# Interpreting $R^2$, F-statistic & p-value of a model

From p. 277 of R Cookbook:

Let's say I have a R model lm(formula = y ~ u + v + w) and the Summary() shows:

Multiple R-Squared: 0.4981, Adjust R-Squared: 0.4402 F-statistic:
8.603 on 3 and 26 DF, p-value: 0.0003915


Using Adjusted r-Squared I can say that my model explains 44.02% of the variance of y with the remaining 55.98 unexplained.

Question: Does the associated F-statistic (with the p-value being < .05) tell me:

1. the model, in general, is significant (not taking into account other values from Summary)
2. the model is significant in explaining the 44.02% variance (adjusted r-squared)
• How did you arrive at 3, and 26 d.f.?
– user10619
Dec 3, 2017 at 12:43

The F-statistics tells you if the model fits the data better than the mean. Or, in other words, if $H_0:\;R^2=0$ should be rejected.

See: Wikipedia

To illustrate that the formula given in the link is indeed used by summary.lm:

x1 <- 1:10
set.seed(42)
x2 <- rnorm(10)
y <- x1+2*x2+rnorm(10)

fit0 <- lm(y~1)
fit1 <- lm(y~x1+x2)

summary(fit1)
#F-statistic:  14.1 on 2 and 7 DF,  p-value: 0.003507

#14.10014
pf(Fvalue,2,7,lower.tail=FALSE)
#0.00350697

• The question concerns about the chisquared test of variability and does not ask for the significance of mean.
– user10619
Jun 28, 2019 at 23:06
• My answer does not say anything about significance of the mean. Jun 29, 2019 at 11:17
• The two formulations are different. I think, one (Ist) imbibes - so called random effects and second is based on on (true )(fi xed- effects model.itmaybe noted that the two shoud yield same results. To my knowlege, first one model acts on assumption of random effects assumption and second one under fixed effeccts assumption. Therefore, I am not conviced your structure and proof.
– user10619
Jun 30, 2019 at 0:55
• Sorry, but both OP's model and the so-called Null model y ~ 1 are pure fixed effects models. Jun 30, 2019 at 4:19
• fit1 <- lm(y~x1+x2) please explain briefly. thanks for your valuable response. I am a halfbaked statistics worker.
– user10619
Jun 30, 2019 at 10:57