Well,since the intercept is the same as mean of group1,I take it for granted
that the 1-sample ttest must test based on group==1...

If the intercept is estimated from the whole sample,why does the intercept
is the same as mean of group1?




2011/8/17 Eik Vettorazzi <e.vettora...@uke.uni-hamburg.de>

> Hi Lao,
> thats not the same test. The concept of linear regression applies here
> (and you might take any introductory at your hand to refresh that
> concept). The intercept is estimated from the whole sample not just
> group==1, dfs are 20-2, not sum(group==1)-1!
>
> best regards
>
> Am 17.08.2011 09:57, schrieb Lao Meng:
> > Thanks Eik.
> > As to your words:"The intercept in lm is tested against 0 (one sample
> > t-test)"
> >
> > So, I perform the following test:
> > t.test(extra[group==1],mu=0)
> >
> > Since goup1 is regarded as reference,I do the 1-sample ttest based on
> > group1's mean vs 0.
> > But the result:
> > t value= 1.3257
> > p-value = 0.2176
> >
> > And t value and p value of s1 is:
> > t value= 1.249
> > p value= 0.2276
> >
> > So the t value and p value are different between 1-sample ttest of
> > group1'mean vs 0  and s1(lm's result).
> >
> > What's the reason for the difference then?
> >
> > Thanks a lot for your help.
> >
> > My best.
> >
> >
> > 2011/8/16 Eik Vettorazzi <e.vettora...@uke.uni-hamburg.de
> > <mailto:e.vettora...@uke.uni-hamburg.de>>
> >
> >     Hi,
> >     you may have noticed, that your t-test and lm had not the same
> p-values
> >     for the difference in means, which is calculated for group2 when you
> use
> >     treatment contrasts and that is what R does by default (see
> >     ?contr.treatment). This is because R uses Welsh test by default. Pros
> >     and cons are beyond this post, but look at
> >
> >     (t1<-t.test(extra~group,data=sleep,var.equal=T))
> >     (s1<-summary(lm(extra~group,data=sleep)))
> >     all.equal(s1$coef["group2","Pr(>|t|)"],t1$p.value)
> >
> >     The intercept in lm is tested against 0 (one sample t-test),
> >     so the t-statistic is (mean-0)/sd, having n-k (sample size - number
> of
> >     parameters) degrees of freedom.
> >
> >     cc<-s1$coef["(Intercept)",1:2]
> >     2*(1-pt(cc[1]/cc[2],df=18))
> >
> >
> >     hth.
> >
> >     Am 16.08.2011 07:25, schrieb Lao Meng:
> >     > Hi all:
> >     > I have a question about lm on t-test.
> >     >
> >     > data(sleep)
> >     >
> >     > I wanna perform t-test to test the difference between the 2 groups:
> >     >
> >     > I can use:
> >     > t.test(extra~group)
> >     >
> >     > The t.test result shows that:t = -1.8608; mean1=0.75,mean2=2.33
> >     >
> >     >
> >     > But I still wanna use:
> >     > summary(lm(extra~group))
> >     >
> >     > Intercept=0.75,which is mean1,just the same as t.test.
> >     > group2=1.58 means the difference of the 2 groups,so
> >     > mean2=1.58+0.75=2.33,just the same as t.test.
> >     > And some parameters of group2(t value,Pr) are the same as
> t.test,since
> >     > group2 is the difference of the 2 groups.
> >     >
> >     > My question is:
> >     > How the "t value" of Intercept(group1 acturally) is calculated?
> >     >
> >     >
> >     > Thanks a lot.
> >     >
> >     > My best
> >     >
> >     >       [[alternative HTML version deleted]]
> >     >
> >     > ______________________________________________
> >     > R-help@r-project.org <mailto:R-help@r-project.org> mailing list
> >     > https://stat.ethz.ch/mailman/listinfo/r-help
> >     > PLEASE do read the posting guide
> >     http://www.R-project.org/posting-guide.html
> >     > and provide commented, minimal, self-contained, reproducible code.
> >
> >     --
> >     Eik Vettorazzi
> >     Institut für Medizinische Biometrie und Epidemiologie
> >     Universitätsklinikum Hamburg-Eppendorf
> >
> >     Martinistr. 52
> >     20246 Hamburg
> >
> >     T ++49/40/7410-58243
> >     F ++49/40/7410-57790
> >
> >
>
> --
> Eik Vettorazzi
> Institut für Medizinische Biometrie und Epidemiologie
> Universitätsklinikum Hamburg-Eppendorf
>
> Martinistr. 52
> 20246 Hamburg
>
> T ++49/40/7410-58243
> F ++49/40/7410-57790
>

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