a question about lm on t-test.

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]]

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.Vettorazzi at uke.uni-hamburg.de
<mailto:E.Vettorazzi at 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 at r-project.org <mailto:R-help at r-project.org> mailing list
    > https://stat.ethz.ch/mailman/listinfo/r-help
    > PLEASE do read the posting guide

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]]

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.Vettorazzi at uke.uni-hamburg.de
<mailto:E.Vettorazzi at 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.Vettorazzi at uke.uni-hamburg.de

    <mailto:E.Vettorazzi at uke.uni-hamburg.de>

    > <mailto:E.Vettorazzi at uke.uni-hamburg.de

    <mailto:E.Vettorazzi at 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 at r-project.org <mailto:R-help at r-project.org>

    <mailto:R-help at r-project.org <mailto:R-help at 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

    code.

    >
    >     --
    >     Eik Vettorazzi
    >     Institut f?r Medizinische Biometrie und Epidemiologie
    >     Universit?tsklinikum Hamburg-Eppendorf
    >
    >     Martinistr. 52
    >     20246 Hamburg
    >
    >     T ++49/40/7410-58243
    >     F ++49/40/7410-57790
    >
    >

    --
    Eik Vettorazzi
    Institut f?r Medizinische Biometrie und Epidemiologie
    Universit?tsklinikum Hamburg-Eppendorf

    Martinistr. 52
    20246 Hamburg

    T ++49/40/7410-58243
    F ++49/40/7410-57790

Pick up a book or the like on ordinary least squares regression, which is
what lm() in its plain vanilla application does. The t-value is the
estimated coefficient divided by the standard error. The standard errors
of the coefficients are the diagonal entries of the variance-covariance
matrix. 

x<-rnorm(100)
y<-2+x+rnorm(100)
reg<-lm(y~x)
summary(reg)$coefficients
sqrt(diag(vcov(reg)))

See also: http://en.wikipedia.org/wiki/Ordinary_least_squares

HTH,
Daniel


Lao Meng wrote:

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]]