Chromatography Forum

I have a table of n=65 experimental vs predicted data points. The data points can be classified into 1 of 3 groups, A, B, C. The regression equations for these three groups, along with a regression equation for the entire set is shown. (95% confidence interval shown in parenthesis)

How can I determine if the three regression equations, A, B, C are coincident or statistically different? Alternately, how can I determine if the "all" regression equation is different from A, B, C?

A y=1.62(±0.07)x-4.30(±0.27) R^2 = 0.992 n=19 F=2007 SE=0.20
B y=1.69(±0.11)x-4.71(±0.33) R^2 = 0.972 n=27 F=853 SE=0.21
C y=1.63(±0.13)x-4.30(±0.48) R^2 = 0.972 n=19 F=590 SE=0.30

(All) y=1.67(±0.06)x-4.55(±0.20) R^2 = 0.981 n=65 F=3190 SE=0.25

You could do this with a t-test to compare slope A with slope B, A with C, and B with C and do a bonferroni correction on your uncertainity.
But for a t-test you need the error of your slope and not the Residual SE.
You could back calculate this from your confidence intervals of your slope, but this isn't really great.

So I think you need the data from all 65 points to calculate the right numbers.

But if you have your data, you could do ANOVA instead of a t-test, which is better to compare more than 2 groups.

HTH

Ace

I'll vote for aceto_81 on this too; you need to look at the original numbers. As well as anova, there's a thing called generalised least squares (and anova is basically a subset of generalised least squares). It can tease out the correlations of a set of numbers with a whole set of factors, including non-numerical ones (A, B, C) and numerical correlations (i.e if A, B, and C were three calibration curves, for example, each containing measurements at different concentration levels; the concentration would be one of the factors). It's offered by Genstat (and probably other similar packages). It will tell you which factors explain significant variation in the data. It has one advantage over anova, which is that it can handle unbalanced data-sets, which yours is: you have more measurements in B than A and C.