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Consult a analysis problem.Thank you!

2 messages · Ivy_Li, Brian Ripley

Ivy_Li
# 
Hello everybody,
	Could I consult you a question?
	I am doing an analysis about some data. I used Anova analysis. Its PValue returned is about 0.275, no signal. But through the box-chart, I think it exist discrepancy between A and B. And then I tried to use the ks.test, fisher.test and var.test to do analysis. Their PValue returned are all imperfect. If we think the PValue below 0.05 means it exist significant. The all test result are all bigger than 0.1. I don't know why the anova and other tests can not find out the issue? And could you help me to find out which analysis method fit for this case? Thank you very much!

##################R script
	#### -creat a data frame
	Value <- c(0.01592016, 0.05034839, 0.01810571, 0.05129173, 0.01557562, 0.04321186,
 		0.01851016, 0.05214449, 0.01912795, 0.02081264, 0.05580136, 0.03097065,
		0.01706546, 0.01534989, 0.01367946, 0.01734044, 0.02419865, 0.04541759,
		0.08735891, 0.03297321, 0.02311511, 0.05972912, 0.04356657, 0.02234764,
		0.01291197, 0.02203159, 0.17550784, 0.08726857, 0.01557562, 0.04486457,
		0.01498870)
	Group <- c("A", "A", "B", "B", "B", "A", "B", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A",
		"A", "A", "A", "A", "A", "A", "A", "A", "A", "B", "B", "B")
	df.new <- data.frame(Value=Value,Group=Group)
	#### -plot the boxchart
	plot(as.factor(df.new$Group),df.new$Value)
	points(as.factor(df.new$Group), df.new$Value, pch=16,col=2)
	#### -anova test
	anova(lm(df.new$Value~as.factor(df.new$Group)))
###############end

Thank you very much for your kindly help!

BR
Ivy
Brian Ripley
#
On Mon, 5 Dec 2005, Ivy_Li wrote:

            

      
      
      
    

        
It you want free statistical consultancy, please use an informative 
signature giving your affiliation and credentials.

            

      
      
      
    

        
What was the aim of the analysis?

            

      
      
      
    

        
You need to think about transforming your data.  However, a one-way 
two-class ANOVA is the same as a t-test with equal variances:

            

      
      
      
    

        
and clearly your two samples have different variances.

            

      
      
      
    

        
[1] 0.04595329

so

            

      
      
      
    

        
would be better.

So, you tested for a difference in mean assuming equal variances, but 
there exists a marginally difference in variances, and testing for a 
difference in mean no assuming equal variances is not significant.

Using 1/Value makes sense to equalize the variances.

You are doing things in R in a convoluted way.  Try simply

            

      
      
      
    

        
Do also be aware of the dangers of multiple testing: it is invalid to 
choose the one you like out of several tests applied to a set of data. 
The bottom line is to collect more data.