What null hypothesis does the overall F-test in a linear regression output test?

## What is being compared The overall F reported at the foot of a regression summary is a comparison of two models on the same rows. The **restricted** model contains only an intercept: it predicts `y_bar` for every observation, and its residual sum of squares is the total sum of squares `TSS`. The **unrestricted** model is the one you fitted, with `k` slopes plus the intercept, and residual sum of squares `RSS`. The null hypothesis is the joint statement `H0: beta_1 = beta_2 = ... = beta_k = 0` with the intercept left free. The alternative is not that all of them are non-zero - it is that **at least one** is. ## The statistic `F = ((TSS - RSS)/k) / (RSS/(n - k - 1))` which, dividing top and bottom by TSS, can be written in terms of the fit statistic as `F = (R2/k) / ((1 - R2)/(n - k - 1))` The numerator is the variation the slopes explain, per slope. The denominator is the leftover variation, per residual degree of freedom. Under the null and the usual linear-model assumptions, this ratio follows an F distribution with `k` numerator and `n - k - 1` denominator degrees of freedom. Large values are evidence against the null; the p-value is the upper tail. ## Joint versus individual tests This is where most interview follow-ups live. Each t-statistic in the coefficient table tests one coefficient **given the others are in the model**. The F tests all of them at once. The two can disagree in both directions: - **Significant F, no significant t.** The typical cause is correlated predictors. Several columns carry overlapping information, so the model as a whole explains real variation, but the shared information inflates every individual standard error and no single coefficient can be separated from zero. The joint test does not suffer from this because it never has to attribute the explained variation to a particular column. - **A significant t, unremarkable F.** With many predictors and weak signal, one t-statistic crossing the 5 percent line is unsurprising by chance alone, and one small contribution among many need not move the joint statistic. With a single predictor there is no room for disagreement: `F = t^2` exactly, and the two p-values coincide. ## What significance here does and does not buy The alternative hypothesis is extremely weak - the model beats the mean by some non-zero amount. The denominator degrees of freedom grow with the sample, so on a large dataset even a tiny R-squared clears the threshold comfortably. With `n = 50000` and `k = 5`, an R-squared of 0.002 already gives a large F. So on modern data volumes a significant overall F is close to guaranteed and carries almost no information; the interesting questions are about effect sizes and specification, not about whether the model beats a flat line. Significance also says nothing about whether the model is **correctly specified**. A curved relationship fitted with a straight line, a model missing its main confounder, and a model contaminated by a predictor derived from the outcome will all produce a hugely significant F. The test is a floor to clear, not a certificate. ## Assumptions worth naming The exact F distribution follows from the usual linear-model conditions: correctly specified mean, independent observations, constant residual variance, and approximately normal errors (or a large enough sample for the approximation to hold). Heavy dependence between observations, such as repeated measurements on the same unit, inflates the statistic because the effective sample size is smaller than `n`. ## How to answer crisply Name the null as all slopes simultaneously zero, name the comparison model as intercept-only, name both degrees of freedom, and finish with the joint-versus-individual point. Adding the `F = t^2` identity for the single-predictor case reliably signals that you understand the two tests are the same machinery at different scopes.