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I'm running a non-parametric ANCOVA in R. I'm using the non-parametric ANCOVA from the fANCOVA package. https://cran.r-project.org/web/packages/fANCOVA/fANCOVA.pdf I'm getting the results below. Now I need to run pair-wise comparisons, similar to Dunn test in Kruskal-Wallis test and to Tukey HSD in ANOVA -- even though the results below are not significant, I want to do the pairwise comparisons, in the case other analyses shows significance. Therefore I have the following question: How do I run pairwise comparisons in a non-parametric ANCOVA in R/SPSS? I greatly appreciate your help.

I'm applying the formula from fANCOVA:

T.aov(x, y, group, B = 200, degree = 1, criterion = c("aicc", "gcv"), family = c("gaussian", "symmetric"), tstat = c("DN", "YB"), user.span = NULL, ...)

T.aov(subsetframe$IriSum1, subsetframe$EmDiffSum1,subsetframe$Condition2), IriSum1 is the covariate, EmDiffSum1 is the dependent variable, and Condition2 is the grouping variable.

Test the equality of curves based on an ANOVA-type statistic

Comparing 5 nonparametric regression curves Local polynomial regression with automatic smoothing parameter selection via AICC >is used for curve fitting. Wide-bootstrap algorithm is applied to obtain the null distribution.

Null hypothesis: there is no difference between the 5 curves. T = 4.793 p-value = 0.209

I also tried the loess.ancova (Fit a semiparametric ANCOVA model with a local polynomial smoother) from the fANCOVA package. The result is:

loess.ancova(x, spdata.2$EmDiffSum1,spdata.2$Condition, plot=TRUE, data.points = > TRUE) $linear.fit [,1] (Intercept) 4.510333 groupB 1.447953 groupC 1.373840 groupD 1.450809 groupE 1.047067

$smooth.fit Call: loess(formula = lm.res ~ x1 + x2, span = span1, degree = degree, family = family)

Number of Observations: 132 Equivalent Number of Parameters: 8.28 Residual Standard Error: 1.957

Thank you very much for your help. I use both R and SPSS. While these tests have been run in R, if anybody knows the method for running non-parametric ANCOVA with pairwise comparisons in SPSS, I'd be very grateful to hear that too.

Shimano
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  • *"Null hypothesis: there is no difference between the 5 curves."* If there is no difference among any of the curves, there will be no significant individual difference, especially after multiple testing correction. Non-parametric statistics are less powerful than parametric statistics if the assumptions of parametric statistics hold. Can you include a bit more about the type of data and the research question? Non-parametric statistics is not generally the first thing to resort to. – Frans Rodenburg Oct 31 '17 at 05:29
  • Thank you for your response. I'm using non-parametric tests because the assumptions for ANCOVA are not met: the data are not normally distributed (Shapiro-Wilks test) and the variances are not homogenous (Levene's test). Therefore, non-parametric tests have to be used. Yes, I know that the result I shared doesn't have statistically significant differences. However, I need to be able to do pairwise comparisons, because some other variables may have statistical significance. But if I can't run pairwise comparisons, I will not be able to identify where the differences are. – Shimano Oct 31 '17 at 16:47
  • So to clarify: My question is: How to run pairwise comparisons in non-parametric ANCOVA in R/SPSS? I know how to run them after a non-parametric ANOVA (Kruskal-Wallis test) in R/SPSS: Dunn post hoc test with bh corrections. Thank you for your help. – Shimano Oct 31 '17 at 16:49
  • Why did you assume normality in the first place? Can a different distribution (e.g. Poisson for independent counts, binomial for successes out of $n$ trials) be assumed in the context of a GLM? Even if normality seemed a reasonable assumption, a Shapiro-Wilks test is hardly a reason to use non-parametric methods. See for example this question: https://stats.stackexchange.com/q/2492/176202. Lastly, in my previous comment I tried to explain that if there is no significance in the omnibus test of the ANOVA, there will also be no individual difference that remains significant after correction. – Frans Rodenburg Nov 01 '17 at 06:13

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