wald test and score test, normal or chi square?

I learnt from section 10.3 of statistical inference that both Wald test statistic $\frac{W_n-\theta_0}{S_n}\approx\frac{W_n-\theta_0}{\sqrt{\hat I_n(W_n)}}$ and score test statistic $\frac{S(\theta_0)}{\sqrt{I_n(\theta_0)}}$, where $S(\theta)$ is the score and $I_n(W_n)$ is the expected/observed Fisher information, have standard normal distributions.

However, in this tutorial, it says both Wald and Score tests have chi-square distributions. How to explain the difference?

My guess is that since the square of the standard normal variable has chi square distribution with d.o.f. 1, simply squaring the test statistics mentioned above leads to chi square distributions. Then the d.o.f. will be one. But in the tuorial, it states explictly that the d.o.f. of the score test is two.

it is distributed chi-squared, with degrees of freedom equal to the number of variables being added to the model, so in our example, 2.

So, any better explanation? Thanks!

• Casella/Berger, Sec 10.3, deals with testing a restriction on a single parameter $\theta$, so 1 df. The other link tests whether the coefficients on math and science are jointly zero, so two restrictions, so 2 df. Indeed, that does not explain why the statistic is $\chi^2$ for more restrictions, may be more on that later. – Christoph Hanck Apr 13 '15 at 3:31