R: t.test and pairwise.t.test give different results?

I tried to do a t-test with R over the following dataframe. ``` df <- structure(list(freq = c(9, 11, 14, 12, 10, 9, 16, 10, 11, 15, 13, 12, 12, 13, 13, 9, 16, 14, 12, 15, 16, 10, 11, 13, 14, 14, 14, 16, 8, 10, 14, 14, 11, 11, 11, 11, 13, 7, 12, 13, 14, 11, 11, 13, 10, 14, 10, 10, 12, 8, 9, 12, 14, 11, 12, 12, 14, 14, 14, 15, 12, 13, 14, 8, 9, 11, 10, 14, 12, 12, 9, 10, 8, 14, 11, 14, 9, 13, 13, 13, 10, 9, 13, 10, 13, 10, 13, 12, 11, 12, 10, 12, 8, 11, 12, 15, 12, 12, 11, 13, 12, 10, 13, 9, 11, 9, 11, 8, 12, 12, 12, 10, 11, 12, 9, 13, 14, 11, 11, 14, 13, 12, 14, 15, 12, 12, 12, 14), class = structure(c(3L, 3L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 3L, 4L, 4L, 4L, 4L, 3L, 2L, 3L, 2L, 1L, 4L, 1L, 4L, 1L, 4L, 2L, 2L, 3L, 3L, 2L, 4L, 1L, 4L, 4L, 4L, 3L, 3L, 3L, 2L, 1L, 4L, 3L, 3L, 1L, 4L, 1L, 2L, 2L, 3L, 3L, 4L, 2L, 2L, 3L, 3L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 4L, 1L, 1L, 1L, 2L, 2L, 3L, 2L, 3L, 2L, 3L, 3L, 4L, 2L, 1L, 4L, 1L, 1L, 3L, 2L, 2L, 2L, 3L, 1L, 1L, 1L, 1L, 3L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 3L, 3L, 4L, 4L, 3L, 4L, 4L, 4L, 4L, 3L, 3L, 1L, 4L, 4L, 1L, 4L, 4L, 1L, 3L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 3L, 3L, 2L, 1L), .Label = c("ending", "mobile", "stem.first", "stem.second"), class = "factor")), .Names = c("freq", "class"), row.names = c(NA, -128L), class = "data.frame") ``` As I read in a previous post there is more than one way to do this in R. I tried both with using the `t.test`-function and with using the `pairwise.t.test`-function. For using `t.test` I subsetted the dataframe by the classes to be compared and ran subsequent t-tests over the subsets. ``` ending.vs.mobile <- df[df$class=="ending"|df$class=="mobile",] ending.vs.first <- df[df$class=="ending"|df$class=="stem.first",] ending.vs.second <- df[df$class=="ending"|df$class=="stem.second",] mobile.vs.first <- df[df$class=="mobile"|df$class=="stem.first",] mobile.vs.second <- df[df$class=="mobile"|df$class=="stem.second",] first.vs.second <- df[df$class=="stem.first"|df$class=="stem.second",] t.test(ending.vs.mobile$freq ~ ending.vs.mobile$class, var.equal=T) t.test(ending.vs.first$freq ~ ending.vs.first$class, var.equal=T) t.test(ending.vs.second$freq ~ ending.vs.second$class, var.equal=T) t.test(mobile.vs.first$freq ~ mobile.vs.first$class, var.equal=T) t.test(mobile.vs.second$freq ~ mobile.vs.second$class, var.equal=T) t.test(first.vs.second$freq ~ first.vs.second$class, var.equal=T) ``` As far as I have understood it (here I might be wrong) the `pairwise.t.test` would be more convenient here, as I don't need to create all the subsets and can run it over the original dataframe. ``` pairwise.t.test(df$freq, df$class, p.adjust.method="none", paired=FALSE, pooled.sd=FALSE) ``` However I get different results here, most pronounced for the comparison ending vs. stem.second: p=0.7 using `t.test` and p=0.1 using `pairwise.t.test`. What's wrong here? Where have I done sth. wrong? Although the problem itself is solved, I think the reason why it occurred, makes me a little paranoid (not trusting myself anymore): Just by typing `pooled.sd` instead of `pool.sd` I do not get the results I expect. Isn't this very prone to errors? In many other cases you can type variants, e.g. `bonf` or `bonferroni`, `fa()` or `factor()`, and so on. But here `pooled.sd` is completely ignored although "pooled sd" is actually intended. Ok, if you thoroughly read the headline of the output you can guess that `pooled.sd` wasn't recognized as it still says "t tests with pooled SD" but what if I don't even print this, e.g. when piping the output to a self-written function? There are chances that this error will never be recognized. Should I write to some developers of R, that in future releases of R both spelling variants should be valid?