Hi, I am trying to reconcile anova table in R (summary(lm)) with individual
t.test.
datafilename="http://personality-project.org/R/datasets/R.appendix1.data"
data.ex1=read.table(datafilename,header=T) #read the data into a table
summary(lm(Alertness~Dosage,data=data.ex1))
gives:
Call:
lm(formula = Alertness ~ Dosage, data = data.ex1)
Residuals:
Min 1Q Median 3Q Max
-8.500 -2.437 0.250 2.687 8.500
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 32.500 2.010 16.166 6.72e-11 ***
Dosageb -4.250 2.659 -1.598 0.130880
Dosagec -13.250 3.179 -4.168 0.000824 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.924 on 15 degrees of freedom
Multiple R-squared: 0.5396, Adjusted R-squared: 0.4782
F-statistic: 8.789 on 2 and 15 DF, p-value: 0.002977
As far as I understand it the lines "Dosageb" and "DosageC"
represent the difference between DosageA and the other two dosages.
My question is this: are these differences and the p-values associated with them
the same as a t.test or pairwise.t.test on these groups? If I do t.tests, I get
different values for t and p-value from those in the anova table above.
Can someone please explain what the discrepancy is?
Thanks
[[alternative HTML version deleted]]
On Mon, 20 Sep 2010, Jabez Wilson wrote:> Hi, I am trying to reconcile anova table in R (summary(lm)) with individual t.test. > datafilename="http://personality-project.org/R/datasets/R.appendix1.data" > data.ex1=read.table(datafilename,header=T) #read the data into a table > summary(lm(Alertness~Dosage,data=data.ex1)) > > gives: > > Call: > lm(formula = Alertness ~ Dosage, data = data.ex1) > > Residuals: > ???? Min???????? 1Q Median???????? 3Q?????? Max > -8.500 -2.437?? 0.250?? 2.687?? 8.500 > > Coefficients: > ?????????????????????? Estimate Std. Error t value Pr(>|t|)?????? > (Intercept)???? 32.500?????????? 2.010?? 16.166 6.72e-11 *** > Dosageb???????????? -4.250?????????? 2.659?? -1.598 0.130880?????? > Dosagec?????????? -13.250?????????? 3.179?? -4.168 0.000824 *** > --- > Signif. codes:?? 0 ???***??? 0.001 ???**??? 0.01 ???*??? 0.05 ???.??? 0.1 ??? ??? 1 > > Residual standard error: 4.924 on 15 degrees of freedom > Multiple R-squared: 0.5396,???????? Adjusted R-squared: 0.4782 > F-statistic: 8.789 on 2 and 15 DF,?? p-value: 0.002977 > > As far as I understand it the lines "Dosageb" and "DosageC" represent the difference between DosageA and the other two dosages. > My question is this: are these differences and the p-values associated with them the same as a t.test or pairwise.t.test on these groups? If I do t.tests, I get different values for t and p-value from those in the anova table above.The t tests in the table use all the data to estimate a residual variance that is used to calculate the standard error, whereas t.test() uses just two groups. If you use pairwise.t.test() with pool.sd=TRUE you will get the same p-values. There is another reason for the difference from t.test(), which is that t.test() does not assume equal variance in the outcome across groups. If you were to use vcovHC(,type="HC0") from the 'sandwich' package to estimate the standard errors in the linear model these would also not assume equal variance. The tests would still not be identical, although they would be very similar in large samples. The reasons they would not be identical are - some differences in using n vs n-p vs n1+n2-2 in denominators - different approximations to the denominator degrees of freedom. -thomas Thomas Lumley Professor of Biostatistics University of Washington, Seattle
On Sep 20, 2010, at 11:53 AM, Jabez Wilson wrote:> Hi, I am trying to reconcile anova table in R (summary(lm)) with > individual t.test. > datafilename="http://personality-project.org/R/datasets/R.appendix1.data > " > data.ex1=read.table(datafilename,header=T) #read the data into a > table > summary(lm(Alertness~Dosage,data=data.ex1))The quick answer is that in the ANOVA situation where you are interpreting individual level parameters, you are testing for the difference of a particular group from a shared mean (the intercept) across all three groups, whereas with the t-test you are only considering two groups at a time. (The default treatment contrasts do not actually result in the intercept estimate being a global mean, however.) -- David.> > gives: > > Call: > lm(formula = Alertness ~ Dosage, data = data.ex1) > > Residuals: > Min 1Q Median 3Q Max > -8.500 -2.437 0.250 2.687 8.500 > > Coefficients: > Estimate Std. Error t value Pr(>|t|) > (Intercept) 32.500 2.010 16.166 6.72e-11 *** > Dosageb -4.250 2.659 -1.598 0.130880 > Dosagec -13.250 3.179 -4.168 0.000824 *** > --- > Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 > > Residual standard error: 4.924 on 15 degrees of freedom > Multiple R-squared: 0.5396, Adjusted R-squared: 0.4782 > F-statistic: 8.789 on 2 and 15 DF, p-value: 0.002977 > > As far as I understand it the lines "Dosageb" and "DosageC" > represent the difference between DosageA and the other two dosages. > My question is this: are these differences and the p-values > associated with them the same as a t.test or pairwise.t.test on > these groups? If I do t.tests, I get different values for t and p- > value from those in the anova table above. > Can someone please explain what the discrepancy is? > Thanks > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > 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.David Winsemius, MD West Hartford, CT