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Different slopes between lin. regression and main axis of density elipse: why?

Thierry_S

Hi JMP Community,

This may be a silly question but I cannot figure out why the slope of a linear regression does not match that of the main axis of the distribution elipse for the same data set (see below)?

DOT PLOT LIN REG + DENSITY ELIPSE.png

Thanks for your help.

Sincerely,

Accepted answers

Dan_Obermiller

Because the regression line assumes there is no error/variability in the X-values. All of the variability is in the Y. A density ellipse does not make that assumption. To see this, from your Fit Y by X plot, choose to Fit Orthogonal > Univariate Variances, Prin Comp. That line will be the major axis of the density ellipse.

All comments

Dan_Obermiller

Pisit_Seagate

Hi Dan

I have some question about Fit Orthogonal between Univariate Variances, Prin Comp and Fit X and Y.

From my understand. When we used Fit X and Y to plot Orthogonal line with no error. For Univariate Variances, Prin Comp It will add error variance ratio to calculatetion and finding the equetion symmetry between Upper and Lower Othogonal line.

My question is How Jmp can calculated variance ration, UpperCL and LowerCL ?. I still stuck this long time.

I very appreciate to you.

Thanks you

2019-11-15 15_55_48-Evans_at99 - Fit Y by X of CTQ_WRT_WDTH by PWT95 - JMP.png

Mark_Bailey

You could start with Help > Books > Basic Analysis > Bivariate. The Statistical Details at the end of this chapter provides this explanation:

Fit Orthogonal
Standard least square fitting assumes that the X variable is fixed and the Y variable is a
function of X plus error. If there is random variation in the measurement of X, you should fit a
line that minimizes the sum of the squared perpendicular differences (Figure 5.26). However,
the perpendicular distance depends on how X and Y are scaled, and the scaling for the
perpendicular is reserved as a statistical issue, not a graphical one.

Figure 5.26 Line Perpendicular to the Line of Fit

The fit requires that you specify the ratio of the variance of the error in Y to the error in X. This
is the variance of the error, not the variance of the sample points, so you must choose carefully.
The ratio is infinite in standard least squares because is zero. If you do an
orthogonal fit with a large error ratio, the fitted line approaches the standard least squares line
of fit. If you specify a ratio of zero, the fit is equivalent to the regression of X on Y, instead of Y
on X.

The most common use of this technique is in comparing two measurement systems that both
have errors in measuring the same value. Thus, the Y response error and the X measurement
error are both the same type of measurement error. Where do you get the measurement error
variances? You cannot get them from bivariate data because you cannot tell which
measurement system produces what proportion of the error. So, you either must blindly
assume some ratio like 1, or you must rely on separate repeated measurements of the same
unit by the two measurement systems.

An advantage to this approach is that the computations give you predicted values for both Y
and X; the predicted values are the point on the line that is closest to the data point, where
closeness is relative to the variance ratio.

Confidence limits are calculated as described in Tan and Iglewicz (1999).

Pisit_Seagate

Hi Mark,

Thanks you for you comment. After i try to understand a concept of Fit Orthogonal with variance ration. From Jmp Document, I already knows concept and how to calculate a Intercept , Slope and variance ratio when variance ratio = 0.

Anyway, Jmp Document still not mention a theory of How to calculate a slope and intercept when variance ration not equal 0. Do you have any knowleadge about this ?

2019-12-12 10_53_28-.png Quote