R help - Nov 2009 - The equivalence of t.test and the hypothesis testing of one way ANOVA

I read somewhere that t.test is equivalent to a hypothesis testing for
one way ANOVA. But I'm wondering how they are equivalent. In the
following code, the p-value by t.test() is not the same from the value
in the last command. Could somebody let me know where I am wrong?
> set.seed(0)
> N1=10
> N2=10
> x=rnorm(N1)
> y=rnorm(N2)
> t.test(x,y)
	Welch Two Sample t-test

data:  x and y
t = 1.6491, df = 14.188, p-value = 0.1211
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.2156863  1.6584968
sample estimates:
 mean of x  mean of y
 0.3589240 -0.3624813
>
> A = c(rep('x',N1),rep('y',N2))
> Y = c(x,y)
> fr = data.frame(Y=Y,A=as.factor(A))
> afit=aov(Y ~ A,fr)
>
> X=model.matrix(afit)
> B=afit$coefficients
> V=solve(t(X) %*% X)
>
> mse=tail(summary(afit)[[1]]$'Mean Sq',1)
> df=tail(summary(afit)[[1]]$'Df',1)
> t_statisitic=(B/(mse * sqrt(diag(V))))[[2]]
> 2*(1-pt(abs(t_statisitic),df))#the p-value from aov
[1] 0.1090802
>

compare

t.test(x, y, var.equal=T)

with

summary(afit)

b

On Nov 5, 2009, at 12:21 PM, Peng Yu wrote:
> I read somewhere that t.test is equivalent to a hypothesis testing for
> one way ANOVA. But I'm wondering how they are equivalent. In the
> following code, the p-value by t.test() is not the same from the value
> in the last command. Could somebody let me know where I am wrong?
>
>> set.seed(0)
>> N1=10
>> N2=10
>> x=rnorm(N1)
>> y=rnorm(N2)
>> t.test(x,y)
>
>        Welch Two Sample t-test
>
> data:  x and y
> t = 1.6491, df = 14.188, p-value = 0.1211
> alternative hypothesis: true difference in means is not equal to 0
> 95 percent confidence interval:
> -0.2156863  1.6584968
> sample estimates:
> mean of x  mean of y
> 0.3589240 -0.3624813
>
>>
>> A = c(rep('x',N1),rep('y',N2))
>> Y = c(x,y)
>> fr = data.frame(Y=Y,A=as.factor(A))
>> afit=aov(Y ~ A,fr)
>>
>> X=model.matrix(afit)
>> B=afit$coefficients
>> V=solve(t(X) %*% X)
>>
>> mse=tail(summary(afit)[[1]]$'Mean Sq',1)
>> df=tail(summary(afit)[[1]]$'Df',1)
>> t_statisitic=(B/(mse * sqrt(diag(V))))[[2]]
>> 2*(1-pt(abs(t_statisitic),df))#the p-value from aov
> [1] 0.1090802
>>
>
> ______________________________________________
> 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
> and provide commented, minimal, self-contained, reproducible code.

Notice also
> t.stat <- t.test(x,y, var.equal=TRUE)$statistic
> F.stat <- summary(afit)[[1]][1,4]
> t.stat^2 == F.stat 
   t 
TRUE 


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of Peng Yu
Sent: Thursday, November 05, 2009 9:21 AM
To: r-help at stat.math.ethz.ch
Subject: [R] The equivalence of t.test and the hypothesis testing of one
way ANOVA

I read somewhere that t.test is equivalent to a hypothesis testing for
one way ANOVA. But I'm wondering how they are equivalent. In the
following code, the p-value by t.test() is not the same from the value
in the last command. Could somebody let me know where I am wrong?
> set.seed(0)
> N1=10
> N2=10
> x=rnorm(N1)
> y=rnorm(N2)
> t.test(x,y)
	Welch Two Sample t-test

data:  x and y
t = 1.6491, df = 14.188, p-value = 0.1211 alternative hypothesis: true
difference in means is not equal to 0
95 percent confidence interval:
 -0.2156863  1.6584968
sample estimates:
 mean of x  mean of y
 0.3589240 -0.3624813
>
> A = c(rep('x',N1),rep('y',N2))
> Y = c(x,y)
> fr = data.frame(Y=Y,A=as.factor(A))
> afit=aov(Y ~ A,fr)
>
> X=model.matrix(afit)
> B=afit$coefficients
> V=solve(t(X) %*% X)
>
> mse=tail(summary(afit)[[1]]$'Mean Sq',1)
> df=tail(summary(afit)[[1]]$'Df',1)
> t_statisitic=(B/(mse * sqrt(diag(V))))[[2]] 
> 2*(1-pt(abs(t_statisitic),df))#the p-value from aov
[1] 0.1090802
>
______________________________________________
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
and provide commented, minimal, self-contained, reproducible code.


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Hi, 

Student's T-test is a test that can be used to test ONE SINGLE linear
restriction - which serves the as alternative hypothesis - on a linear model -
which serves as the null hypothesis - AT THE SAME TIME.

Fisher's F test is an extension of the T test. The F test can be used to
test ONE OR MORE linear restriction(s) on a linear model AT THE SAME TIME.

So, to test a single restriction on a linear model one can use both the F test
and the T test. When multiple restrictions are tested at the same time one needs
to apply the F test.

Both the F and the T test actually require equal variances. However, using a
transformation matrix one can transform a model assuming unequal variances into
an equivalent model assuming equal variances. On such a transformed model the F
test or T test can be applied. The untransformed models are usually called
general linear models. In R they can be handled using the glm() function. (See
?glm)

A (one-way) Anova model is a specific type of general linear model (glm). So
hypotheses on an Anova model are tested in exactly the same way as any other
restrictions on a glm should be tested.

