Question about reading an output for using ANOVA to compare two linear models I tried to compare the following two models using "anova.lm()" in R:
Model 1: score ~ gpa + class 
Model 2: score ~ gpa   
  Res.Df     RSS Df   Sum of Sq      F    Pr(>F) 
1     90     213                              
2     91     201  1          96  14.07  <2.2e-16 

Since there is only one DF, how should I report the result of this F test?
Is "F(df=1) = 14.07, p-value < 2.2e-16" correct?
 A: F distributions have two degrees of freedom, one for the numerator and one for the denominator. 
http://en.wikipedia.org/wiki/F-distribution#Definition
In an ANOVA calculation, the numerator df is the df for the thing your null hypothesis relates to, the denominator df are the degrees of freedom for the residual. 
You appear to be testing whether the coefficient for the 'class' effect differs from zero, and it looks like it has 1df, so your numerator df is 1.
While it looks like the denominator df being applied here is 90, neither 90 nor 91 come anywhere near producing the observed p-value from that F... so I don't know what's going on. 
Indeed, to get that p-value with 1 and 90 (or 1 and 91) df would require an F of about 100.
Further, your sums of squares are inconsistent with the F-ratio. They suggest an F of around 40, not 14.
This doesn't add up anywhere. Is it not a standard situation? Is this made up output? Or is it real data with something edited in the output? 
Can you give more details to clarify what we're dealing with so more sensible answers can be given?
--
In the absence of consistent results (there are a number of additional problems there that I can't resolve), let me give a real data example (one that comes with R):
 a0b <- lm(count~0, data=InsectSprays[1:18,])
 ab <- lm(count~spray, data=InsectSprays[1:18,])
 anova(a0b,ab)

Analysis of Variance Table

Model 1: count ~ 0
Model 2: count ~ spray
  Res.Df  RSS Df Sum of Sq      F    Pr(>F)    
1     18 4192                                  
2     16  319  2      3873 97.129 1.124e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

This I can do something with!The d.f. for this F are (2,16). We can see this as follows:
 pf(97.129, 2, 16, lower.tail=FALSE)

[1] 1.124449e-09

