Calculate the time delay introduced by group delay for IIR-Filters First of all, let me show what I mean by a simple example of an FIR-Filter:

I designed an FIR-Filter with a constant group delay of 323 samples and my sampling rate is 500 Hz. Now I can calculate the time delay by this filter with 323/(500 Hz) = 0.646s.

An IIR notch filter has a frequency dependent group delay and I am interested in the time delay around 50 Hz.

We know the group delay is around 2.5 samples at 50 Hz and we sample with 500 Hz. Can I do the same calculation as before? This means the time delay for 50 Hz is 2.5 samples/500 Hz = 5ms.

I am not sure but I think the correct calculation is with fs/2, that means it's 2.5/250 = 10ms.
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 $(1)$.