Why would I use ANOVA instead of a Rank-Sum test? A colleague of mine with little statistics experience is trying to perform an experimental evaluation of a computer program. He created a between subjects design and solicited test subjects.
11 people were given his "new and improved" computer program to use. 10 others got the "old and boring" computer program to use, for the same task. 
He asked me, and several other people around the lab how to analyze his data.
I told him he should examine the data for normality. If it was normally distributed, he should use a t-test. If it was not, he should use a Wilcoxon rank sum test.
One of my colleagues told him he should use ANOVA, even though he only has 2 groups. Apparently using ANOVA on non-normal data in R produces a new degree of freedom measure which can be put into a t-test.
I've never heard of such a thing. Is this true? Is it statistically valid? Why would anyone use it instead of just doing a rank-sum test?
 A: First, the data should not be checked for "normality". This is a useless test. The t test is quite robust to departures from this assumption anyway.
Second, the t test is just a special case of ANOVA for two groups.
If you believe the data to likely be close to a normal distribution (based on experience of the variable in question, visualization of a histogram, but NOT using a normality test), just use the t test. If not, use the rank sum test.
A: The Wilcoxon is vastly superior to the ANOVA in most cases, especially with non-normal data (read anything by Cliff Blair or Shlomo Sawiliowsky for references). The ANOVA is somewhat robust to non-normality, but not very. Additionally the Wilcoxon (as with most rank based statistical tests) are more powerful and increase in power with non-normal data. However, some fields tend to have a bias against distribution free statistical tests. This bias is founded in the tradition and NOT the math. Both statistical tests are the same in R, SPSS, SAS, Minitab, etc.
