Non-parametric methods

#1
Hi everyone.

This is my first time on this forum … thank you for accepting me!

I have run a Quade's rank analysis of covariance (Quade, 1967) using SPSS.

The reason I did not use a straight ANCOVA was because three of the five assumptions underpinning parametric ANCOVA were not confirmed in my data. As the non-parametric method (Quade's) identifies, you are running the ANOVA part of the ANCOVA analysis on the rank-ordered dependent variable with the residuals saved in the OLS regression part of the ANCOVA analysis as the factor. The residuals are clearly the differences between the rank-ordered DV scores minus the predicted scores from the OLS regression model.

All I am able to do currently is, based on the F-ratio (Quade's index), is accept or reject the null.

Any post-hoc follow up is based on rank ordered data. I need to be able to refer the data back to the original units of measurement as a post-hoc procedure otherwise the results will make no sense!

Does anybody have some advice about how I might resolve this?
 

Karabiner

TS Contributor
#2
Could you tell us something about your study (research questions, variables included, measurement of the variables, sample size), and why you assume that ANCOVA assumptions are violated?

With kind regards

Karabiner
 
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#3
Thanks for coming back to me on this Karabiner.
I have three groups of males (n1 = 89, n2 = 295 and n3 = 66). The dependent variable (DV) is body mass (kg) and the covariate (CV) is age (yrs). I wanted to use ANCOVA to test the H0: M1 = M2 = M3 on the group masses holding the covariate stable.
Only three of the five assumptions underpinning parametric ANCOVA were met in my data.
Linearity between the DV and the CV - r448 = 0.061, P = 0.097 - based on this the assumption is confirmed. However, there is also an issue related to significant differences in the mean ages between the three groups (G1 = 19.1 ± 6.0 yrs, G2 = 24.8 ± 8.4 yrs, G3 = 27.1 ± 5.7 yrs; F2,447 = 25.4, P < 0.001). I concluded therefore that parametric ANCOVA would be inappropriate.
Homogeneity of error variances Levene’s test: L2,447 = 7.76, P < 0.001. This assumption is violated.
Independence of residuals – the scatterplot is orthogonal: r448 = -0.063, P = 0.093, R2 = 0.004, R2adj = 0.002. The assumption is confirmed.
Normality of residuals – Kolmogorov-Smirnov and Shapiro-Wilk tests: K-S450 = 0.061, P < 0.001; S-W450 = 0.957, P < 0.001; Skewness = 0.02 and Kurtosis = -1.19. The residuals are not badly skewed (i.e. S is within ±1.96). I am less concerned about the K index, which is still within acceptable limits. Indeed, it is not at all unusual to see issues with the peaks of normal curves in biological data. I was inclined to assume that a parametric ANCOVA is robust enough to apply to these scores, based on these outcomes, i.e. the assumption is confirmed.
Homogeneity of regression slopes – the slopes of the DV vs CV OLS regression lines for the three groups are not equivalent – this assumption is violated.
Consequently I ran Quade's rank ANCOVA:
1. I rank ordered the DV (mass) and the CV (age).
2. I ran an OLS regression as: Rank Mass = a + (b x Rank Age), and I saved the unstandardised residuals in the process.
3. I ran a one-way ANOVA with the unstandardized residuals saved in 2 as the DV and the grouping variable as the Factor
Results: F2,447 = 24.28 (P < 0.001) which allowed me to reject the H0 in favour of the H1: M1 ≠ M2 ≠ M3.
My issue is how can I run a post-hoc test to identify which of the group means differ, but report the results related to the original units of measurement of the DV (kg)?
Any advice would be most welcome.
Steve
 

Karabiner

TS Contributor
#4
Normality of residuals is irrelevant, if sample size is large enough (as is here the case, cf. central limit theorem).

You should include covariate*dactor interaction in the analysis anyway, I have rarely seen that the covariate is independent from the factor level.

As to homogeneity of variances, that could be a problem, but it certainly depends on the degree of inhomogeneity. p-values just tell you that the differnce is not 0.00000000... in the population.

With kind regards

Karabiner