Calculate the time delay introduced by group delay for IIR-Filters
If you want to compute the delay experienced by a sine wave at a given frequency $\omega_0$, then it's not the group delay but the phase delay that you need to consider. In the case of linear phase FIR filters the phase delay and the group delay are identical, but in the general case they aren't.

The phase delay is defined by

$$\tau_p(\omega)=-\frac{\phi(\omega)}{\omega}\tag{1}$$

where $\phi(\omega)$ is the frequency dependent (unwrapped) phase of the system's frequency response. Clearly, if $\phi(\omega)=c\cdot \omega$ with some constant $c$, then the phase delay $(1)$ equals the negative derivative of the phase, i.e., group delay and phase delay are equal.

Of course, the (de-)normalization by the sampling frequency is always done in the same way. If $\omega$ is normalized, then you get the phase delay in samples, otherwise in seconds.

EDIT:

The response of an LTI system with frequency response $H(\omega)=M(\omega)e^{j\phi(\omega)}$ to a sinusoidal input signal $x(t)=A\sin(\omega_0t+\theta)$ is given by

$$\begin{align}y(t)&amp;=AM(\omega_0)\sin(\omega_0t+\theta+\phi(\omega_0))\\&amp;=AM(\omega_0)\sin\left[\omega_0\left(t+\frac{\phi(\omega_0)}{\omega_0}\right)+\theta\right]\end{align}\tag{2}$$

So the delay experienced by $x(t)$ is given by $\tau_p(\omega_0)$ as defined in $(1)$.