On May 11, 2013, at 06:10 , meng wrote:

> Hi all:
> 
> I have a question about one-way anova.
> 
> The data is ¡°sleep¡±which belongs to R.
> 
> 
> 
> code1:
> 
> summary(lm(extra~group))
> 
> 
> 
> Estimate of group2(1.58) is the difference between mean of group1 and 
> group2,and t value(1.861) and p value(0.0792) is the same as 2 sample t 
> test,and the code is ¡° t.test(extra~group)¡±.
> 
> 
> 
> My question:
> 
> The p value of Intercept(0.2276)  tests for what?
> 
> 
> 
> My guess is test the difference between mean of group1 vs 0:
> 
> dat1<-extra[group==1]
> 
> t.test(dat1,mu=0)
> 
> 
> 
> But the p value is 0.2176,which is different from 0.2276 of lm result.
> 
> 
> 
> 
> 
> 
> 
> code2:
> 
> summary(lm(extra~group+0))
> 
> 
> 
> The pvalue of group1(0.2276) and group2(0.0011).
> 
> 
> 
> My question:
> 
> The p value of group1 and group2 tests for what?
> 
> 
> 
> My guess is test the difference between mean of group1 vs 0 and group2 vs 0:
> 
> 
> 
> dat1<-extra[group==1]
> 
> t.test(dat1,mu=0)
> 
> 
> 
> The p value is 0.2176,which is different from 0.2276 of lm result.
> 
> 
> 
> dat2<-extra[group==2]
> 
> t.test(dat2,mu=0)
> 
> 
> 
> The p value is 0.005076, which is different from 0.0011 of lm result.
> 
> 
> 
> 
> 
> So,what¡¯s wrong with my guess?
> 

You need to pay more attention to the degrees of freedom for error.

-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd....@cbs.dk  Priv: pda...@gmail.com

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