Why does the Chow test use a single tailed test? Suppose we want to test for differential intercepts using the Chow test. Suppose the variables we have are Y, the dependent variable and X, the independent variable. The sample is divided into two subsamples and the following regressions are run. Model 1:
\begin{equation} Y = \beta_0 + \beta_1 X + \epsilon \end{equation}
and, model 2:
$$ Y = \beta_0 + \beta_1 X + d + \epsilon $$
Where $ d $ is a dummy, taking the value 0 for one sub sample and 1 for another. We get two RSS values, $RSS_1$ and $RSS_2$ respectively from running these two regressions. The F-test for differential intercepts is then given by:
\begin{equation} F = \frac{\text{RSS}_1 - \text{RSS}_2}{\text{RSS}_2 /(n-k-1)} \stackrel{}{\sim} F(1, n-k-1) \end{equation}
This is from Jack Johnston & John DiNardo's "Econometric methods". In a later example, they use this F-test to test for differential intercepts. However, the critical value they take at the 5% significance level is the critical value for a single tailed test! (The F score that gives 0.05 on the upper tail). 
The hypotheses are:
$$H_0: \beta_1 = \beta_2 $$
$$H_a: \beta_1 \ne \beta_2 $$
Should we not be using the critical value as the F-score that gives 0.025% significance level instead? (Reference: Example in section 4.5.5 - "Econometric Methods" by Jack Johnston, John DiNardo)
 A: Motivated by @whuber's comments, I revise my previous answer. The book by Johnston and Dinardo is freely available and they actually give an explanation themselve in one of their previous chapters in a slightly different context (p.30):

The $F$ statistic [...] is seen to be the ratio of the mean square due
  to $X$ to the residual mean suqare. The latter may be regarded as a
  measure of the "noise" in the system, and thus an $X$ effect is only
  detected if it is greater than the inherent noise level. The
  significance of $X$ is thus tested by examning whether the sample $F$
  exceeds the appropriate critical value of $F$ taken from the upper
  tail of the $F$ distribution.

In other words, if $F$ is smaller than 1, this indicates that the additional feature explains less than a totally uncorrelated feature (i.e. noise) would explain. This might hint towards problems in the model specification, but is not helpful in assessing whether the new predictor adds some explanational power to the model. Therefore, we are usually not interested in values for $F$ smaller than 1. Values greater than 1 indicate evidence against the null hypothesis in both directions. Therefore, although the $F$-test itself is one-tailed, it will detect deviations of the new coefficient from $0$ in both directions. 
Edit: Maybe it also helps to take a closer look at the formula.
Johnston and Dinardo write that the degrees of freedom are $df_1=n-k$ and $df_2=n-k-1$ for the two models. 
Therefore, $\frac{RSS_1 - RSS_2}{RSS_2/(n-k-1)} $ can be rewritten as 
$\frac{(RSS_1 - RSS_2)/(df_1-df_2)}{RSS_2/df_2} $. 
So, you have the decrease in $RSS$ relative to the decrease in degrees of freedom (basically the gain of sacrificing one degree of freedom) in the nominator and in the denominator $RSS_2/df_2$, which is nothing else than the variance (or random noise) of the second model. If with the second model we gain less than random noise, i.e. if $F<1$, then something went wrong. Hence for the hypothesis test, we only care for $F>1$.
