p-value of a test statistic on a two-sided test
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
New contributor
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
New contributor
@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
2 days ago
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
2 days ago
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
2 days ago
could you explicitly add what likelihood was in the denominator?
– LinAlg
2 days ago
@LinAlg see above
– LucasMation
yesterday
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
New contributor
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
New contributor
New contributor
edited yesterday
New contributor
asked Nov 15 at 16:08
LucasMation
1012
1012
New contributor
New contributor
@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
2 days ago
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
2 days ago
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
2 days ago
could you explicitly add what likelihood was in the denominator?
– LinAlg
2 days ago
@LinAlg see above
– LucasMation
yesterday
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
2 days ago
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
2 days ago
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
2 days ago
could you explicitly add what likelihood was in the denominator?
– LinAlg
2 days ago
@LinAlg see above
– LucasMation
yesterday
@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
2 days ago
@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
2 days ago
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
2 days ago
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
2 days ago
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
2 days ago
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
2 days ago
could you explicitly add what likelihood was in the denominator?
– LinAlg
2 days ago
could you explicitly add what likelihood was in the denominator?
– LinAlg
2 days ago
@LinAlg see above
– LucasMation
yesterday
@LinAlg see above
– LucasMation
yesterday
add a comment |
1 Answer
1
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.
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.
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.
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.
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.
edited yesterday
answered Nov 15 at 18:14
LinAlg
7,5741520
7,5741520
add a comment |
add a comment |
LucasMation is a new contributor. Be nice, and check out our Code of Conduct.
LucasMation is a new contributor. Be nice, and check out our Code of Conduct.
LucasMation is a new contributor. Be nice, and check out our Code of Conduct.
LucasMation is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999881%2fp-value-of-a-test-statistic-on-a-two-sided-test%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
@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
2 days ago
By the way, I forgot to mention that -2ln(LR) is what has Chi^2 distribution, not LR itself. Will edit that now.
– LucasMation
2 days ago
@LinAlg, please incorporate my point avobe into your answer so I can mark it as a solution to the question.
– LucasMation
2 days ago
could you explicitly add what likelihood was in the denominator?
– LinAlg
2 days ago
@LinAlg see above
– LucasMation
yesterday