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I am working on R, and have multiple data.frames. They all look like this one:

 head(Stim.Adjs)
      LogFreq    Word   PhonCV    FreqDev  LengthDev
6316    2.673  enkele VC-CV-CV -0.1794029 -1.0608453
12638   2.415 laatste CVVC-CCV -0.2207496 -0.7113483
12639   2.415 laatste CVVC-CCV -0.2207496 -0.7113483
1633    2.248    bang      CVC -0.2475127 -1.7598394
3167    2.243   bezig  CVV-CVC -0.2483140 -1.4103424
25467   2.243 vroeger CCVV-CVC -0.2483140 -0.7113483

These are list of words extracted from a corpus, and the columns 4 and 5 represent measures of standard deviation of Frequency and Word length respectively. What I want to do is to have all lists with a similar frequency deviation and word length deviation (two separate controls). So far I had been doing this by comparing each dimension in pairs of list using t.tests, and then removing values from the lists so that the t.tests showed no significant difference in means.

Now that I have more than two lists, I'm a bit confused as to how to test if the means are equal or not. At first I though of an ANOVA, but I don't have a predictor variable, just 3 independent means (for each of the two values I want to control).

The question is, then, what alternative to the t.test could I use that allows me to compare the means of independent vectors. Help with the function for this in R is much appreciated.

Thanks in advance.

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Welcome to our site! I am struggling to understand what you mean by "have all lists with a similar frequency deviation and word length deviation (two separate controls)." Could you expand on this, perhaps by showing the output you would like to see for the example data? Out of curiosity: what "measure of standard deviation" are you using? It is strange for such measures to be negative, and yours are all negative. Finally, the repetition of "laatste" in your example seems inconsistent with your description of the data as describing word frequencies: how, then, could a word ever be repeated? – whuber♦ Oct 25 '12 at 14:16
@whuber The column FreqDev stands for Frequency Deviation. Basically what I have there are the z-scores of each word (That is why there are negative values there), and thus represents how much the frequency of each words deviates from the mean frequency of the whole list. What I want is to control for frequency both within lists (by removing all entries which deviate more than .8 z-scores from the mean in either direction) and between lists. The later is what I don't know how to do, as I have to compare more than two means, and so a t.test is not feasible. – Hernan_L Oct 29 '12 at 15:07

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