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.










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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
    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















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.










share|cite|improve this question









New contributor




LucasMation is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • @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













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.










share|cite|improve this question









New contributor




LucasMation is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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






share|cite|improve this question









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LucasMation is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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LucasMation is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited yesterday





















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asked Nov 15 at 16:08









LucasMation

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1012




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New contributor





LucasMation is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






LucasMation is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • @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










  • 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










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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.






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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.






    share|cite|improve this answer



























      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.






      share|cite|improve this answer

























        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.






        share|cite|improve this answer














        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered Nov 15 at 18:14









        LinAlg

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