cancel
Showing results for 
Show  only  | Search instead for 
Did you mean: 
  • Learn how to build custom Python data connectors and further customize JMP’s Data Connector Framework with the Python Data Connector Demo, available now in the JMP Marketplace!
  • See how to use Accelerated Life Testing (ALT) to evaluate reliability. Register for June 5 webinar, 2pm US Eastern Time.

Discussions

Solve problems, and share tips and tricks with other JMP users.
%3CLINGO-SUB%20id%3D%22lingo-sub-307974%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ER%C3%A9sum%C3%A9%20des%20effets%20et%20le%20Logworth%20avec%20une%20valeur%20p%20de%200%26nbsp%3B%3F%3F%3F%3F%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-307974%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3E%3CP%3ESalut%20quelqu'un%20peut-il%20m'aider%3F%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3EJ'ex%C3%A9cute%20une%20r%C3%A9gression%20logistique%20avec%20deux%20variables%20x%20ci-dessous%20montre%20mon%20r%C3%A9sum%C3%A9%20d'effet%20avec%20le%20logworth.%20Maintenant%2C%20je%20comprends%20que%20le%20logworth%20prend%20essentiellement%20le%20-log(p-value)%20pour%20chaque%20variable%20x%20et%20cela%20fonctionne%20pour%20la%20p-value%200.00219.%20Ce%20qui%20m'embrouille%2C%20c'est%20le%20logworth%20de%20la%20valeur%20p%200%2C0000%2C%20comment%20obtient-il%2056%2C125%26nbsp%3B%3F%20De%20plus%2C%20dans%20d'autres%20mod%C3%A8les%20de%20r%C3%A9gression%2C%20j'ai%20rencontr%C3%A9%20diff%C3%A9rents%20logworths%20pour%20la%20valeur%20p%20de%200.%20Quelqu'un%20peut-il%20m'expliquer%20s'il%20vous%20pla%C3%AEt%20afin%20que%20je%20puisse%20continuer%20ma%20vie%3F%3CSPAN%20class%3D%22lia-inline-image-display-wrapper%20lia-image-align-inline%22%20image-alt%3D%22jMOP.PNG%22%20style%3D%22width%3A%20400px%3B%22%3E%3CSPAN%20class%3D%22lia-inline-image-display-wrapper%22%20image-alt%3D%22jMOP.PNG%22%20style%3D%22width%3A%20400px%3B%22%3E%3Cspan%20class%3D%22lia-inline-image-display-wrapper%22%20image-alt%3D%22jMOP.PNG%22%20style%3D%22width%3A%20400px%3B%22%3E%3Cimg%20src%3D%22https%3A%2F%2Fcommunity.jmp.com%2Ft5%2Fimage%2Fserverpage%2Fimage-id%2F26780i1759E01849E47FB5%2Fimage-size%2Fmedium%3Fv%3Dv2%26amp%3Bpx%3D400%22%20role%3D%22button%22%20title%3D%22jMOP.PNG%22%20alt%3D%22jMOP.PNG%22%20%2F%3E%3C%2Fspan%3E%3C%2FSPAN%3E%3C%2FSPAN%3E%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3EMerci!!!%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3ERachel%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3E%23statistiques%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-LABS%20id%3D%22lingo-labs-307974%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3E%3CLINGO-LABEL%3E%C3%89tudes%20de%20consommation%20et%20de%20march%C3%A9%3C%2FLINGO-LABEL%3E%3C%2FLINGO-LABS%3E%3CLINGO-SUB%20id%3D%22lingo-sub-472249%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ERe%26nbsp%3B%3A%20R%C3%A9sum%C3%A9%20des%20effets%20et%20le%20Logworth%20avec%20une%20valeur%20p%20de%200%26nbsp%3B%3F%3F%3F%3F%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-472249%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3E%3CP%3Evous%20avez%20probablement%20utilis%C3%A9%20%22e%22%20dans%20la%20calculatrice%20qui%20est%202%2C71828.%20Essayez%2010%5E-57%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-308039%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ERe%26nbsp%3B%3A%20R%C3%A9sum%C3%A9%20des%20effets%20et%20le%20Logworth%20avec%20une%20valeur%20de%20p%20de%200%26nbsp%3B%3F%3F%3F%3F%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-308039%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3EEh%20bien%20merci%2C%20je%20l'ai%20fait%20avec%20une%20calculatrice%20et%20c'est%20l%C3%A0%20que%20j'ai%20obtenu%20la%20r%C3%A9ponse%2049.%20Merci%20de%20l'avoir%20fait%20!%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-308038%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ERe%26nbsp%3B%3A%20R%C3%A9sum%C3%A9%20des%20effets%20et%20le%20Logworth%20avec%20une%20valeur%20p%20de%200%26nbsp%3B%3F%3F%3F%3F%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-308038%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ERevoyons%20%3A%20Log10%20(7.5e-57)%20%3D%20Log10%20(7.5)%20%2B%20Log10%20(e-57)%20%3D%20Log10%20(7.5)%20-%2057%20%3D%200.875%20-%2057%20%3D%20-56.125%3CBR%20%2F%3E%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-308037%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ERe%26nbsp%3B%3A%20R%C3%A9sum%C3%A9%20des%20effets%20et%20le%20Logworth%20avec%20une%20valeur%20p%20de%200%26nbsp%3B%3F%3F%3F%3F%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-308037%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3E%3CP%3E-log10%20renvoie%2049%2C8785%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-308036%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ERe%26nbsp%3B%3A%20R%C3%A9sum%C3%A9%20des%20effets%20et%20le%20Logworth%20avec%20une%20valeur%20p%20de%200%26nbsp%3B%3F%3F%3F%3F%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-308036%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3E%C3%8Ates-vous%20s%C3%BBr%20d'avoir%20utilis%C3%A9%20-Log10%26nbsp%3B%3F%3CBR%20%2F%3E%20Meilleur%2C%3CBR%20%2F%3E%20TS%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-307997%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ERe%26nbsp%3B%3A%20R%C3%A9sum%C3%A9%20des%20effets%20et%20le%20Logworth%20avec%20une%20valeur%20p%20de%200%26nbsp%3B%3F%3F%3F%3F%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-307997%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3E%3CP%3E%3CA%20href%3D%22https%3A%2F%2Fcommunity.jmp.com%2Ft5%2Fuser%2Fviewprofilepage%2Fuser-id%2F2687%22%20target%3D%22_blank%22%3E%40txnelson%3C%2FA%3E%20eh%20bien%20merci%20parce%20que%20je%20ne%20savais%20pas%20que%20je%20pouvais%20faire%20%C3%A7a.%20Cela%20%C3%A9tant%20dit%2C%20la%20valeur%20est%20de%207%2C5e-57%20mais%20quand%20je%20prends%20le%20-log%20de%20cela%2C%20cela%20me%20donne%3CSPAN%3E%2049.8784920133.%20Je%20suis%20donc%20plus%20proche%20mais%20toujours%20pas%20au%20bon%20endroit.%3C%2FSPAN%3E%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-307995%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3ERe%26nbsp%3B%3A%20R%C3%A9sum%C3%A9%20des%20effets%20et%20le%20Logworth%20avec%20une%20valeur%20de%20p%20de%200%26nbsp%3B%3F%3F%3F%3F%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-307995%22%20slang%3D%22en-US%22%20mode%3D%22NONE%22%3E%3CP%3Eessayez%20de%20double-cliquer%20sur%20la%20valeur%20p%20et%20modifiez%20le%20format%20pour%20afficher%20plus%20de%20d%C3%A9cimales%20pour%20voir%20quelle%20est%20la%20valeur%20p%20r%C3%A9elle.%3C%2FP%3E%3C%2FLINGO-BODY%3E
Choose Language Hide Translation Bar
rlw268
Level III

