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

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  1. POR: t.test and pairwise.t.test give different results?
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    <p>I tried to do a t-test with R over the following dataframe. </p> <pre><code>df &lt;- 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") </code></pre> <p>As I read in a <a href="https://stackoverflow.com/questions/9661469/r-t-test-over-all-columns">previous post</a> there is more than one way to do this in R. I tried both with using the <code>t.test</code>-function and with using the <code>pairwise.t.test</code>-function.</p> <p>For using <code>t.test</code> I subsetted the dataframe by the classes to be compared and ran subsequent t-tests over the subsets.</p> <pre><code>ending.vs.mobile &lt;- df[df$class=="ending"|df$class=="mobile",] ending.vs.first &lt;- df[df$class=="ending"|df$class=="stem.first",] ending.vs.second &lt;- df[df$class=="ending"|df$class=="stem.second",] mobile.vs.first &lt;- df[df$class=="mobile"|df$class=="stem.first",] mobile.vs.second &lt;- df[df$class=="mobile"|df$class=="stem.second",] first.vs.second &lt;- 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) </code></pre> <p>As far as I have understood it (here I might be wrong) the <code>pairwise.t.test</code> would be more convenient here, as I don't need to create all the subsets and can run it over the original dataframe.</p> <pre><code>pairwise.t.test(df$freq, df$class, p.adjust.method="none", paired=FALSE, pooled.sd=FALSE) </code></pre> <p>However I get different results here, most pronounced for the comparison ending vs. stem.second: p=0.7 using <code>t.test</code> and p=0.1 using <code>pairwise.t.test</code>.</p> <p>What's wrong here? Where have I done sth. wrong?</p> <hr> <p>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 <code>pooled.sd</code> instead of <code>pool.sd</code> I do not get the results I expect. Isn't this very prone to errors?</p> <p>In many other cases you can type variants, e.g. <code>bonf</code> or <code>bonferroni</code>, <code>fa()</code> or <code>factor()</code>, and so on. But here <code>pooled.sd</code> is completely ignored although "pooled sd" is actually intended. Ok, if you thoroughly read the headline of the output you can guess that <code>pooled.sd</code> 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.</p> <p>Should I write to some developers of R, that in future releases of R both spelling variants should be valid?</p>
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