
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
Nov 17 at 20:40
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
Nov 17 at 20:40
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
Nov 17 at 20:41
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
Nov 17 at 20:41
could you explicitly add what likelihood was in the denominator?
– LinAlg
Nov 17 at 21:05
could you explicitly add what likelihood was in the denominator?
– LinAlg
Nov 17 at 21:05
@LinAlg see above
– LucasMation
Nov 19 at 2:48
@LinAlg see above
– LucasMation
Nov 19 at 2:48
add a comment |
1 Answer
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1
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The $p$ value is related to the rejection region, so the answer depends on how you divide $alpha$ over both sides for the test. If you reject H0 when the test statistic is less than $F(alpha/2)$ or more than $F(1-alpha/2)$, then $2alpha$ is the right answer. However, you can use any rejection region $(-infty,a] cup [b,infty)$ as long as $a$ and $b$ satisfy $F^{-1}(b) - F^{-1}(a) = 1-alpha$. For skewed unimodal distributions you can impose $f(a)=f(b)$, which leads to a different answer.
In your example of a generalized likelihood ratio test, the rejection region is an interval $[b,infty)$, even though the test is two sided. So $alpha$ is the p-value.
answered Nov 15 at 18:14
LinAlg
7,6491520
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The $p$ value is related to the rejection region, so the answer depends on how you divide $alpha$ over both sides for the test. If you reject H0 when the test statistic is less than $F(alpha/2)$ or more than $F(1-alpha/2)$, then $2alpha$ is the right answer. However, you can use any rejection region $(-infty,a] cup [b,infty)$ as long as $a$ and $b$ satisfy $F^{-1}(b) - F^{-1}(a) = 1-alpha$. For skewed unimodal distributions you can impose $f(a)=f(b)$, which leads to a different answer.
In your example of a generalized likelihood ratio test, the rejection region is an interval $[b,infty)$, even though the test is two sided. So $alpha$ is the p-value.
answered Nov 15 at 18:14
LinAlg
7,6491520
add a comment |
up vote
1
down vote
The $p$ value is related to the rejection region, so the answer depends on how you divide $alpha$ over both sides for the test. If you reject H0 when the test statistic is less than $F(alpha/2)$ or more than $F(1-alpha/2)$, then $2alpha$ is the right answer. However, you can use any rejection region $(-infty,a] cup [b,infty)$ as long as $a$ and $b$ satisfy $F^{-1}(b) - F^{-1}(a) = 1-alpha$. For skewed unimodal distributions you can impose $f(a)=f(b)$, which leads to a different answer.
In your example of a generalized likelihood ratio test, the rejection region is an interval $[b,infty)$, even though the test is two sided. So $alpha$ is the p-value.
answered Nov 15 at 18:14
LinAlg
7,6491520
add a comment |
up vote
1
down vote
up vote
1
down vote
The $p$ value is related to the rejection region, so the answer depends on how you divide $alpha$ over both sides for the test. If you reject H0 when the test statistic is less than $F(alpha/2)$ or more than $F(1-alpha/2)$, then $2alpha$ is the right answer. However, you can use any rejection region $(-infty,a] cup [b,infty)$ as long as $a$ and $b$ satisfy $F^{-1}(b) - F^{-1}(a) = 1-alpha$. For skewed unimodal distributions you can impose $f(a)=f(b)$, which leads to a different answer.
In your example of a generalized likelihood ratio test, the rejection region is an interval $[b,infty)$, even though the test is two sided. So $alpha$ is the p-value.
answered Nov 15 at 18:14
LinAlg
7,6491520
The $p$ value is related to the rejection region, so the answer depends on how you divide $alpha$ over both sides for the test. If you reject H0 when the test statistic is less than $F(alpha/2)$ or more than $F(1-alpha/2)$, then $2alpha$ is the right answer. However, you can use any rejection region $(-infty,a] cup [b,infty)$ as long as $a$ and $b$ satisfy $F^{-1}(b) - F^{-1}(a) = 1-alpha$. For skewed unimodal distributions you can impose $f(a)=f(b)$, which leads to a different answer.
In your example of a generalized likelihood ratio test, the rejection region is an interval $[b,infty)$, even though the test is two sided. So $alpha$ is the p-value.
answered Nov 15 at 18:14
LinAlg
7,6491520
edited Nov 19 at 13:50
answered Nov 15 at 18:14
LinAlg
7,6491520
answered Nov 15 at 18:14
LinAlg
7,6491520
answered Nov 15 at 18:14
LinAlg
7,6491520
7,6491520
add a comment |
add a comment |
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@LinAlg, tks. I realized that in general tests (of a transformation, standardization of X) we need to look at both sides. But in this case, we have a likelihood ratio of a constrained likelihood (in the numerator) divided by an unconstrained likelihood (in the denominator). Thus the smaller the value the greatest the evidence against H0.
– LucasMation
Nov 17 at 20:39
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
Nov 17 at 20:40
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
Nov 17 at 20:41
could you explicitly add what likelihood was in the denominator?
– LinAlg
Nov 17 at 21:05
@LinAlg see above
– LucasMation
Nov 19 at 2:48