Best wishes, 

Guido 
> ------------------------------
> 
> Message: 11
> Date: Fri, 6 Nov 2009 09:48:18 -0500
> From: JLucke at ria.buffalo.edu
> Subject: Re: [R] The equivalence of t.test and the
> hypothesis testing
> ??? of one??? way ANOVA
> To: Peng Yu <pengyu.ut at gmail.com>
> Cc: r-help-bounces at r-project.org,
> r-help at stat.math.ethz.ch
> Message-ID:
> ??? <OFC71A4670.65D468B9-ON85257666.0050EE14-85257666.005175E1 at
ria.buffalo.edu>
> ??? 
> Content-Type: text/plain
> 
> There extensions to aov for without assuming equal
> variances.
> 
> Reed, James F., I. & Stark, D. B. (1988), 'Robust
> alternatives to 
> traditional analyses of variance: Welch $W^*$, James
> $J_I^*$, James 
> $J_II^*$, and Brown-Forsythe $BF^*$', Computer Methods and
> Programs in 
> Biomedicine 26, 233--238.
> 
> 
> I don't???know whether they are implemented
> in R.
> 
> 
> 
> 
> Peng Yu <pengyu.ut at gmail.com>
> 
> Sent by: r-help-bounces at r-project.org
> 11/06/2009 07:59 AM
> 
> To
> r-help at stat.math.ethz.ch
> cc
> 
> Subject
> Re: [R] The equivalence of t.test and the hypothesis
> testing of one way 
> ANOVA
> 
> 
> 
> 
> 
> 
> Is it possible to use aov() to compute the same p-value
> that is
> generated by t.test() with var.equal=F. An assumption of
> ANOVA is
> equal variance, I'm wondering how to relax such assumption
> to allow
> non equal variance?
> 
> On Thu, Nov 5, 2009 at 8:31 AM, Benilton Carvalho <bcarvalh at
jhsph.edu>
> 
> wrote:
> > compare
> >
> > t.test(x, y, var.equal=T)
> >
> > with
> >
> > summary(afit)
> >
> > b
> >
> > On Nov 5, 2009, at 12:21 PM, Peng Yu wrote:
> >
> >> I read somewhere that t.test is equivalent to a
> hypothesis testing for
> >> one way ANOVA. But I'm wondering how they are
> equivalent. In the
> >> following code, the p-value by t.test() is not the
> same from the value
> >> in the last command. Could somebody let me know
> where I am wrong?
> >>
> >>> set.seed(0)
> >>> N1=10
> >>> N2=10
> >>> x=rnorm(N1)
> >>> y=rnorm(N2)
> >>> t.test(x,y)
> >>
> >>? ? ???Welch Two Sample
> t-test
> >>
> >> data:? x and y
> >> t = 1.6491, df = 14.188, p-value = 0.1211
> >> alternative hypothesis: true difference in means
> is not equal to 0
> >> 95 percent confidence interval:
> >> -0.2156863? 1.6584968
> >> sample estimates:
> >> mean of x? mean of y
> >> 0.3589240 -0.3624813
> >>
> >>>
> >>> A = c(rep('x',N1),rep('y',N2))
> >>> Y = c(x,y)
> >>> fr = data.frame(Y=Y,A=as.factor(A))
> >>> afit=aov(Y ~ A,fr)
> >>>
> >>> X=model.matrix(afit)
> >>> B=afit$coefficients
> >>> V=solve(t(X) %*% X)
> >>>
> >>> mse=tail(summary(afit)[[1]]$'Mean Sq',1)
> >>> df=tail(summary(afit)[[1]]$'Df',1)
> >>> t_statisitic=(B/(mse * sqrt(diag(V))))[[2]]
> >>> 2*(1-pt(abs(t_statisitic),df))#the p-value
> from aov
> >>
> >> [1] 0.1090802
> >>>
> >>
> >> ______________________________________________
> >> 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
> >> and provide commented, minimal, self-contained,
> reproducible code.
> >
> >
> 
> ______________________________________________
> 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
> and provide commented, minimal, self-contained,
> reproducible code.
> 
> 
> ??? [[alternative HTML version deleted]]
> 
> 
> 


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Hi Joe, 

You are right about the Behrens-Fisher problem. I was merely referring to
situations where the distribution of error terms is - assumed to be - known, and
not necessarily equal for all observations.

Thanks for pointing this out. 

Best wishes, 

Guido 

--- On Mon, 9/11/09, JLucke at ria.buffalo.edu <JLucke at ria.buffalo.edu>
wrote:
> From: JLucke at ria.buffalo.edu <JLucke at ria.buffalo.edu>
> Subject: Re: [R] The equivalence of t.test and the hypothesis testing of
one way ANOVA
> To: "Guido van Steen" <gvsteen at yahoo.com>
> Cc: "Peng Yu" <pengyu.ut at gmail.com>, r-help at
r-project.org
> Date: Monday, 9 November, 2009, 8:44 PM
> 
> 
> Guido wrote
> 
> "However, using a transformation
> matrix one can
> transform a model assuming unequal variances into an
> equivalent model assuming
> equal variances. On such a transformed model the F test or
> T test can be
> applied."
> 
> 
> 
> This is indeed news to me. ?I
> thought such transformations
> for unequal variances applied only to cases where the
> variances were known.
> ?Unknown, unequal variances leads to the
> Behrens-Fisher problem and
> its generalizations, a problem not resolved by mere linear
> transformation.
> ?Correct me and point me to the literature if I've
> misunderstood.
> 
> 
> 
> Joe
> 

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