From: klebyn
Hello all R-users
_question 1_
I need to make a statistical model and respective ANOVA table
but I get distinct results for
the T-test (in summary(lm.object) function) and
the F-test (in anova(lm.object) )
shouldn't this two approach give me the same result, i.e
to indicate the same significants terms in both tests???
No, because they are not the same tests. The t-tests in summary.lm() test
whether the coefficient is zero, when all other terms are present in the
model. The F-tests in anova.lm() test the terms by sequentially adding them
into the model. Here's an example:
set.seed(1)
d - data.frame(x1=runif(20), x2=runif(20), y=rnorm(20))
fm - lm(y ~ ., d)
summary(fm)$coef
Estimate Std. Errort value Pr(|t|)
(Intercept) 1.0187254 0.5534310 1.8407452 0.08318123
x1 -1.6914784 0.6377065 -2.6524404 0.01675543
x2 -0.1817831 0.6618875 -0.2746435 0.78689983
anova(fm)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(F)
x1 1 4.2341 4.2341 7.0936 0.01638 *
x2 1 0.0450 0.0450 0.0754 0.78690
Residuals 17 10.1472 0.5969
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(fm2 - lm(y ~ x2 + x1, d))
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(F)
x2 1 0.0797 0.0797 0.1336 0.71928
x1 1 4.1994 4.1994 7.0354 0.01676 *
Residuals 17 10.1472 0.5969
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Notice how the p-value for x1 in the last output matches that of the t-test:
because both are testing if the coefficient for x1 is 0 given that x2 is
already in the model. (It's the same reason that the p-value for x2 in the
first anova() output matches that of the summary.lm(), but not the second
anova() output.)
I may be off, but I do not think the restrictions you mentioned have any
bearing on the analysis. If x + z is restricted to something _for each
case_ then you do have to worry, but not the way you have it. You can
choose the independent variables to take on any value you like (as in
designed experiments), so such restrictions should not matter.
Andy
obs.
The system has two restrictions:
1) sum( x_i ) = 1
2) sum( z_j ) = 1
*output below*
_question 2_
Has I to considerate a SST in ANOVA table with:
1) N-2 d.f. because of 2 restrictions?
or
2) N-1 d.f. because of 1 global restriction: sum( x ) + sum( z ) = 2 ?
I don't find any paper, book or another reference,
if someone may to indicate references for this type model (with 2
restrictions),
I would be very grateful.
Thanks a lot.
Regards
Cleber N. Borges
###
# OUTPUT
###
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(|t|)
(Intercept) 15.5000 0.5270 29.409 2.97e-10 ***
z1:x1-5. 0.7454 -6.708 8.77e-05 ***
z1:x2 0.5000 0.7454 0.671 0.519177
z1:x3-3. 0.7454 -4.025 0.002996 **
z2:x1-6. 0.7454 -8.050 2.11e-05 ***
z2:x2-5. 0.7454 -6.708 8.77e-05 ***
z2:x3-4.5000 0.7454 -6.037 0.000193 ***
z3:x1 1. 0.7454 1.342 0.212580
z3:x2 1.5000 0.7454 2.012 0.075029 .
z3:x3 NA NA NA NA
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F valuePr(F)
z1:x1 1 16.674 16.674 30.0125 0.0003910 ***
z1:x2 1 13.580 13.580 24.4446 0.0007977 ***
z1:x3 1 1.190 1.190 2.1429 0.1772677
z2:x1 1 35.267 35.267 63.4800 2.287e-05 ***
z2:x2 1 32.400 32.400 58.3200 3.202e-05 ***
z2:x3 1 42.667 42.667 76.8000 1.061e-05 ***
z3:x1 1 0.083 0.083 0.1500 0.7075349
z3:x2 1 2.250 2.250 4.0500 0.0750295 .
Residuals 9 5.000 0.556
---
###
# DATA
###
z1 z2 z3 x1 x2 x3 y
1 0 0 1 0 0 10
1 0 0 0 1 0 15
1 0 0 0 0 1 12
0 1 0 1 0 0 10
0 1 0 0 1 0 11
0 1 0 0 0 1 11
0 0 1 1 0 0 16
0 0 1 0 1 0 17
0 0 1 0 0 1 15
1 0 0 1 0 0 11
1 0 0 0 1 0 17
1 0 0 0 0 1 13
0 1 0 1 0 0 9
0 1 0 0 1 0 10
0 1 0 0 0 1 11
0 0 1 1 0 0 17
0 0 1 0 1 0 17
0 0 1 0 0 1 16
###
# CODE
###
x = read.table(file(clipboard),h=T)
## NOT a Scheffé Model:
x.lm - lm( y ~ (z1+z2+z3):(x1+x2+x3), data=x)
summary(x.lm)
anova(x.lm)
## Scheffé Model: - IS CORRECT the analysis below?
x.lm - lm( y ~ -1 + (z1+z2+z3):(x1+x2+x3), data=x)
summary(x.lm)
x.aov - aov( y ~
