Linearity - R-sq evaluation - Where can we find R-sq evaluation criteria?

V

Vlad12

#1
Hello everybody,

During linearity check of tyre uniformity automatic test line
we have received the following results:
line "bias=0" is in confidence limit; ta=0,60246;
tb=0,59391; t value=2,0017; R-sq=0,6 %.
I am sure that R-sq is well caiculated.
1. Can we consider linearity of such system acceptable? Why?
2. Who can comment the result of R-sq? Where can we find R-sq evaluation criteria?

Thank you in advance.
 
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Wesley Richardson

Wes R
Trusted Information Resource
#2
During linearity check of tyre uniformity automatic test line
we have received the following results:
line "bias=0" is in confidence limit; ta=0,60246;
tb=0,59391; t value=2,0017; R-sq=0,6 %.
I am sure that R-sq is well caiculated.
1. Can we consider linearity of such system acceptable? Why?
2. Who can comment the result of R-sq? Where can we find R-sq evaluation criteria?
Hi Vlad,

Linearity of the gage:
http://www.itl.nist.gov/div898/handbook/mpc/section4/mpc452.htm

Least squares fit:
http://www.itl.nist.gov/div898/handbook/pmd/section4/pmd431.htm

R^2 is the coefficient of determination. R^2 = 1 - (SSE/(S_Y^2))
R^2 is the proportion of variation in the dependent variable that is explained by the variation in the independent variable.

You state R-sq = 0.6%. Did you possibly mean R-sq = 0.6 or 60%? If so, then 60% of the variation is explained using your model for the relationship. If it really is 0.6%, then that is a very poor model of the relationship.

A low R^2 value means that either the relationship is nonlinear, or there are other factors contributing to the variation in the dependent variable. The acceptable amount of explained variation is up to the user.

For the t value, you did not state the number of degrees of freedom or the confidence level. The t value can be used to estimate the confidence interval for the slope of the line.

You also did not state what ta and tb represent. These are possibly the intercept and slope of your line, but I do not know if that is what you meant.

Whether the linearity of the system is acceptable is again dependent on the user to make a determination. I have seen data in which an R^2 = 0.65 was considered very good, and other data in which R^2 = 0.87 was not acceptable.

Wes R.
 

Bev D

Heretical Statistician
Staff member
Super Moderator
#3
if you post your data, we can provide better responses to your questions. summary data leads to many assumptions and you may get answers that you like - but they will be wrong becuase our assumptions will were wrong. Or you will only get general responses that may not help you
 

Steve Prevette

Deming Disciple
Staff member
Super Moderator
#4
I very much distrust R-squared. There is no direct correlation between R-squared and level of significance. You can perform such a calculation, but it is difficult. Even more difficult, but worthwhile, is to plot the prediction and confidence limit curves (yes, they are curves) on the chart. Another good option is to look at the ANOVA analysis that goes with the regression, and it will let you know if the slope is significantly different than zero, and if the intercept is significantly different than zero.
 
V

Vlad12

#5
Hello everybody!

First of all thanks to all who answered my letter. Thanks for the answers.
Linearity study report of this measurement process is attached. There you can find all basic data.
It turns out, that if coefficient of determination R-sq value calculated during linearity study is too low (in my case R-sq=0,6 % not 60%, but 0,6%)), then relation between y and x is not linear, or there are some other factors, that influence dependent variable y. Third edition of MSA asserts during linearity study only double linear regression model (y=ax+b).
May be I should pay less attention to R-sq value?
Can I use obtained data (graphic and numeric analysis) to conclude about linearity acceptability ignoring small values of R-sq or should I chose another regression equation structure and repeat calculation?
 

Attachments

Miner

Forum Moderator
Staff member
Admin
#6
Attached is a Minitab analysis of your data. The reults of it match those in your spreadsheet, so your calculations are correct.

A few observations about your model:
  • The low p-value for the 91.7 reference part indicates that you have a problem with that part
  • The low R-squared value is probably caused by the poor repeatability of measurements for each part.
  • The poor repeatability invalidates the conclusion that linearity is acceptable even though the line falls within the acceptance limits.
  • The reason for this is because the regression model itself is not significant. The graph alone should raise questions about this.

I recommend performing a Repeatability and Reproducibility study.
 

Attachments

V

Vlad12

#7
Hello, Miner!

1. We have investigated our system for Repeatability and Reproducibility and have received the following results:
R&R=14,14 %, ndc=10.

2. If we correct the obtained data we get the following results of linearity analysis:
- If we increase the reproducibility of values for each part it makes R-sq value rise till about 1,1 %.
- If we correct the pert 91.7 values it makes R-sq value rise till about 2 %.
The R-sq value depends much on the regression slope; the larger the slope is, the more the value of the R-sq is. When we increased the slope and at the same time provided that ta and tb < t value, we managed to get the R-sq value about 6 %.
If we make R-sq value equal to 75% then the range of values with linearity in tolerance becomes very limited and ta and tb become larger then t value.
Have you ever received R-sq high values while graphic and numerical analysis of linearity study results showed it was acceptable?
 

Miner

Forum Moderator
Staff member
Admin
#8
After reviewing your feedback on the modeling that you did, I rethought my reponse. In a perfectly linear gage, the slope of the line would be zero. If there is no bias, the y-intercept would also be zero. The equation(model) becomes y = 0. In this situation, R^2 would be driven to zero also. It would become a meaningless indicator because the value of X would have no impact on Y.

My recommendations would be to use the criteria below to assess linearity:
- All part means within the prediction limits or P > 0.05
- Prediction line within the limits

Since I could not see the equations that you used in your worksheet, I am presuming that you are performing a t-test on the measured bias versus the repeatability variation to determine significance of the bias. If your calculations were made correctly, this should be a valid approach. Though you are using the same study to assess linearity and bias, evaluate bias separately from linearity. A gage may be perfectly linear, yet highly biased. And vice-versa.
 
A

Atul Khandekar

#9
Since I could not see the equations that you used in your worksheet, I am presuming that you are performing a t-test on the measured bias versus the repeatability variation to determine significance of the bias. If your calculations were made correctly, this should be a valid approach. Though you are using the same study to assess linearity and bias, evaluate bias separately from linearity. A gage may be perfectly linear, yet highly biased. And vice-versa.
The method and equations used are as per the AIAG MSA manual. I think these are standard linear regression equations.

ta is the t-statistic for test of hypothesis that slope=0
tb is the t-statistic for test of hypothesis that bias(intercept)=0
 
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