Consult a analysis problem.Thank you!
On Mon, 5 Dec 2005, Ivy_Li wrote:
Hello everybody, Could I consult you a question?
It you want free statistical consultancy, please use an informative signature giving your affiliation and credentials.
I am doing an analysis about some data.
What was the aim of the analysis?
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
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:
t.test(Value ~ Group, data=df.new, var.equal=TRUE)
and clearly your two samples have different variances.
var.test(Value ~ Group, data=df.new)$p.value
[1] 0.04595329 so
t.test(Value ~ Group, data=df.new)
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
boxplot(1/Value ~ Group, data=df.new) summary(aov(1/Value ~ Group, data=df.new))
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.
Brian D. Ripley, ripley at stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595