Effect Summary and the Logworth with p-value of 0????

Hi can someone help me?

 

I am running logistic regression with two x variables below shows my effect summary with the logworth. Now I understand the logworth basically takes the -log(p-value) for each x variable and that works for the p-value 0.00219. What i am confused about is the logworth of for the p-value 0.0000, how is it getting 56.125? Moreover in other regression models I've run into different logworths for p-value  of 0. Can someone please explain this to me so I can move on with my life?jMOP.PNG

 

Thank you!!!

 

Rachel

 

#statistics

3 ACCEPTED SOLUTIONS

Accepted Solutions
txnelson
Super User

Re: Effect Summary and the Logworth with p-value of 0????

try double clicking on the p-value and change the format to show more decimal places to see what the actual p-value is.

Jim

View solution in original post

Thierry_S
Super User

Re: Effect Summary and the Logworth with p-value of 0????

Are you sure you used -Log10?
Best,
TS
Thierry R. Sornasse

View solution in original post

Thierry_S
Super User

Re: Effect Summary and the Logworth with p-value of 0????

Let's review: Log10 (7.5e-57) = Log10 (7.5) + Log10 (e-57) = Log10 (7.5) - 57 = 0.875 - 57 = -56.125
Thierry R. Sornasse

View solution in original post

7 REPLIES 7
txnelson
Super User

Re: Effect Summary and the Logworth with p-value of 0????

try double clicking on the p-value and change the format to show more decimal places to see what the actual p-value is.

Jim
rlw268
Level III

Re: Effect Summary and the Logworth with p-value of 0????

@txnelson well thank you cause i didn't realize I could do that. With that being said the value is 7.5e-57 but when I take the -log of that it gives me 49.8784920133. So I'm closer but still not spot on. 

Thierry_S
Super User

Re: Effect Summary and the Logworth with p-value of 0????

Are you sure you used -Log10?
Best,
TS
Thierry R. Sornasse
rlw268
Level III

Re: Effect Summary and the Logworth with p-value of 0????

-log10 returns 49.8785

Thierry_S
Super User

Re: Effect Summary and the Logworth with p-value of 0????

Let's review: Log10 (7.5e-57) = Log10 (7.5) + Log10 (e-57) = Log10 (7.5) - 57 = 0.875 - 57 = -56.125
Thierry R. Sornasse
rlw268
Level III

Re: Effect Summary and the Logworth with p-value of 0????

Well thank you, I did it with a calculator and thats where I got the 49 answer. Thank you for doing it out!
Metin_Tueluemen
Level III

Re: Effect Summary and the Logworth with p-value of 0????

probably you used "e" in calculator which is 2,71828. Try 10^-57

Recommended Articles