anovat-test
I have 7 groups, one healthy group and 6 treatment groups and 5 traits. I would like to test for each trait if there are any significant difference between each treatment group and healthy group.
Would it be more appropriate to perform a T-test for each trait or ANOVA?
In case T-test is the answer, should I consider adjustment for 6×5=30 tests (i.e. 0.05/30)?
In general, you wouldn't necessarily expect one way ANOVA and the Kruskal-Wallis to be similar, sometimes they can give quite different p-values. See here for a little partial motivation for why you might expect a difference. [When samples are reasonably normal-looking and with means not too many standard errors apart, they often tend to give similar p-values. Outside that, they frequently don't.]
However, in this case the reason is more prosaic: Your Kruskal-Wallis p-value is wrong.
Here's a summary of results in R (details below).
p-value
Welch t-test: 0.001287
Equal-var. t-test: 8.552e-05
One way anova: 8.55e-05
Wilcoxon test: 0.004847
Kruskal-Wallis: 0.003761
(Neither of the last two p-values are exact; if they were, you'd get the same p-value for the two-group comparison.)
Your problem is you're treating the second group's data as a factor (see the end of this answer).
Here's what I get in R with your data:
frh <- data.frame(group1 = c(103.56, 103.32, 103.32, 104.27, 103.56, 103.8),
group2 = c( 97.16, 97.16, 96.69, 98.58, 90.76, 97.64))
# strip chart:
# Welch t-test:
> with(frh,t.test(group1,group2))
Welch Two Sample t-test
data: group1 and group2
t = 6.3316, df = 5.163, p-value = 0.001287
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
4.368147 10.245186
sample estimates:
mean of x mean of y
103.63833 96.33167
$\,$
# equal-variance t-test:
> with(frh,t.test(group1,group2,var.equal=TRUE))
Two Sample t-test
data: group1 and group2
t = 6.3316, df = 10, p-value = 8.552e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
4.735411 9.877922
sample estimates:
mean of x mean of y
103.63833 96.33167
$\,$
#one way anova:
summary(aov(values~ind,stack(frh)))
Df Sum Sq Mean Sq F value Pr(>F)
ind 1 160.16 160.2 40.09 8.55e-05 ***
Residuals 10 39.95 4.0
$\,$
# Wilcoxon-Mann-Whitney:
> with(frh,wilcox.test(group1,group2))
Wilcoxon rank sum test with continuity correction
data: group1 and group2
W = 36, p-value = 0.004847
alternative hypothesis: true location shift is not equal to 0
Warning message:
In wilcox.test.default(group1, group2) :
cannot compute exact p-value with ties
$\,$
# Kruskal-Wallis test:
> kruskal.test(frh)
Kruskal-Wallis rank sum test
data: frh
Kruskal-Wallis chi-squared = 8.3958, df = 1, p-value = 0.003761
Those are all about as consistent with each other as I would expect on that data.
Now, here's how to get what you got for the Kruskal-Wallis:
with(frh,kruskal.test(group1,group2))
Kruskal-Wallis rank sum test
data: group1 and group2
Kruskal-Wallis chi-squared = 4.3939, df = 4, p-value = 0.3553
The problem is, if you're getting this, you're using it wrong. That's not how the function works - group2 is being treated as a factor defining different groups for data in group1.
So the main reason the Kruskal Wallis isn't giving you a roughly similar p-value to ANOVA is you didn't call it correctly.
Without looking at your data, I can only guess. However it is likely that the t test appears significant because there is no multiplicity penalization. Conversely, the Tukey correction inflates (appropriately) the p value in light of the multiple comparisons. Check attentively the software output and you should be able to recognize this.
Best Answer
The usual approach is to perform ANOVA and, if ANOVA shows significant differences, perform group comparisons making adjustments for multiple comparisons.
The adjustment you propose would be Bonferroni correction which is widely used (because it's simple) but rather conservative. You might be interested in some alternatives.