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Tagged: Regression
This topic contains 14 replies, has 6 voices, and was last updated by MBBinWI 1 week, 2 days ago.
Hi guys,
I need some advice if possible. I work in the automotive industry making valve train components. Long story short, we’ve found what we think is a quadratic relationship between a Stem Ø and Gauge Height.
We carried out a trial to find the data’s appearance did resemble that of a quadratic curve. Also, the R-Squared is 97%.
The problem I have is: the quadratic equation that it has given us does not seem to match the displayed line. For example, I’ve substituted 6.96 in place of the ‘Stem Ø’ in the equation. The result I got was -0.257mm. If you take this 6.96mm value and plot it on the fitted line plot, the value is around +0.04mm (which is what I’d expect it to me).
Anyone able to shed any light on this?
Thanks,
James
I don’t see how you are getting +.04 for a value of 6.96 for that fitted line. A quick eyeballing with two rulers on my computer screen suggests the intercept is below 0 which would mean a negative number.
I ran a quick check with your equation as printed on the graph and I agree, the equation does not match the plot of the fitted line. I played around with the coefficients of the equation as stated and I suspect you will find that the issue is one of too much rounding of the equation coefficients. For example if you take the coefficient of 359.3 and change it to 359.2458 (this would round up to 359.3)and plug in 6.96 you will get a value of -.039 for the predicted value. This value has the correct sign and magnitude of the fitted equation.
A couple of additional items
1. If you are going to use this equation to model your process you will need to include the 95% confidence intervals for INDIVIDUALS on that graph.
2. The region between what amounts to two big, fuzzy data points (approximately 6.93 – 6.97)needs to have some actual measurements to confirm the predicted curvature. As it stands I suspect the significance of the quadratic term is highly dependent on the extreme measurement on the right. It would be interesting to see what happens to your model if that point is removed and the analysis run without it.
My mistake guys. I have no idea where I got the statement; ” If you take this 6.96mm value and plot it on the fitted line plot, the value is around +0.04mm (which is what I’d expect it to me).”
Thanks for the replies. I’m currently digesting what you’ve raised Robert.
I had the same initial thought as @rbutler regarding rounded coefficients.
To get more precise estimates in Minitab, I would store the coefficients. In case this isn’t something you’ve done before (or someone reading this post is familiar with), here are the steps:
1. Choose Stat > Regression > Fitted Line Plot > Storage to select the Coefficients > OK.
2. In your worksheet, click anywhere in the stored coefficients ‘COEF1’ column > right-click > Format Column > Fixed decimal to increase the number of decimals > OK.
Once you use more decimals, do the calculated values match the plot?
Hi Robert,
Again, thanks for your response. Although, I can see you’ve changed the coefficient to give one the near-correct answer, I cannot see the logic in having to consciously change numbers.
Hi M.Paret,
Thanks for taking the time out to respond.
I have extended the coefficients out to 6 decimal places, which I hadn’t previously considered.
The coefficients I get are:
1240.475692
-359.339160
26.021060
Submitting those values into our equation:
GH = 1240.475692 – (359.339160 x 6.96) + (26.021060 x (6.96^2))
GH = -1260.5248616 + 1260.501780096
GH = -0.023mm
This makes total sense now. I thank you for pointing that out!
No problem!
@jamesm the logic was to test the hypothesis that the excessive rounding of your coefficients was the cause of your problem. Of the coefficients provided 359.3 had the fewest significant digits after the decimal point so it was the logical place to start looking.
I built a quick program with the equation you initially reported and had it generate predictions for 6.92, 6.94, and 6.96. I also output each term of the model to see how they were impacting the final answer. The term with 359.3 looked to be the culprit so I just played around with a string of digits after the decimal point to see how many decimal places I would need to generate an answer that was in the ballpark of what was illustrated in your graph.
That quick test suggested excessive rounding was the issue and, as we can see from your most recent post, that was indeed the problem.
I’d highly recommend that you fit this model in Minitab using Fit Regression Model. That way you can see the predicted R-squared value. This statistic helps determine how well the model fits data points that are removed from the model estimation.
Again big thanks to you both for your help on this.
For anyone else reading, I thought it’d be decent of me to give an update. I’ve attached what my model have evolved into. I have this pinned on my wall by my desk with around 100 additional manual handwritten points that include random samples that I’ve been taking everyday to challenge the model. I’m trying to prove it wrong but have not been able to yet which is positive.
Jim Frost;
Are you the same Jim Frost who’s active on the blogs on the Minitab help site, particularly on regression? I’ve been reading many of your articles and they’re very informative!
I’ve never used “those dots as a background”….if using Minitab, what is that option called and where is it?
Hi Chris,
If you click your model in Minitab and select ‘edit region’ I think. You can then select your desired pattern. I’ve attached a screenshot of it.
I like to use it particular when axis are far away from points.
James
always nice to see collaboration and learning on here. i’ve been using it way too long and always want to see tips–especially on tools i know
thanks @jamesm
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