[R] The equivalence of t.test and the hypothesis testing of one way ANOVA

[Peng Yu]

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
>>>
>>
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>
>