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How to add parabola equation on JMP(Ver.12)?

MissPN

New Contributor

Joined:

Sep 17, 2017

Dear JMP community members,

 

Could you help to explain which how to plot parabola on JMP while using below equation?

Ex. Based on formular -->> y = 0.7471x^2 + 444.96x - 41913 (:y=0.7471(x-166.2148)^2+11405.54)

Note: Excel plot as in the attachment

 

Thank you in advance

Best regards,

PN.

4 ACCEPTED SOLUTIONS

Accepted Solutions
ian_jmp

Staff

Joined:

Jun 23, 2011

Solution

I made this JMP table by reading in your .XLSX (using 'File > Open'). I then used 'Graph > Graph Builder' to look at the data, then 'Analyze > Fit Y By X' to fit the parabola. You can run the scripts saved to the table to reproduce these. From this fit, I saved the predicted values to make the new column 'Predicted Y', and if you click on the '+' sign next to this column in the table columns panel you can inspect the formula of the parabola that was fit. If you need more help with the steps just ask again.

 

However, having seen the data, I am left wondering why you think a parabola is a useful model to fit? I guess the answer depends on why you are interested in this data, and what you intend to do with the results of any analysis.

Craige_Hales

Staff

Joined:

Mar 21, 2013

Solution

Ian's answer is probably the one you want. Here's another answer, using Help->Scripting Index to look up the YFunction and insert your function. I scrolled the axes until I got a nice picture.

Y Function plotting a parabolaY Function plotting a parabola

Craige
MissPN

New Contributor

Joined:

Sep 17, 2017

Solution

 Hi ian_jmp,

 

Many Thanks for your advicing. Following your question, based on my data from 2 groups(Pass Hd & Fail Hd).

I would like to find the equation that could be cover on fail Hd while impacting Pass Hd as less than as we can (Please refer on Pass+Fail.xlsx).So, I think Parabola model could be useful to analysis.

Based on your JMP file as attached, may I ask about column "predicted y" ? from the formula, how to calculate Parabola fit model?? It's the better than formula as I provided? 

 

Appriciate for your help.

Best Regards,

Miss PN.

 

ian_jmp

Staff

Joined:

Jun 23, 2011

Solution

With the 'pass-fail' data you have provided, then this becomes a 'classification problem' in which you would like to find the best separation between the two groups. There are many ways that you could do this in JMP. Using your data I used Graph Builder to get:

 

Screen Shot 2017-09-19 at 11.09.06.png

I used your 'No. Hd' variable in the 'Wrap' role. I assume this is the number of heads in the drive. This might suggest that a linear boundary could work OK, but (although I'm sure you have good reason for looking at 'A' and 'B') it might be that other variables could be useful in making this classification too.

There is lots of great content on the JMP website that could help you progress further, and you might consider signinging up for a live webcast. 

6 REPLIES
ian_jmp

Staff

Joined:

Jun 23, 2011

Solution

I made this JMP table by reading in your .XLSX (using 'File > Open'). I then used 'Graph > Graph Builder' to look at the data, then 'Analyze > Fit Y By X' to fit the parabola. You can run the scripts saved to the table to reproduce these. From this fit, I saved the predicted values to make the new column 'Predicted Y', and if you click on the '+' sign next to this column in the table columns panel you can inspect the formula of the parabola that was fit. If you need more help with the steps just ask again.

 

However, having seen the data, I am left wondering why you think a parabola is a useful model to fit? I guess the answer depends on why you are interested in this data, and what you intend to do with the results of any analysis.

Craige_Hales

Staff

Joined:

Mar 21, 2013

Solution

Ian's answer is probably the one you want. Here's another answer, using Help->Scripting Index to look up the YFunction and insert your function. I scrolled the axes until I got a nice picture.

Y Function plotting a parabolaY Function plotting a parabola

Craige
MissPN

New Contributor

Joined:

Sep 17, 2017

Hi Craige_Hales,

 

Your JMP script is very useful, I could help my job to be esier.

Thank you very much for your help.

 

Best Regards,

Miss PN.

 

MissPN

New Contributor

Joined:

Sep 17, 2017

Solution

 Hi ian_jmp,

 

Many Thanks for your advicing. Following your question, based on my data from 2 groups(Pass Hd & Fail Hd).

I would like to find the equation that could be cover on fail Hd while impacting Pass Hd as less than as we can (Please refer on Pass+Fail.xlsx).So, I think Parabola model could be useful to analysis.

Based on your JMP file as attached, may I ask about column "predicted y" ? from the formula, how to calculate Parabola fit model?? It's the better than formula as I provided? 

 

Appriciate for your help.

Best Regards,

Miss PN.

 

ian_jmp

Staff

Joined:

Jun 23, 2011

Solution

With the 'pass-fail' data you have provided, then this becomes a 'classification problem' in which you would like to find the best separation between the two groups. There are many ways that you could do this in JMP. Using your data I used Graph Builder to get:

 

Screen Shot 2017-09-19 at 11.09.06.png

I used your 'No. Hd' variable in the 'Wrap' role. I assume this is the number of heads in the drive. This might suggest that a linear boundary could work OK, but (although I'm sure you have good reason for looking at 'A' and 'B') it might be that other variables could be useful in making this classification too.

There is lots of great content on the JMP website that could help you progress further, and you might consider signinging up for a live webcast. 

MissPN

New Contributor

Joined:

Sep 17, 2017

Hi ian_jmp,

Thank you a lot for your kindly, your suggestion could help my JMP skill to grow up.

Thanks again.
Best Regards,
Miss PN.