Contradicting p-values for Anova and Kruskal-Wallis on same data: Which is right? I have a data file with task duration values for three groups, and I want to determine group effect on task duration (tasks were executed by individuals; each group had 7 different individuals; each individual executed the same three tasks; and the data for one individual in group B was not recorded because of a setup problem during the experiment).
I have created from the data file the following box plot (red dots are the means, and "n" is the number of time values in each group):

and also the following histogram (duration given in "min:sec"):

My data sample per group is small, and "Shapiro-Wilk normality test" tells me that group A does not come from a normal distribution, and that groups B and C come from a normal distribution. Because groups are small and one group is non-normal, I decided to run Kruskal-Wallis one-way analysis of variance (non-parametric) and its result was:
Kruskal-Wallis rank sum test
data:  Duration by Group 
Kruskal-Wallis chi-squared = 4.2811, df = 2, p-value = 0.1176

so I should accept that the effect of the groups was not significant (p-value > 0.05).
However, when I run one-way Anova (sanity check just in case Kruskal-Wallis was not the correct choice), Anova's result was:
ANOVA Duration ~ Group 
            Df    Sum Sq   Mean Sq F value  Pr(>F)   
Group        2 0.0003692 1.846e-04   6.473 0.00293 **
Residuals   57 0.0016257 2.852e-05                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Tukey multiple comparisons of means
   95% family-wise confidence level

             diff          lwr           upr     p adj
B-A -0.0055763154 -0.009704328 -0.0014483027 0.0054132
C-A -0.0048032407 -0.008769307 -0.0008371744 0.0138750
C-B  0.0007730747 -0.003354938  0.0049010874 0.8943085

That is, Anova returns p-value < 0.05, that is, it is telling that the group effect is significant (in this case, group A was significantly different regarding B and C).
However, because of a small number of samples per group and the fact that one group is not normally distributed, I tend to accept Kruskal-Wallis result, but I am not sure.
So my questions are: 
Should I accept Kruskal-Wallis result because this test is more adequate than Anova for this particular case?
How should I have used the box plot and the histogram to decide for the most adequate test?
Thanks
 A: The boxplot and histogram tell you all by themselves that your data are skewed, especially in group A.  The Shapiro-Wilk test is kind of pointless.  With data thusly skewed the ANOVA isn't really appropriate.  The Kruskal-Wallis rank sum test is based on the ranks, not the absolute values and doesn't require normality, either in the measures or residuals.  It is the more appropriate test.
A quick Google search will tell you one requires normality and one does not.
One thing you might consider is that durations are an arbitrary representation of time.  For example, you can indicate the duration of an event as 2 s or you can say the event has a rate 0.5 events/s.  It's the exact same thing and both numbers can arbitrarily be interchanged for representation.  However, rates tend to be much less skewed and more appropriate for statistical analysis.  It's possible your rates are normally distributed and you can use ANOVA in that case.
If you do decide to look at rates, keep in mind that the direction of magnitude changes, a higher duration values = a lower rate value.  Some people use a negative rate just to avoid that confusion.
A: The Kruskal-Wallis test and the Anova test are testing different hypotheses, both could be correct, the answers differ because they are answering different questions.
A: The distributions all overlap very much.  The Kruskal Wallis test seems to be indicating that the centers of the distributions are nearly the same.  The distribution for group is highly skewed due to several very extremely high values.  That is what causes the distribution to fail the Shapiro-Wilk test.  The anova F test wrongly interprets group A to have a significantly larger mean because it "ignores" the skewness. The Kruskal Wallis test is giving the appropriate answer while the F test is not.
A: Two things to keep in mind: first, ANOVA is robust in the face of non-normality if sample sizes are equal - the larger the difference in sample sizes, the less reliable it is; second, the K-W test is not a test of means or medians - it is really a test of similarity of distributions and, if the distributions are similar, it can be interpreted as a test of location. In my experience, most people ignore that both the Mann-Whitney and K-W test expect (require) the groups being compared to have similar distributions.
Several options are available for your problem.  You might try a data transformation (e.g., log) to put the data on a scale that yields normal distributions in each group.  Or, you might try running standard ANOVA after replacing the data with their ranks.  Both approaches are effective when the assumptions for ANOVA are violated. 
