no, because the formula for the test statistics ( even assuming that
variances are equal ) of the two different tests are different. in the
pairwise t test, the pairwise differences are
viewed as one sample so it turns into a one sample test. any intro stat book
will have the formulas.
mark
----- Original Message -----
From: "Li,Qinghong,ST.LOUIS,Molecular Biology" <[EMAIL PROTECTED]>
To: <r-help@stat.math.ethz.ch>
Sent: Thursday, September 07, 2006 5:07 PM
Subject: [R] pairwise.t.test vs. t. test
> Hi,
>
> If I set the p.adjust="none", does it meant that the output p values from
> the pairwise.t.test will be the same as those from individual t.tests (set
> var.equal=T, alternative="t")?
>
> I actually got different p values from the two tests. See below. Is it
> supposed to be this way?
>
> Thanks
> Johnny
>
>> x
> [1] 61.6 52.7 61.3 65.2 62.8 63.7 64.8 58.7 44.9 57.0 64.3 55.1 50.0 41.0
> [15] 43.0 45.9 52.2 45.5 46.9 31.6 40.6 44.8 39.4 31.0 37.5 32.6 23.2 34.6
> [29] 38.3 38.1 19.5 21.2 15.8 33.3 28.6 25.8
>> Grp
> [1] Yng Yng Yng Yng Yng Yng Yng Yng Yng Yng Yng Yng Med Med Med Med Med
> Med
> [19] Med Med Med Med Med Med Old Old Old Old Old Old Old Old Old Old Old
> Old
> Levels: Yng Med Old
>> pairwise.t.test(x=x,g=Grp,p.adjust.method="none")
>
> Pairwise comparisons using t tests with pooled SD
>
> data: x and Grp
>
> Yng Med
> Med 1.0e-06 -
> Old 2.0e-12 2.6e-05
>
> P value adjustment method: none
>
>
>> t.test(x=x[1:12],y=x[25:36],var.equal=T, alternative="t")
>
> Two Sample t-test
>
> data: x[1:12] and x[25:36]
> t = 10.5986, df = 22, p-value = 4.149e-10
> alternative hypothesis: true difference in means is not equal to 0
> 95 percent confidence interval:
> 24.37106 36.22894
> sample estimates:
> mean of x mean of y
> 59.34167 29.04167
>
> [[alternative HTML version deleted]]
>
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