Solved – Degrees of freedom in quadratic regression simultaneous confidence bands

confidence intervaldegrees of freedomregression

I have been working on an inverse regression addon for SPSS and using this paper heavily:
I. Lavagnini, F. Magno, A statistical overview on univariate calibration, inverse regression, and detection limits: Application to gas chromatography/mass spectrometry technique., Mass spectrometry reviews 26, 1-18 (2007).

They do a good job of explaining the difference between the simultaneous and non-simultaneous bands for linear regression, i.e. using $t_{1-{\frac\alpha 2},n-2}$ for non-simultaneous vs $\sqrt{2F^\alpha _{2,n-2}}$ for simultaneous as follows:
$$
y^{\pm} = b_0+b_1x\pm t_{1-{\frac\alpha 2},n-2}s_R\sqrt{{\frac 1 N}+{\frac {(x-\bar x)^2} {\sum (x_i-\bar x)^2}}}
$$

and

$$
y^{\pm} = b_0+b_1x\pm \sqrt{2F^\alpha _{2,n-2}}s_R\sqrt{{\frac 1 N}+{\frac {(x-\bar x)^2} {\sum (x_i-\bar x)^2}}}
$$

…But they do not give the analogous simultaneous multiplier for quadratic regression. Do the two degrees of freedom in the $\sqrt{2F^\alpha _{2,n-2}}$ term come from having 2 regression coefficients in linear regression or from having two variables, the independent and dependent?

My question stated briefly:

Should my simultaneous multiplier in quadratic regression be $\sqrt{2F^\alpha _{2,n-3}}$ or $\sqrt{2F^\alpha _{3,n-3}}$ ?

Best Answer

Neither. (But they are good guesses.)

This is called Scheffe's Method. The multiplier (see formula 1.2 in the link) is $\sqrt{p F^\alpha_{p, n-p}}$ for $p$ parameters and $n$ independent observations. For linear regression there are $p=2$ parameters (intercept and slope); for quadratic regression there are $p=3$ parameters (intercept, slope, and quadratic term): see Section E in the reference ("Quadratic Calibration Curve").

However, I'm not so sure you can derive the same formula for quadratic regression: those bands are really the tops of tubes in three dimensions and surely they must also include terms involving the squares of the $x_i$ (and consequently their fourth powers) in addition to the $x_i$ themselves.

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