On page 606, of Lock et al “Statistics: Unlocking the Power of Data”, the authors state in item D “The p-value from the ANOVA table is 0.000 so the model as a whole is effective at predicting grade point average.” Ah no.

library(data.table)
library(mvtnorm)
rho <- 0.5
n <- 10^5
Sigma <- diag(2)
Sigma[1,2] <- Sigma[2,1] <- rho
X <- rmvnorm(n, mean=c(0,0), sigma=Sigma)
colnames(X) <- c("X1", "X2")
beta <- c(0.01, -0.02)
y <- X%*%beta + rnorm(n)
fit <- lm(y ~ X)
summary(fit)
## 
## Call:
## lm(formula = y ~ X)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -5.0500 -0.6750  0.0003  0.6730  4.8276 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  0.002480   0.003158   0.785  0.43232    
## XX1          0.011213   0.003651   3.071  0.00213 ** 
## XX2         -0.019416   0.003651  -5.319 1.05e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.9987 on 99997 degrees of freedom
## Multiple R-squared:  0.0002849,  Adjusted R-squared:  0.0002649 
## F-statistic: 14.25 on 2 and 99997 DF,  p-value: 6.489e-07

A p-value is not a measure of the predictive capacity of a model because the p-value is a function of 1) signal, 2) noise (unmodeled error), and 3) sample size while predictive capacity is a function of the signal:noise ratio. If the signal:noise ratio is tiny, the predictive capacity is small but the p-value can be tiny if the sample size is large.