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 
## -4.4908 -0.6749 -0.0011  0.6736  4.2441 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.006924   0.003165  -2.188   0.0287 *  
## XX1          0.016228   0.003642   4.456 8.35e-06 ***
## XX2         -0.019160   0.003641  -5.263 1.42e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.001 on 99997 degrees of freedom
## Multiple R-squared:  0.0003215,  Adjusted R-squared:  0.0003015 
## F-statistic: 16.08 on 2 and 99997 DF,  p-value: 1.04e-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.